NCERT Exemplar Class 12 Maths Solutions Chapter 6 Applications Of Derivatives

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NCERT Exemplar Class 12 Maths Solutions Chapter 6 Applications Of Derivatives

Edited By Ravindra Pindel | Updated on Sep 15, 2022 04:59 PM IST | #CBSE Class 12th

NCERT exemplar solutions for Class 12 Maths chapter 6 Applications Of Derivatives- For those who want to learn more about Calculus Mathematics, the best way to do it is by solving the NCERT exemplar Class 12 Maths solutions chapter 6. Students, who want to learn the topic of applications of derivatives, should attempt all the questions, mentioned in the chapter. Solving questions will not only help in understanding the elements but will also help in solving questions in the exam to score well. However, many tend to find it difficult to solve the questions and thus, require timely guidance.

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Question:1

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Answer:

Given: When a spherical ball salt is dissolved, the rate of decrease of the volume at any instant is proportional to the surface
To prove: The rate of decrease of radius is constant at any given time
Explanation: Take the radius of the spherical ball at any time t be ‘r’
Assume S as the surface area of the spherical ball
Then, $S = 4\pi r^2$ ……….(i)
Take the volume of the spherical ball be V
Then, $\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$$
According to the given criteria,
$-\frac{\mathrm{dV}}{\mathrm{dt}} \propto \mathrm{S}$$
The rate of decrease of volume is indicated by the negative sign It can also be written as
$-\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\mathrm{kS}$$
Here K is the proportional constant
After substitution of values from equation (i) and (ii), we get
$\Rightarrow-\frac{\mathrm{d}\left(\frac{4}{3} \pi r^{3}\right)}{\mathrm{dt}}=\mathrm{k} 4 \pi \mathrm{r}^{2}$$
When the constant term is taken outside the LHS, we get
$\Rightarrow-\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dt}}=\mathrm{k} 4 \pi \mathrm{r}^{2}$$
After the derivatives are applied with respect to t, we get
$\Rightarrow-\frac{4}{3} \pi \times 3 \mathrm{r}^{2} \times \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k} 4 \pi \mathrm{r}^{2}$$
After cancelling of the like terms, we get
$\\ \Rightarrow-\frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k} \\ \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=-\mathrm{k}$
Hence the rate of decrease of the radius of the spherical ball is constant.
Hence Proved

Question:2

If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

Answer:

Given: A circle with uniformly increasing area rate
To prove: relation between perimeter and radius is inversely proportional
Explanation: Take the radius of circle ‘r’
Let the area of the circle be A
Then $A = \pi r^2$ ……..(i)
After considering the give criteria of area increasing at a uniform rate,
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\mathrm{k}$$
Substitute the value of equation (i) into above equation,
$\frac{\mathrm{d}\left(\pi r^{2}\right)}{d t}=k$$
Differentiating with respect to t results in
$\\ \Rightarrow \pi \times 2 \mathrm{r} \times \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dt}}=\mathrm{k}$\\ $\Rightarrow \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dt}}=\frac{\mathrm{k}}{2 \pi \mathrm{r}} \ldots$(ii)$
Let P as the perimeter of the circle,
P = 2πr
Now differentiate the perimeter with respect to t,
$\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{d}(2 \pi \mathrm{r})}{\mathrm{dt}}$$
Apply all the derivatives,
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=2 \pi \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dt}}$$
Substituting of equation (ii) in the above equation,
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=2 \pi\left(\frac{\mathrm{k}}{2 \pi \mathrm{r}}\right)$$
Cancelling out of the like terms results into
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\left(\frac{\mathrm{k}}{\mathrm{r}}\right)$$
Now covert it into proportionality,
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}} \propto \frac{1}{\mathrm{r}}$$
Hence in the given conditions, the relation between perimeter of circle and radius is inversely proportional.
Hence Proved

Question:3

A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10 m/s, how fast is the string being let out; when the kite is 250 m away from the boy who is flying the kite? The height of boy is 1.5 m.

Answer:

Given: a boy of n height 1.5 m is flying a kite at a height of 151.5 m. The kite is moving with a speed of 10m/s. And the kite is 250 m away from the boy.
To find: the letting out speed of the string
Explanation: the above situation is explained by the figure,

Referring to the above figure,
Height of the kite, H = AD = 151.5 m
Height of the boy, b = BC = 1.5 m
x = CD = BE
Distance between kite and boy = AB = y =250
So, we need to calculate the increasing rate of the string
From figure, h = AE
= AD-ED
= 151.5-1.5
= 150m
The figure implies that the ΔABE is a right-angled triangle
Applying of the Pythagoras theorem results into,
$\\AB\textsuperscript{2} = BE\textsuperscript{2}+AE\textsuperscript{2}\\ $ {Or y\textsuperscript{2} = x\textsuperscript{2}+h\textsuperscript{2}$ \ldots $ $ \ldots $ $ \ldots $ ..(i)}\\ {Substitute the corresponding values}\\ {y\textsuperscript{2} = (x)\textsuperscript{2}+(150)\textsuperscript{2}}$
Let’s differentiate the equation (i) with respect to time,
$\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{h}^{2}\right)}{\mathrm{dt}}$$
After using the differentiation sum rule, we get
$\Rightarrow \frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+\frac{\mathrm{d}\left(\mathrm{h}^{2}\right)}{\mathrm{dt}}$$
since the height is not increasing, it indicates that it is constant, thus
$\Rightarrow \frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+0$$
Let's apply the derivative with respect to t
$\Rightarrow 2 \mathrm{y} \cdot \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=2 \mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}$$
$\Rightarrow \mathrm{y} \cdot \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}} \ldots$(iii)$
since the speed of the kite is 10 m/s so
$\frac{d x}{d t}=10 \mathrm{~m} / \mathrm{s}$$
When y = 250
${250\textsuperscript{2} = (x)\textsuperscript{2}+(150)\textsuperscript{2}}$
$x=200$
After substituting the corresponding values in equation (iii), we get
$\Rightarrow(250) \cdot \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=(200).(10)$
$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\frac{(200).(10)}{250}=8$
Therefore, the letting out speed of the string is 8 m/s

Question:4

Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being separated.

Answer:

Given: two men A and B start with velocities v at the same time from the junction of the two roads inclined at 45° to each other
To find: The rate of separation of the two men
Explanation:

The distance x travelled by A and B on any given time t will be same as they have velocity.
Hence
$\begin{aligned} &\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \ldots \ldots (1)\\ &\text { In the } \triangle A O B \text { , after applying the cosine rule, we get }\\ &y^{2}=x^{2}+x^{2}-2 x \cdot x \cdot \cos 45^{\circ} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}-2 \mathrm{x}^{2} \cdot \frac{1}{\sqrt{2}}\\ &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}\left(1-\frac{1}{\sqrt{2}}\right)\\ &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\\ &\text { Lets multiple and divide by } \sqrt{2} \text { , }\\ &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \times \frac{\sqrt{2}}{\sqrt{2}}\\ &\Rightarrow \mathrm{y}^{2}=\sqrt{2} \mathrm{x}^{2}(\sqrt{2}-1)\\ &\Rightarrow \mathrm{y}=\sqrt{\sqrt{2} \mathrm{x}^{2}(\sqrt{2}-1)}\\ &\Rightarrow \mathrm{y}= \mathrm{x} \cdot \sqrt{(2-\sqrt{2})} \end{aligned}$
Apply the derivatives with respect to t,
$\Rightarrow \frac{d y}{d t}=\frac{d( x \cdot \sqrt{(2-\sqrt{2}})}{d t}$$
Now take out the constant terms, we get
$\Rightarrow \frac{d y}{d t}= \sqrt{(2-\sqrt{2})} \frac{d(x)}{d t}$$
After value substitution for equation (i) we get
$\Rightarrow \frac{d y}{d t}= \sqrt{(2-\sqrt{2})}(v) \mathrm{m} / \mathrm{s}$$
Thus, the above given amount is the rate of separation of the two roads.

Question:5

Find an angle $\theta, 0<\theta<\frac{\pi}{2}$ which increases twice as fast as its sine.

Answer:

Given: a condition $\theta, 0<\theta<\frac{\pi}{2}$
To find: the angle θ such that it increases twice as fast as its sine.
Explanation: Let x = sin θ
Let’s differentiate with respect to t,
$\frac{d x}{d t}=\frac{d(\sin \theta)}{d t}$$
Now applying the derivative results into,
$\frac{d x}{d t}=\cos \theta \cdot \frac{d(\theta)}{d t} \ldots . .(i)$$
As this is given in the question
$\frac{\mathrm{d}(\theta)}{\mathrm{dt}}=2 \cdot \frac{\mathrm{d} x}{\mathrm{dt}}$$
Let's substitute this value in equation (i)
$\frac{d x}{d t}=\cos \theta \cdot 2 \cdot \frac{d x}{d t}$$
After cancelling the like terms, we get $1=2 \cos \theta$$
$\Rightarrow \cos \theta=\frac{1}{2}$$
But given $0<\theta<\frac{\pi}{2},$ this possibility occurs only when $\cos \theta=\frac{\pi}{3}$$
Hence the angle $\theta$$ is $\frac{\pi}{3}$.$

Question:6

Find the approximate value of $(1.999)^5$ .

Answer:

Given: $(1.999)^5$
And as the nearest integer to 1.999 is 2 , $\mathrm{So}, 1.999=2-0.001$$
$\therefore a=2$ and $h=-0.001$$
Hence, $(1.999)^{5}=(2+(-0.001))^{5}$
Therefore, the function becomes, $f(x)=x^{5} \ldots \ldots \ldots(i)$
After applying of first derivative, we get $$f^{\prime}(x)=5 x^{4} \ldots \ldots \ldots .$ (ii)$
Now let $f(a+h)=(1.999)^{5}$
Now we know,
$f(a+h)=f(a)+h f^{\prime}(a)$$
From equations (i) and (ii), substituting of functions results in,
$f(a+h)=a^{5}+h\left(5 a^{4}\right)$$
Substitution of values of a and h, we get
$f(2+(-0.001))=2^{5}+(-0.001)\left(5\left(2^{4}\right)\right)$$
$\\\Rightarrow f(1.999)=32+(-0.001)(5(16))$ \\\Rightarrow(1.999)^{5}=32+(-0.001)(80)$$
$\\ \Rightarrow(1.999)^{5}=32-0.08 \\ \Rightarrow(1.999)^{5}=31.92$
So, the approximate value of $(1.999)^5$ = 31.92.

Question:7

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

Answer:

Given: A hollow spherical shell having an internal radius of 3cm and external radii of 3.0005 cm
To find: The amount of metal used in the formation of the spherical shell.
Explanation: Let the r and R be the internal and external radii respectively.
So, it is given,
R = 3.0005 and r = 3
Let V be the volume of the hollow shell.
So according to question,
$\begin{aligned} &\mathrm{V}=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right)\\ &\mathrm{R} \text { and } \mathrm{r} \text { values are substituted, we get }\\ &\mathrm{V}=\frac{4}{3} \pi\left((3.0005)^{3}-3^{3}\right) \ldots (i) \end{aligned}$
To get the approximate value of $(3.0005)^3$ , differentiation is carried out
But the integer nearest to 3.0005 is 3,
So 3.0005 = 3+0.0005
So let a = 3 and h = 0.0005
Hence, $(3.0005)\textsuperscript{3} = (3+0.0005)\textsuperscript{3}$
Let the function becomes,
$f(x) = x\textsuperscript{3}$ \ldots $ $ \ldots $ $ \ldots $ (ii)$
Now applying first derivative, we get
$f'(x) = 2x\textsuperscript{2}$ \ldots $ $ \ldots $ $ \ldots $ .(iii)$
Now let $f(a+h) = (3.0005)\textsuperscript{3}$
Now we know,
$f(a+h) = f(a)+hf'(a)$
Now substituting the function from (ii) and (iii), we get
$f(a+h) = a\textsuperscript{3}+h(3a\textsuperscript{2})$
Substituting the values of a and h, we get
$\begin{aligned} &f(3+0.0005)=3^{3}+(0.0005)\left(3\left(3^{2}\right)\right)\\ &\Rightarrow f(3.0005)=27+(0.0005)(3(9))\\ &\Rightarrow(3.0005)^{3}=27+(0.0005)(27)\\ &\Rightarrow(3.0005)^{3}=27+0.0135\\ &\Rightarrow(3.0005)^{3}=27.0135\\ &\text { Thus, the approximate value of }(3.0005)^{3}=27.0135 \text { . } \end{aligned}$
So, substituting of the value in equation (i),
$\begin{aligned} \mathrm{V} &=\frac{4}{3} \pi(27.0135-27) \\ \mathrm{V} &=\frac{4}{3} \pi(0.0135) \\ \mathrm{V} &=4 \pi(0.0045) \\ \mathrm{V} &=0.018 \pi \mathrm{cm}^{3} \end{aligned}$
Hence, the approximate volume of the metal in the hollow spherical shell is $0.018\pi cm^3$ .

Question:8

A man, 2m tall, walks at the rate of $1\frac{2}{3}$ m/s towards a street light which is $5\frac{1}{3}$ m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is $3\frac{1}{3}$ m from the base of the light?

Answer:

Given: a 2m tall man walks at the rate of $1\frac{2}{3}$ m/s towards a $5\frac{1}{3}$ m tall street light.
To find: calculate the rate of movement of the tip of the shadow and also the rate of change in the length of the shadow when he is $3\frac{1}{3}$ m from the base of the light.
Explanation:

Here the street light is AB = $5\frac{1}{3}$
And man is DC = 2m
Let BC = x m and CE = y m
The rate of the man’s walk towards the streetlight is $1\frac{2}{3}m/s$ , and as the man is moving towards the street light, the entity carries a negative charge
Hence, $\frac{\mathrm{dx}}{\mathrm{dt}}=-1 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ ....(i)
Now consider ΔABE and ΔDCE
∠DEC = ∠AEB (same angle)
∠DCE = ∠ABD = 90°
Hence by AA similarity,
ΔABE≅ΔDCE
Hence by CPCT,
$\frac{AB}{DC}=\frac{BE}{CE}$
$\\ \begin{aligned} &\text { After substituting the values mentioned in the figure, we get }\\ &\Rightarrow \frac{\left(5 \frac{1}{3}\right)}{2}=\frac{x+y}{y}\\ &\Rightarrow\left(\frac{16}{3}\right) \mathrm{y}=2(\mathrm{x}+\mathrm{y})\\ &\Rightarrow 16 y=6(x+y)\\ &\Rightarrow 16 y=6 x+6 y\\ &\Rightarrow 16 y-6 y=6 x\\ &\Rightarrow 10 y=6 x\\ &\Rightarrow \mathrm{y}=\frac{6}{10} \mathrm{x}\\ &\Rightarrow \mathrm{y}=\frac{3}{5} \mathrm{x} \end{aligned}$
Apply the first derivative with respect to t,
$\\ \Rightarrow \frac{d y}{d t}=\frac{d\left(\frac{3}{5} x\right)}{d t}$\\ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{3 \mathrm{~d}(\mathrm{x})}{5}$$
After substituting the value in the above equation from equation (i), we get
$\\\Rightarrow \frac{d y}{d t}=\frac{3}{5}\left(-1 \frac{2}{3}\right)$\\ $\Rightarrow \frac{d y}{d t}=\frac{3}{5}\left(-\frac{5}{3}\right)$\\ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-1 \mathrm{~m} / \mathrm{s}$$
So, the rate of movement of tip of the shadow is -1m/s, i.e., the length of the shadow is decreasing at the rate of 1m/s.
Let BE = z
So from fig,
z = x+y
Let’s apply the first derivative with respect to t on the above equation.
$\\ \begin{aligned} &\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}\\ &\text { Substituting oft eh corresponding values result in, }\\ &\frac{d z}{d t}=-1 \frac{2}{3}-1\\ &\Rightarrow \frac{d z}{d t}=-\frac{5}{3}-1\\ &\Rightarrow \frac{\mathrm{dz}}{\mathrm{dt}}=\frac{-5-3}{3}\\ &\Rightarrow \frac{\mathrm{dz}}{\mathrm{dt}}=\frac{-8}{3}\\ &\Rightarrow \frac{\mathrm{dz}}{\mathrm{dt}}=-2 \frac{2}{3} \mathrm{~m} / \mathrm{s} \end{aligned}$
So, the rate of tip of the shadow moving towards the light source is of $2 \frac{2}{3}$ m/s.

Question:9

A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and $L = 200 (10 - t)^2$ . How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

Answer:

Given: L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and
$L = 200 (10 - t)^2$
To find: the rate of water running out of the pool at the last 5s. and also the average rate of water flowing at during the first 5s.
Explanation: Take the rate of Water running out given by $-\frac{dL}{dt}$
Given $L = 200 (10 - t)^2$
Differentiation of the above given equation results in,
$\\\Rightarrow-\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}=-\frac{\mathrm{d}\left(200(10-\mathrm{t})^{2}\right)}{\mathrm{dt}}$ \\Removing all the constant terms, we get \\$\Rightarrow-\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}=-200 \cdot \frac{\mathrm{d}\left((10-\mathrm{t})^{2}\right)}{\mathrm{dt}}$\\ Using the power rule of differentiation, we get \\$\Rightarrow-\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}=-200.2(10-\mathrm{t}) \cdot \frac{\mathrm{d}(10-\mathrm{t})}{\mathrm{dt}}$\\ $\Rightarrow-\frac{\mathrm{dL}}{\mathrm{dt}}=-400(10-\mathrm{t}).(-1)$\\ $\Rightarrow-\frac{\mathrm{dL}}{\mathrm{dt}}=400(10-\mathrm{t}) \ldots$(i)$
Now, to find the speed of water running out at the end of 5s, we need to find the value of equation (i) with t=5
$\\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=(400(10-t))_{t=5} \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=400(10-5) \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=400(5) \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=2000 \frac{L}{s} \ldots . .(i i)$
Therefore, 2000 L/s is the rate of water running out at the end of 5s
To calculate the initial rate we need to take t=o in equation (i)
$\\ \Rightarrow\left(-\frac{\mathrm{dL}}{\mathrm{dt}}\right)_{\mathrm{t}=0}=(400(10-\mathrm{t})) \mathrm{t}=0 \\ \Rightarrow\left(-\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}\right)_{\mathrm{t}=0}=400(10-0) \\ \Rightarrow\left(-\frac{\mathrm{d} L}{\mathrm{dt}}\right)_{\mathrm{t}=0}=4000 \mathrm{~L} / \mathrm{s}$
Equation (ii) tells about the final rate of water flowing whereas the equation (iii) is the initial rate.
Thus, the average rate during 5s is
$\\=\frac{\text { initial rate }+\text { final rate }}{2}$ Substituting the corresponding values, we get $=\frac{4000+2000}{2}=3000 \mathrm{~L} / \mathrm{s}$$
After calculation, the final rate of water flowing out of the pool in 5s is 3000L/s.

Question:10

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Answer:

Given: a cube with volume increasing at a constant rate
To prove: the relation between the increase int eh surface area with the length of the side is inversely proportional.
Explanation: Let ‘a’ the length of the side of the cube.
Take the volume of the cube ‘V’
Then $V=a^3........(i)$
As mentioned in the question, the rate of volume increase is constant, then
$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\mathrm{k}$$
After substituting in the above equation, the values from equation (i) we get
$\frac{\mathrm{d}\left(\mathrm{a}^{3}\right)}{\mathrm{dt}}=\mathrm{k}$$
Differentiating the equation with respect to t,
$\\ \Rightarrow 3 a^{2} \times \frac{d(a)}{d t}=k$\\ $\Rightarrow \frac{\mathrm{d}(\mathrm{a})}{\mathrm{dt}}=\frac{\mathrm{k}}{3 \mathrm{a}^{2}} \ldots \ldots(\mathrm{ii})$$
Take S as the surface area of the cube, then $\mathrm{S}=6 \mathrm{a}^{2}$$
After differentiating the surface area with respect to t, we get
$\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}\left(6 \mathrm{a}^{2}\right)}{\mathrm{dt}}$$
Applying the derivatives, we get
$\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=6 \times 2 \mathrm{a} \times \frac{\mathrm{d}(\mathrm{a})}{\mathrm{dt}}$$
After substituting value from equation (ii) in the given equation we get
$\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=12 \mathrm{a}\left(\frac{\mathrm{k}}{3 \mathrm{a}^{2}}\right)$$
After taking out the like terms,
$\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=4\left(\frac{\mathrm{k}}{\mathrm{a}}\right)$$
Now converting it to proportional, we get
$\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}} \propto \frac{1}{\mathrm{a}}$$
Therefore, the relation between the length and the side of the cube is inversely proportional in the given condition.
Hence Proved.

Question:11

x and y are the sides of two squares such that $y = x - x^2$ . Find the rate of change of the area of the second square with respect to the area of the first square.

Answer:

Given: two squares of sides x and y, such that $y = x - x^2$
To find: The rate of change of area of both the squares with respect to each other
Explanation: Take $A_1$ and $A_2$ as the area of first and 2nd square respectively
Thus, the area of the 1st square will be
$A_1 = x^2$
Differentiating the equation with respect to time, we get
$\\ \frac{\mathrm{d} \mathrm{A}_{1}}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{d} \mathrm{A}_{1}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}} \ldots \ldots \text { (i) }$
And the area of the second square is
$A\textsubscript{2} = y\textsuperscript{2}$ \ldots $ $ \ldots $ .(ii)$
But given, $y = x-x\textsuperscript{2}$
Now substituting the known value in equation (ii),
$A\textsubscript{2} = (x-x\textsuperscript{2}) ^{2} $ \ldots $ $ \ldots $ .(iii)$
After differentiating equation (iii) with respect to time it results into,
$\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=\frac{\mathrm{d}\left(\left(\mathrm{x}-\mathrm{x}^{2}\right)^{2}\right)}{\mathrm{dt}}$
Apply the power rule of differentiation to get,
$\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2\left(\mathrm{x}-\mathrm{x}^{2}\right) \times \frac{\mathrm{d}\left(\mathrm{x}-\mathrm{x}^{2}\right)}{\mathrm{dt}}$
Applying the sum rule of differentiation, we get
$\begin{aligned} &\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2\left(\mathrm{x}-\mathrm{x}^{2}\right)\left(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}-\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}\right)\\ &\text { Applying the derivative, we get }\\ &\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2\left(\mathrm{x}-\mathrm{x}^{2}\right)\left(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}-2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}\right)\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2 \mathrm{x}(1-\mathrm{x})\left(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}(1-2 \mathrm{x})\right)\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2 \mathrm{x}(1-\mathrm{x})(1-2 \mathrm{x}) \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}} \ldots \ldots(\mathrm{iv}) \end{aligned}$
Since we need to find the rate of change of area of both the squares with respect to each other, which is
$\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=\frac{\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}}{\frac{\mathrm{d} \mathrm{A}_{1}}{\mathrm{dt}}}$
Substituting the known values from equation (i) and (iv), we get
$\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=\frac{2 \mathrm{x}(1-\mathrm{x})(1-2 \mathrm{x}) \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}}{2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}}$$
By cancelling the like terms, we get
$\begin{aligned} &\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=(1-\mathrm{x})(1-2 \mathrm{x})\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=1(1-2 \mathrm{x})-\mathrm{x}(1-2 \mathrm{x})\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=1-2 \mathrm{x}-\mathrm{x}+2 \mathrm{x}^{2}\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=2 \mathrm{x}^{2}-3 \mathrm{x}+1\\ &\text { Therefore, the rate of change of area of second square with respect to area of first square is }\\ &2 x^{2}-3 x+1 \end{aligned}$

Question:12

Find the condition that the curves $2x = y^2$ and 2xy = k intersect orthogonally.

Answer:

Given: two curves $2x = y^2$ and 2xy = k
To find: to track the condition where both the curves intersect orthogonally
Explanation: Given 2xy = k
$\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{2 \mathrm{x}} \ldots \ldots$ (i)$
Put in the value of y in another curve equation, i.e., $2 x=y^{2},$ we get
$2 \mathrm{x}=\left(\frac{\mathrm{k}}{2 \mathrm{x}}\right)^{2}$$
$\\\Rightarrow 2 \mathrm{x}=\frac{\mathrm{k}^{2}}{4 \mathrm{x}^{2}}$ \\$\Rightarrow \mathrm{x}^{3}=\frac{\mathrm{k}^{2}}{8}$$
Putting both the sides under cube root, we get
$\Rightarrow \mathrm{x}=\frac{\mathrm{k}^{\frac{2}{3}}}{2} \ldots \ldots$ (ii)$
Substituting equation (ii) in equation (i), we get
$\Rightarrow y=\frac{k}{2\left ( \frac{k^{\frac{2}{3}}}{2} \right )}$
$\begin{aligned} &\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{\mathrm{k}^{\frac{2}{3}}}\\ &\Rightarrow \mathrm{y}=\mathrm{k}^{1-\frac{2}{3}}\\ &\Rightarrow \mathrm{y}=\mathrm{k}^{\frac{1}{3}}\\ & \end{aligned}$
$\text { Thus, } \left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$ is the point of intersection of two curves

Now given $2x = y^2$
After differentiating the equation with respect to x, we get
$\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}$ $\Rightarrow 2=2 \mathrm{y} \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{y}}$$
After tracking the value of differentiation at the point of intersection i.e., at $\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$ we
get
$\left ( \frac{dy}{dx} \right )$ $\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$ $=\frac{1}{k^{\frac{1}{3}}}=m_1...(ii)$
Also given 2xy = k
Differentiating this with respect to x, we get
$\begin{aligned} &\frac{\mathrm{d}(2 \mathrm{xy})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{k})}{\mathrm{dx}}\\ &\Rightarrow 2 \frac{\mathrm{d}(\mathrm{xy})}{\mathrm{dx}}=0\\ &\text { When applying the product rule of differentiation, it results into }\\ &\Rightarrow 2\left(x \frac{d y}{d x}+y \frac{d x}{d x}\right)=0\\ &\Rightarrow \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0\\ &\Rightarrow x \frac{d y}{d x}=-y\\ &\Rightarrow \frac{d y}{d x}=-\frac{y}{x} \end{aligned}$
Then again finding the about given differentiation value at the point of intersection i.e., at $\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$ , we get
$\left ( \frac{dy}{dx} \right )$ $\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$
$\\=-\frac{k^{\frac{1}{3}}}{\frac{k^{\frac{2}{3}}}{2}} \\ =-2 k^{\frac{1}{3}-\frac{2}{3}} \\ =-2 k^{-\frac{1}{3}}=m_{2} \ldots \text { (iv) }$
But the orthogonal interaction of two curves occurs if
m1.m2 = -1
Then Substituting the values from equation (iii) and equation (iv), we get
$\begin{aligned} &\frac{1}{k^{\frac{1}{3}}} \cdot\left(-2 k^{-\frac{1}{3}}\right)=-1\\ &\Rightarrow \mathrm{k}^{-\frac{1}{3}} \cdot\left(-2 \mathrm{k}^{-\frac{1}{3}}\right)=-1\\ &\Rightarrow-2 k^{-\frac{1}{3}-\frac{1}{3}}=-1\\ &\Rightarrow 2 k^{-\frac{2}{3}}=1\\ &\Rightarrow \frac{2}{k^{\frac{2}{3}}}=1\\ &\Rightarrow \mathrm{k}^ \frac{2}{3}=2\\ &\text { Now taking cube on both sides, we get }\\ &\Rightarrow \mathrm{k}^{2}=2^{3}=8\\ &\Rightarrow \mathrm{k}=\pm2 \sqrt 2 \end{aligned}$
This condition proves to fulfill the orthogonal interaction point for the two curves.

Question:13

Prove that the curves $xy = 4$ and $x^2+y^2 = 8$ touch each other.
Answer:

Given: two curves $xy = 4$ and $x^2+y^2 = 8$
To prove: two curves meet each other at a point
Explanation:
Now given $x^2+y^2 = 8$
Differentiating this with respect to x, we get
$\begin{aligned} &\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}\\ &\text { By using the sum rule of differentiation, we get }\\ &\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0\\ &\Rightarrow 2 x+2 y \frac{d y}{d x}=0\\ &\Rightarrow 2 x=-2 y \frac{d y}{d x}\\ &\Rightarrow \frac{d y}{d x}=-\frac{x}{y}=m_{1} \ldots \ldots(i) \end{aligned}$

Also given xy = 4
Differentiating this with respect to x, we get
$\frac{\mathrm{d}(\mathrm{xy})}{\mathrm{dx}}=\frac{\mathrm{d}(4)}{\mathrm{dx}}$$
The using the product rule of differentiation, we get
$\Rightarrow\left(x \frac{d y}{d x}+y \frac{d x}{d x}\right)=0$ $\Rightarrow x \frac{d y}{d x}+y=0$$
$\\\Rightarrow \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}$\\ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{m}_{2} \ldots \ldots$ (ii)$
But the touch of 2 curves is possible if
$\mathrm{m}_{1}=\mathrm{m}_{2}$$
Now substituting the values from equation (ii) and equation (ii), we get
$\\-\frac{y}{x}=-\frac{x}{y}$\\ $\Rightarrow \mathrm{y}^{2}=\mathrm{x}^{2}$\\ $\Rightarrow \mathrm{x}=\mathrm{y}$$
Now substituting $x=y$ in $x^{2}+y^{2}=8,$ we get $y^{2}+y^{2}=8$$
$\\$\Rightarrow 2 \mathrm{y}^{2}=8$ \\$\Rightarrow y^{2}=4$ \\$\Rightarrow y=\pm 2$$
When $y=2$ $x y=4$ becomes $x(2)=4 \Rightarrow x=2$$
when $y=-2$ $x y=4$ becomes $$ x(-2)=4 \Rightarrow x=-2 $$$
Therefore, (2,2) and (-2, -2) is the intersection point of the two curve
Substituting these points of intersection equation (i) and equation (ii), we get
For (2,2),
$\\ \mathrm{m}_{1}=-\frac{\mathrm{x}}{\mathrm{y}}=-\frac{2}{2}=-1 \\ \mathrm{~m}_{2}=-\frac{\mathrm{y}}{\mathrm{x}}=-\frac{2}{2}=-1 \\ \therefore \mathrm{m}_{1}=\mathrm{m}_{2} \\ \text { For }(-2,-2) \\ \mathrm{m}_{1}=-\frac{\mathrm{x}}{\mathrm{y}}=-\frac{-2}{-2}=-1 \\ \mathrm{~m}_{2}=-\frac{\mathrm{y}}{\mathrm{x}}=-\frac{-2}{-2}=-1 \\ \therefore \mathrm{m}_{1}=\mathrm{m}_{2}$
Thus, the condition for both the curves to touch is possible if that they have same slope
Hence the two given curves touch each other.
Hence proved

Question:14

Find the co-ordinates of the point on the curve $\sqrt x +\sqrt y =4$ at which the tangent is equally inclined to the axes.

Answer:

Given: curve $\sqrt x +\sqrt y =4$
To find: point coordinates on which tangent is equally inclined to the axis on the curve
Explanation: given $\sqrt x +\sqrt y =4$
After differentiating with respect to,
$\frac{\mathrm{d}(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(4)}{\mathrm{dx}}$$
Now using the sum rule of differentiation
$\\ \frac{\mathrm{d}(\sqrt{\mathrm{x}})}{\mathrm{d} \mathrm{x}}+\frac{\mathrm{d}(\sqrt{\mathrm{y}})}{\mathrm{d} \mathrm{x}}=0$ \\$\frac{\mathrm{d}\left(\mathrm{x}^{\frac{1}{2}}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{\frac{1}{2}}\right)}{\mathrm{d} \mathrm{x}}=0$$
Then differentiating the equation, we get
$\\ \Rightarrow \frac{1}{2} x^{\frac{1}{2}-1}+\frac{1}{2} y^{\frac{1}{2}-1} \frac{d(y)}{d x}=0 \\ \Rightarrow \frac{1}{2} x^{\frac{-1}{2}}+\frac{1}{2} y^{\frac{-1}{2}} \frac{d(y)}{d x}=0 \\ \Rightarrow \frac{1}{2} x^{\frac{-1}{2}}=-\frac{1}{2} y^{\frac{-1}{2}} \frac{d(y)}{d x} \\ \Rightarrow \frac{d y}{d x}=-\frac{x^{\frac{-1}{2}}}{y^{\frac{-1}{2}}} \\ \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{y}{x}} \ldots . . \text { (i) }$
The given curve has this tangent
As mentioned in the question tangent is equally inclined to the axis,
$\\ \frac{d y}{d x}=\pm 1 \\\\ -\sqrt{\frac{y}{x}}=\pm 1 \\\\ \frac{y}{x}=1 \Rightarrow y=x$
Substituting values in the curve equation from equation (ii)
$\\$\sqrt y+\sqrt y=4$ \\$2 \sqrt y=4$ \\$ \sqrt y=2$ \\$\Rightarrow y=4$$
When y = 4, then x = 4 from equation (ii)
Show the points on the curve $\sqrt{x}+\sqrt {y}=4$ at which the tangent equally inclined to the axis has the coordinates (4,4).

Question:15

Find the angle of intersection of the curves $y = 4-x^2$ and $y = x^2.$

Answer:

Given: the curves $y = 4-x^2$ and $y = x^2.$
To find: the interaction angle between two curves
Explanation: acknowledging first curve
$y = 4-x^2$
when the above curve is differentiated with respect to X
$\begin{aligned} &\frac{d y}{d x}=\frac{d\left(4-x^{2}\right)}{d x}\\ &\frac{d y}{d x}=-2 x=m_{1} \ldots\\ &\text { Considering the second curve }\\ &y=x^{2} \end{aligned}$
second curve differentiated with respect to X
$\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}$$
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}=\mathrm{m}_{2} \ldots$ (ii)$
Given $y=x^{2}$ Substituting the other curve equation with this
$\\$x^{2}=4-x^{2}$ \\$\Rightarrow 2 x^{2}=4$ \\$\Rightarrow x^{2}=2$ \\$\Rightarrow x=\pm \sqrt{2}$$
When $$x=\sqrt{2}$ , we get $y=(\sqrt{2})^{2} \Rightarrow y=2$$
When $$x=-\sqrt{2}$ we get $y=(-\sqrt{2})^{2} \Rightarrow y=2$
Hence the intersection points are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$ since angle of intersection can be found using the formula
i.e., $\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$$
Substituting the values from equation (i) and equation (ii), we get
$\Rightarrow \tan \theta=\left|\frac{-2 x-2 x}{1+(-2 x)(2 x)}\right|$$
$\Rightarrow \tan \theta=\left|\frac{-4 x}{1-4 x^{2}}\right|$$
For $$(\sqrt{2}, 2)$,$ the equation gets converted into,
$\\ \Rightarrow \tan \theta=\left|\frac{-4(\sqrt{2})}{1-4(\sqrt{2})^{2}}\right|$ \\$\Rightarrow \tan \theta=\left|\frac{-4(\sqrt{2})}{-7}\right|$ \\$\Rightarrow \theta=\tan ^{-1}\left|\frac{4 \sqrt{2}}{7}\right|$$
Hence, the angle at which the curve intersect at $y=4-x^{2}$ and $y=x^{2}$ is $\tan ^{-1}\left|\frac{4 \sqrt{2}}{7}\right|$$

Question:16

Prove that the curves $y^2 = 4x$ and $x^2 + y^2 - 6x + 1 = 0$ touch each other at the point (1, 2).

Answer:

Given: two curves $y^2 = 4x$ and $x^2 + y^2 - 6x + 1 = 0$
To prove: Two curves have the possibilities of meeting at a point(1,2)
Explanation:
Now given $x^2 + y^2 - 6x + 1 = 0$
Differentiating the above equation with respect to x
$\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}+1\right)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(0)}{\mathrm{dx}}$
Now using the sum rule of differentiation
$\begin{aligned} &\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(6 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(1)}{\mathrm{dx}}=0\\ &\Rightarrow 2 x+2 y \frac{d y}{d x}-6+0=0\\ &\Rightarrow 2 \mathrm{x}-6=-2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\\ &\Rightarrow 6-2 \mathrm{x}=2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\\ &\Rightarrow \frac{d y}{d x}=\frac{6-2 x}{2 y}\\ &\Rightarrow \frac{d y}{d x}=\frac{3-x}{y}\\ &\text { Finding the solution of above equation at point }(1,2) \text { , we get }\\ &\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,2)}=\frac{3-1}{2}=1=\mathrm{m}_{1} \ldots \ldots(i) \end{aligned}$
Also given $y^2 = 4x$
Differentiating the value with respect to x, we get
$\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(4 \mathrm{x})}{\mathrm{dx}}$$
Using a product rule of differentiation, we get
$\\\Rightarrow 2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=4$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\mathrm{y}}=\mathrm{m}_{2}$$
Finding the solution of the above equation at point (1,2), we get
$\Rightarrow\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}=1=m_{2} \ldots \ldots$ (ii)$
From equation (i) and (ii),
∴ $m_1 = m_2$
Therefore it is possible for both the curves to touch each other at point(1,2).
Hence proved

Question:17

Find the equation of the normal lines to the curve $3x^2 - y^2 = 8$ which are parallel to the line $x + 3y = 4$ .

Answer:

Given: equation of the curve $3x^2 - y^2 = 8$ equation of line $x + 3y = 4$
To find: the equation of the normal lines to the given curve which are parallel to the given line
Explanation:
Now given equation of curve as $3x^2 - y^2 = 8$
Differentiating the equation with respect to X
$\\\frac{\mathrm{d}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}$ \\$\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{d} \mathrm{x}}=0$ \\$\Rightarrow 3 \times 2 \mathrm{x}-2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$ \\$\Rightarrow 6 \mathrm{x}=2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6 \mathrm{x}}{2 \mathrm{y}}=\frac{3 \mathrm{x}}{\mathrm{y}}=\mathrm{m}_{1} \ldots \ldots(\mathrm{i})$$
Assume the slope of the normal curve to be m₂ is given by
$\Rightarrow \mathrm{m}_{2}=\frac{-1}{\mathrm{~m}_{1}}$$
Substituting value from equation (i), we get
$\\\Rightarrow \mathrm{m}_{2}=\frac{-1}{\frac{3 \mathrm{x}}{\mathrm{y}}} \\\Rightarrow \mathrm{m}_{2}=\frac{-\mathrm{y}}{3 \mathrm{x}}$
The known equation of the line is
$\\$x+3 y=4$\\ $\Rightarrow 3 y=4-x$$
$\\\Rightarrow \mathrm{y}=\frac{4}{3}-\frac{\mathrm{x}}{3}$\\ $\Rightarrow \mathrm{y}=-\frac{\mathrm{x}}{3}+\frac{4}{3}$$
$\therefore$ the slope of this line will be
$\mathrm{m}_{3}=-\frac{1}{3} \ldots \ldots$ (iii)$
since, slope of normal to the curve should be equal to the slope of the line which is parallel to the curve,
$\therefore \mathrm{m}_{2}=\mathrm{m}_{3}$$
Substituting values from equation (ii) and (iii), we get
$\\ \Rightarrow-\frac{y}{3 x}=-\frac{1}{3}$ \\$\Rightarrow 3 y=3 x$ \\$\Rightarrow y=x_{\cdots \cdots}(i v)$$
After putting y=x in the equation of the curve, we get
$\\3 x^{2}-y^{2}=8$ \\$\Rightarrow 3 x^{2}-x^{2}=8$ \\$\Rightarrow 2 x^{2}=8$ \\$\Rightarrow x^{2}=4$ \\$\Rightarrow x=\pm 2$$
But from equation (iv)
y=x
$\Rightarrow y=\pm 2$
Therefore, the points at which normal to the given curve is parallel to the given line are (2, 2) and (-2, -2)
Thus, the equations of the normal can be calculated by
$\begin{aligned} &y-2=m_{2}(x-2) \text { and } y+2=m_{2}(x+2)\\ &\Rightarrow \mathrm{y}-2=-\frac{1}{3}(\mathrm{x}-2) \quad \text{And }\mathrm{y}+2=-\frac{1}{3}(\mathrm{x}+2)\\ &3 y-6=-x+2 \text { and } 3 y+6=-x-2\\ &3 y+x=8 \text { and } 3 y+x=-8\\ &\text { Hence the equation of the normal lines to the curve } 3 x^{2}-y^{2}=8 \text { which are parallel to the line }\\ &x+3 y=8 \text { and } 3 y+y=- 8 \end{aligned}$

Question:18

At what points on the curve $x^2 + y^2 - 2x - 4y + 1 = 0$ , the tangents are parallel to the y-axis?

Answer:

Given: equation of a curve $x^2 + y^2 - 2x - 4y + 1 = 0$
To find: the points on the curve $x^2 + y^2 - 2x - 4y + 1 = 0$ , the tangents are parallel to the y-axis
Explanation: the given equation of curve as $x^2 + y^2 - 2x - 4y + 1 = 0$
Differentiating the equation with respect to X
$\begin{aligned} &\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}+1\right)}{\mathrm{dx}}=\frac{\mathrm{d}(0)}{\mathrm{dx}}\\ &\text { Now applying the sum rule of differentiation }\\ &\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(4 \mathrm{y})}{\mathrm{dx}}+\frac{\mathrm{d}(1)}{\mathrm{dx}}=0\\ &\Rightarrow 2 x+2 y \frac{d(y)}{d x}-2-4 \frac{d y}{d x}+0=0\\ &\Rightarrow(2 y-4) \frac{d(y)}{d x}=2-2 x\\ &\Rightarrow \frac{d y}{d x}=\frac{2(1-x)}{2(y-2)} \end{aligned}$
Sense the tangents are parallel to the axis as mentioned in the question
Thus,
$\begin{aligned} &\tan \theta=\tan 90^{\circ}=\frac{\mathrm{dy}}{\mathrm{dx}}\\ &\text { Value from equation (i) can be substituted equation }\\ &\frac{1}{0}=\frac{(1-x)}{(y-2)} \end{aligned}$
$\\ { \Rightarrow $ y-2 = 0}\\ {$ \Rightarrow $ y = 2}\\$
Putting y = 2 in curve equation, we get
$\\x\textsuperscript2 + y^2 - 2x - 4y + 1 = 0\\$
$\\ \Rightarrow x\textsuperscript2 + 2^2 - 2x - 4(2) + 1 = 0\\$
$\\\Rightarrow x\textsuperscript2 + 4 - 2x - 8 + 1 = 0\\ \Rightarrow x\textsuperscript2- 2x-3 = 0\\$
After the splitting of the middle term,
$\\ \Rightarrow x\textsuperscript{2}-3x+x-3 = 0\\ { \Rightarrow x(x-3)+1(x-3) = 0}\\ { \Rightarrow (x+1)(x-3) = 0}\\ { \Rightarrow x+1 = 0 or x-3 = 0}\\ { \Rightarrow x = -1 or x = 3}\\$
Thus, the needed points are(-1, 2) and (3, 2).
Hence the points on the curve $x^2 + y^2 - 2x - 4y + 1 = 0$ , the tangents are parallel to the y-axis are (-1, 2) and (3, 2).

Question:19

Show that the line $\frac{x}{a}+\frac{y}{b}=1$ touches the curve $\mathrm{y}=\mathrm{b.e}^{-\frac{x}{a}}$ at the point where the curve intersects the axis of y.

Answer:

Given: equation of line $\frac{x}{a}+\frac{y}{b}=1$ the curve $\mathrm{y}=\mathrm{b.e}^{-\frac{x}{a}}$ intersects the y-axis
To show: the line touches the curve at the point where the curve intersects the axis of y
Explanation: given the curve $\mathrm{y}=\mathrm{b.e}^{-\frac{x}{a}}$ intersects the y-axis, i.e., at x = 0
Now differentiate the given curve equation with respect to x, i.e.,
$\frac{d y}{d x}=\frac{d\left(b \cdot e^{-\frac{x}{a}}\right)}{d x}$ \\Removing all the constant term, \\$\Rightarrow \frac{d y}{d x}=b \frac{d\left(e^{-\frac{x}{a}}\right)}{d x}$ \\After differentiate in the equation we get \\$\Rightarrow \frac{d y}{d x}=b \cdot e^{-\frac{x}{a}} \frac{d\left(-\frac{x}{a}\right)}{d x}$ \\$\Rightarrow \frac{d y}{d x}=b \cdot e^{-\frac{x}{a}} \cdot\left(-\frac{1}{a}\right)$ \\Now put the value of $x=0$, \\$\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}=b \cdot e^{-\frac{0}{a}} \cdot\left(-\frac{1}{a}\right)$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}=$ b. $1 \cdot\left(-\frac{1}{a}\right)$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}=\left(-\frac{b}{a}\right)=m_{1} \ldots \ldots(i)$$
Then considering the line equation,
$\frac{x}{a}+\frac{y}{b}=1$ \\Now differentiate it with respect to X \\$\frac{\mathrm{d}\left(\frac{\mathrm{X}}{\mathrm{a}}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{b}}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(1)}{\mathrm{dx}}$ \\Removing all the constant terms \\$\Rightarrow \frac{1}{a}+\frac{1}{b} \frac{d y}{d x}=0$ \\$\Rightarrow \frac{1}{\mathrm{~b}} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{\mathrm{a}}$ \\$\Rightarrow \frac{d y}{d x}=-\frac{b}{a}=m_{2} \ldots .$$
Line touches the curve only if their slopes are equal
From equation (i) and (ii), we see that
$m_1 = m_2$
Hence, the line touches the curve at the point where the curve intersects the axis of y.

Question:20

Show that $f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^{2}}-x\right)$ is increasing in R.

Answer:

Given: $f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^{2}}-x\right)$
To show: the mentioned function increases in R
Explanation: Given $f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^{2}}-x\right)$
Substituting the first derivative in respect to x
$\\\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(2 \mathrm{x}+\cot ^{-1} \mathrm{x}+\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}$\\ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\cot ^{-1} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}$$
But the derivative of 2x is 2,so
$\mathrm{f}^{\prime}(\mathrm{x})=2+\frac{\mathrm{d}\left(\cot ^{-1} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}$$
But the derivative of $\cot ^{-1} \mathrm{x}=-\frac{1}{\mathrm{x}^{2}+1}$$ ,so
$\\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{\mathrm{d}\left(\log \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)}{\mathrm{dx}}$\\ $f^{\prime}(x)=2+\left(-\frac{1}{x^{2}+1}\right)+\frac{1}{\sqrt{1+x^{2}}-x} \cdot \frac{d\left(\sqrt{1+x^{2}}-x\right)}{d x}$$
When the sum rule is applied to the last part we get,
$\\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{d}\left(\sqrt{1+\mathrm{x}^{2}}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{d}\left(\left(1+\mathrm{x}^{2}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}-1\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{1}{2} \cdot\left(1+\mathrm{x}^{2}\right)^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}\left(1+\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}-1\right) \\$
$\\\mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{1}{2} \cdot\left(1+\mathrm{x}^{2}\right)^{-\frac{1}{2}} \cdot\left[\frac{\mathrm{d}(1)}{\mathrm{d} \mathrm{x}}+\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}\right]-1\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{1}{2 \sqrt{1+\mathrm{x}^{2}}} \cdot[0+2 \mathrm{x}]-1\right) \\ \mathrm{f}^{\prime}(\mathrm{x})=2+\left(-\frac{1}{\mathrm{x}^{2}+1}\right)+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{2 \mathrm{x}}{2 \sqrt{1+\mathrm{x}^{2}}}-1\right)$
$\\ \mathrm{f}^{\prime}(\mathrm{x})=2-\frac{1}{\mathrm{x}^{2}+1}+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}-1\right)$ $\\\\\mathrm{f}^{\prime}(\mathrm{x})=2-\frac{1}{\mathrm{x}^{2}+1}+\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}}\left(\frac{\mathrm{x}-\sqrt{1+\mathrm{x}^{2}}}{\sqrt{1+\mathrm{x}^{2}}}\right)$ \\Cutting out all the like terms, $\mathrm{f}^{\prime}(\mathrm{x})=2-\frac{1}{\mathrm{x}^{2}+1}+\frac{-1}{\sqrt{1+\mathrm{x}^{2}}}$ \\Then, adding by taking the LCM, we get \\$\mathrm{f}^{\prime}(\mathrm{x})=\frac{2+2 \mathrm{x}^{2}-1-\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}^{2}+1}$\\ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}^{2}+1-\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}^{2}+1}$$
Then to calculate any real value x, the above value of f(x) is larger than or equal to zero
Hence $\frac{2 x^{2}+1-\sqrt{1+x^{2}}}{x^{2}+1} \forall x \in R$
And we know, if $f'(x)\geq 0$ , then f(x) is increasing function.
Hence, the given function is an increasing function in R.

Question:21

Show that for $a \geq 1, f (x) = \sqrt 3 \sin x - \cos x - 2ax + b$ is decreasing in R.

Answer:

Given: $a \geq 1, f (x) = \sqrt 3 \sin x - \cos x - 2ax + b$
To show: the above function is decreasing in R.
Explanation: Given $f (x) = \sqrt 3 \sin x - cos x - 2ax + b$
The first derivative is applied with respect to x,
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sqrt{3} \sin \mathrm{x}-\cos \mathrm{x}-2 \mathrm{ax}+\mathrm{b})}{\mathrm{dx}}$$
By using the sum rule of differentiation, we get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sqrt{3} \sin \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(2 \mathrm{ax})}{\mathrm{dx}}+0$$
Removing all the constant terms, we get
$\mathrm{f}^{\prime}(\mathrm{x})=\sqrt{3} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}-2 \mathrm{a} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}$$
But the derivative of sin X = cos x and that of cos x = -sin x, so
$\\ {f' (x) = \sqrt 3cos x-(-sin x)-2a}\\ {f' (x) = \sqrt 3\cos x+ \sin x-2a}\\$
Multiplying and dividing RHS by 2,
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=2\left[\frac{\sqrt{3}}{2} \cos \mathrm{x}+\frac{1}{2} \sin \mathrm{x}\right]-2 \mathrm{a}\\ &\sin \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \text { and } \sin \frac{\pi}{6}=\frac{1}{2} \text { , putting these values in the above equation, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=2\left[\cos \frac{\pi}{6} \cos \mathrm{x}+\sin \frac{\pi}{6} \cdot \sin \mathrm{x}\right]-2 \mathrm{a}\\ &\text { But } \cos (A-B)=\cos A \cos B+\sin A \cdot \sin B \text { , placing the values in the above given equation, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=2\left[\cos \left(\frac{\pi}{6}-\mathrm{x}\right)\right]-2 \mathrm{a}\\ &\text { Now, we know cos } \mathrm{x} \text { always belong to }[-1,1] \text { for } \mathrm{a} \geq 1\\ &2\left[\cos \left(\frac{\pi}{6}-x\right)\right]-2 a \leq 0 \end{aligned}$
And we know, if $f'(x)\leq0$ , then f(x) is decreasing function.
Therefore, the given function is decreasing function in R.

Question:22

Show that $f (x) = tan^{-1}(\sin x + \cos x)$ is an increasing function in $\left(0,\frac{\pi}{4}\right)$

Answer:

Given: $f (x) = tan^{-1}(\sin x + \cos x)$
To show: the given function is increasing in $\left(0,\frac{\pi}{4}\right)$
Explanation: Given
$f (x) = tan^{-1}(\sin x + \cos x)$
First derivative is applied with respect to x,
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\tan ^{-1}(\sin \mathrm{x}+\cos \mathrm{x})\right)}{\mathrm{dx}}$$
Using the differentiation rule for $tan $^{-1}$$ , results into
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{(\sin \mathrm{x}+\cos \mathrm{x})^{2}+1} \cdot \frac{\mathrm{d}(\sin \mathrm{x}+\cos \mathrm{x})}{\mathrm{dx}}$$
Now use the sum rule of differentiation,
$\begin{aligned} &f^{\prime}(x)=\frac{1}{(\sin x+\cos x)^{2}+1}\left[\frac{d(\sin x)}{d x}+\frac{d(\cos x)}{d x}\right]\\ &\text { But the derivative of } \sin \mathrm{X}=\cos \mathrm{x} \text { and that } \text { of } \cos \mathrm{x}=-\sin \mathrm{x}, \mathrm{s} \circ\\ &\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{(\sin \mathrm{x}+\cos \mathrm{x})^{2}+1}[\cos \mathrm{x}+(-\sin \mathrm{x})]\\ &\text { Expanding }(\sin x+\cos x)^{2}, \text { we get }\\ &f^{\prime}(x)=\frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x+1}\\ &\text { But } \sin ^{2} x+\cos ^{2} x=1 \text { and } 2 \sin x \cos x=\sin 2 x, \text { thus the equation given above gets converted }\\ &\text { into, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos \mathrm{x}-\sin \mathrm{x}}{1+\sin 2 \mathrm{x}+1}\\ &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos \mathrm{x}-\sin \mathrm{x}}{\sin 2 \mathrm{x}+2} \end{aligned}$
To make f(x) to be increasing function,
$\\f^{\prime}(x) \geq 0$ \\$\Rightarrow \frac{\cos x-\sin x}{\sin 2 x+2} \geq 0$ \\$\Rightarrow \cos x-\sin x \geq 0$ \\$\Rightarrow \cos x \geq \sin x$$
But this is possible only when $\mathrm{x} \in\left(0, \frac{\pi}{4}\right)$$
Hence, the given function is increasing function in $\left(0, \frac{\pi}{4}\right)$$

Question:23

At what point, the slope of the curve $y = -x^3+3x^2+9x-27$ is maximum? Also find the maximum slope.

Answer:

Given: $y = -x^3+3x^2+9x-27$
To find: the point in curve where the slope is maximum and the maximum value of the slope.
Explanation: given $y = -x^3+3x^2+9x-27$
The slope of the curve can be found by calculating the first derivative of the known curve equation,
Thus, slope of the curve is
$\frac{d y}{d x}=\frac{d\left(-x^{3}\right)}{d x}+\frac{d\left(3 x^{2}\right)}{d x}+\frac{d(9 x)}{d x}-\frac{d(27)}{d x}$$
Then using the derivative,
$\\ \frac{d y}{d x}=-3 x^{2}+3.2 x+9-0$ $\\\frac{d y}{d x}=-3 x^{2}+6 x+9$$
To know the critical point, it is necessary to have the value of the slope,
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(-3 \mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(6 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(9)}{\mathrm{dx}}$$
Using the derivative leads to,
$\\ \frac{d^{2} y}{d x^{2}}=-3 \cdot 2 x+6+0$\\ $\frac{d^{2} y}{d x^{2}}=-6 x+6$$
When the second derivative is equated to 0 it gives the critical point, i.e.,
$\\ \frac{d^{2} y}{d x^{2}}=0$\\ $\Rightarrow-6 x+6=0$\\ $\Rightarrow 6 x=6$\\ $\Rightarrow x=1 \ldots \ldots .0(\mathrm{i})$$
Then finding the third derivative of the curve,
i.e.,
$\frac{d^{3} y}{d x^{3}}=\frac{d(-6 x)}{d x}+\frac{d(6)}{d x}$$
Putting the values of the derivative results in
$\frac{d^{3} y}{d x^{3}}=-6<0$$
As the third derivative is less than 0, so the maximum slope of the given curve is at x=1.
Equating the first derivative with x=1 leads to the maximum value of the slope, i.e.,
$\\ \left(\frac{d y}{d x}\right)_{x=1}=-3 x^{2}+6 x+9$ \\$\left(\frac{d y}{d x}\right)_{x=1}=-3(1)^{2}+6(1)+9$ \\$\left(\frac{d y}{d x}\right)_{x=1}=-3+6+9$ \\$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=1}=12$$
Therefore, the slope of the curve is maximum at x=1, and the maximum value of the slope is 12 .

Question:24

Prove that $f(x)=\sin x+\sqrt{3} \cos x$ has maximum value at $x=\frac{\pi}{6}$ .

Answer:

Given: $f(x)=\sin x+\sqrt{3} \cos x$
To prove: the given function has maximum value at $x=\frac{\pi}{6}$
Explanation: given $f(x)=\sin x+\sqrt{3} \cos x$
While calculating the first derivative we get,
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(\sqrt{3} \cos \mathrm{x})}{\mathrm{d} \mathrm{x}}$
Then putting the derivative, we get
$f'(x) = cos x- \sqrt 3sin x$
Critical point ca be calculated by equating the derivative with 0,
$f'(x) = 0$
$\Rightarrow cos x- \sqrt 3sin x =0 \Rightarrow \sqrt 3sin x = cos x$
$\\ \Rightarrow \frac{\sin x}{\cos x}=\frac{1}{\sqrt{3}}$ \\$\Rightarrow \tan x=\frac{1}{\sqrt{3}}$ \\$x=\frac{\pi}{6}$$
This is possible only when
The second derivative if the c=function can be calculated by,
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}-\sqrt{3} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}$$
Putting the value of derivative, we get
$\\ f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x$$
Then, we will substitute $\\x=\frac{\pi}{6}$$ in the above equation, we get
$\\ f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x$ \\$f^{\prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6}$$
Putting the corresponding value, we get
$\\ \mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-\frac{1}{2}-\sqrt{3} \times \frac{\sqrt{3}}{2} \\ \mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-\frac{1}{2}-\frac{3}{2} \\ \mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-2<0$
Hence f(x) has a maximum value at $x=\frac{\pi}{6}$ .
Hence proved.

Question:25

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{\pi}{3}$ .

Answer:

Given: a right-angled triangle with the sum of the lengths of its hypotenuse and side.
To show: at this angle $\frac{\pi}{3}$ the area of the triangle is maximum
Explanation:

Let ΔABC be the right-angled triangle,
Let hypotenuse, AC = y,
side, BC = x, AB = h
then the calculation of sum of the side and hypotenuse is done using,
⇒ x+y = k, where k is any constant value
⇒ y = k-x………..(i)
Take A as the area of the triangle, as we know
$\mathrm{A}=\frac{1}{2} \mathrm{~h} \cdot \mathrm{x} \ldots$(ii)$
Then using the Pythagoras theorem, we get
$y^{2}=x^{2}+h^{2}$$
$\Rightarrow \mathrm{h}=\sqrt{\mathrm{y}^{2}-\mathrm{x}^{2}}$$
Putting the value from equation (i) in above equation, we get
$\Rightarrow \mathrm{h}=\sqrt{(\mathrm{k}-\mathrm{x})^{2}-\mathrm{x}^{2}}$$
$\\\Rightarrow \mathrm{h}=\sqrt{\mathrm{k}^{2}+\mathrm{x}^{2}-2 \mathrm{kx}-\mathrm{x}^{2}}$\\ $\Rightarrow \mathrm{h}=\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}} \ldots \ldots \ldots$(iii)$
Applying the values from equation (iii) into equation (ii), we get
$A=\frac{1}{2}\left(\sqrt{k^{2}-2 k x}\right) \cdot x$$
The above equation is differentiated with respect to x,
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\frac{1}{2}\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right) \cdot \mathrm{x}\right)}{\mathrm{dx}}$$
Then the constant terms are taken out,
$\frac{d A}{d x}=\frac{1}{2} \frac{d\left(\left(\sqrt{k^{2}-2 k x}\right) \cdot x\right)}{d x}$$
$\begin{aligned} &\text { By using the product rule, }\\ &\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right) \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+(\mathrm{x}) \frac{\mathrm{d}\left(\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)\right)}{\mathrm{dx}}\right]\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)(1)+(\mathrm{x}) \frac{\mathrm{d}\left(\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}\right] \end{aligned}$
Power rule pf differentiation is applied on the second part of the above equation,
$\\ \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)+(\mathrm{x})\left[\frac{1}{2} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)}{\mathrm{dx}}\right]\right] \\ \Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)+\frac{1}{2}(\mathrm{x})\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\left[\frac{\mathrm{d}\left(\mathrm{k}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}(2 \mathrm{kx})}{\mathrm{d} \mathrm{x}}\right]\right] \\ \Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)+\frac{(\mathrm{x})}{2 \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}[0-2 \mathrm{k}]\right] \\$
$\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{2 \mathrm{k}(\mathrm{x})}{2 \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]$
$\begin{aligned} &\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right] \ldots . \text { (iv) }\\ &\text { When the first derivative is equated with } 0, \text { it gives critical point, }\\ &\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=0 \end{aligned}$
$\\ \Rightarrow \frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]=0 \\ \Rightarrow\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}=0 \\ \Rightarrow\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)=\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}} \\ \Rightarrow\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)=\mathrm{kx} \\ \Rightarrow \mathrm{k}^{2}-2 \mathrm{kx}=\mathrm{kx} \\ \Rightarrow \mathrm{k}^{2}=2 \mathrm{kx}+\mathrm{kx} \\ \Rightarrow \mathrm{k}^{2}=3 \mathrm{kx} \\ \Rightarrow \mathrm{k}=3 \mathrm{x} \\ \Rightarrow \mathrm{x}=\frac{\mathrm{k}}{3}$
Once again, differentiating equation (iv) with respect to x, we get
$\begin{aligned} &\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]\right)}{\mathrm{dx}}\\ &\text { Taking out the constant term and taking the LCM, we get }\\ &\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} \frac{\mathrm{d}\left(\left[\frac{\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)-\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]\right)}{\mathrm{dx}}\\ &\Rightarrow \frac{d^{2} A}{d x^{2}}=\frac{1}{2} \frac{d\left(\left[\frac{\left(k^{2}-3 k x\right)}{\sqrt{k^{2}-2 k x}}\right]\right)}{d x}\\ &\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} \frac{\mathrm{d}\left(\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\right)}{\mathrm{dx}} \end{aligned}$
Using the product rule of differentiation,
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2}\left[\left(\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}} \frac{\mathrm{d}\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)}{\mathrm{dx}}\right)-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right) \frac{\mathrm{d}\left(\left[\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\right]\right)}{\mathrm{dx}}\right]$$
Then using the power rule of differentiation,
$\begin{aligned} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} &\left[\left(\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\left(\frac{\mathrm{d}\left(\mathrm{k}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}(3 \mathrm{kx})}{\mathrm{d} \mathrm{x}}\right)\right)\right.\\ &\left.-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(-\frac{1}{2} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right) \frac{-1}{2}-1 \cdot \frac{\mathrm{d}\left(\left[\mathrm{k}^{2}-2 \mathrm{kx}\right]\right)}{\mathrm{dx}}\right)\right] \\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} &\left[\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}(-3 \mathrm{k})\right) \\ &\left.-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(-\frac{1}{2} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{-1} \cdot(-2 \mathrm{k})\right)\right] \end{aligned}$
$\\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(\frac{\mathrm{k}}{\left(\mathrm{k}^{2}-2 \mathrm{kx}\right) \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right)\right] \\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}-\frac{\mathrm{k}\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)}{\left(\mathrm{k}^{2}-2 \mathrm{kx}\right) \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]$
Putting $\mathrm{x}=\frac{\mathrm{k}}{3}$$ , in above equation, we get
$\\\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)}}-\frac{\mathrm{k}\left(\mathrm{k}^{2}-3 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)\right)}{\left(\mathrm{k}^{2}-2 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)\right) \sqrt{\mathrm{k}^{2}-2 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)}}\right]$ \\$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}\left(1-\frac{2}{3}\right)}}-\frac{\mathrm{k}\left(\mathrm{k}^{2}-\mathrm{k}^{2}\right)}{\left(\mathrm{k}^{2}\left(1-\frac{2}{3}\right)\right) \sqrt{\mathrm{k}^{2}\left(1-\frac{2}{3}\right)}}\right]$$
$\\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\mathrm{k} \sqrt{\left(\frac{1}{3}\right)}}-0\right] \\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3}{\left.\sqrt{\left(\frac{1}{3}\right)}\right]}\right. \\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}[-3 \sqrt{3}] \\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{-3 \sqrt{3}}{2}<0$
Hence the maximum value of A is at $X=\frac{k}{3}$
We know,
$\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}$$
Then from figure,
$\Rightarrow \cos \theta=\frac{x}{y}$$
Applying the value of y=k -x from equation (i), we get
$\Rightarrow \cos \theta=\frac{x}{k-x}$$
Putting the value of $x=\frac{k}{3},$$ we get
$\\\Rightarrow \cos \theta=\frac{\frac{\mathrm{k}}{3}}{\mathrm{k}-\frac{\mathrm{k}}{3}}$ \\$\Rightarrow \cos \theta=\frac{1}{2}$$
This possibility is present when $\theta=\frac{\pi}{3}$$
Therefore, the area of the triangle is maximum only when the angle between them is $\frac{\pi}{3}$$

Question:26

Find the points of local maxima, local minima and the points of inflection of the function $f (x) = x^5 - 5x^4 + 5x^3 -1$ . Also find the corresponding local maximum and local minimum values.

Answer:

Given: function $f (x) = x^5 - 5x^4 + 5x^3 -1$
To find: the points of local maxima, local minima and the points of inflection of f(x) and also to find the corresponding local maximum and local minimum values.
Explanation: given $f (x) = x^5 - 5x^4 + 5x^3 -1$
Calculating the first derivative of f(x), i.e.,
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\right)}{\mathrm{d} \mathrm{x}}$ \\When the sum rule of differentiation is applied followed by taking out all the constant term, we get \\$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{x}^{5}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(5 \mathrm{x}^{4}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(5 \mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(1)}{\mathrm{dx}}$$
$f' (x) = 5x\textsuperscript{4}-5.4x\textsuperscript{3}+5.3x\textsuperscript{2}-0$
$\Rightarrow f' (x) = 5x\textsuperscript{4}-20x\textsuperscript{3}+15x\textsuperscript{2}$
Equating the first derivative with 0 to find out the critical point,
$\\ {f'(x) = 0}\\ { \Rightarrow 5x\textsuperscript{4}-20x\textsuperscript{3}+15x\textsuperscript{2} = 0}\\ { \Rightarrow 5x\textsuperscript{2}(x\textsuperscript{2}-4x+3) = 0}\\ { \Rightarrow 5x\textsuperscript{2}(x\textsuperscript{2}-4x+3) = 0}\\$
Then splitting the middle term, we get $\Rightarrow 5x\textsuperscript{2}(x^2-3x-x+3) = 0$
$\\{ \Rightarrow 5x\textsuperscript{2}[ x(x-3)-1(x-3)] = 0}\\ { \Rightarrow 5x\textsuperscript{2}(x-1)(x-3) = 0}\\ \Rightarrow 5x\textsuperscript{2} = 0 \ or\ (x-1) = 0 \ or\ (x-3) = 0\\ { \Rightarrow x = 0 \ or\ x = 1 \ or\ x = 3\\$
Now we will find the corresponding y value by putting the numerous values of x in given function
$f (x) = x^5 - 5x^4 + 5x^3 -1$
$$When x = 0, f(0) = 0\textsuperscript{5} - 5(0)\textsuperscript{4} + 5(0)\textsuperscript{3} - 1 = -1$
Hence the point is (0,-1)
$$When x = 1, f(1) = 1\textsuperscript{5} - 5(1)\textsuperscript{4} + 5(1)\textsuperscript{3} - 1 = 1-5+5-1 = 0$
Hence the point is (1,0)
$$When x = 3, f(3) = 3\textsuperscript{5} - 5(3)\textsuperscript{4} + 5(3)\textsuperscript{3} - 1 = 243-405+135-1 = -28\\$
Hence the point is (3,-28)
Therefore, we see that
At x = 3, y has minimum value = -28. Hence x = 3 is point of local minima.
At x = 1, y has maximum value = 0. Hence x = 1 is point of local maxima.
And at x = 0, y has neither maximum nor minimum value, hence this is point of inflection.

Question:27

A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

Answer:

Given: a telephone company in a town has 500 subscribers and collects fixed charges of Rs 300/- per subscriber per year, company increase the annual subscription and for every increase of Re 1/- one subscriber will discontinue the service
To find: the best increase amount for the company to earn maximum profit
Explanation: company has 500 subscribers, and collects 300 per subscriber per year.
Let x as the increase in annual subscription by the company
As per the question, the number of subscribers to discontinue the service will be x
The total revenue earned after the increment would be calculated by,
$\\{R(x) = (500-x)(300+x)}\\ { \Rightarrow R(x) = 500(300+x)-x(300+x)}\\ { \Rightarrow R(x) = 150000+500x-300x-x\textsuperscript{2}}\\ { \Rightarrow R(x) = 150000+200x-x\textsuperscript{2}}\\$
We need to calculate the first derivative of the above equation,
$\\ \mathrm{R}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(150000+200 \mathrm{x}-\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}} \\ \mathrm{R}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(150000)}{\mathrm{dx}}+\frac{\mathrm{d}(200 \mathrm{x})}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}$
$R'(x) = 0+200-2x \ldots ..(i)$
The critical point is calculated by equating the first derivative with 0,
$\\{R'(x) = 0}\\ { \Rightarrow 200-2x = 0}\\ { \Rightarrow 2x = 200}\\ { \Rightarrow x = 100 \ldots \ldots (ii)}\\$
Then we calculate the second derivative of the total revenue function, i.e., again differentiate equation (i), i.e.,
$\\ \mathrm{R}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(200-2 \mathrm{x})}{\mathrm{dx}} \\ \mathrm{R}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(200)}{\mathrm{dx}}-\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{d} \mathrm{x}} \\ \mathrm{R}^{\prime \prime}(\mathrm{x})=0-2<0$
Hence R’’(100) is also less than 0,
Therefore, R(x) is maximum at x = 0, i.e.,
Thus, the required increase on the subscription fee for the company to make profit is by Rs 100.

Question:28

If the straight-line $x \cos\alpha + y \sin\alpha = p$ touches the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ then prove that $a^2 \cos2\alpha + b^2 \sin2\alpha = p^2.$

Answer:

Given: equation of straight-line $x \cos\alpha + y \sin\alpha = p$ , equation of curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the straight line touches the curve
To prove: $a^2 cos2\alpha + b^2 sin2\alpha = p^2.$
Explanation: the know line equation is,
$x \cos\alpha + y \sin\alpha = p$
$\Rightarrow y sin \alpha = p-x cos \alpha$
$\\\Rightarrow y=\frac{p-x \cos \alpha}{\sin \alpha}$\\\\ $\Rightarrow y=\frac{p}{\sin \alpha}-\frac{x \cos \alpha}{\sin \alpha}$ \\\\$\Rightarrow \mathrm{y}=\frac{\mathrm{p}}{\sin \alpha}-\mathrm{x}\left(\frac{\cos \alpha}{\sin \alpha}\right)$ \\\\$\Rightarrow \mathrm{y}=-\mathrm{x}\left(\frac{\cos \alpha}{\sin \alpha}\right)+\frac{\mathrm{p}}{\sin \alpha}$ \\\\Comparing this with the equation $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ the slope and intercept of the given line can be seen as \\\\$\mathrm{m}=-\frac{\cos \alpha}{\sin \alpha}, \mathrm{C}=\frac{\mathrm{p}}{\sin \alpha}$$
We know that, if a line y = mx+c touches the eclipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , then required condition is
$c\textsuperscript{2} = a\textsuperscript{2}m\textsuperscript{2}+b\textsuperscript{2}$
Then putting the corresponding values, we get
$\\\left(\frac{p}{\sin \alpha}\right)^{2}=a^{2}\left(-\frac{\cos \alpha}{\sin \alpha}\right)^{2}+b^{2}\\\\$ $\frac{p^{2}}{\sin ^{2} \alpha}=\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}\left(a^{2}\right)+b^{2}\\\\$ $\frac{\mathrm{p}^{2}}{\sin ^{2} \alpha}=\frac{\mathrm{a}^{2} \cos ^{2} \alpha+\mathrm{b}^{2} \sin ^{2} \alpha}{\sin ^{2} \alpha}$$
Removing the like terms we get,
$\mathrm{p}^{2}=\mathrm{b}^{2} \sin ^{2} \alpha+\mathrm{a}^{2} \cos ^{2} \alpha$
Hence, proved.

Question:29

An open box with square base is to be made of a given quantity of card board of area c2. Show that the maximum volume of the box is $\frac{c^3}{6\sqrt3}$ cubic units.

Answer:

Given: a cardboard box that is open and square in shape has $c^2$ area
To show: $\frac{c^3}{6\sqrt3}$ cubic units is the maximum volume of the box.
Explanation:

Take the side of the square be x cm and
Take the height the box be y cm.
So, the total area of the cardboard used is
A = area of square base + 4x area of rectangle
$\Rightarrow A = x\textsuperscript{2}+4xy$
But it is given this is equal to $c\textsuperscript{2}$ , hence
$\\{c\textsuperscript{2} = x\textsuperscript{2}+4xy}\\ { \Rightarrow 4xy = c\textsuperscript{2}-x\textsuperscript{2}}\\$
$\Rightarrow \mathrm{y}=\frac{\mathrm{c}^{2}-\mathrm{x}^{2}}{4 \mathrm{x}} \ldots \ldots(i)$
According to the given condition the area of the square base will be
V = base × height
Since the base is square, the volume is
$V = x\textsuperscript{2}y \ldots \ldots (ii)$
Then putting the values of equation (i) in equation (ii), we get
$\\ \mathrm{V}=\mathrm{x}^{2}\left(\frac{\mathrm{c}^{2}-\mathrm{x}^{2}}{4 \mathrm{x}}\right) \\ \mathrm{V}=\frac{\mathrm{x}}{4}\left(\mathrm{c}^{2}-\mathrm{x}^{2}\right) \\ \mathrm{V}=\frac{1}{4}\left(\mathrm{xc}^{2}-\mathrm{x}^{3}\right)$
Calculation of the first derivative of the equation,
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\frac{1}{4}\left(\mathrm{xc}^{2}-\mathrm{x}^{3}\right)\right)}{\mathrm{dx}}$$
Removing all the constant terms
$\mathrm{V}^{\prime}=\frac{1}{4} \frac{\mathrm{d}\left(\mathrm{xc}^{2}-\mathrm{x}^{3}\right)}{\mathrm{dx}}$
Using the sum rule of differentiation, we get
$\mathrm{V}^{\prime}=\frac{1}{4}\left[\frac{\mathrm{d}\left(\mathrm{xc}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}\right]$$
Removing all the constant terms, we get
$\mathrm{V}^{\prime}=\frac{1}{4}\left[\mathrm{c}^{2} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{d} \mathrm{x}}\right]$$
After differentiating the equation, we get
$\mathrm{V}^{\prime}=\frac{1}{4}\left[\mathrm{c}^{2}-3 \mathrm{x}^{2}\right] \ldots \ldots$ (iii)$
We need to calculate the second derivative to find out the maximum value of x , so for that let $$\mathrm{V}^{\prime}=0,$ so$ equating above equation with 0, we get
$\frac{1}{4}\left[c^{2}-3 x^{2}\right]=0$$
$\\ \Rightarrow c^{2}-3 x^{2}=0$ \\\\$\Rightarrow 3 x^{2}=c^{2}$ \\\\$\Rightarrow \mathrm{x}^{2}=\frac{\mathrm{c}^{2}}{3}$ \\\\$\Rightarrow \mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}$$
Differentiating equation (iii) again with respect to x, we get
$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(\frac{1}{4}\left[\mathrm{c}^{2}-3 \mathrm{x}^{2}\right]\right)}{\mathrm{dx}}$$
Removing all the constant terms results into
$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{1}{4} \frac{\mathrm{d}\left(\mathrm{c}^{2}-3 \mathrm{x}^{2}\right)}{\mathrm{dx}}$$
Using the differentiation rule of sum, we get
$\\\Rightarrow \mathrm{V}^{\prime \prime}=\frac{1}{4}\left[\frac{\mathrm{d}\left(\mathrm{c}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}\right]$ \\$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{1}{4}[0-3(2 \mathrm{x})]$ \\$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{-6 \mathrm{x}}{4}$ \\$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{-3 \mathrm{x}}{2}$$
$x=\frac{c}{\sqrt{3}}$ , the above equation becomes,
$\\\Rightarrow\left(\mathrm{V}^{\prime \prime}\right)_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{-3\left(\frac{\mathrm{c}}{\sqrt{3}}\right)}{2}$ \\\\$\Rightarrow\left(\mathrm{V}^{\prime \prime}\right)_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{-3 \mathrm{c}}{2 \sqrt{3}}<0$$
Thus, the volume (V) is maximum at $x=\frac{c}{\sqrt{3}}$
∴ The box has a maximum value of
$\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\left(\frac{\mathrm{c}}{\sqrt{3}}\right) \mathrm{c}^{2}-\left(\frac{\mathrm{c}}{\sqrt{3}}\right)^{3}\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\left(\frac{\mathrm{c}^{3}}{\sqrt{3}}\right)-\frac{\mathrm{c}^{3}}{3 \sqrt{3}}\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\frac{\mathrm{c}^{3}}{\sqrt{3}}\left(1-\frac{1}{3}\right)\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\frac{\mathrm{c}^{3}}{\sqrt{3}}\left(\frac{2}{3}\right)\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{\mathrm{c}^{3}}{6 \sqrt{3}} \mathrm{units}$
So, the box has a maximum value of is $\frac{\mathrm{c}^{3}}{6 \sqrt{3}}$ cubic units.
Hence, proved.

Question:30

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also, find the maximum volume.

Answer:

Given: rectangle of perimeter 36cm
To find: to estimate the dimensions of a rectangle in a way that it can sweep out maximum amount of volume when resolved to about one of its sides. Also, to find the maximum volume
Explanation: x and y can be the length and the breadth of the rectangle
The known perimeter of the rectangle is 36cm
$\\ { \Rightarrow 2x+2y = 36}\\ { \Rightarrow x+y = 18}\\ { \Rightarrow y = 18-x \ldots \ldots \ldots (i)}\\$
Now when the rectangle revolve about side y it will form a cylinder with y as the height and x as the radius, then if the volume of the cylinder is V, then we know
$V = \pi x\textsuperscript{2}y$
Applying the value from equation (i) in above equation we get
$V = \pi x\textsuperscript{2}(18-x)$
$\Rightarrow V = \pi (18x\textsuperscript{2}-x\textsuperscript{3})$
Then find out the first derivative of the given equation,
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\pi\left(18 \mathrm{x}^{2}-\mathrm{x}^{3}\right)\right)}{\mathrm{dx}}$$
Taking out the constant terms from equation followed by using the sum rule of differentiation,
$\\ \mathrm{V}^{\prime}=\pi\left[\frac{\mathrm{d}\left(18 \mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{d} \mathrm{x}}\right]$ \\$\mathrm{v}^{\prime}=\pi\left[18(2 \mathrm{x})-3 \mathrm{x}^{2}\right]$ \\$\mathrm{V}^{\prime}=\pi\left[36 \mathrm{x}-3 \mathrm{x}^{2}\right] \ldots \ldots$ (ii)$
To calculate critical point, we will equate the first derivative to 0, i.e.,
$\\ {V' = 0}\\ { \Rightarrow \pi (36x-3x\textsuperscript{2}) = 0}\\ { \Rightarrow 36x-3x\textsuperscript{2} = 0}\\ { \Rightarrow 36x = 3x\textsuperscript{2}}\\ { \Rightarrow 3x = 36}\\ { \Rightarrow x = 12 \ldots \ldots ..(iii)}\\$
By differentiating the second equation, second derivative of the volume equations can be easily calculated,
$\mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(\pi\left(36 \mathrm{x}-3 \mathrm{x}^{2}\right)\right)}{\mathrm{dx}}$$
Taking out the constant terms from equation followed by using the sum rule of differentiation,
$\\\mathrm{V}^{\prime \prime}=\pi\left[\frac{\mathrm{d}(36 \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}\right]$ \\$\mathrm{V}^{\prime \prime}=\pi[36-3(2 \mathrm{x})]$ \\$V^{\prime \prime}=\pi[36-6 x]$$
Now substituting x = 12 (from equation (iii)), we get
$\\ V''\textsubscript{x} = 12 = \pi [36-6(12)]\\ V''\textsubscript{x} = 12 = \pi [36-72]\\ V''\textsubscript{x} = 12 = -36 \pi <0\\$
Hence at x = 12, V will have maximum value.
The maximum value of V can be found by substituting x = 12 in
$V = \pi (18x\textsuperscript{2}-x^3),$ i. e $V _x = 12 = \pi (18(12)^2-(12)^3)$
$\\ V\textsubscript{x} = 12 = \pi (18(144)-(1728))\\$
$V\textsubscript{x} = 12 = \pi (2592-(1728))\\$
$V\textsubscript{x} = 12 = 864 \pi cm \textsuperscript{3}$
Therefore, the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides equal to 12cm.
And the maximum volume is $864 \pi cm \textsuperscript{3}$ .

Question:31

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum?

Answer:

Given: The combined surface area of a cube and sphere are constant
To find: the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum
Explanation: Let ‘a’ be the side of the cube
Then surface area of the cube = $6a^2$ ….(i)
Take ‘r’ as the radius of the sphere
Then the surface area of the sphere = $4\pi r^2$ …(ii)
According to the question, the surface area of both the figures is added, thus adding the equation (i) and (ii), we get
$\\ {6a\textsuperscript{2}+4 \pi r\textsuperscript{2} = k ($where k is the constant)}\\ { \Rightarrow 6a\textsuperscript{2} = k-4 \pi r\textsuperscript{2}}\\$
$\\\Rightarrow \mathrm{a}^{2}=\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}$ \\$\Rightarrow \mathrm{a}=\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}} \ldots$ (iii)$
As the formula of volume of cube is
$\mathrm{v}_{c}=\mathrm{a}^{3}$
Plus the volume of a sphere is
$\mathrm{V}_{\mathrm{s}}=\frac{4}{3} \pi \mathrm{r}^{3}$$
Hence adding both the volumes will result into,
$\mathrm{V}=\mathrm{a}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}$$
Then putting the values from equation (iii) in above equation,
$$$ \begin{aligned} \mathrm{V} &=\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)^{3}+\frac{4}{3} \pi \mathrm{r}^{3} \\ \mathrm{~V} &=\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}+\frac{4}{3} \pi \mathrm{r}^{3} \end{aligned}$
After finding the first derivative of the volume, we get
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}+\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}}$
After taking out the constant terms along with using the sum rule of differentiating,
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}\right)}{\mathrm{dr}}+\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dr}}$
Using the power rule of differentiation,
$\begin{aligned} \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}-1} \cdot \frac{\mathrm{d}\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{\mathrm{dr}}+\frac{4}{3} \pi\left(3 \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[\frac{1}{6} \cdot \frac{\mathrm{d}\left(\mathrm{k}-4 \pi \mathrm{r}^{2}\right)}{\mathrm{dr}}\right]+\left(4 \pi \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[\frac{1}{6} \cdot(-4 \pi(2 \mathrm{r}))\right]+\left(4 \pi \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[-\frac{4}{3} \pi r\right]+\left(4 \pi r^{2}\right) \\ \end{aligned}$
$\mathrm{V}^{\prime} =-2 \pi \mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}+\left(4 \pi \mathrm{r}^{2}\right)$
$\begin{aligned} &\mathrm{V}^{\prime}=-2 \pi \mathrm{r}\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right] \ldots . \text { (iv) }\\ &\text { Critical point can be calculated by equating the first derivative with } 0 \text { , }\\ &V^{\prime}=0\\ &\Rightarrow-2 \pi r\left[\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]=0\\ &\Rightarrow-2 \pi r=0 \text { or }\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]=0\\ &\Rightarrow \mathrm{r}=0 \text { or }\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}=2 \mathrm{r}\\ &\Rightarrow \mathrm{r}=0 \text { or } \frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}=4 \mathrm{r}^{2} \end{aligned}$
$\\ { \Rightarrow r = 0\ or\ (k-4 \pi r\textsuperscript{2}) = 24r\textsuperscript{2}}\\ { \Rightarrow r = 0\ or\ k = 4 \pi r\textsuperscript{2}+24r\textsuperscript{2}}\\ { \Rightarrow r = 0\ or\ (4 \pi +24)r\textsuperscript{2} = k}\\$
$\\\Rightarrow \mathrm{r}=0$ or $\mathrm{r}^{2}=\frac{\mathrm{k}}{(4 \pi+24)}$ \\$\Rightarrow \mathrm{r}=0$ or $\mathrm{r}=\pm \sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}$$
Now we know, $r \neq 0$
Hence
$\mathrm{r}=\sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}$
To find the second derivative of this volume equation, we cam simply differentiate the equation (ii),
$\mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(-2 \pi \mathrm{r}\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]\right)}{\mathrm{dx}}$
$\mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(-2 \pi \mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}+\left(4 \pi \mathrm{r}^{2}\right)\right)}{\mathrm{dx}}$
After removing the constant terms, we apply the sum rule of differentiation,
$\mathrm{V}^{\prime \prime}=-2 \pi \frac{\mathrm{d}\left(\mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dx}}$
Using the product rule of differentiation,
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\mathrm{r} \frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}} \cdot \frac{\mathrm{dr}}{\mathrm{dr}}\right]+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dx}}$
Again, the power rule of differentiation is used,
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\mathrm{r} \cdot \frac{1}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{\mathrm{dx}}+\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right]+4 \pi(2 \mathrm{r})$
Differentiating the equation, we get
$V"=-2\pi\left [ r.\frac{1}{2}\left ( \frac{k-4\pi r^2}{6} \right )^\frac{-1}{2} \left ( \frac{1}{6}\left ( -4\pi(2r) \right ) \right )+\left ( \frac{k-4\pi r^2}{6} \right )^\frac{1}{2} \right ]+8\pi r$
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{-2 \pi r^{2}}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}+\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right]+8 \pi r$
$\\ \mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{-2 \pi r^{2}+3\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \mathrm{\pi} \\ \left.\mathrm{~V}^{\prime \prime}=-2 \pi \left[ \frac{-2 \pi \mathrm{r}^{2}+\left(\frac{\mathrm{k}-4 \pi r^{2}}{2}\right)}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]\right]+8 \pi \mathrm{r}$
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}-4 \pi \mathrm{r}^{2}}{2}}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \pi \mathrm{r}$
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{\mathrm{k}-8 \pi \mathrm{r}^{2}}{6\left(\left(\frac{\mathrm{k}-4 \mathrm{\pi r}^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \mathrm{\pi r}$
$\mathrm{V}^{\prime \prime}=-\pi\left[\frac{\mathrm{k}-8 \mathrm{\pi r}^{2}}{6\left(\left(\frac{\mathrm{k}-4 \mathrm{\pi r}^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \pi \mathrm{r}>0$
$r=\sqrt{\frac{k}{(4 \pi+24)}}$$
Hence for
The substituting $r=\sqrt{\frac{k}{(4 \pi+24)}}$ , in equation (iii), we get
$\Rightarrow \mathrm{a}=\left(\frac{\mathrm{k}-4 \pi\left(\sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}\right)^{2}}{6}\right)^{\frac{1}{2}}$$
$\\ \Rightarrow a=\left(\frac{k-4 \pi\left(\frac{k}{4(\pi+6)}\right)}{6}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k-\left(\frac{k \pi}{(\pi+6)}\right)}{6}\right)^\frac{1}{2} \\ \Rightarrow a=\left(\frac{k(\pi+6)-k \pi}{6(\pi+6)}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k \pi+6 k-k \pi}{6(\pi+6)}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k}{(\pi+6)}\right)^{\frac{1}{2}}$
Now we will find the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum, i.e.,
a:2r
$\\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\left(\frac{\mathrm{k}}{(4 \pi+24)}\right)^{\frac{1}{2}}\right) \\ \\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\frac{1}{4^{\frac{1}{2}}}\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right) \\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\frac{1}{2}\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right)$
$\Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}:\left(\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right)$
Hence the required ratio is
a:2r = 1:1

Question:32

AB is a diameter of a circle and C is any point on the circle. Show that the area of Δ ABC is maximum when it is isosceles.

Answer:

Given: a circle with AB as diameter and C is any point on the circle
To show: area of Δ ABC is maximum, when it is isosceles
Explanation: If we take r as the radius of the circle, the diameter will become 2r= AB

This indicates that any angle in a semicircle is 90°.
Hence ∠ACB = 90°
Now let AC = x and BC = y
Using the Pythagoras theorem in this right-angled triangle ABC,
$\\{(2r)\textsuperscript{2} = x\textsuperscript{2}+y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4r\textsuperscript{2}-x\textsuperscript{2}}\\$
$\begin{aligned} &\Rightarrow \mathrm{y}=\sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}} \ldots \ldots\\ &\text { Now we know area of } \triangle A B C \text { is }\\ &A=\frac{1}{2} x y \end{aligned}$
Then putting the values from the equation (i), we get
$\mathrm{A}=\frac{1}{2} \mathrm{x} \sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}} \ldots \ldots$(ii)$
By finding the first derivative of the area,
$\mathrm{A}^{\prime}=\frac{\mathrm{d}\left(\frac{1}{2} \mathrm{x} \sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}}\right)}{\mathrm{dx}}$
Simultaneously using the product rule of differentiation and also taking out the constant terms,
$\mathrm{A}^{\prime}=\frac{1}{2}\left[\mathrm{x} \frac{\mathrm{d}\left(\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dx}}\right]$
Using the power rule of differentiation,
$\\ A^{\prime}=\frac{1}{2}\left[x \cdot \frac{1}{2} \cdot\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}-1} \frac{d\left(4 r^{2}-x^{2}\right)}{d x}+\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}\right] \\ A^{\prime}=\frac{1}{2}\left[x \cdot \frac{1}{2} \cdot\left(4 r^{2}-x^{2}\right)^{\frac{-1}{2}}(-2 x)+\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}\right] \\ A^{\prime}=\frac{1}{2}\left[\frac{-x^{2}}{\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}}+\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}\right] \\ A^{\prime}=\frac{1}{2}\left[\frac{-x^{2}+\left(4 r^{2}-x^{2}\right)}{\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}}\right] \\ A^{\prime}=\left[\frac{\left(2 r^{2}-x^{2}\right)}{\sqrt{4 r^{2}-x^{2}}}\right] \ldots \ldots \text { (iii) }$
Critical point can be calculated by putting the first derivative equal to 0
$\begin{aligned} &\mathrm{A}^{\prime}=0\\ &\Rightarrow \frac{\left(2 r^{2}-x^{2}\right)}{\sqrt{4 r^{2}-x^{2}}}=0\\ &\Rightarrow 2 r^{2}-x^{2}=0\\ &\Rightarrow 2 r^{2}=x^{2}\\ &\Rightarrow \mathrm{r}^{2}=\frac{\mathrm{x}^{2}}{2}\\ &\Rightarrow \mathrm{r}=\pm \sqrt{\frac{\mathrm{x}^{2}}{2}}\\ &\Rightarrow \mathrm{r}=\pm \frac{\mathrm{x}}{\sqrt{2}}\\ &0 \Rightarrow x=\pm r \sqrt{2}\\ &\text { Now we know, } r \neq 0 \end{aligned}$
Hence $r=\frac{x}{\sqrt{2}}$ and $x=r \sqrt{2}$$
Differentiating the equation (ii) will give us the second derivative of the equation
$\\\mathrm{A}^{\prime \prime}=\frac{\mathrm{d}\left(\left[\frac{\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right)}{\sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}}}\right]\right)}{\mathrm{dx}}$ \\$\mathrm{A}^{\prime \prime}=\frac{\mathrm{d}\left(\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right)\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}}\right)}{\mathrm{dx}}$$
Removing all the constant terms and then using the product rule of differentiation,
$\\\mathrm{A}^{\prime \prime}=\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right) \frac{\mathrm{d}\left(\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}}\right)}{\mathrm{dx}}+\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}} \frac{\mathrm{d}\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right)}{\mathrm{dx}}$$
After using the power rule of differentiation,
$\\ A^{\prime \prime}=\left(2 r^{2}-x^{2}\right) \cdot\left(-\frac{1}{2}\right) \cdot\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}-1} \frac{d\left(4 r^{2}-x^{2}\right)}{d x}+\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}}(-2 x) \\ A^{\prime \prime}=\left(2 r^{2}-x^{2}\right) \cdot\left(-\frac{1}{2}\right) \cdot\left(4 r^{2}-x^{2}\right)^{-\frac{3}{2}}(-2 x)+\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}}(-2 x) \\ A^{\prime \prime}=-\left(2 r^{2}-x^{2}\right) \cdot\left(4 r^{2}-x^{2}\right)^{-\frac{3}{2}}-2 x\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}} \\ A^{\prime \prime}=-\frac{\left(2 r^{2}-x^{2}\right)}{\left(4 r^{2}-x^{2}\right) \sqrt{4 r^{2}-x^{2}}}-\frac{2 x}{\sqrt{4 r^{2}-x^{2}}} \\ A^{\prime \prime}=\frac{-\left(2 r^{2}-x^{2}\right)-2 x\left(\left(4 r^{2}-x^{2}\right)\right)}{\left(4 r^{2}-x^{2}\right) \sqrt{4 r^{2}-x^{2}}} \\ A^{\prime \prime}=\frac{-\left(2 r^{2}-x^{2}\right)-8 x r^{2}+2 x^{3}}{\left(4 r^{2}-x^{2}\right) \sqrt{4 r^{2}-x^{2}}}$
For $x=r\sqrt 2$ in above equation, we get
$\\\mathrm{A}_{\mathrm{x}=\mathrm{r} \sqrt{2}}^{\prime \prime}=\frac{\left.-\left(2 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}\right)-8(\mathrm{r} \sqrt{2}) \mathrm{r}^{2}+2(\mathrm{r} \sqrt{2})^{3}\right)}{\left(4 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}\right) \sqrt{4 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}}}$ \\$A_{x=r \sqrt{2}}^{\prime \prime}=\frac{-\left(2 r^{2}-2 r^{2}\right)-8 \sqrt{2} r^{3}+4 \sqrt{2} r^{3}}{\left(4 r^{2}-2 r^{2}\right) \sqrt{4 r^{2}-2 r^{2}}}$ \\$\mathrm{A}_{\mathrm{x}=\mathrm{r} \sqrt{2}}^{\prime \prime}=\frac{-(0)-4 \sqrt{2} \mathrm{r}^{3}}{\left(2 \mathrm{r}^{2}\right) \sqrt{2 \mathrm{r}^{2}}}$ \\$A_{x=r \sqrt{2}}^{\prime \prime}=\frac{-4 \sqrt{2} r^{3}}{\left(2 r^{3}\right) \sqrt{2}}<0$$
Thus, for $\mathrm{x}=\mathrm{r} \sqrt{2}$ , the area of $\triangle \mathrm{ABC}$ is maximum
The maximum value can be calculated by substituting $\mathrm{x}=\mathrm{r} \sqrt{2}$ in equation (i),
$\Rightarrow \mathrm{y}=\sqrt{4 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}}$$
$\\ \Rightarrow y=\sqrt{4 r^{2}-2 r^{2}} \\ \Rightarrow y=\sqrt{2 r^{2}} \\ \Rightarrow y=r \sqrt 2=x$
i.e., the two sides of the $\triangle \mathrm{ABC}$ are equal
Hence, the area of $\triangle \mathrm{ABC}$ is maximum, when it is isosceles
Hence, proved.

Question:33

A metal box with a square base and vertical sides is to contain $1024 cm^3$ . The material for the top and bottom costs Rs $5/cm^2$ and the material for the sides costs Rs $2.50/cm^2$ . Find the least cost of the box.

Answer:

Given: A metal box with a square base and vertical sides is to contain $1024 cm^3$ . The material for the top and bottom costs Rs $5/cm^2$ and the material for the sides costs Rs $2.50/cm^2$
To find: the minimum cost of the box

.
Take x cm as the side of the square
Take y cm as the vertical side of the metal box
According to the given information in the question, the formula used volume for square base is
V=base × height
Due to its square base, the formula of the volume is
$V=x^2y$
This is equal to $$1024 \mathrm{~cm}^{3}$ . So, volume becomes
$\\1024 \mathrm{~cm}^{3}=\mathrm{x}^{2} \mathrm{y}$\\ $\mathrm{y}=\frac{1024}{\mathrm{x}^{2}} \ldots \ldots$ (i)$
Then we need to calculate the total area of the metal box.
Area of top and bottom $= 2x\textsuperscript{2}cm\textsuperscript{2}$
The mentioned material for the top and bottom costs Rs $5/cm\textsuperscript{2}$ , thus, the material cost for top and bottom becomes
Cost of top and bottom =Rs. 5( $2x^2$ )
Area of one side of the metal box = xy cm
The total sides present in the metal box are 4, so
Thus, the total area of all the sides of the metal box = 4xy
The cost of the material for sides is Rs $2.50/cm^2$
∴ Cost of all the sides of the metal box =Rs. 2.50(4xy)
The overall area of the metal box will be
$A=2x\textsuperscript{2}+4xy$
This will make the cost of the box to be
$C=5(2x\textsuperscript{2}) + 2.50(4xy)$
Putting the value of y from equation (i) in the above equation,
$\\C=5\left(2 x^{2}\right)+2.50\left(4 x\left(\frac{1024}{x^{2}}\right)\right)$\\ $\Rightarrow C=10 x^{2}+\frac{10240}{x}$$
Both the sides are differentiated with respect to x
$\Rightarrow C^{\prime}=\frac{d\left(10 x^{2}+\frac{10240}{x}\right)}{d x}$$
Using differentiation rule of sum, we get
$\Rightarrow C^{\prime}=\frac{d\left(10 x^{2}\right)}{d x}+\frac{d\left(\frac{10240}{x}\right)}{d x}$$
$\Rightarrow C^{\prime}=10 \frac{d\left(x^{2}\right)}{d x}+10240 \frac{d\left(x^{-1}\right)}{d x}$$
Then using the derivative, we get
$\\\Rightarrow C^{\prime}=10 \times 2 x+10240 \times\left(-x^{-2}\right)$ \\$\Rightarrow C^{\prime}=20 x-\frac{10240}{x^{2}} \ldots$ (ii)$
Take c=0 to find the minimum value of x by apply second derivative test, so the above equation is equated with 0
$\\20 x-\frac{10240}{x^{2}}=0$ \\\\$\Rightarrow 20 \mathrm{x}=\frac{10240}{\mathrm{x}^{2}}$ \\\\$\Rightarrow \mathrm{x}^{3}=\frac{10240}{20}$$
$\Rightarrow x^{3}=512$
Solving this we get
$\Rightarrow x=8$
Again, differentiating equation (ii) with respect to x,
$\Rightarrow C^{\prime \prime}=\frac{d\left(20 x-\frac{10240}{x^{2}}\right)}{d x}$$
Using the differentiation rule of sum,
$\\\Rightarrow C^{\prime \prime}=\frac{d(20 x)}{d x}-\frac{d\left(\frac{10240}{x^{2}}\right)}{d x}$ \\$\Rightarrow \mathrm{C}^{\prime \prime}=20 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}-10240 \frac{\mathrm{d}\left(\mathrm{x}^{-2}\right)}{\mathrm{dx}}$ \\$\Rightarrow \mathrm{C}^{\prime \prime}=20+20480 \times\left(-\mathrm{x}^{-3}\right)$ \\$\Rightarrow C^{\prime \prime}=20+\frac{20480}{x^{3}}$$
At x=8, the above equation becomes,
$\\\Rightarrow\left(C^{\prime \prime}\right)_{x=8}=20+\frac{20480}{(8)^{3}}$ \\$\Rightarrow\left(C^{\prime \prime}\right)_{x=8}=20+\frac{20480}{512}>0$$
Now at x=8, $C^{\prime}(8)=0$ and $C^{\prime \prime}(8)>0$ , so as per the second derivative test, x is a point of local minima and $\mathrm{C}(8)$ will be minimum value of C.
Hence least cost becomes
$\\C_{x=8}=10(8)^{2}+\frac{10240}{8}$ \\$\Rightarrow C_{y=8}=640+1280=RS.1920$
Hence the least cost of the metal box is Rs. 1920

Question:34

The sum of the surface areas of a rectangular parallelepiped with sides x, 2x and $\frac{x}{3}$ and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
Answer:

Given: x, 2x and $\frac{x}{3}$ are the sides of a rectangular parallelepiped. The sum of the surface areas of a sphere and a rectangular parallelepiped is given to be constant.
To prove: if x is equal to three times the radius of the sphere, the sum of their volumes is minimum and find the minimum value of the sum of their volumes
Surface area of rectangular parallelepiped:
$\\ S_{\mathrm{rp}}=2\left(\mathrm{x}(2 \mathrm{x})+2 \mathrm{x}\left(\frac{\mathrm{x}}{3}\right)+\left(\frac{\mathrm{x}}{3}\right) \mathrm{x}\right) \\ \Rightarrow \mathrm{S}_{\mathrm{rp}}=2\left(2 \mathrm{x}^{2}+\frac{2 \mathrm{x}^{2}}{3}+\frac{\mathrm{x}^{2}}{3}\right)$
Let radius of sphere be r cm, then surface area is $S_s=4\pi r^2$
Now sum of the surface areas is,
$\mathrm{S}=\mathrm{S}_{\mathrm{rp}}+\mathrm{S}_{\mathrm{s}}=2\left(2 \mathrm{x}^{2}+\frac{2 \mathrm{x}^{2}}{3}+\frac{\mathrm{x}^{2}}{3}\right)+4 \pi \mathrm{r}^{2}$
$\Rightarrow \mathrm{S}=2\left(\frac{6 \mathrm{x}^{2}+2 \mathrm{x}^{2}+\mathrm{x}^{2}}{3}\right)+4 \mathrm{\pi}r^{2}$
$$\Rightarrow \mathrm{S}=6 \mathrm{x}^{2}+4 \mathrm{\pi}r^{2} \ldots$(i)$
Now given that the sum of the surface areas is constant, so
$\frac{\mathrm{dS}}{\mathrm{dr}}=0$$
Now, differentiate (i) with respect to r and get
$\frac{d S}{d r}=\frac{d\left(6 x^{2}+4 \pi r^{2}\right)}{d r}=0$$
Apply differentiation rule of sum and get
$\Rightarrow \frac{\mathrm{d}\left(6 \mathrm{x}^{2}\right)}{\mathrm{dr}}+\frac{\mathrm{d}\left(4 \pi \mathrm{r}^{2}\right)}{\mathrm{dr}}=0$$
Take the constant terms out and get
$\Rightarrow 6 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dr}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dr}}=0$$
Apply derivative and get
$\\\Rightarrow 6 \times 2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dr}}+4 \pi \times 2 \mathrm{r} \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dr}}=0$ \\\\$\Rightarrow 12 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dr}}+8 \mathrm{\pi r}=0$\\ \\$\Rightarrow 12 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dr}}=-8 \pi \mathrm{r}$ \\\\$\Rightarrow \frac{d x}{d r}=-\frac{8 \pi r}{12 x}$ \\\\$\Rightarrow \frac{d x}{d r}=-\frac{2 \pi r}{3 x} \ldots \ldots . .$ (ii)$
Let V denote the sum of volumes of both the shapes, so
$\mathrm{V}=\mathrm{x} \times 2 \mathrm{x} \times \frac{\mathrm{x}}{3}+\frac{4}{3} \pi \mathrm{r}^{3}$$
$\Rightarrow \mathrm{V}=\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi \mathrm{r}^{3} \ldots \ldots$ (iii)$
The first derivative of volume must be equal to 0 for minima or maxima
$\frac{\mathrm{dV}}{\mathrm{dr}}=0$$
Differentiate (iii) with respect to r and get
$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dr}}=\frac{\mathrm{d}\left(\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}}=0$$
Apply differentiation rule of sum and get
$\Rightarrow \frac{\mathrm{d}\left(\frac{2}{3} \mathrm{x}^{3}\right)}{\mathrm{dr}}+\frac{\mathrm{d}\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}}=0$$
Take constant terms out and get
$\Rightarrow \frac{2}{3} \frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dr}}+\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dr}}=0$$
Apply derivative and get
$\Rightarrow \frac{2}{3} \times 3 \mathrm{x}^{2} \times \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dr}}+\frac{4}{3} \pi \times 3 \mathrm{r}^{2} \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dr}}=0$$
$\Rightarrow 2 \mathrm{x}^{2} \frac{\mathrm{dx}}{\mathrm{dr}}+4 \mathrm{\pi}r^{2}=0$ $$
Substitute value of $\frac{d x}{d r}$$ from (ii) and get
$\\\Rightarrow 2 \mathrm{x}^{2}\left(-\frac{2 \pi r}{3 \mathrm{x}}\right)+4 \pi r^{2}=0$ \\$\Rightarrow-\frac{4 \pi r x}{3}+4 \pi r^{2}=0 \ldots . .(i v)$ \\$\Rightarrow 4 \pi r^{2}=\frac{4 \pi r x}{3}$ \\$\Rightarrow \mathrm{r}=\frac{\mathrm{x}}{3}$ $\Rightarrow x=3 r$$
i.e., the radius of the sphere is 1/3 of x.
Hence proved
Now let’s find the second derivative value at x=3r.
Now, apply derivative with respect to r to (iv) and get
$\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=\frac{\mathrm{d}\left(-\frac{4 \pi \mathrm{r} \mathrm{x}}{3}+4 \mathrm{\pi r}^{2}\right)}{\mathrm{dr}}$$
Apply differentiation rule of sum and get
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=\frac{\mathrm{d}\left(-\frac{4 \pi \mathrm{rx}}{3}\right)}{\mathrm{dr}}+\frac{\mathrm{d}\left(4 \mathrm{\pi r}^{2}\right)}{\mathrm{dr}}$$
Take constant terms out and get
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3} \frac{\mathrm{d}(\mathrm{rx})}{\mathrm{dr}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dr}}$$
The first part is applied the differentiation rule of product, so
$\\\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(\mathrm{r} \frac{\mathrm{dx}}{\mathrm{dr}}+\mathrm{x} \frac{\mathrm{dr}}{\mathrm{dr}}\right)+4 \pi \times 2 \mathrm{r} \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dr}}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(\mathrm{r} \frac{\mathrm{d} \mathrm{x}}{\mathrm{dr}}+\mathrm{x}\right)+4 \pi \times 2 \mathrm{r}$$
Substitute value of $\frac{dr}{dx}$ from (ii) and get
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(\mathrm{r}\left(-\frac{2 \pi \mathrm{r}}{3 \mathrm{x}}\right)+\mathrm{x}\right)+4 \pi \times 2 \mathrm{r}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(-\frac{2 \pi \mathrm{r}^{2}}{3 \mathrm{x}}+\mathrm{x}\right)+8 \mathrm{\pi r}$$
Substitutex=3r and get
$\\\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=-\frac{4 \pi}{3}\left(-\frac{2 \pi \mathrm{r}^{2}}{3(3 \mathrm{r})}+3 \mathrm{r}\right)+8 \mathrm{\pi r}$ \\$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=\frac{4 \pi}{3}\left(\frac{2\pi \mathrm{r}}{9}-3 \mathrm{r}\right)+8 \mathrm{\pi r}$ \\$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=\frac{4 \pi}{3}\left(\frac{2\pi \mathrm{r}}{9}-3 \mathrm{r}+6 \mathrm{r}\right)$$
$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=\frac{4 \pi}{3}\left(\frac{\pi \mathrm{r}}{3}+3 \mathrm{r}\right)$$
It is positive;so V is minimum when $\mathrm{x}=3 \mathrm{r}$ or $\mathrm{r}=\frac{\mathrm{x}}{3}$ , and the minimum value of Volume
can be obtained by substituting $r=\frac{x}{3}$$ in equation (iii), we get
$\\\mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}$ \\$\Rightarrow \mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2}{3}(\mathrm{x})^{3}+\frac{4}{3} \pi\left(\frac{\mathrm{x}}{3}\right)^{3}$ \\$\Rightarrow \mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi\left(\frac{\mathrm{x}^{3}}{27}\right)$ \\$\Rightarrow \mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2 \mathrm{x}^{3}}{3}\left[1+\frac{2 \pi}{27}\right]$$
Therefore, it Is the minimum value of the sum of their volumes.

Question:35

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:
A. 10 $cm^2/s$
B. $\sqrt3$ $cm^2/s$
C. $10\sqrt3$ $cm^2/s$
D. $\frac{10}{3}$ $cm^2/s$

Answer:

Let x cm be the side of the equilateral triangle, then the area of the triangle is
$\\A=\frac{\sqrt{3}}{4} x^{2}$ \\$\frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{sec}$$
Also, the rate of side increasing at instant of time t is
Differentiate area with respect to time t and get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{d}\left(\frac{\sqrt{3}}{4} \mathrm{x}^{2}\right)}{\mathrm{dt}}$$
Take the constants out and get,
$\\ \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{4} \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}$$
Apply the derivative and get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{4} \times 2 \mathrm{x} \times \frac{\mathrm{dx}}{\mathrm{dt}}$$
Substitute given value of $\frac{\mathrm{dx}}{\mathrm{dt}}$$ and get
$\\\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{4} \times 2 \mathrm{x} \times 2$ \\$\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\sqrt{3} \mathrm{x}$$
Now, put side $\mathrm{x}=10 \mathrm{~cm}$$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}_{\mathrm{x}=10}}=\sqrt{3} \times 10$$
Hence, the rate at which the area increases is $10 \sqrt 3 \mathrm{~cm}^{2} / \mathrm{s}$.$
So the correct answer is option C.

Question:36

A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is:
A. $\frac{1}{10}$ radian/sec
B. $\frac{1}{20}$ radian/sec
C. 20 radian/sec
D. 10 radian/sec

Answer:

Let 5m/500cm be the length of the ladder, which is the hypotenuse of the right triangle formed in the above figure.
Now let the angle between the ladder and the floor be β, so
$\\\sin \beta=\frac{\text { opposite side }}{\text { hypotenuse }}$ \\\\$\Rightarrow \sin \beta=\frac{\mathrm{x}}{500}$$
Differentiate both sides with respect to time t and get
$\\\Rightarrow \frac{\mathrm{d}(\sin \beta)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\frac{\mathrm{x}}{500}\right)}{\mathrm{dt}}$$
Apply derivatives and get
$\\\Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{500} \frac{\mathrm{dx}}{\mathrm{dt}}$$
Now, the top of the ladder slides downwards at the rate of
$10 \mathrm{~cm} / sec$
$\frac{d x}{d t}=10 \mathrm{~cm} / \mathrm{sec}$$
So the equation is
$\\ \Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{500} \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{500} \times 10$ \\$\Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{50} \ldots \ldots$ (ii)$
Now,
$\cos \beta=\frac{\text { adacent side }}{\text { hypotenuse }}$$
$\begin{aligned} &\text { Substitute (iii) in (ii) and get }\\ &\Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{50}\\ &\Rightarrow \frac{y}{500} \times \frac{d \beta}{d t}=\frac{1}{50}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{50} \times \frac{500}{\mathrm{y}}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{10}{\mathrm{y}}\\ &\begin{array}{l} \text { Now when } y=200 \mathrm{~cm} \text { , we get } \end{array}\\ &\Rightarrow \frac{\mathrm {d \beta}}{\mathrm{dt}}_{\mathrm{y}=200}=\frac{10}{\mathrm{y}}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}_{\mathrm{y}=200}=\frac{10}{200}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}_{y=200}=\frac{1}{20} \end{aligned}$
Hence the rate at which the angle between the floor and the ladder is decreasing is $\frac{1}{20}$ radian/sec
So the correct answer is option B.

Question:37

The curve $y=x^{\frac{1}{5}}$ has at (0, 0)
A. a vertical tangent (parallel to y-axis)
B. a horizontal tangent (parallel to x-axis)
C. an oblique tangent
D. no tangent

Answer:

Given $y=x^{\frac{1}{5}}$
Differentiate both sides with x and get
$\frac{d y}{d x}=\frac{d\left(x^{\frac{1}{5}}\right)}{d x}$$
Apply power rule and get
$\\\Rightarrow \frac{d y}{d x}=\frac{1}{5} \times x^{\frac{1}{5}-1}$ \\$\Rightarrow \frac{d y}{d x}=\frac{1}{5} \times x^{-\frac{4}{5}}$ \\$\Rightarrow \frac{d y}{d x}=\frac{1}{5 x^{\frac{4}{5}}}$$
Now at (0,0)
$\\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,0)}=\frac{1}{5 \mathrm{x}^{\frac{4}{5}}}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{(0,0)}=\frac{1}{5(0)^{\frac{4}{5}}}$\\ $\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,0)}=\infty$$
So the curve $\mathrm{y}=\mathrm{x}^{\frac{1}{5}}$$ at (0,0) has vertical tangent parallel to Y-axis.
Hence the correct answer is option A.

Question:38

The equation of normal to the curve $3x\textsuperscript{2} - y^2 = 8$ which is parallel to the line $x + 3y = 8$ is
$\\A. 3x - y = 8\\ B. 3x + y + 8 = 0\\ C. x + 3y \pm 8 = 0 \\D. x + 3y = 0$

Answer:

Given the equation of the line is $3x\textsuperscript{2} - y^2 = 8$
Differentiate both sides with x and get
$\frac{\mathrm{d}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}$$
Apply sum rule and 0 is the differentiation of constant, so
$\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0$$
Take the constants out and get
$\Rightarrow 3 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0$$
Apply power rule and get
$\\\Rightarrow 3 \times 2\left(\mathrm{x}^{2-1}\right) \times \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}-2\left(\mathrm{y}^{2-1}\right) \times \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}=0$ \\$\Rightarrow 6 \mathrm{x}-2 \mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}=0$ \\$\Rightarrow 6 \mathrm{x}=2 \mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} \\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6 \mathrm{x}}{2 \mathrm{y}}=\frac{3 \mathrm{x}}{\mathrm{y}}$$
Hence, the slope of the given curve is provided.
Also, the slope of the normal to the curve is
$=-\frac{1}{\frac{d y}{d x}}$ \\$\Rightarrow=-\frac{1}{\frac{3 \mathrm{x}}{\mathrm{y}}}=\left(-\frac{\mathrm{y}}{3 \mathrm{x}}\right) \ldots \ldots$ (i)$
Now, $x+3 y=8$
$\Rightarrow 3 y=8-x$$
After differentiating with respect to x
$\\\Rightarrow \frac{\mathrm{d}(3 \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}(8-\mathrm{x})}{\mathrm{dx}}$ \\$\Rightarrow 3 \frac{\mathrm{dy}}{\mathrm{dx}}=-1$$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{3}$$
Therefore, the slope is $-\frac{1}{3}$$
Now, because the normal to the curve is parallel to this line, that means the slope of the line must be equal to slope of the normal to the given curve,
$\therefore\left(-\frac{y}{3 x}\right)=-\frac{1}{3}$$
$\\ { \Rightarrow 3y=3x}\\ { \Rightarrow y=x}\\$
Substitute the value of the given equation
$\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3x\textsuperscript{2} - (x)\textsuperscript{2} = 8}\\ { \Rightarrow 2x\textsuperscript{2} = 8}\\ { \Rightarrow x\textsuperscript{2} = 4}\\ { \Rightarrow x= \pm 2}\\$
When x=2, the equation is
$\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(2)\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(4) - y\textsuperscript{2} = 8}\\ { \Rightarrow 12-8= y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4}\\ { \Rightarrow y= \pm 2}\\$
When x=-2, the equation is
$\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(-2)\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(4) - y\textsuperscript{2} = 8}\\ { \Rightarrow 12-8= y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4}\\ { \Rightarrow y= \pm 2}\\$
So, the points are $( \pm 2, \pm 2)$ at which normal is parallel to the given line.
And required equation at $( \pm 2, \pm 2)$ is
$\\ y-(\pm 2)=-\frac{1}{3}[x-(\pm 2)] \\ \Rightarrow 3(y-(\pm 2))=-(x-(\pm 2)) \\ \Rightarrow 3 y-(\pm 6)=-x+(\pm 2) \\ \Rightarrow x+3 y-(\pm 6)-(\pm 2)=0 \\ \Rightarrow x+3 y+(\pm \quad 8)=0$
Hence the equation of normal to the curve is $x+3 y+(\pm \quad 8)=0$
So the correct answer is option C

Question:39

If the curve $ay + x^2 = 7$ and $x^3 = y$ , cut orthogonally at (1, 1), then the value of a is:
A. 1
B. 0
C. – 6
D. 6

Answer:

Given the fact that curve $ay + x^2 = 7$ and $x^3 = y$ , cut orthogonally at (1, 1)
Differentiate on both sides with x and get
$\frac{\mathrm{d}\left(\mathrm{ay}+\mathrm{x}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(7)}{\mathrm{dx}}$$
Apply sum rule and also 0 is the derivative of the constant, so
$\Rightarrow \frac{\mathrm{d}(\mathrm{ay})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=0$$
Apply power rule and get
$\\\Rightarrow \mathrm{a} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{x}=0$ \\$\Rightarrow \frac{d y}{d x}=-\frac{2 x}{a}$$
Putting (1,1)
$\\\Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{2 x}{a}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=-\frac{2(1)}{\mathrm{a}}=-\frac{2}{\mathrm{a}} \ldots \ldots(\mathrm{i})$ \\$x^{3}=y$$
Differentiate on both sides with x and get
$\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}$$
Apply power rule and get
$\Rightarrow 3 \mathrm{x}^{2}=\frac{\mathrm{dy}}{\mathrm{dx}}$$
Putting (1,1)
$\\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=3 \mathrm{x}^{2}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=3(1)^{2}=3 \ldots \ldots$(ii)$
Both curves cut orthogonally at (1,1), 50
$\left(\frac{d y}{d x}\right)_{(1,1)} \times\left(\frac{d y}{d x}\right)_{(1,1)}=-1$$
So from (i) and (ii), we get
$\\\left(-\frac{2}{a}\right) \times 3=-1$ \\$\Rightarrow-\frac{6}{a}=-1$ \\$\Rightarrow a=6$$
Hence when the curves cut orthogonally at (1, 1), then the value of a is 6.
So the correct answer is option D.

Question:40

If $y = x^4 - 10$ and if x changes from 2 to 1.99, what is the change in y
A. 0.32
B. 0.032
C. 5.68
D. 5.968

Answer:

A)
Given $y = x^4 - 10$
Differentiate on both sides with x and get
$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{4}-10\right)}{\mathrm{dx}}$$
Apply power rule and get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{x}^{3}$$
Now, value of x changes from 2 to 1.99, so the change in x is
$\Delta x=2-1.99=0.01$$
So the change in y is,
$\Rightarrow \Delta y=\frac{d y}{d x} \times \Delta x$$
Substitute corresponding values and get
$\Rightarrow \Delta y=4x\textsuperscript{3} \times (0.01)$
Now at x=2, the change in y becomes
$\\ { \Rightarrow \Delta y\textsubscript{x=2}=4(2)\textsuperscript{3} \times (0.01)}\\ { \Rightarrow \Delta y\textsubscript{x=2}=4 \times 8 \times (0.01)}\\ { \Rightarrow \Delta y\textsubscript{x=2}=0.32}\\$
Therefore, change in y is 0.32.

Question:41

The equation of tangent to the curve $y(1 + x^2) = 2 - x,$ where it crosses x-axis is:
$\\A. x + 5y = 2 \\B. x - 5y = 2 \\C. 5x - y = 2 \\D. 5x + y = 2$

Answer:

A)
Given the equation of the curve is
$y(1 + x^2) = 2 - x,$
Both the sides are differentiated with respect to x,
$\frac{d\left(y\left(1+x^{2}\right)\right)}{d x}=\frac{d(2-x)}{d x}$$
Using the power rule
$\Rightarrow y \cdot \frac{d\left(1+x^{2}\right)}{d x}+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=\frac{d(2-x)}{d x}$$
As the derivative of a constant is always 0 we get
$\Rightarrow y \cdot \frac{d\left(x^{2}\right)}{d x}+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=\frac{d(-x)}{d x}$$
Again, using the power rule
$\\\Rightarrow y \cdot 2 x+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=-1$ \\$\Rightarrow\left(1+x^{2}\right) \cdot \frac{d y}{d x}=2 x y-1$ \\$\Rightarrow \frac{d y}{d x}=\frac{2 x y-1}{1+x^{2}} \ldots \ldots$(i)$
The mentioned curve passes through the x -axis, i.e., y=0
Thus, the curve equation becomes
$\\ {y(1+x\textsuperscript{2})=2-x}\\ { \Rightarrow 0(1+x\textsuperscript{2})=2-x}\\ { \Rightarrow 0=2-x}\\ { \Rightarrow x=2}\\$
As the point of passing for the given curve is (2,0)
So the equation (i) at point (2,0) is,
$\\\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{2 x y-1}{1+x^{2}}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{2(2)(0)-1}{1+(2)^{2}}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{0-1}{1+4}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,0)}=-\frac{1}{5}$$
So, the slope of tangent to the curve is $-\frac{1}{5}$$
Therefore, the equation of tangent of the curve passing through (2,0) is given by
$y-0=-\frac{1}{5}(x-2)$$
$\\ { \Rightarrow 5y=-x+2}\\ { \Rightarrow x+5y=2}\\$
Thus, the equation of tangent to the curve $y(1 + x^2) = 2 - x,$ , where it crosses x-axis is $x+5y=2$ .
Hence, the correct option is option A

Question:42

The points at which the tangents to the curve $y = x^3 - 12x + 18$ are parallel to x-axis are:
A. (2, -2), (-2, -34)
B. (2, 34), (-2, 0)
C. (0, 34), (-2, 0)
D. (2, 2), (-2, 34)

Answer:

D)
Given the equation of the curve is
$y = x^3 - 12x + 18$
Differentiating on both sides with respect to x, we get
$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}-12 \mathrm{x}+18\right)}{\mathrm{dx}}$$
Applying the sum rule of differentiation, we get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-12 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(18)}{\mathrm{dx}}$$
We know derivative of a constant is 0,so above equation becomes
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-12 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+0$$
Applying the power rule we get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-12 \ldots$(i)$
Thus, the slope of line parallel to the x -axis is given by
$\frac{\mathrm{dy}}{\mathrm{dx}}=0$$
So equating equation (i) to 0 we get
$3 x^{2}-12=0$$
$\\ { \Rightarrow 3x\textsuperscript{2}=12}\\ { \Rightarrow x\textsuperscript{2}=4}\\ { \Rightarrow x= \pm 2}\\$
When x=2, the given equation of curve becomes,
$\\ {y = x\textsuperscript{3} - 12x + 18}\\ { \Rightarrow y = (2)\textsuperscript{3} - 12(2) + 18}\\ { \Rightarrow y = 8- 24 + 18}\\ { \Rightarrow y = 2}\\$
When x=-2, the given equation of curve becomes,
$\\ {y = x\textsuperscript{3} - 12x + 18}\\ { \Rightarrow y = (-2)\textsuperscript{3} - 12(-2) + 18}\\ { \Rightarrow y = -8+24 + 18}\\ { \Rightarrow y = 34}\\$
Hence, the points at which the tangents to the curve $y = x^3 - 12x + 18$ are parallel to x-axis are (2, 2) and (-2, 34).
So, the correct option is option D.

Question:43

The tangent to the curve $y = e^{2x}$ at the point (0, 1) meets x-axis at:
A. (0, 1)
B. $\left ( -\frac{1}{2},0 \right )$
C. (2, 0)
D. (0, 2)

Answer:

Given the equation of the curve is
$y = e^{2x}$
Differentiating on both sides with respect to x, we get
$\frac{d(y)}{d x}=\frac{d\left(e^{2 x}\right)}{d x}$$
Applying the exponential rule of differentiation, we get
$\\\Rightarrow \frac{d y}{d x}=e^{2 x} \frac{d(2 x)}{d x}$ \\$\Rightarrow \frac{d y}{d x}=2 e^{2 x} \ldots . .$(i)$
As it is given the curve has tangent at (0,1), so the curve passes through the point (0,1), so above equation at (0,1), becomes
$\\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0.1)}=2 \mathrm{e}^{2 \mathrm{x}}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,1)}=2 \mathrm{e}^{2(0)}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,1)}=2$$
So, the slope of the tangent to the curve at point (0,1) is 2
Hence the equation of the tangent is given by
$\\y-1=2(x-0) \\\Rightarrow y-1=2x \\\Rightarrow y=2x+1$
It is given that the tangent to the curve $y=e\textsuperscript{2x}$ at the point (0,1) meet x-axis i.e., y=0
So the equation on tangent becomes,
$\\ { \Rightarrow y=2x+1}\\ { \Rightarrow 0=2x+1}\\ { \Rightarrow 2x=-1}\\$
$\Rightarrow \mathrm{x}=-\frac{1}{2}$$
Hence, the required point is $\left(-\frac{1}{2}, 0\right)$$
Therefore, the tangent to the curve $y=e^{2 x}$$ at the point (0,1) meets x -axis at $\left(-\frac{1}{2}, 0\right)$
So, the correct option is option B.

Question:44

The slope of tangent to the curve $x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point (2, -1) is:
A. $\frac{22}{7}$
B. $\frac{6}{7}$
C. $-\frac{6}{7}$
D. $-6$

Answer:

Curve of the given equation is $x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$
With respect to t, while differentiating on both sides, we get
$\frac{d(x)}{d t}=\frac{d\left(t^{2}+3 t-8\right)}{d t}$$
After application of the sum rule of differentiation, we get
$\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{t}^{2}\right)}{\mathrm{dt}}+\frac{\mathrm{d}(3 \mathrm{t})}{\mathrm{dt}}+\frac{\mathrm{d}(-8)}{\mathrm{dt}}$$
Constant's derivative is 0, so above equation becomes
$\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{t}^{2}\right)}{\mathrm{dt}}+3 \frac{\mathrm{d}(\mathrm{t})}{\mathrm{dt}}+0$$
Power Rule application leads to
$\\ \frac{d(x)}{d t}=2 t+3 \ldots \ldots$(i) \\$y=2 t^{2}-2 t-5$$
With respect to t, we differentiate on both side and get
$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\frac{\mathrm{d}\left(2 \mathrm{t}^{2}-2 \mathrm{t}-5\right)}{\mathrm{dt}}$$
Sum Rule application leads to
$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\frac{\mathrm{d}\left(2 \mathrm{t}^{2}\right)}{\mathrm{dt}}-\frac{\mathrm{d}(2 \mathrm{t})}{\mathrm{dt}}+\frac{\mathrm{d}(-5)}{\mathrm{dt}}$$
The Constant's derivative is 0, so the equation becomes
$\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=2 \frac{\mathrm{d}\left(\mathrm{t}^{2}\right)}{\mathrm{dt}}-2 \frac{\mathrm{d}(\mathrm{t})}{\mathrm{dt}}+0$$
Applying power rule
$\frac{d(y)}{d t}=4 t-2 \ldots$(ii)$
We know,
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
Substitute values from equation (i) and (ii)
$\frac{d y}{d x}=\frac{4 t-2}{2 t+3}$$
The point through which the curve passes is (2,-1), now, substitute the same and get
$\\ {x = t\textsuperscript{2} + 3t - 8}\\ { \Rightarrow 2= t\textsuperscript{2} + 3t - 8}\\ { \Rightarrow t\textsuperscript{2} + 3t - 8-2=0}\\ { \Rightarrow t\textsuperscript{2} + 3t - 10=0}\\$
Split the middle term
$\\ { \Rightarrow t\textsuperscript{2} + 5t-2t - 10=0}\\ { \Rightarrow t(t+ 5) -2(t+5)=0}\\ { \Rightarrow (t+ 5) (t-2)=0}\\ { \Rightarrow t+5=0 or t-2=0}\\ { \Rightarrow t=-5or t=2 \ldots \ldots \ldots .(iii)}\\ {y = 2t\textsuperscript{2} - 2t - 5}\\ { \Rightarrow -1=2t\textsuperscript{2} - 2t - 5}\\ { \Rightarrow 2t\textsuperscript{2} - 2t - 5+1=0}\\ { \Rightarrow 2t\textsuperscript{2} - 2t - 4=0}\\$
Take 2 as common
$\Rightarrow t\textsuperscript{2} - t - 2=0$
Split the middle term again
$\\ { \Rightarrow t\textsuperscript{2} - 2t +t- 2=0}\\ { \Rightarrow t(t- 2)+1(t- 2)=0}\\ { \Rightarrow (t- 2) (t+1) = 0}\\ { \Rightarrow (t- 2) =0 or (t+1) = 0}\\ { \Rightarrow t=2 or t=-1 \ldots \ldots .. (iv)}\\$
In equation (iii) and (iv), 2 is common
So, t=2
So, the slope of the tangent at t=2 is as follows
$\\ \left(\frac{d y}{d x}\right)_{t=2}=\frac{4 t-2}{2 t+3}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{t}=2}=\frac{4(2)-2}{2(2)+3}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{8-2}{4+3}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{6}{7}$$
Therefore, the slope of tangent at the point (2,-1) is $\frac{6}{7}$$
So, the correct answer is option B.

Question:45

The two curves $x^3 - 3xy^2 + 2 = 0$ and $3x^2y - y^3 - 2 = 0$ intersect at an angle of
A. $\frac{\pi}{4}$
B. $\frac{\pi}{3}$
C. $\frac{\pi}{2}$
D. $\frac{\pi}{6}$

Answer:

Given the curve $x^3 - 3xy^2 + 2 = 0$ and $3x^2y - y^3 - 2 = 0$
Differentiate on both the sides with respect to x
$\frac{\mathrm{d}\left(\mathrm{x}^{3}-3 \mathrm{xy}^{2}+2\right)}{\mathrm{dx}}=\frac{\mathrm{d}(0)}{\mathrm{dx}}$$
Apply the sum rule and also 0 is the the derivative of the constant, so it becomes
$\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(3 \mathrm{xy}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(2)}{\mathrm{dx}}=0$$
Apply power rule and get
$\Rightarrow 3 x^{2}-3 \frac{d\left(x y^{2}\right)}{d x}+0=0$$
Apply product rule and get
$\begin{aligned} &\Rightarrow 3 x^{2}-3\left(x \frac{d\left(y^{2}\right)}{d x}+y^{2} \frac{d x}{d x}\right)=0\\ &\Rightarrow 3 x^{2}-3\left(x .2 y \frac{d(y)}{d x}+y^{2}\right)=0\\ &\Rightarrow 3 x^{2}-6 x y \frac{d y}{d x}-3 y^{2}=0\\ &\Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\\ &\Rightarrow \frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y}=\frac{3\left(x^{2}-y^{2}\right)}{6 x y}\\ &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}\\ &\text { Let it be equal to } \mathrm{m}_{\mathrm{d}_{k}} \mathrm{so}\\ &\Rightarrow \mathrm{m}_{1}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} . . \text { (i) }\\ &3 x^{2} y-y^{3}-2=0 \end{aligned}$
Differentiate on both the sides with respect to x and get
$\frac{d\left(3 x^{2} y-y^{3}-2\right)}{d x}=\frac{d(0)}{d x}$$
Apply the sum rule and also 0 is the derivative of the constant, so
$\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2} \mathrm{y}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(2)}{\mathrm{dx}}=0$$
Apply power rule and get
$\Rightarrow 3 \frac{d\left(x^{2} y\right)}{d x}-3 y^{2} \frac{d(y)}{d x}+0=0$$
Apply product rule and get
$\\\Rightarrow 3\left(x^{2} \frac{d(y)}{d x}+y \frac{d\left(x^{2}\right)}{d x}\right)-3 y^{2} \frac{d(y)}{d x}=0$ \\$\Rightarrow 3 x^{2} \frac{d y}{d x}+3 y(2 x)-3 y^{2} \frac{d y}{d x}=0$$
$\begin{aligned} &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)+6 \mathrm{xy}=0\\ &\Rightarrow \frac{d y}{d x}\left(3 x^{2}-3 y^{2}\right)=-6 x y\\ &\Rightarrow \frac{d y}{d x}=-\frac{6 x y}{\left(3 x^{2}-3 y^{2}\right)}\\ &\text { Let it be equal to } \mathrm{m}_{2} \mathrm{z}_{\mathrm{r}}\\ &\Rightarrow \mathrm{m}_{2}=-\frac{6 \mathrm{xy}}{\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)} \ldots \ldots .\\ &\text { The multiply equation (i) and (ii) and get }\\ &\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\left(\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}\right) \cdot\left(-\frac{6 \mathrm{xy}}{\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}\right)\\ &\Rightarrow \mathrm{m}_{1} \cdot \mathrm{m}_{2}=\left(-\frac{3\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}\right)\\ &\Rightarrow m_{1} \cdot \mathrm{m}_{2}=-1 \end{aligned}$
Since the product of the slopes is -1, it means that both the curves are intersecting at right angle i.., they are making $\frac{\pi}{2}$ angle with each other.
So, the correct answer is option C

Question:46

The interval on which the function $f (x) = 2x^3 + 9x^2 + 12x - 1$ is decreasing is:
A. [-1, ∞)
B. [-2, -1]
C. (-∞, -2]
D. [-1, 1]

Answer:

Given $f (x) = 2x^3 + 9x^2 + 12x - 1$
Apply first derivative and get
$\begin{aligned} &f(x)=\frac{d\left(2 x^{3}+9 x^{2}+12 x-1\right)}{d x}\\ &\text { Apply the sum rule of the differentiation and } 0 \text { is the derivative of the constant, so }\\ &\Rightarrow f(x)=\frac{d\left(2 x^{3}\right)}{d x}+\frac{d\left(9 x^{2}\right)}{d x}+\frac{d(12 x)}{d x}-\frac{d(1)}{d x} \end{aligned}$
Apply power rule and get
$\\ { \Rightarrow f'(x)=6x\textsuperscript{2}+18x+12-0}\\ { \Rightarrow f'(x)=6(x\textsuperscript{2}+3x+2)}\\$
Split the middle term and get
$\\ { \Rightarrow f'(x)=6(x\textsuperscript{2}+2x+x+2)}\\ { \Rightarrow f'(x)=6(x(x+2)+1(x+2))}\\ { \Rightarrow f'(x)=6((x+2) (x+1))}\\$
Now f'(x)=0 gives
x=-1, -2
Three intervals are made when these points divide the real number line
$\\ {(- \infty , -2), [-2,-1] and (-1, \infty )}\\ ${(i) in the interval $(- \infty , -2), f'(x)>0}\\ { \therefore $ f(x) is increasing in (- \infty ,-2)}\\ ${(ii) $in the interval [-2,-1], f'(x) \leq 0}\\ ${ \therefore $ f(x) is decreasing in [-2, -1]}$ \\ $(iii) in the interval$ (-1, \infty ), f'(x)>0\\ { \therefore $ f(x) is increasing in (-1, \infty )}\\$
So, the interval on which the function decreases is [-2, -1].
So, the correct answer is option B.

Question:47

Let the $f :R\rightarrow R$ be defined by $f (x) = 2x + cosx$ , then f(x) :
A. has a minimum at x = π
B. has a maximum, at x = 0
C. is a decreasing function
D. is an increasing function

Answer:

Given $f (x) = 2x + cosx$ if $f :R\rightarrow R$
Apply the first derivative and get
$\begin{aligned} &f(x)=\frac{d(2 x+\cos x)}{d x}\\ &\text { Apply the sum rule and } 0 \text { is the derivative of the constant, so }\\ &\Rightarrow f(x)=\frac{d(2 x)}{d x}+\frac{d(\cos x)}{d x} \end{aligned}$
Apply power rule and get
$\Rightarrow f'(x)=2-sin x$
Now, 1 is the maximum value of sin x.
$So f'(x)>0, \forall x$
So, function f is an increasing function.
So the correct answer is option D.

Question:48

y = $x (x - 3)^2$ decreases for the values of x given by :
A. $1 < x < 3$
B. $x < 0$
C. $x > 0$
D. $0<x<\frac{3}{2}$

Answer:

Given $y = x (x - 3)\textsuperscript{2}$
$\\ { \Rightarrow y=x(x\textsuperscript{2}-6x+9)}\\ { \Rightarrow y=x\textsuperscript{3}-6x\textsuperscript{2}+9x}\\$
Apply first derivative and get
$\frac{d y}{d x}=\frac{d\left(x^{3}-6 x^{2}+9 x\right)}{d x}$$
Apply sum rule of differentiation and get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(6 \mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(9 \mathrm{x})}{\mathrm{dx}}$$
Apply power rule and get
$\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-12 \mathrm{x}+9$ \\$\Rightarrow \frac{d y}{d x}=3\left(x^{2}-4 x+3\right)$$
Now, split middle term and get
$\\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3\left(\mathrm{x}^{2}-3 \mathrm{x}-\mathrm{x}+3\right)$ \\$\Rightarrow \frac{d y}{d x}=3(x(x-3)-1(x-3))$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3((\mathrm{x}-3)(\mathrm{x}-1))$$
Now, $\frac{\mathrm{d} y}{\mathrm{dx}}=0$ gives us
x=1, 3
The points divide this real number line into three intervals
$\\ {(- \infty , 1), (1,3) and (3, \infty )}\\ {(i)$ in the interval (- \infty , 1), f'(x)>0}\\ { \therefore \text{ f(x) is increasing in} (- \infty ,1)}\\ {(ii)$ \text{in the interval} (1,3), f'(x) \leq 0}\\ { \therefore $f(x) is decreasing in (1,3)}\\ {(iii) in the interval (3, \infty ), f'(x)>0}\\ { \therefore \text{f(x) is increasing in} (3, \infty )}\\$
So, the interval on which the function decreases is (1,3) i.e., 1<x<3
So the correct answer is option A.

Question:

The function $f (x) = 4 sin^3x - 6 sin^2x + 12 sinx + 100$ is strictly
A. increasing in $\left ( \pi,\frac{3\pi}{2} \right )$
B. decreasing in $\left ( \frac{\pi}{2},\pi \right )$
C. decreasing in $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
D. decreasing in $\left ( 0,\frac{\pi}{2} \right )$

Answer:

Given $f (x) = 4 sin^3x - 6 sin^2x + 12 sinx + 100$
Apply the first derivative and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(4 \sin ^{3} \mathrm{x}-6 \sin ^{2} \mathrm{x}+12 \sin \mathrm{x}+100\right)}{\mathrm{dx}}$$
Apply sum rule and get
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(4 \sin ^{3} \mathrm{x}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(6 \sin ^{2} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(12 \sin \mathrm{x})}{\mathrm{dx}}+0$$
Then apply power rule and get
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=12 \sin ^{2} \mathrm{x} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}-12 \sin \mathrm{x} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}+12 \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}$$
Now apply the derivative,
$\\ { \Rightarrow f' (x)=12 sin\textsuperscript{2}x (cos x)-12sin x (cos x)+12(cos x)}\\ { \Rightarrow f' (x)=12 sin\textsuperscript{2}x cos x-12sin x cos x +12 \cos x}\\ { \Rightarrow f' (x)=12 cos x(sin\textsuperscript{2}x -sin x +1)}\\$
Now, $1-sin x \geq 0$ and $sin\textsuperscript{2}x \geq 0$
Hence $sin\textsuperscript{2}x -sin x +1 \geq 0$
Therefore, $\\f^{\prime}(x)>0,$ when $\cos x>0,$ i.e $_{x}$ X $\in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ \\$
Hence f(x) is increasing when $\mathrm{x} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$$
and $f^{\prime}(x)<0$ , when $\cos x<0,$ i.e. $\mathrm{x} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$$
Hence f(x) is decreasing when $\mathrm{X} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$$
Now $\left(\frac{\pi}{2}, \pi\right) \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$$
Hence, f(x) is decreasing in $\left(\frac{\pi}{2}, \pi\right)$$
So the correct answer is option B

Question:50

Which of the following functions is decreasing on $\left ( 0,\frac{\pi}{2} \right )$ .
A. sin2x
B. tan x
C. cos x
D. cos 3x

Answer:

(i) Let f(x)=sin 2x
Apply first derivative and get
f’(x)=2cos 2x
Put f’(x)=0, and get
2cos 2x =0
⇒ cos 2x=0
It is possible when
0≤x≤2π
Thus, sin 2x does not decrease or increase on $x\in \left ( 0,\frac{\pi}{2} \right )$
(ii) Let f(x)=tan x
Apply first derivative and get
f’(x)= $sec^2 x$
Now. square of every number is always positive,
So, tan x is increasing function in $x\in \left ( 0,\frac{\pi}{2} \right )$
(iii) Let f(x)=cos x
Apply first derivative and get
f’(x)=-sin x
But, sin x>0 for $x\in \left ( 0,\frac{\pi}{2} \right )$
And -sin x<0 for $x\in \left ( 0,\frac{\pi}{2} \right )$
Hence f’(x)<0 for $x\in \left ( 0,\frac{\pi}{2} \right )$
⇒ cos x is strictly decreasing on $x\in \left ( 0,\frac{\pi}{2} \right )$
(iv) Let f(x)=cos 3x
Apply first derivative and get
f’(x)=-3sin 3x
Put f’(x)=0, we get
-3sin 3x=0
⇒ sin 3x=0
Because sin θ=0 if θ=0, π, 2π, 3π
⇒ 3x=0,π, 2π, 3π
$\begin{aligned} &\Rightarrow x=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi\\ &x \in\left(0, \frac{\pi}{2}\right)\\ &\Rightarrow x=\frac{\pi}{3}\\ &\text { since }\\ &\mathrm{x} \in\left(0, \frac{\pi}{2}\right) \end{aligned}$
so we write it on number line as

Now, this point $x=\frac{\pi}{3}$ divides the interval $\left(0, \frac{\pi}{2}\right)$$ into 2 disjoint intervals.
i.e. $\left(0, \frac{\pi}{3}\right)$ and $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$$
case 1 : for $\mathrm{x} \in\left(0, \frac{\pi}{3}\right)$$
$\\0<x<\frac{\pi}{3}$\\ $\Rightarrow 3 \times 0<3 x<\frac{\pi}{3} \times 3$ \\$0<3 x<\pi$$
So wher $\mathrm{x} \in\left(0, \frac{\pi}{3}\right), 3 \mathrm{x} \in(0, \pi) \ldots . .(\mathrm{a})$$
Also,
$\\ \sin \theta>0$ for $\theta \in(0, \pi)$ \\$\sin 3 x>0$ for $3 x \in(0, \pi)$$
From equation (a), we get
$\sin 3 x>0$ for $x \in\left(0, \frac{\pi}{3}\right)$$
sin 3x <0 for $x \in\left(0, \frac{\pi}{3}\right)$$
$\\\Rightarrow f^{\prime}(x)<0$ for \\$x \in\left(0, \frac{\pi}{3}\right)$ \\$\Rightarrow f(x)$ is strictly decreasing $\left(0, \frac{\pi}{3}\right)$$
case 2: for $x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$$
Now $\frac{\pi}{3}<x<\frac{\pi}{2}$$
$\\\Rightarrow 3 \times \frac{\pi}{3}<3 x<\frac{\pi}{2} \times 3$ \\$\Rightarrow \pi<3 x<\frac{3 \pi}{2}$ \\$\mathrm{x} \in\left(\frac{\pi}{3}, \frac{\pi}{3}\right), 3 \mathrm{x} \in\left(\pi, \frac{3 \pi}{2}\right) \ldots$(b)$
Also,
$\\ \sin \theta<0$ in $3^{\text {rd }}$ quadrant \\$\sin \theta<0$ for $\theta \in(0, \pi)$ \\$\sin \theta<0$ for \\$\theta \in\left(\pi, \frac{3 \pi}{2}\right)$ \\$\sin 3 x<0$ for \\$3 \mathrm{x} \in\left(\pi, \frac{3 \pi}{2}\right)$$
Equation (b) gives
$\\ \sin 3 x<0$ for $x \in\left(\frac{\pi}{3}, \frac{3 \pi}{2 \times 3}\right)$ \\$-\sin 3 x>0$ for $x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ \\$\Rightarrow f^{\prime}(x)<0$ for $x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ \\$\Rightarrow f(x)$ is strictly increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$$
Hence, cos 3 x does not decrease or increase on $\mathrm{x} \in\left(0, \frac{\pi}{2}\right)$$
So, the correct answer is option C i.e., $\cos \mathrm{x}$$ is decreasing in $\left(0, \frac{\pi}{2}\right)$$

Question:51

The function $f (x) = tanx - x$
A. always increases
B. always decreases
C. never increases
D. sometimes increases and sometimes decreases.

Answer:

Given $f (x) = tanx - x$
Apply first derivative and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\tan \mathrm{x}-\mathrm{x})}{\mathrm{dx}}$$
Apply sum rule and get
$\Rightarrow f^{\prime}(x)=\frac{d(\tan x)}{d x}-\frac{d(x)}{d x}$$
Apply derivative,
$\Rightarrow f^{\prime}(x)=\sec ^{2} x-1$$
Square of every number is always positive,
So $f^{\prime}(x)>0 \forall x \in R$$
So $f (x) = tanx - x$ always increases.
So the correct answer is option A

Question:52

If x is real, the minimum value of $x^2 - 8x + 17$ is
A. -1
B. 0
C. 1
D. 2

Answer:

Let $f(x)=x^2 - 8x + 17$
Apply first derivative and get
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{x}^{2}-8 \mathrm{x}+17\right)}{\mathrm{dx}}\\ &\text { Apply sum rule and get }\\ &\Rightarrow f^{\prime}(x)=\frac{d\left(x^{2}\right)}{d x}-8 \frac{d(x)}{d x}+0 \end{aligned}$
Apply derivative,
$\\ { \Rightarrow f' (x)=2x-8}\\ {$Put f'(x)=0 and get}\\ {2x-8=0}\\ { \Rightarrow 2x=8}\\ { \Rightarrow x=4}\\$
Hence the minimum value of f(x) at x=4 is given by
$\\ {f(x)= x\textsuperscript{2} - 8x + 17}\\ {f(4)=4\textsuperscript{2}-8(4)+17}\\ { \Rightarrow f(4)=16-32+17}\\ { \Rightarrow f(4)=1}\\$
So, if x is real, 1 is the minimum value of $x\textsuperscript{2} - 8x + 17$
So the correct answer is option C.

Question:53

The smallest value of the polynomial $x^3 - 18x^2 + 96x$ in [0, 9] is
A. 126
B. 0
C. 135
D. 160

Answer:

Let $f(x)=x^3 - 18x^2 + 96x$
Apply first derivative and get
$\begin{aligned} &f^{\prime}(x)=\frac{d\left(x^{3}-18 x^{2}+96 x\right)}{d x}\\ &\text { Apply sum rule and get }\\ &\Rightarrow f^{\prime}(x)=\frac{d\left(x^{3}\right)}{d x}-18 \frac{d\left(x^{2}\right)}{d x}+96 \frac{d(x)}{d x} \end{aligned}$
Apply derivative,
$\\ { \Rightarrow f' (x)=3x\textsuperscript{2}-36x+96}\\ {Put f'(x)=0, $and get critical points}\\ {3x\textsuperscript{2}-36x+96=0}\\ { \Rightarrow 3(x\textsuperscript{2}-12x+32)=0}\\ { \Rightarrow x\textsuperscript{2}-12x+32=0}\\$
Split middle term and get
$\\ { \Rightarrow x\textsuperscript{2}-8x-4x+32=0}\\ { \Rightarrow x(x-8)-4(x-8)=0}\\ { \Rightarrow (x-8)(x-4)=0}\\ { \Rightarrow x-8=0 or x-4=0}\\ { \Rightarrow x=8 or x=4}\\ { \Rightarrow x \in [0,9]}\\$
Now we find the values of f(x) at x=0, 4, 8, 9
$\\ {f(x)= x\textsuperscript{3} - 18x\textsuperscript{2} + 96x}\\ {f(0)= 0\textsuperscript{3} - 18(0)\textsuperscript{2} + 96(0)=0}\\ {f(4)= 4\textsuperscript{3} - 18(4)\textsuperscript{2} + 96(4)=64-288+384=160}\\ {f(8)= 8\textsuperscript{3} - 18(8)\textsuperscript{2} + 96(8)=512-1152+768=128}\\ {f(9)= 9\textsuperscript{3} - 18(9)\textsuperscript{2} + 96(9)=729-1458+864=135}\\$
Hence we find that 0 is the absolute minimum value of f(x) in [0,9] at x=0.
So the correct answer is option B.

Question:54

The function $f (x) = 2x^3 - 3x^2 - 12x + 4$ , has
A. two points of local maximum
B. two points of local minimum
C. one maxima and one minima
D. no maxima or minima
Answer:

Let $f (x) = 2x^3 - 3x^2 - 12x + 4$
Apply first derivative and get
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+4\right)}{\mathrm{dx}}\\ &\text { Apply sum rule and get }\\ &\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(2 \mathrm{x}^{3}\right)}{\mathrm{dx}}-3 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}-12 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+0 \end{aligned}$
Apply derivative,
$\\ { \Rightarrow f' (x)=6x\textsuperscript{2}-6x-12}\\$
Put f'(x)=0, and get
$\\{6x\textsuperscript{2}-6x-12=0}\\ { \Rightarrow 6(x\textsuperscript{2}-x-2)=80}\\ { \Rightarrow x\textsuperscript{2}-x-2=0}\\$
Split middle term and get
$\\ { \Rightarrow x\textsuperscript{2}-2x+x-2=0}\\ { \Rightarrow x(x-2)+1(x-2)=0}\\ { \Rightarrow (x-2)(x+1)=0}\\ { \Rightarrow x-2=0 or x+1=0}\\ { \Rightarrow x=2 or x=-1}\\$
Now we find the values of f(x) at x=-1, 2
$\\ {f(x)= 2x\textsuperscript{3} - 3x\textsuperscript{2} - 12x + 4}\\ {f(-1)= 2(-1)\textsuperscript{3} - 3(-1)\textsuperscript{2} - 12(-1) + 4=-2-3+12+4=11}\\ {f(2)= 2(2)\textsuperscript{3} - 3(2)\textsuperscript{2} - 12(2) + 4 =16-12-24+4=-16}\\$
Hence from above we find that the point of local maxima is x=-1 and 11 is the maximum value of f(x).
Whereas the point of local minima is x=2 and -16 is the minimum value of f(x).
So, the correct answer is option C.
Hence, the given function has 1 minima and 1 maxima.

Question:55

The maximum value of sin x cos x is
A. $\frac{1}{4}$
B. $\frac{1}{2}$
C. $\sqrt 2$
D. $2\sqrt 2$

Answer:

Let f(x)= sin x cos x
sin2x=2sin x cos x
$\Rightarrow f(x)=\frac{1}{2} \sin 2 x$$
Apply first derivative and get
$\\ \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\frac{1}{2} \sin 2 \mathrm{x}\right)}{\mathrm{dx}}$\\ $\Rightarrow f^{\prime}(x)=\frac{1}{2} \frac{d(\sin 2 x)}{d x}$$
Apply derivative,
$\Rightarrow f^{\prime}(x)=\frac{1}{2} \cdot \cos 2 x \cdot \frac{d(2 x)}{d x}$$
$\\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cdot \cos 2 \mathrm{x} \cdot 2$ $\\\Rightarrow f(x)=\cos 2 x \ldots \ldots(i)$$
Put $f^{\prime}(x)=0$ and get
$\\\cos 2 x=0$ \\$\Rightarrow \cos 2 x=\cos \frac{\pi}{2}$$
Equate the angles and get
$\\\Rightarrow 2 x=\frac{\pi}{2}$ \\$\Rightarrow x=\frac{\pi}{4}$$
Now we find second derivative by deriving equation (i) and get
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\cos 2 \mathrm{x})}{\mathrm{dx}}$$
Apply derivative,
$\\\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin 2 \mathrm{x} \cdot \frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}$ \\$\Rightarrow f^{\prime}(x)=-\sin 2 x .2$ \\$\Rightarrow f^{\prime}(x)=-2 \sin 2 x$$
Now we find the value of $f^{\prime \prime}(x)$ at $x=\frac{\pi}{4},$$ we get
$\\\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2 x$ \\$\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2\left(\frac{\pi}{4}\right)$ \\$\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin \left(\frac{\pi}{2}\right)$$
But $\sin \left(\frac{\pi}{2}\right)=1$$ so above equation becomes
$\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \times 1=-2<0$$
Hencee at $\mathrm{x}=\frac{\pi}{4}$ , $\mathrm{f}(\mathrm{x})$$ is maximum and $\frac{\pi}{4}$$ is the point of maxima.
Now we will find the maximum value of $\sin \mathrm{x} \cos \mathrm{x}$$ by substituting $\mathrm{x}=\frac{\pi}{4},$$ in $\mathrm{f}(\mathrm{x}),$$ we get
$\\ f(x)=\sin x \cos x$ \\$f\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)$ \\$\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}$ \\$\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{2}$$
So, maximum value of $\sin \mathrm{x} \cos \mathrm{x}$$ is $\frac{1}{2}$
So, the correct answer is option B

Question:56

At $x=\frac{5\pi}{6}$ , $f (x) = 2 sin3x + 3 cos3x$ is:
A. maximum
B. minimum
C. zero
D. neither maximum nor minimum

Answer:

Given $f (x) = 2 sin3x + 3 cos3x$
Apply first derivative and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(2 \sin 3 \mathrm{x}+3 \cos 3 \mathrm{x})}{\mathrm{dx}}$$
Apply sum rule and take the constant terms out and get
$\Rightarrow f^{\prime}(x)=2 \frac{d(\sin 3 x)}{d x}+3 \frac{d(\cos 3 x)}{d x}$$
Apply derivative,
$\Rightarrow f^{\prime}(x)=2 . \cos 3 x \cdot \frac{d(3 x)}{d x}+3 \cdot(-\sin 3 x) \cdot \frac{d(3 x)}{d x}$$
$\begin{aligned} &\Rightarrow f^{\prime}(x)=2 \cdot \cos 3 x \cdot 3-3 \cdot \sin 3 x \cdot 3\\ &\Rightarrow f'(x)=6 \cos 3 x-9 \sin 3 x \ldots \ldots(i)\\ &x=\frac{5 \pi}{6}\\ &\text { Now we find the value of } f^{\prime}(x) \text { at } \frac{5\pi}{6}\text { and get }\\ &\Rightarrow(f'(x))_{x=\frac{5 \pi}{6}}=6 \cos 3 x-9 \sin 3 x\\ &\Rightarrow(f'(x))_{x=\frac{5 \pi}{6}}=6 \cos \left(\frac{5 \pi}{2}\right)-9 \sin \left(\frac{5 \pi}{2}\right)\\ &\text { Splitting } \frac{5 \pi}{2} \text { as } 2 \pi+\frac{\pi}{2}\\ &\Rightarrow(\mathrm{f'}(\mathrm{x}))_{\mathrm{x}=\frac{5 \pi}{6}}=6 \cos \left(2 \pi+\frac{\pi}{2}\right)-9 \sin \left(2 \pi+\frac{\pi}{2}\right)\\ &\text { Also, } \cos (2 \pi+\theta)=\cos \theta \text { and } \sin (2 \pi+\theta)=\sin \theta\\ &\Rightarrow(\mathrm{f'}(\mathrm{x}))_{\mathrm{x}=\frac{5 \pi}{6}}=6 \cos \left(\frac{\pi}{2}\right)-9 \sin \left(\frac{\pi}{2}\right)\\ \end{aligned}$
$\\\cos \left(\frac{\pi}{2}\right)=0 \quad \sin \left(\frac{\pi}{2}\right)=1\\ \\\Rightarrow(f'(x))_{x=\frac{5 \pi}{6}}=6(0)-9(1)=-9$
And we found f’(x) at $x=\frac{5\pi}{6}$ not equal to 0.
So $x=\frac{5\pi}{6}$ cannot be point of minima or maxima.
Hence, $f (x) = 2 sin3x + 3 cos3x$ at $x=\frac{5\pi}{6}$ is not minima nor maxima.
So, the correct answer is option D.

Question:57

Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is:
A. 0
B. 12
C. 16
D. 32

Answer:

Given equation of curve is $y = -x^3 + 3x^2 + 9x - 27$
Apply first derivative and get
$\frac{d y}{d x}=\frac{d\left(-x^{3}+3 x^{2}+9 x-27\right)}{d x}$$
Apply sum rule and $\varrho$ is the differentiation of the constant term, so
$\frac{d y}{d x}=-\frac{d\left(x^{3}\right)}{d x}+3 \frac{d\left(x^{2}\right)}{d x}+9 \frac{d(x)}{d x}-0$$
Apply power rule and get
$\\\frac{d y}{d x}=-3 x^{2}+3 \cdot(2 x)+9.1$ $\\\\\frac{\mathrm{dy}}{\mathrm{dx}}=-3 \mathrm{x}^{2}+6 \mathrm{x}+9 \ldots \ldots \ldots$(i)$
Hence, it is the slope of the curve.
Now to find out the second derivative of the given curve, we will differentiate equation (i) once again
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(-3 \mathrm{x}^{2}+6 \mathrm{x}+9\right)}{\mathrm{dx}}$$
Apply sum rule and 0 is the differentiation of the constant term so
$\frac{d^{2} y}{d x^{2}}=-3 \frac{d\left(x^{2}\right)}{d x}+6 \frac{d(x)}{d x}+0$$
Apply power rule and get
$\\ \frac{d^{2} y}{d x^{2}}=-3(2 x)+6.1$ \\$\frac{d^{2} y}{d x^{2}}=-6 x+6=-6(x-1) \ldots$(ii)$
Now we will find the critical point by equating the second derivative to 0, we get
-6(x-1) =0
⇒ x-1=0
⇒ x=1
Now, to find out the third derivative of the given curve, we will differentiate equation (ii) once again
$\frac{d^{3} y}{d x^{3}}=\frac{d(-6 x+6)}{d x}$$
Apply sum rule and 0 is the differentiation of the constant term, so
$\frac{d^{3} y}{d x^{3}}=-6 \frac{d(x)}{d x}+0$$
Apply power rule and get
$\frac{d^{3} y}{d x^{3}}=-6<0$$
Hence, maximum slope is at $\mathrm{x}=1$$
Now, substitute $\mathrm{x}=1$$ in (i), and get
$\\ \left(\frac{d y}{d x}\right)_{x=1}=-3 x^{2}+6 x+9$ \\$\left(\frac{d y}{d x}\right)_{x=1}=-3(1)^{2}+6(1)+9=-3+6+9=12$$
Therefore, 12 is the maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ .
So, the correct answer is option B.

Question:58

$f (x) = x^x$ has a stationary point at
A. x = e
B. $x=\frac{1}{e}$
C. x = 1
D. $x=\sqrt e$

Answer:

Given equation is $f (x) = x^x$
Let $y= x^x$ ………(i)
Take logarithm on both side
$\log y=\log (x^x)$
⇒ log y=x log x
Apply first derivative and get
$\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(\mathrm{x} \log \mathrm{x})}{\mathrm{d} \mathrm{x}}$$
Apply product rule and get
$\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\mathrm{x} \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{d} \mathrm{x}}+\log \mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{d} \mathrm{x}}$$
Apply first derivative and get
$\\ \frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{x}+\log x \cdot 1$ \\$\Rightarrow \frac{d y}{d x}=y(1+\log x)$$
Substitute value of y from (i) and get
$\Rightarrow \frac{d y}{d x}=x^{x}(1+\log x)$$
Now we find the critical point by equating (i) to 0 and get
$\\ {x\textsuperscript{x} (1+log x)=0}\\ { \Rightarrow 1+log x =0 \: \: as \: \: x\textsuperscript{x}$ cannot be equal to 0}\\$
$\\ { \Rightarrow log x=-1}\\ {But -1=log e\textsuperscript{-1}}\\ { \Rightarrow log x=log e\textsuperscript{-1}}\\$
Equate the terms and get
$x= e\textsuperscript{-1}$
$\Rightarrow X=\frac{1}{e}$
Therefore f(x) has a stationary point at $X=\frac{1}{e}$
So, the correct answer is option B.

Question:59

The maximum value of $\left ( \frac{1}{x} \right )^x$ is:
A. $e$
B. $e^e$
C. $e^{\frac{1}{e}}$
D. $\left ( \frac{1}{e} \right )^{\frac{1}{e}}$

Answer:

Let $y=\left(\frac{1}{x}\right)^{x}$$
Take logarithm on both side
$\\ \log y=\log \left(\frac{1}{x}\right)^{x}$ \\$\log y=x \log \frac{1}{x}$$
Applying first derivative and get
$\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x} \log \frac{1}{\mathrm{x}}\right)}{\mathrm{dx}}$$
Apply product rule and get
$\\\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\mathrm{x} \frac{\mathrm{d}\left(\log \frac{1}{\mathrm{x}}\right)}{\mathrm{dx}}+\log \frac{1}{\mathrm{x}} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}$$
Applying first derivative and get
$\begin{aligned} &\frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{\frac{1}{x}} \cdot \frac{d\left(\frac{1}{x}\right)}{d x}+\log \frac{1}{x}, 1\\ &\frac{1}{y} \cdot \frac{d y}{d x}=x^{2} \cdot \frac{d\left(x^{-1}\right)}{d x}+\log \frac{1}{x}\\ &\frac{1}{y} \cdot \frac{d y}{d x}=x^{2} \cdot(-1) \cdot x^{-1-1}+\log \frac{1}{x}\\ &\frac{1}{y} \cdot \frac{d y}{d x}=x^{2} \cdot(-1) \cdot x^{-2}+\log \frac{1}{x}\\ &\frac{1}{y} \cdot \frac{d y}{d x}=-1+\log \frac{1}{x}\\ &\Rightarrow \frac{d y}{d x}=y\left(-1+\log \frac{1}{x}\right)\\ &\text { Substitute value of } y \text { from (i) and get }\\ &\Rightarrow \frac{d y}{d x}=\left(\frac{1}{x}\right)^{x} \cdot\left(-1+\log \frac{1}{x}\right) \end{aligned}$
Now we find critical point by equating (i) to 0
$\begin{aligned} &\left(\frac{1}{x}\right)^{x} \cdot\left(-1+\log \frac{1}{x}\right)=0\\ &\Rightarrow\left(-1+\log \frac{1}{x}\right)=0 \quad\left(\frac{1}{x}\right)^{x} \text { can't be equal to } 0\\ &\Rightarrow\left(\log \frac{1}{x}\right)=1\\ &\text { But } 1=\log \mathrm{e}^{1}\\ &\Rightarrow\left(\log \frac{1}{x}\right)=\log e\\ &\text { Equate the terms }\\ &\Rightarrow \frac{1}{\mathrm{x}}=e\\ &\Rightarrow x=\frac{1}{e} \end{aligned}$
Therefore f(x) has a stationary point at $x=\frac{1}{e}$ .
i.e the maximum value of $f\left (\frac{1}{e} \right )=e^{\frac{1}{e}}$
So, the correct answer is option C.

Question:60

Fill in the blanks in each of the following
The curves $y = 4x^2 + 2x - 8$ and $y = x^3 - x + 13$ touch each other at the point_____.

Answer:

Given the first curve is $y = 4x^2 + 2x - 8$
Applying first derivative and get
$\frac{d y}{d x}=\frac{d\left(4 x^{2}+2 x-8\right)}{d x}$$
Apply sum rule and 0 is the differentiation of the constant term is 0,so
$\frac{d y}{d x}=4 \frac{d\left(x^{2}\right)}{d x}+2 \frac{d(x)}{d x}+0$$
Apply power rule and get
$\frac{d y}{d x}=8 x+2$$
This is the slope of the first curve; let $m_{1}$ be equal to this.
$\Rightarrow m_{1}=8 x+2 \ldots \ldots(0)$$
The second curve is $y=x^{3}-x+13$$
Applying first derivative and get
$\frac{d y}{d x}=\frac{d\left(x^{3}-x+13\right)}{d x}$$
Apply sum rule and 0 is the differentiation of the constant term, so
$\begin{aligned} &\frac{d y}{d x}=\frac{d\left(x^{3}\right)}{d x}-\frac{d(x)}{d x}+0\\ &\text { Apply power rule and get }\\ &\frac{d y}{d x}=3 x^{2}-1 \end{aligned}$
This is the slope of the second curve; let $m_2$ be equal to this.
$\Rightarrow m\textsubscript{2}=3x\textsuperscript{2}-1 \ldots \ldots (ii)$
Now the slopes must be equal, because they touch each other, i.e.,
$m\textsubscript{1}=m\textsubscript{2}$
$\\ { \Rightarrow 8x+2=3x\textsuperscript{2}-1}\\ { \Rightarrow 3x\textsuperscript{2}-8x-1-2=0}\\ { \Rightarrow 3x\textsuperscript{2}-8x-3=0}\\$
Split middle term and get
$\\ { \Rightarrow 3x\textsuperscript{2}+x-9x -3=0}\\ { \Rightarrow x(3x+1)-3(3x+1)=0}\\ { \Rightarrow (3x+1)(x-3)=0}\\ { \Rightarrow (3x+1)=0 or (x-3)=0}\\$
$\\ \Rightarrow x=-\frac{1}{3}$ or $x=3$ \\$
Substitute $x=-\frac{1}{3}$$ in both the equations and get
For first curve, $y=4 x^{2}+2 x-8$$
$\\ \Rightarrow y=4\left(-\frac{1}{3}\right)^{2}+2\left(-\frac{1}{3}\right)-8$ \\$\Rightarrow y=4\left(\frac{1}{9}\right)-\frac{2}{3}-8$ \\$\Rightarrow \mathrm{y}=\frac{4-2 \times 3-8 \times 9}{9}=-\frac{74}{9}$$
For second curve, $y=x^{3}-x+13$$
$\Rightarrow y=\left(-\frac{1}{3}\right)^{3}-\left(-\frac{1}{3}\right)+13$$
$\begin{aligned} &\Rightarrow \mathrm{y}=-\frac{1}{27}+\frac{1}{3}+13\\ &\Rightarrow \mathrm{y}=\frac{-1+1 \times 9+13 \times 27}{27}=\frac{8}{27}+13\\ &\text { Thus at }\\ &\mathrm{x}=-\frac{1}{3^{\prime}} \text { curves don't touch } \end{aligned}$
Substitute x=3 in both equations
For first curve, $y=4 x^{2}+2 x-8$$
$\\ { \Rightarrow y=4(3)\textsuperscript{2}+2(3)-8}\\ { \Rightarrow y=4(9)+6-8=34}\\$
For second curve, $y=x^{3}-x+13$$
$\\ { \Rightarrow y= (3)\textsuperscript{3} - (3) + 13}\\ { \Rightarrow y=27-3+13=37}\\$
Hence at x=3 both curves don't touch
So, the curves $y = 4x\textsuperscript{2} + 2x - 8$ and $y = x\textsuperscript{3} - x + 13$ do not touch each other.

Question:61

Fill in the blanks in each of the following
The equation of normal to the curve $y = tanx$ at (0, 0) is ________.

Answer:

Given curve is $y = tanx$
Apply first derivative and get
$\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}(\tan \mathrm{x})}{\mathrm{dx}}$ \\$\Rightarrow \frac{d y}{d x}=\sec ^{2} x$$
It is the slope of tangent
Substitute (0,0) in slope and get
$\\\left(\frac{d y}{d x}\right)_{(0,0)}=\sec ^{2} x$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,0)}=\sec ^{2} 0=1$$
Hence, -1 is the slope of the normal to the curve at (0,0)
Hence the equation is
$\\$y-0=-1(x-0)$ \\$\Rightarrow y=-x$ \\$or $x+y=0$$

Question:62

Fill in the blanks in each of the following
The values of a for which the function $f (x) = sinx - ax + b$ increases on R are ______.

Answer:

Given $f (x) = sinx - ax + b$
Apply first derivative and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sin \mathrm{x}-\mathrm{ax}+\mathrm{b})}{\mathrm{dx}}$$
Apply sum rule and 0 is the differentiation of the constant term, so
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}-\mathrm{a} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+0$$
Apply first derivative and get
$f^{\prime}(x)=\cos x-a$$
Also f(x) increases on R
$\\ $\Rightarrow \mathrm{f}^{\prime}(x) \geq 0 \ \ \forall x \epsilon R$ \\$\Rightarrow \cos x-a \geq 0\ \ \forall x \epsilon R \\$\Rightarrow \cos x \geq a \ \ \forall x \epsilon R$
This is possible when $a\leq-1$
Hence $a \in(-\infty,-1]$$
The values of a increases on $\mathrm{R}$ are $(-\infty,-1] .$$

Question:63

Fill in the blanks in each of the following
The function $f(x)=\frac{2x^2-1}{x^4},x>0$ decreases in the interval _______.

Answer:

Given $f(x)=\frac{2x^2-1}{x^4},x>0$
After applying derivative, we get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\frac{2 \mathrm{x}^{2}-1}{\mathrm{x}^{4}}\right)}{\mathrm{d} \mathrm{x}}$
Apply quotient rule and 0 is the differentiation of the constant term, so
$\\ f^{\prime}(x)=\frac{x^{4} \cdot \frac{d\left(2 x^{2}-1\right)}{d x}-\left(2 x^{2}-1\right) \cdot \frac{d\left(x^{4}\right)}{d x}}{\left(x^{4}\right)^{2}} \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(x^{4} \cdot \frac{d\left(2 x^{2}-1\right)}{d x}-\left(2 x^{2}-1\right) \cdot \frac{d\left(x^{4}\right)}{d x}\right) \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(x^{4} \cdot(4 x)-\left(2 x^{2}-1\right) \cdot\left(4 x^{3}\right)\right) \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(4 x^{5}-\left(2 x^{2}\right) \cdot\left(4 x^{3}\right)-\left(4 x^{3}\right)\right) \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(4 x^{5}-8 x^{5}-\left(4 x^{3}\right)\right) \\ f^{\prime}(x)=\frac{x^{3}}{x^{8}}\left(4 x^{2}-8 x^{2}-4\right) \\ f^{\prime}(x)=\frac{1}{x^{5}}\left(-4 x^{2}-4\right) \\ f^{\prime}(x)=\frac{-4}{x^{5}}\left(x^{2}-1\right)$
Equate this with 0 and get
$\\ f^{\prime}(x)=0$ \\$\Rightarrow \frac{-4}{x^{5}}\left(x^{2}-1\right)=0$ \\$\Rightarrow x^{2}-1=0$ \\$\Rightarrow x^{2}=1$ \\$\Rightarrow x=\pm 1$$
$(-\infty,-1),(-1,0),(0,1)$$ and $(1, \infty)$$ are the intervals formed by these two critical numbers
(i) in the interval $(-\infty,-1), f^{\prime}(x)>0$$
$\therefore f(x)$ is increasing in $(-\infty,-1)$$
(ii) in the interval (-1,0), $\mathrm{f}^{\prime}(\mathrm{x}) \leq 0$$
$\therefore \mathrm{f}(\mathrm{x})$ is decreasing in (-1,0)
(iii) in the interval (0,1), $\mathrm{f'}(x)>0$$
$\therefore \mathrm{f}(\mathrm{x})$ is increasing in $(0,1)$
(iii) in the interval $(1, \infty), f^{\prime}(x)<0$
$\therefore \mathrm{f}(\mathrm{x})$ is decreasing in $(1, \infty)$
$f(x)=\frac{2 x^{2}-1}{x^{4}}, x>0$$
Therefore, the function $f(x)=\frac{2 x^{2}-1}{x^{4}}, x>0$$ decreases in the interval $(1, \infty)$ .

Question:64

Fill in the blanks in each of the following
The least value of the function $f(x)=ax+\frac{b}{x}$ $(a > 0, b > 0, x > 0)$ is ______.

Answer:

Given $f(x)=ax+\frac{b}{x}$
After applying the derivative
$f^{\prime}(x)=\frac{d\left(a x+\frac{b}{x}\right)}{d x}$$
Apply sum rule and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\mathrm{ax})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\frac{\mathrm{b}}{\mathrm{x}}\right)}{\mathrm{dx}}$$
Apply quotient rule on second part and get
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a}+\frac{\mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{b})}{\mathrm{d} \mathrm{x}}-\mathrm{b} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{d} \mathrm{x}}}{(\mathrm{x})^{2}}$ \\$f^{\prime}(x)=a+\frac{x \cdot 0-b \cdot 1}{x^{2}}$ \\$f^{\prime}(x)=a-\frac{b}{x^{2}}$$
Equate it with 0 and get
$\\f^{\prime}(x)=0$ \\$\Rightarrow \mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}=0$ \\$\Rightarrow \mathrm{a}=\frac{\mathrm{b}}{\mathrm{x}^{2}}$ \\$\Rightarrow x^{2}=\frac{b}{a}$ \\$\Rightarrow x=\sqrt{\frac{b}{a}}($ as $x>0)$$
Now given by second derivative,
Apply derivative and get
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}\right)}{\mathrm{dx}}$$
Apply sum rule and get
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\mathrm{a})}{\mathrm{dx}}-\frac{\mathrm{d}\left(\frac{\mathrm{b}}{\mathrm{x}^{2}}\right)}{\mathrm{dx}}$$
Apply quotient rule on the second part and get
$\\ \mathrm{f}^{\prime \prime}(\mathrm{x})=0-\frac{\mathrm{x}^{2} \cdot \frac{\mathrm{d}(\mathrm{b})}{\mathrm{d} \mathrm{x}}-\mathrm{b} \cdot \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}}{\left(\mathrm{x}^{2}\right)^{2}}$ \\$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{x}^{2} \cdot 0-\mathrm{b} \cdot(2 \mathrm{x})}{\mathrm{x}^{4}}$ \\$f^{\prime \prime}(x)=\frac{2 x b}{x^{4}}=\frac{2 b}{x^{3}}$$
Now, equate it with $x=\sqrt{\frac{b}{a}}$$
$\\\left(\mathrm{f}^{\prime \prime}(\mathrm{x})\right) \mathrm{x}=\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}=\frac{2 \mathrm{~b}}{\mathrm{x}^{3}}$ \\$\left(\mathrm{f}^{\prime \prime}(\mathrm{x})\right)_{\mathrm{x}=\sqrt{\frac{\mathrm{b}}{2}}}=\frac{2 \mathrm{~b}}{\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)^{3}}$ \\$\left(\mathrm{f}^{\prime \prime}(\mathrm{x})\right)_{\mathrm{x}=\sqrt{\frac{\mathrm{b}}{2}}}=\frac{2 \mathrm{ab}}{\mathrm{b}}=2 \mathrm{a}>0$$
The least value of f(x) is
$\begin{array}{l} \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\mathrm{a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)+\frac{\mathrm{b}}{\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\mathrm{a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)+\mathrm{b} \sqrt{\frac{\mathrm{a}}{\mathrm{b}}} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{\mathrm{ab}+\mathrm{ab}}{\sqrt{\mathrm{ab}}} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{ab}}{\sqrt{\mathrm{ab}}} \end{array}$
Multiply and divide by $\sqrt{\mathrm{ab}}$$ and get
$\\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{ab}}{\sqrt{\mathrm{ab}}} \times \frac{\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{ab}}}$ \\$\Rightarrow f\left(\sqrt{\frac{b}{a}}\right)=2 \sqrt{a b}$$
Therefore, the least value of function $f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$$ is $2 \sqrt{a b}$$

With the NCERT exemplar Class 12 Maths solutions chapter 6, you will have a better grasp of the topics and how to solve the questions. Our team of experts will help in expanding the questions in a way that will make it easier for the students to make sense of. Also, the solutions provided are following CBSE standards and guidelines.

You can easily download the pdf copy of NCERT exemplar Class 12 Maths solutions chapter 6 by using the Maths NCERT exemplar Class 12 solutions chapters 6 pdf download feature which includes the solutions for the in-chapter questions and final exercise questions. Our guidance team and teachers have solved the questions in a way that is easy to understand and simple to grasp.

The language is simple and the process is shown in a step-by-step manner while being thorough enough for the students to understand every question in detail. Solving the questions before exams during the preparatory phase can help in getting an idea of what will be asked and how to manage the score well in mathematics.

Subtopics in NCERT Exemplar Class 12 Maths Solutions Chapter 6

The sub-topics that are covered in this chapter are:

Introduction
Rate of change of quantities
Increasing and decreasing functions
Normals and Tangents
Approximations
Maxima and minima
Minimum and maximum value of the function is a closed interval

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What you will learn from NCERT Exemplar Class 12 Maths Solutions Chapter 6?

In NCERT exemplar Class 12 Mathematics chapter 6 Applications of Derivatives, the students will come to know in detail about derivatives of functions and numbers. Also, one will know about the applications and uses of these derivatives. One will learn about the most crucial part of the derivatives topic, and that is the rate of change. One will learn about how one or two quantities change due to change in some other quantity.

Students will get a clear picture of how derivatives affect functions after referring to NCERT exemplar Class 12 Maths solutions chapter 6. It will teach whether the function is increasing, decreasing or strictly increasing/decreasing. This chapter will also cover Rolle's Theorem, LaGrange's Theorem of Mean Value, finding tangent line equations, and all its cases.

They will learn in-depth about approximations and how to find approximate value to some of the quantities by reading NCERT exemplar Class 12 Maths solutions chapter 6. One of the most crucial things that the students will cover is details about finding maxima and minima of functions, points of local, etc. They will also learn about the closed function's absolute maxima and minima.

NCERT Exemplar Class 12 Maths Solutions

Chapter 1 Relations and Functions

Chapter 2 Inverse Trigonometric Functions

Chapter 3 Matrices

Chapter 4 Determinants

Chapter 5 Continuity and Differentiability

Chapter 7 Integrals

Chapter 8 Applications of Integrals

Chapter 9 Differential Equations

Chapter 10 Vector Algebra

Chapter 11 Three dimensional Geometry

Chapter 12 Linear Programming

Chapter 13 Probability

Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 6

In NCERT Exemplar Class 12 Mathematics solutions chapter 6, the students will learn in detail about the derivatives and their fundamentals and applications. Several topics are covered in this, which has significance in higher calculus and even other subjects as it is a very common topic in exams.

In NCERT exemplar Class 12 Maths solutions chapter 6, the students will come face to face with several topics like decreasing and increasing functions, minima and maxima, approximations, normal, tangents, change of quantities and its rate which are very commonly asked in the exams.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Also, check NCERT Solutions for questions given in the book:

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

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Frequently Asked Question (FAQs)

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First of all, you should read the chapters well for the exam. You must also go through the questions after the end of each chapter.

2. Who has prepared the solutions?

The Class 12 Maths NCERT exemplar solutions chapters 6 are prepared by us who are experts in mathematics. The solutions are prepared after understanding and reference from professional books.

3. Can I download the solutions for this chapter?

Yes, you can click the NCERT exemplar Class 10 Maths solutions chapter 1 pdf download in the given link on the page.

4. Are these solutions helpful for board examinations?

Yes, the NCERT exemplar Class 12 solutions for Mathematics chapter 6 will provide you with the ability to ace the board exams if you utilise it properly.

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I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

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Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

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Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Maths Solutions Chapter 6 Applications Of Derivatives

Subtopics in NCERT Exemplar Class 12 Maths Solutions Chapter 6

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