NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

Question:10: A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer:

Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given that $\frac{dx}{dt} = 2$ cm/s
We need to find the rate at which the height of the ladder decreases $(\frac{dh}{dt})$
length of ladder(L) = 5m and x = 4m (given)
By Pythagoras' theorem, we can say that
$h^{2}+x^{2} = L^{2}$
$⇒h^{2} = L^{2} - x^{2}$
$⇒h= \sqrt{L^{2} - x^{2}}$
Differentiate on both sides with respect to t,
$\frac{dh}{dt}=\frac{d(\sqrt{L^{2}-x^{2}})}{dx}.\frac{dx}{dt} = \frac{1}{2}\frac{-2x}{\sqrt{5^{2}-x^{2}}}.\frac{dx}{dt} = \frac{-x}{\sqrt{25-x^{2}}}\frac{dx}{dt}$
at x = 4
$\frac{dh}{dt} = \frac{-4}{\sqrt{25-16}}\times 2 = \frac{-4}{3} \times 2 =-\frac{8}{3}$ cm/s
Hence, the rate at which the height of the ladder decreases is $\frac{8}{3}$ cm/s.

Question:11: A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

We need to find the point at which $\frac{dy}{dt} = 8\frac{dx}{dt}$
Given the equation of curve = $6y = x^3 + 2$
Differentiate both sides w.r.t. t
$6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0$
$= 3x^{2}.\frac{dx}{dt}$
$\frac{dy}{dt} = 8\frac{dx}{dt}$ (required condition)
$⇒6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}$
$⇒3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt}$
$\Rightarrow x^{2} = \frac{48}{3} = 16$
$⇒x = \pm 4$
when x = 4, $y=\frac{4^3+2}{6}=\frac{64+2}{6}=\frac{66}{6}=11$

and
when x = -4, $y=\frac{(-4)^3+2}{6}=\frac{-64+2}{6}=\frac{-62}{6}=-\frac{31}{3}$

So, the coordinates are $(4,11)$ and $(-4,\frac{-31}{3})$.

Question:12: The radius of an air bubble is increasing at the rate of $\frac12$ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

It is given that $\frac{dr}{dt} = \frac{1}{2}$ cm/s
We know that the shape of the air bubble is spherical
So, volume(V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi$ cm3/s
Hence, the rate of change in volume is $2\pi$ cm3/s.

Question:13 A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}( 2x +1)$. Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere(V) = $\frac{4}{3}\pi r^{3}$
Diameter = $\frac{3}{2}(2x+1)$
So, radius(r) = $\frac{3}{4}(2x+1)$
$\frac{dV}{dx} = \frac{d(\frac{4}{3}\pi r^{3})}{dx} = \frac{d(\frac{4}{3}\pi (\frac{3}{4}(2x+1))^{3})}{dx} = \frac{4}{3}\pi\times 3\times\frac{27}{64}(2x+1)^{2}\times 2$
$= \frac{27}{8}\pi (2x+1)^{2}$

Question:14: Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer:

Given = $\frac{dV}{dt} = 12$ cm3/s and $h = \frac{1}{6}r$
To find = $\frac{dh}{dt}$ at h = 4 cm
Volume of cone(V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt} = \frac{d(\frac{1}{3}\pi (6h)^{2}h)}{dh}.\frac{dh}{dt} = \frac{1}{3}\pi\times36\times3h^{2}.\frac{dh}{dt} = 36\pi \times(4)^{2}.\frac{dh}{dt}$
$⇒\frac{dV}{dt} = 576\pi.\frac{dh}{dt}$

Question:15 The total cost C(x) in Rupees associated with the production of x units of an item is given by $C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$. Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = $\frac{dC}{dx}$
$C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$
$\frac{dC}{dx} = \frac{d(0.007x^{3} - 0.003x^{2}+15x+400)}{dx} = 3\times 0.007x^{2} - 2\times0.003x+15$
$= .021x^{2} - 0.006x + 15$
Now, at x = 17
MC $= 0.021(17)^{2} - 0.006(17) + 15$
$= 6.069 - 0.102 + 15$
$= 20.967$
Hence, the marginal cost when 17 units are produced is 20.967.

Question:16 The total revenue in Rupees received from the sale of x units of a product is given by $R ( x) = 13 x^2 + 26 x + 15$. Find the marginal revenue when x = 7

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R (x) = 13 x^2 + 26 x + 15$
$\frac{dR}{dx} = \frac{d(13x^{2}+26x+15)}{dx} = 13\times2x+ 26 = 26(x+1)$
at x = 7
$\frac{dR}{dx} = 26(7+1) = 26\times8 = 208$
Hence, marginal revenue when x = 7 is 208.

Question:17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is (A) 10π (B) 12π (C) 8π (D) 11π

Answer:

Area of circle(A) = $\pi r^{2}$
$\frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2\pi r$
Now, at r = 6 cm
$\frac{dA}{dr}= 2\pi \times 6 = 12\pi$ cm2/s
Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is $12\pi$ cm2/s
Hence, the correct answer is B.

Question:18 The total revenue in Rupees received from the sale of x units of a product is given by $R(x) = 3x^2 + 36x + 5$. The marginal revenue, when x = 15, is (A) 116 (B) 96 (C) 90 (D) 126

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 3 x^2 + 36 x + 5$
$\frac{dR}{dx} = \frac{d(3x^{2}+36x+5)}{dx} = 3\times2x+ 36 = 6(x+6)$
at x = 15
$\frac{dR}{dx} = 6(15+6) = 6\times21 = 126$
Hence, marginal revenue when x = 15 is 126.
Hence, the correct answer is D.

Question:1. Show that the function given by f (x) = 3x + 17 is increasing on R.

Answer:
Let $x_1$ and $x_2$ be two numbers in R.
$x_1 < x_2$
$ \Rightarrow 3x_1 < 3 x_2$
$ \Rightarrow 3x_1 + 17 < 3x_2+17$
$ \Rightarrow f(x_1)< f(x_2)$
Hence, f is strictly increasing on R.

Question:2. Show that the function given by $f(x) = e^{2x}$ is increasing on R.

Answer:
Let $x_1$ and $ x_2$ be two numbers in R.
$x_1 \ < \ x_2$
$ \Rightarrow 2x_1 < 2x_2$
$ \Rightarrow e^{2x_1} < e^{2x_2}$
$ \Rightarrow f(x_1) < f(x_2)$
Hence, the function $f(x) = e^{2x}$ is strictly increasing in R.

Question:3(a) Show that the function given by f (x) = $\sin x$ is increasing in $\left ( 0 , \frac{\pi}2 \right )$

Answer:
Given $f(x) = \sin x$
$f^{'}(x) = \cos x$
Since, $\cos x > 0$ for each x $\in \left ( 0,\frac{\pi}{2} \right )$
$f^{'}(x) > 0$
Hence, $f(x) = \sin x$ is strictly increasing in $\left ( 0,\frac{\pi}{2} \right )$.

Question:3(b) Show that the function given by f (x) = $\sin x$ is decreasing in $\left ( \frac{\pi}{2},\pi \right )$

Answer:
$f(x) = \sin x$
$f^{'}(x) = \cos x$
Since, $\cos x < 0$ for each $x \ \epsilon \left ( \frac{\pi}{2},\pi \right )$
So, we have $f^{'}(x) < 0$
Hence, $f(x) = \sin x$ is strictly decreasing in $\left ( \frac{\pi}{2},\pi \right )$

Question:3(c) Show that the function given by f (x) = $\sin x$ is neither increasing nor decreasing in $( 0 , \pi )$

Answer:
We know that $\sin x$ is strictly increasing in $\left ( 0,\frac{\pi}{2} \right )$ and strictly decreasing in $\left ( \frac{\pi}{2},\pi \right )$
So, by this, we can say that $f(x) = \sin x$ is neither increasing nor decreasing in the range $\left ( 0,\pi \right )$.

Question:4(a). Find the intervals in which the function f given by $f ( x) = 2x ^2 - 3 x$ is increasing

Answer:
$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
Now,
$f^{'}(x) = 0$
$⇒4x - 3 = 0$
$⇒x = \frac{3}{4}$

So, the range is $\left ( -\infty, \frac{3}{4} \right ) $ and $ \left ( \frac{3}{4}, \infty \right )$
So,
$f(x)< 0$, when $x \ \in \left ( -\infty,\frac{3}{4} \right )$ Hence, f(x) is strictly decreasing in this range
and
$f(x) > 0$, when $x \in \left ( \frac{3}{4},\infty \right )$
Hence, f(x) is strictly increasing in this range.
Hence, $f ( x) = 2x ^2 - 3 x$ is strictly increasing in $x \in \left ( \frac{3}{4},\infty \right )$

Question:4(b) Find the intervals in which the function f given by $f ( x) = 2 x ^2 - 3 x$ is decreasing

Answer:
$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
Now,
$f^{'}(x) = 0$
$⇒4x - 3 = 0$
$⇒x = \frac{3}{4}$

So, the range is $\left ( -\infty, \frac{3}{4} \right )$ and $ \left ( \frac{3}{4}, \infty \right )$.
So,
$f(x)< 0$, when $x \ \in \left ( -\infty,\frac{3}{4} \right )$ Hence, f(x) is strictly decreasing in this range
and
$f(x) > 0$, when $x \in \left ( \frac{3}{4},\infty \right )$ Hence, f(x) is strictly increasing in this range.
Hence, $f ( x) = 2x ^2 - 3 x$ is strictly decreasing in $x \in \left ( -\infty ,\frac{3}{4}\right )$.

Question:5(a) Find the intervals in which the function f given by $f (x) = 2x^3 - 3x ^2 - 36 x + 7$ is increasing

Answer:
It is given that
$f (x) = 2x^3 - 3x ^2 - 36 x + 7$
So,
$f^{'}(x)= 6x^{2} - 6x - 36$
Also, $f^{'}(x)= 0$
$⇒6x^{2} - 6x -36 =0$
$ \Rightarrow 6 (x^{2} - x-6)$
$⇒x^{2} - x-6 = 0$
$⇒x^{2} - 3x+2x-6 = 0$
$⇒x(x-3) + 2(x-3) = 0\\$
$⇒(x+2)(x-3) = 0$
$\therefore x = -2 , x = 3$
So, three ranges are there $(-\infty,-2) , (-2,3) $ and $ (3,\infty)$
Function $f^{'}(x)= 6x^{2} - 6x - 36$ is positive in interval $(-\infty,-2) , (3,\infty)$ and negative in the interval $(-2,3)$.
Hence, $f (x) = 2x^3 - 3x ^2 - 36 x + 7$ is strictly increasing in $(-\infty,-2) \cup (3,\infty)$ and strictly decreasing in the interval (-2, 3).

Question:5(b) Find the intervals in which the function f given by $f ( x) = 2x ^3 - 3x ^2 - 36x + 7$ is decreasing

Answer:
We have $f ( x) = 2x ^3 - 3x ^2 - 36x + 7$

Differentiating the function with respect to x, we get :

$f' ( x) = 6x ^2 - 6x - 36= 6\left ( x-3 \right )\left ( x+2 \right )$

When $f'(x)\ =\ 0$ , we have :

$⇒0\ = 6\left ( x-3 \right )\left ( x+2 \right )$
$⇒\left ( x-3 \right )\left ( x+2 \right )\ =\ 0$
So, three ranges are there $(-\infty,-2) , (-2,3)$ and $ (3,\infty)$
Function $f^{'}(x)= 6x^{2} - 6x - 36$ is positive in the interval $(-\infty,-2) , (3,\infty)$ and negative in the interval (-2,3)

So, f(x) is decreasing in (-2, 3).

Question:6(a) Find the intervals in which the following functions are strictly increasing or
decreasing: $x ^2 + 2x -5$

Answer:

f(x) = $x ^2 + 2x -5$
$f^{'}(x) = 2x + 2 = 2(x+1)$
Now,
$f^{'}(x) = 0 \\ 2(x+1) = 0\\ x = -1$
The range is from $(-\infty,-1) \ and \ (-1,\infty)$.
In interval $(-\infty,-1)$, $f^{'}(x)= 2(x+1)$ is -ve
Hence, function f(x) = $x ^2 + 2x -5$ is strictly decreasing in interval $(-\infty,-1)$.
In interval $(-1,\infty)$, $f^{'}(x)= 2(x+1)$ is +ve.
Hence, function f(x) = $x ^2 + 2x -5$ is strictly increasing in interval $(-1,\infty)$.

Question:6(b) Find the intervals in which the following functions are strictly increasing or
Decreasing: $10 - 6x - 2x^2$

Answer:
Given function is,
$f(x) = 10 - 6x - 2x^2$
$f^{'}(x) = -6 - 4x$
Now,
$f^{'}(x) = 0$
$⇒6+4x= 0$
$⇒x= -\frac{3}{2}$
So, the range is $(-\infty , -\frac{3}{2})$ and $ (-\frac{3}{2},\infty)$.
In interval $(-\infty , -\frac{3}{2})$ , $f^{'}(x) = -6 - 4x$ is +ve.
Hence, $f(x) = 10 - 6x - 2x^2$ is strictly increasing in the interval $(-\infty , -\frac{3}{2})$.
In interval $( -\frac{3}{2},\infty)$ , $f^{'}(x) = -6 - 4x$ is -ve.
Hence, $f(x) = 10 - 6x - 2x^2$ is strictly decreasing in interval $( -\frac{3}{2},\infty)$.

Question:6(c) Find the intervals in which the following functions are strictly increasing or
decreasing: $- 2 x^3 - 9x ^2 - 12 x + 1$

Answer:
Given function is,
$f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}$
$f^{'}(x) = - 6 x^2 - 18x - 12$
Now,
$f^{'}(x) = 0\\ - 6 x^2 - 18x - 12 = 0$
$⇒-6(x^{2}+3x+2) = 0$
$⇒x^{2}+3x+2 = 0 $
$⇒x^{2} + x + 2x + 2 = 0$
$⇒x(x+1) + 2(x+1) = 0$
$⇒ (x+2)(x+1) = 0$
$\therefore x = -2$ and $x = -1$
So, the range is $(-\infty , -2) \ , (-2,-1) $ and $ (-1,\infty)$.
In interval $(-\infty , -2) \cup \ (-1,\infty)$ , $f^{'}(x) = - 6 x^2 - 18x - 12$ is -ve.
Hence, $f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}$ is strictly decreasing in interval $(-\infty , -2) \cup \ (-1,\infty)$.
In interval (-2,-1) , $f^{'}(x) = - 6 x^2 - 18x - 12$ is +ve.
Hence, $f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}$ is strictly increasing in the interval (-2,-1).

Question:6(d) Find the intervals in which the following functions are strictly increasing or
decreasing: $6- 9x - x ^2$

Answer:
Given function is,
$f(x) = 6- 9x - x ^2$
$f^{'}(x) = - 9 - 2x$
Now,
$f^{'}(x) = 0$
$⇒ - 9 - 2x = 0 $
$⇒2x = -9$
$⇒x = -\frac{9}{2}$
So, the range is $(-\infty, - \frac{9}{2} )$ and $ ( - \frac{9}{2}, \infty )$
In interval $(-\infty, - \frac{9}{2} )$ , $f^{'}(x) = - 9 - 2x$ is +ve.
Hence, $f(x) = 6- 9x - x ^2$ is strictly increasing in interval $(-\infty, - \frac{9}{2} )$.
In interval $( - \frac{9}{2},\infty )$ , $f^{'}(x) = - 9 - 2x$ is -ve.
Hence, $f(x) = 6- 9x - x ^2$ is strictly decreasing in interval $( - \frac{9}{2},\infty )$.

Question:6(e) Find the intervals in which the following functions are strictly increasing or
decreasing: $( x+1) ^3 ( x-3) ^3$

Answer:
Given function is,
$f(x) = ( x+1) ^3 ( x-3) ^3$
$f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$
Now,
$f^{'}(x) = 0 $
$⇒ 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^{3} =0$
$⇒3(x+1)^{2}(x-3)^{2}((x-3) + (x+1) ) = 0 $
$⇒ (x+1)(x-3) = 0$ or $(2x-2) = 0$
So, $ x=-1$ and $ x = 3$ or, $x = 1$
So, the intervals are $(-\infty,-1), (-1,1), (1,3)$ and $(3,\infty)$.
Our function $f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$ is +ve in the interval $(1,3)$ and $ (3,\infty)$.
Hence, $f(x) = ( x+1) ^3 ( x-3) ^3$ is strictly increasing in the interval $(1,3) $ and $(3,\infty)$.
Our function $f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$ is -ve in the interval $(-\infty,-1)$ and $(-1,1)$
Hence, $f(x) = ( x+1) ^3 ( x-3) ^3$ is strictly decreasing in interval $(-\infty,-1) $ and $ (-1,1)$.

Question:7 Show that $y = \log( 1+ x ) - \frac{2 x }{2+x } , x > -1$ is an increasing function of x throughout its domain.

Answer:
Given function is,
$f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }$
$f^{'}(x)\Rightarrow \frac{dy}{dx} = \frac{1}{1+x} - \frac{2 (2+x) - (1)(2x)} {(2+x)^{2} } = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^{2}}$
$= \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$= \frac{(2+x)^2 - 4(x+1)}{(x+1)(2+x)^{2}}$
$= \frac{4+x^{2} +4x -4x - 4}{(x+1)(2+x)^{2}} $
$= \frac{x^{2} }{(x+1)(2+x)^{2}}$
So, $f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2}$
Now, for $x > -1$, it is clear that $f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2} > 0$
Hence, $f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }$ strictly increasing when $x > -1$

Question:8 Find the values of x for which $y = [x(x-2)]^{2}$ is an increasing function.

Answer:
Given function is,
$f(x)\Rightarrow y = [x(x-2)]^{2}$
$f^{'}(x)\Rightarrow \frac{dy}{dx} = 2[x(x-2)][(x-2)+x]$
$= 2(x^2-2x)(2x-2)$
$= 4x(x-2)(x-1)$
Now,
$f^{'}(x) = 0$
$⇒ 4x(x-2)(x-1) = 0$
So, $ x=0 , x= 2 $ and $x = 1$
So, the intervals are $(-\infty,0),(0,1),(1,2) $ and $ (2,\infty)$
In interval $(0,1)$ and $(2,\infty)$, $f^{'}(x)> 0$
Hence, $f(x)\Rightarrow y = [x(x-2)]^{2}$ is an increasing function in the interval $(0,1)\cup (2,\infty)$.

Question:9 Prove that $y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta$ is an increasing function of $\theta\: \: \text{in}\: \: [ 0 , \frac{\pi}{2} ]$

Answer:
Given function is,
$f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta$

$f^{'}(x) = \frac{dy}{d\theta} = \frac{4 \cos \theta(2+\cos \theta) - (-\sin \theta)4\sin \theta) }{(2+ \cos \theta )^2} - 1$
$= \frac{8 \cos \theta+4\cos^2 \theta + 4\sin^2 \theta - (2+ \cos \theta )^2 }{(2+ \cos \theta )^2}$
$= \frac{8 \cos \theta+4(\cos^2 \theta + \sin^2 \theta) - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}$
$= \frac{8 \cos \theta+4 - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}$
$= \frac{4 \cos \theta-\cos^2 \theta }{(2+ \cos \theta )^2}$
Now, for $\theta \ \in \ [0,\frac{\pi}{2}]$
$\\ 4 \cos \theta \geq \cos^2 \theta$
$⇒4 \cos \theta - \cos^2 \geq 0$ and $(2+\cos \theta)^2 > 0$
So, $f^{'}(x) > 0$ for $\theta$ in $[0,\frac{\pi}{2}]$
Hence, $f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta$ is increasing function in $\theta \ \in [0,\frac{\pi}{2}]$.

Question:10 Prove that the logarithmic function is increasing on $( 0 , \infty )$

Answer:
Let the logarithmic function be $\log x$.
$f(x) = \log x$
$f^{'}(x) = \frac{1}{x}$
Now, for all values of x in $( 0 , \infty )$ , $f^{'}(x) > 0$
Hence, the logarithmic function $f(x) = \log x$ is increasing in the interval $( 0 , \infty )$.

Question:11: Prove that the function f given by $f ( x) = x ^2 - x + 1$ is neither strictly increasing nor decreasing on (– 1, 1).

Answer:
Given function is,
$f ( x) = x ^2 - x + 1$
$f^{'}(x) = 2x - 1$
Now, for interval $(-1,\frac{1}{2})$ , $f^{'}(x) < 0$ and for interval $(\frac{1}{2},1),f^{'}(x) > 0$
Hence, by this, we can say that $f ( x) = x ^2 - x + 1$ is neither strictly increasing nor decreasing in the interval (-1, 1).

Question:12 Which of the following functions are decreasing on $0, \frac{\pi}2$ $(A) \cos x \\(B) \cos 2x \\ (C) \cos 3x \\ (D) \tan x$

Answer:
(A)
$f(x) = \cos x $
$ f^{'}(x) = -\sin x$
$f^{'}(x) < 0$ for x in $(0,\frac{\pi}{2})$
Hence, $f(x) = \cos x$ is decreasing function in $(0,\frac{\pi}{2})$

(B)
$f(x) = \cos 2x$
$f^{'}(x) = -2\sin2 x$
Now, as
$0 < x < \frac{\pi}{2}$
$⇒0 < 2x < \pi$
$f^{'}(x) < 0$ for 2x in $(0,\pi)$
Hence, $f(x) = \cos 2x$ is decreasing function in $(0,\frac{\pi}{2})$

(C)
$f(x) = \cos 3x $
$f^{'}(x) = -3\sin3 x$
Now, as
$0 < x < \frac{\pi}{2}$
$⇒ 0 < 3x < \frac{3\pi}{2}$
$f^{'}(x) < 0$ for $x \ \epsilon \ \left ( 0,\frac{\pi}{3} \right )$ and $f^{'}(x) > 0 \ x \ \epsilon \ \left ( \frac{\pi}{3} , \frac{\pi}{2}\right )$
Hence, it is clear that $f(x) = \cos 3x$ is neither increasing nor decreasing in $(0,\frac{\pi}{2})$

(D)
$f(x) = \tan x$
$ f^{'}(x) = \sec^{2}x$
$f^{'}(x) > 0$ for x in $(0,\frac{\pi}{2})$
Hence, $f(x) = \tan x$ is strictly increasing function in the interval $(0,\frac{\pi}{2})$.

So, only (A) and (B) are decreasing functions in $(0,\frac{\pi}{2})$.

Question:13: On which of the following intervals is the function f given by $f ( x) = x ^{100} + \sin x - 1$ decreasing?
(A) (0,1) (B) $\frac{\pi}{2},\pi$ (C) $0,\frac{\pi}{2}$ (D) None of these

Answer:

(A) Given function is,
$f ( x) = x ^{100} + \sin x - 1$
$f^{'}(x) = 100x^{99} + \cos x$
Now, in interval (0,1)
$f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$ is an increasing function in the interval (0, 1).

(B) Now, in interval $\left ( \frac{\pi}{2},\pi \right )$
$100x^{99} > 0$ but $\cos x < 0$
So, $100x^{99} > \cos x$
$⇒100x^{99} - \cos x > 0$ , $f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$ is increasing function in interval $\left ( \frac{\pi}{2},\pi \right )$.

(C) Now, in interval $\left ( 0,\frac{\pi}{2} \right )$
$100x^{99} > 0 $ and $ \cos x > 0$
So, $100x^{99} > \cos x $
$⇒ 100x^{99} - \cos x > 0$, $f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$ is increasing function in interval $\left ( 0,\frac{\pi}{2} \right )$.
So, $f ( x) = x ^{100} + \sin x - 1$ is increasing for all cases.
Hence, the correct answer is option (D).

Question:14: For what values of a the function f given by $f (x) = x^2 + ax + 1$ is increasing on [1, 2]?

Answer:
Given function is,
$f (x) = x^2 + ax + 1$
$f^{'}(x) = 2x + a$
Now, we can clearly see that for every value of $a > -2$
$f^{'}(x) = 2x + a$ $> 0$
Hence, $f (x) = x^2 + ax + 1$ is increasing for every value of $a > -2$ in the interval [1, 2].

Question:15 Let I be any interval disjoint from [–1, 1]. Prove that the function f given by $f ( x) = x + \frac1x$ is increasing on I.

Answer:
Given function is,
$f ( x) = x + \frac 1x$
$f^{'}(x) = 1 - \frac{1}{x^2}$
Now,
$f^{'}(x) = 0$
$⇒ 1 - \frac{1}{x^2} = 0$
$⇒ x^{2} = 1$
$⇒ x = \pm1$
So, intervals are from $(-\infty,-1), (-1,1)$ and $(1,\infty)$
In interval $(-\infty,-1), (1,\infty)$ , $\frac{1}{x^2} < 1 \Rightarrow 1 - \frac{1}{x^2} > 0$
$f^{'}(x) > 0$
Hence, $f ( x) = x + \frac 1x$ is increasing in interval $(-\infty,-1)\cup (1,\infty)$
In interval (-1,1), $\frac{1}{x^2} > 1 \Rightarrow 1 - \frac{1}{x^2} < 0$
$f^{'}(x) < 0$
Hence, $f ( x) = x + \frac 1x$ is decreasing in the interval (-1, 1).
Hence, the function f given by $f ( x) = x + \frac 1x$ is increasing on I disjoint from [–1, 1].

Question:16 Prove that the function f given by $f (x) = \log \sin x$ is increasing on $\left ( 0 ,\frac{\pi}2 \right )$ and decreasing on $ \left ( \frac{\pi}2 , \pi \right )$

Answer:
Given function is,
$f (x) = \log \sin x$
$f^{'}(x) = \frac{1}{\sin x}\cos x = \cot x$
Now, we know that cot x is +ve in the interval $\left ( 0 ,\frac{\pi}2 \right )$ and -ve in the interval $\left ( \frac{\pi}2 , \pi \right )$
$f^{'}(x) > 0 $ in $\left ( 0,\frac{\pi}{2} \right )$ and $ f^{'}(x) < 0$ in $ \left ( \frac{\pi}{2} , \pi \right )$
Hence, $f (x) = \log \sin x$ is increasing in the interval $\left ( 0 , \frac{\pi}2 \right )$ and decreasing in interval $\left ( \frac{\pi}2 , \pi \right )$.

Question:17 Prove that the function f given by $f (x) = \log |\cos x|$ is decreasing on $( 0 , \frac{\pi}2)$ and increasing on $( \frac{3\pi}2 , 2\pi )$

Answer:
Given function is,
f(x) = log|cos x|
Value of cos x is always +ve in both these cases
So, we can write log|cos x| = log(cos x)
Now,
$f^{'}(x) = \frac{1}{\cos x}(-\sin x) = -\tan x$
We know that in interval $\left ( 0,\frac{\pi}{2} \right )$ , $\tan x > 0 \Rightarrow -\tan x< 0$
$f^{'}(x) < 0$
Hence, f(x) = log|cos x| is decreasing in interval $\left ( 0,\frac{\pi}{2} \right )$
We know that in interval $\left ( \frac{3\pi}{2},2\pi \right )$, $\tan x < 0$
$ \Rightarrow -\tan x> 0$
$f^{'}(x) > 0$
Hence, f(x) = log|cos x| is increasing in interval $\left ( \frac{3\pi}{2},2\pi \right )$.

Question:18 Prove that the function given by $f (x) = x^3 - 3x^2 + 3x - 100$ is increasing in R.

Answer:
Given function is,
$f (x) = x^3 - 3x^2 + 3x - 100$
$f^{'}(x) = 3x^2 - 6x + 3$
$⇒f^{'}(x)= 3(x^2 - 2x + 1) = 3(x-1)^2$
$⇒f^{'}(x) = 3(x-1)^2$
We can clearly see that for any value of x in R, $f^{'}(x) > 0$
Hence, $f (x) = x^3 - 3x^2 + 3x - 100$ is an increasing function in R.

Question:19 The interval in which $y = x ^2 e ^{-x}$ is increasing is

(A) $( - \infty , \infty )$ (B) $( - 2 , 0 )$ (C) $( - 2 , \infty )$ (D) $( 0, 2 )$

Answer:
Given function is,
$f(x) \Rightarrow y = x ^2 e ^{-x}$
$f^{'}(x) \Rightarrow \frac{dy}{dx} = 2x e ^{-x} + -e^{-x}(x^{2})=xe^{-x}(2-x)$
$f^{'}(x) = xe ^{-x}(2 -x)$
Now, it is clear that $f^{'}(x) > 0$ only in the interval (0, 2).
So, $f(x) \Rightarrow y = x ^2 e ^{-x}$ is an increasing function for the interval (0, 2).
Hence, the correct answer is option (D).

Question:1(i) Find the maximum and minimum values, if any, of the following functions given by ( $f (x) = (2x - 1)^2 + 3$

Answer:
Given function is,
$f (x) = (2x - 1)^2 + 3$
$(2x - 1)^2 \geq 0$
$⇒(2x-1)^2+3\geq 3$
Hence, the minimum value occurs when
$(2x-1)=0$
$⇒ x = \frac{1}{2}$
Hence, the minimum value of function $f (x) = (2x - 1)^2 + 3$ occurs at $x = \frac{1}{2}$
and the minimum value is
$f(\frac{1}{2}) = (2.\frac{1}{2}-1)^2+3\\$
$⇒f(\frac{1}{2})= (1-1)^2+3 =0+3 = 3$
and it is clear that there is no maximum value of $f (x) = (2x - 1)^2 + 3$.

Question:1(ii) Find the maximum and minimum values, if any, of the following functions given by $f (x) = 9x^ 2 + 12x + 2$

Answer:
Given function is,
$f (x) = 9x^ 2 + 12x + 2$
Adding and subtracting 2 in the given equation, we get,
$f (x) = 9x^ 2 + 12x + 2 + 2- 2$
$⇒ f(x)= 9x^2 +12x+4-2$
$⇒ f(x)= (3x+2)^2 - 2$
Now,
$(3x+2)^2 \geq 0$
$⇒ (3x+2)^2-2\geq -2$ for every $x \ \in \ R$
Hence, the minimum value occurs when
$(3x+2)=0$
$⇒x = \frac{-2}{3}$
Hence, the minimum value of function $f (x) = 9x^2+12x+2$ occurs at $x = \frac{-2}{3}$
and the minimum value is
$f(\frac{-2}{3}) = 9(\frac{-2}{3})^2+12(\frac{-2}{3})+2=4-8+2 =-2 \\$
and it is clear that there is no maximum value of $f (x) = 9x^2+12x+2$

Question:1(iii) Find the maximum and minimum values, if any, of the following functions given by $f (x) = - (x -1) ^2 + 10$

Answer:
Given function is,
$f (x) = - (x -1) ^2 + 10$
$-(x-1)^2 \leq 0$
$⇒ -(x-1)^2+10\leq 10$ for every $x \ \in \ R$
Hence, the maximum value occurs when
$(x-1)=0$
$⇒ x = 1$
Hence, maximum value of function $f (x) = - (x -1) ^2 + 10$ occurs at x = 1
and the maximum value is
$f(1) = -(1-1)^2+10=10 \\$
and it is clear that there is no minimum value of $f (x) = 9x^2+12x+2$

Question:1(iv): Find the maximum and minimum values, if any, of the following functions
given by $g(x) = x^3 + 1$

Answer:
Given function is,
$g(x) = x^3 + 1$
value of $x^3$ varies from $-\infty < x^3 < \infty$
Hence, function $g(x) = x^3 + 1$ neither has a maximum or minimum value.

Question:2(i) Find the maximum and minimum values, if any, of the following functions
given by $f (x) = |x + 2| - 1$

Answer:
Given function is
$f (x) = |x + 2| - 1$
$⇒|x+2| \geq 0$
$⇒ |x+2| - 1 \geq -1$ as $x \ \in \ R$
Hence, minimum value occurs when |x + 2| = 0
x = -2
Hence, the minimum value occurs at x = -2
and the minimum value is
$f(-2) = |-2+2| - 1 = -1$
It is clear that there is no maximum value of the given function $x \ \in \ R$.

Question:2(ii) Find the maximum and minimum values, if any, of the following functions
given by
$g(x) = - | x + 1| + 3$

Answer:
Given function is
$g(x) = - | x + 1| + 3$
$-|x+1| \leq 0$
$⇒-|x+1| + 3 \leq 3$ as $x \ \in \ R$
Hence, the maximum value occurs when -|x + 1| = 0
x = -1
Hence, the maximum value occurs at x = -1
and the maximum value is
$g(-1) = -|-1+1| + 3 = 3$
It is clear that there is no minimum value of the given function $x \ \in \ R$.

Question:2(iii) Find the maximum and minimum values, if any, of the following functions
given by $h(x) = \sin(2x) + 5$

Answer:
Given function is
$h(x) = \sin(2x) + 5$
We know that the value of sin 2x varies from
$-1 \leq \sin2x \leq 1$
$⇒-1 + 5 \leq \sin2x +5\leq 1 +5$
$⇒ 4 \leq \sin2x +5\leq 6$
Hence, the maximum value of our function $h(x) = \sin(2x) + 5$ is 6, and the minimum value is 4.

Question:2(iv) Find the maximum and minimum values, if any, of the following functions
given by $f (x) = | \sin 4x + 3|$

Answer:
Given function is
$f (x) = | \sin 4x + 3|$
We know that the value of sin 4x varies from
$-1 \leq \sin4x \leq 1$
$⇒-1 + 3 \leq \sin4x +3\leq 1 +3$
$⇒ 2 \leq \sin4x +3\leq 4$
$⇒ 2\leq | \sin4x +3| \leq 4$
Hence, the maximum value of our function $f (x) = | \sin 4x + 3|$ is 4, and the minimum value is 2.

Question:2(v) Find the maximum and minimum values, if any, of the following functions
given by $h(x) = x + 1 , x \in ( -1,1)$

Answer:
Given function is
$h(x) = x + 1$
It is given that the value of $x \ \in (-1,1)$
So, we can not comment about either maximum or minimum value
Hence, function $h(x) = x + 1$ has neither a maximum nor a minimum value.

Question:3(i) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be: $f ( x) = x^2$

Answer:
Given function is
$f ( x) = x^2$
$ f^{'}(x) = 2x$
Also, $ f^{'}(x) = 0$
$ \Rightarrow 2x = 0 $
$\Rightarrow x = 0$
So, x = 0 is the only critical point of the given function.
$f^{'}(0) = 0\\$
So we find it through the 2nd derivative test.
$f^{''}(x) = 2$
$ f^{''}(0) = 2$
$f^{''}(0)> 0$
Hence, by this, we can say that 0 is a point of minima
and the minimum value is $f(0) = (0)^2 = 0$.

Question:3(ii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be: $g(x) = x ^3 - 3x$

Answer:
Given function is
$g(x) = x ^3 - 3x$
$ g^{'}(x) = 3x^2 - 3$
$ g^{'}(x)=0$
$\Rightarrow 3x^2-3 =0$
$ \Rightarrow x = \pm 1\\$
Hence, the critical points are 1 and - 1
Now, by the second derivative test
$g^{''}(x)=6x$
$g^{''}(1)=6 > 0$
Hence, 1 is the point of minima, and the minimum value is
$g(1) = (1)^3 - 3(1) = 1 - 3 = -2$
$g^{''}(-1)=-6 < 0$
Hence, -1 is the point of maxima, and the maximum value is
$g(1) = (-1)^3 - 3(-1) = -1 + 3 = 2$

Question:3(iii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
$h(x) = \sin x + \cos x,\ 0<x<\frac{\pi}{2}$

Answer:
Given function is
$h(x) = \sin x + \cos x$
$ h^{'}(x)= \cos x - \sin x$
$ h^{'}(x)= 0$
$⇒ \cos x - \sin x = 0$
$⇒ \cos x = \sin x$
$⇒ x = \frac{\pi}{4}$ as $x \ \in \ \left ( 0,\frac{\pi}{2} \right )$
Now, we use the second derivative test.
$h^{''}(x)= -\sin x - \cos x$
$⇒ h^{''}(\frac{\pi}{4}) = -\sin \frac{\pi}{4} - \cos \frac{\pi}{4}$
$⇒ h^{''}(\frac{\pi}{4}) = -\frac{1}{\sqrt2}-\frac{1}{\sqrt2}$
$⇒ h^{''}(\frac{\pi}{4})= -\frac{2}{\sqrt2} = -\sqrt2 < 0$
Hence, $\frac{\pi}{4}$ is the point of maxima and the maximum value is $h\left ( \frac{\pi}{4} \right )$ which is $\sqrt2$.

Question:3(iv): Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $f(x) = \sin x - \cos x$

Answer:
Given function is
$h(x) = \sin x - \cos x$
$ h^{'}(x)= \cos x + \sin x$
$ h^{'}(x)= 0$
$⇒ \cos x + \sin x = 0$
$⇒ \cos x = -\sin x$
$⇒ x = \frac{3\pi}{4}$ as $\ x \ \in \ \left ( 0,2\pi \right )$
Now, we use the second derivative test.
$h^{''}(x)= -\sin x + \cos x$
$⇒ h^{''}(\frac{3\pi}{4}) = -\sin \frac{3\pi}{4} + \cos \frac{3\pi}{4}$
$⇒ h^{''}(\frac{3\pi}{4}) = -(\frac{1}{\sqrt2})-\frac{1}{\sqrt2}$
$⇒ h^{''}(\frac{3\pi}{4})=- \frac{2}{\sqrt2} = -\sqrt2 < 0$
Hence, $\frac{\pi}{4}$ is the point of maxima and maximum value is $h\left ( \frac{3\pi}{4} \right )$, which is $\sqrt2$.

Question:3(v): Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $f (x) = x^3 - 6x^2 + 9x + 15$

Answer:
Given function is:
$f (x) = x^3 - 6x^2 + 9x + 15$
$ f^{'}(x) = 3x^2 - 12x + 9$
$ f^{'}(x)= 0$
$⇒3x^2 - 12x + 9 = 0$
$⇒ 3(x^2-4x+3)=0$
$⇒x^2-4x+3 = 0$
$⇒x^2 - x -3x + 3=0$
$⇒x(x-1)-3(x-1) = 0$
$⇒(x-1)(x-3) = 0$
So, $x=1$ and $ x = 3$
Hence, 1 and 3 are critical points.
Now, we use the second derivative test.
$f^{''}(x) = 6x - 12$
$ f^{''}(1) = 6 - 12 = -6 < 0$
Hence, x = 1 is a point of maxima, and the maximum value is
$f (1) = (1)^3 - 6(1)^2 + 9(1) + 15 = 1-6+9+15 = 19$
$f^{''}(x) = 6x - 12$
$ f^{''}(3) = 18 - 12 = 6 > 0$
Hence, x = 1 is a point of minima, and the minimum value is
$f (3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27-54+27+15 = 15$

Question:3(vi): Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $g ( x) = \frac{x}{2} + \frac{2}{x}, x > 0$

Answer:
Given function is
$g ( x) = \frac{x}{2} + \frac{2}{x}$
$ g^{'}(x) = \frac{1}{2}-\frac{2}{x^2}$
$ g^{'}(x) = 0$
$⇒ \frac{1}{2}-\frac{2}{x^2} = 0$
$⇒ x^2 = 4$
$⇒ x = \pm 2$ ( but as $x > 0$ we only take the positive value of x i.e., x = 2)
Hence, 2 is the only critical point.
Now, we use the second derivative test.
$g^{''}(x) = \frac{4}{x^3}$
$g^{''}(2) = \frac{4}{2^3} =\frac{4}{8} = \frac{1}{2}> 0$
Hence, 2 is the point of minima, and the minimum value is
$g ( x) = \frac{x}{2} + \frac{2}{x} $
$ g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$

Question:3(vii): Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $g (x) = \frac{1}{x^2 + 2}$

Answer:
Given function is
$g (x) = \frac{1}{x^2 + 2}$
$ g^{'}(x) = \frac{-2x}{(x^2+2)^2}$
$ g^{'}(x) = 0$
$ \frac{-2x}{(x^2+2)^2} = 0$
$⇒ x = 0$
Hence, x = 0 is the only critical point.
Now, we use the second derivative test.
$g^{''}(x) = -\frac{-2(x^2+2)^2-(-2x){2(x^2+2)(2x)}}{((x^2+2)^2)^2} $
$g^{''}(0) = \frac{-2\times4}{(2)^4} = \frac{-8}{16} = -\frac{1}{2}< 0$
Hence, 0 is the point of local maxima, and the maximum value is
$g (0) = \frac{1}{0^2 + 2} = \frac{1}{2}$

Question:3(viii): Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $f (x) = x \sqrt{ 1-x }, 0 < x < 1$

Answer:
Given function is
$f (x) = x \sqrt{ 1-x }$
$f ^{'}(x) = \sqrt{1-x} + \frac{x(-1)}{2\sqrt{1-x}}$
$= \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} $
$\Rightarrow \frac{2-3x}{2\sqrt{1-x}}$
$ f^{'}(x) = 0$
$⇒ \frac{2-3x}{2\sqrt{1-x}} = 0$
$⇒ 3x = 2$
$⇒ x = \frac{2}{3}$
Hence, $x = \frac{2}{3}$ is the only critical point.
Now, we use the second derivative test.
$f^{''}(x)= \frac{(-1)(2\sqrt{1-x})-(2-x)(2.\frac{-1}{2\sqrt{1-x}}(-1))}{(2\sqrt{1-x})^2}$
$= \frac{-2\sqrt{1-x}-\frac{2}{\sqrt{1-x}}+\frac{x}{\sqrt{1-x}}}{4(1-x)}$
$= \frac{3x}{4(1-x)\sqrt{1-x}}$
$f^{"}(\frac{2}{3}) > 0$
Hence, it is the point of minima, and the minimum value is
$f (x) = x \sqrt{ 1-x }$
$ f(\frac{2}{3}) = \frac{2}{3}\sqrt{1-\frac{2}{3}}$
$⇒ f(\frac{2}{3}) = \frac{2}{3}\sqrt{\frac{1}{3}}$
$⇒ f(\frac{2}{3}) = \frac{2}{3\sqrt3}$
$⇒ f(\frac{2}{3}) = \frac{2\sqrt3}{9}$

Question:4(i) Prove that the following functions do not have maxima or minima: $f (x) = e ^x$

Answer:
Given function is
$f (x) = e ^x$
$f^{'}(x) = e^x$
$ f^{'}(x) = 0$
$⇒ e^x=0\\$
But exponential can never be 0.
Hence, the function $f (x) = e ^x$ does not have either maxima or minima.

Question:4(ii) Prove that the following functions do not have maxima or minima: $g(x) = \log x$

Answer:
Given function is
$g(x) = \log x$
$g^{'}(x) = \frac{1}{x}$
$ g^{'}(x) = 0$
$ ⇒\frac{1}{x}= 0\\$
Since log x deifne for positive x, i.e. $x > 0$
Hence, by this, we can say that $g^{'}(x)> 0$ for any value of x.
Therefore, there is no $c \ \in \ R$ such that $g^{'}(c) = 0$
Hence, the function $g(x) = \log x$ does not have either maxima or minima.

Question:4(iii) Prove that the following functions do not have maxima or minima: $h(x) = x^3 + x^2 + x +1$

Answer:
Given function is
$h(x) = x^3 + x^2 + x +1$
$h^{'}(x) = 3x^2+2x+1$
$ h^{'}(x) = 0$
$ ⇒3x^2+2x+1 = 0$
$ ⇒2x^2+x^2+2x+1 = 0$
$ ⇒2x^2 + (x+1)^2 = 0\\$
But, it is clear that there is no $c \ \in \ R$ such that $f^{'}(c) = 0$
Hence, the function $h(x) = x^3 + x^2 + x +1$ does not have either maxima or minima.

Question:5(i) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $f (x) = x ^ 3, x \epsilon [- 2, 2]$

Answer:
Given function is
$f(x) = x^3$
$f^{'}(x) = 3x^2$
$ f^{'}(x) = 0$
$⇒ 3x^2 = 0$
$\Rightarrow x = 0$
Hence, 0 is the critical point of the function $f(x) = x^3$
Now, we need to see the value of the function $f(x) = x^3$ at x = 0 and as $x \ \epsilon \ [-2,2]$
We also need to check the value at the end points of the given range, i.e. x = 2 and x = -2
$f(0) = (0)^3 = 0$
$f(2= (2)^3 = 8$
$ f(-2)= (-2)^3 = -8$
Hence, maximum value of function $f(x) = x^3$ occurs at x = 2 and value is 8
and minimum value of function $f(x) = x^3$ occurs at x = -2 and value is -8.

Question:5(ii) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $f (x) = \sin x + \cos x, x \epsilon [0, \pi]$

Answer:
Given function is
$f(x) = \sin x + \cos x$
$f^{'}(x) = \cos x - \sin x$
$ f^{'}(x)= 0$
$⇒ \cos x- \sin x= 0$
$⇒ \cos = \sin x$
$⇒ x = \frac{\pi}{4}$ as $x \ \epsilon \ [0,\pi]$
Hence, $x = \frac{\pi}{4}$ is the critical point of the function $f(x) = \sin x + \cos x$
Now, we need to check the value of function $f(x) = \sin x + \cos x$ at $x = \frac{\pi}{4}$ and at the end points of given range i.e. $x = 0 \ and \ x = \pi$
$f(\frac{\pi}{4}) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4}\\$
$=\frac{1}{\sqrt2}+\frac{1}{\sqrt2} = \frac{2}{\sqrt2} = \sqrt2$
$f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$
$f(\pi) = \sin \pi + \cos \pi = 0 +(-1) = -1$
Hence, the absolute maximum value of function $f(x) = \sin x + \cos x$ occurs at $x = \frac{\pi}{4}$ and value is $\sqrt2$
and absolute minimum value of function $f(x) = \sin x + \cos x$ occurs at $x = \pi$ and value is -1.

Question:5(iii) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $f (x) = 4 x - \frac{1}{2} x^2, x \epsilon \left [ -2, \frac{9}{2} \right ]$

Answer:
Given function is
$f(x) =4x - \frac{1}{2}x^2$
$f^{'}(x) = 4 - x $
$ f^{'}(x)= 0$
$⇒4-x= 0$
$⇒ x=4$
Hence, x = 4 is the critical point of function $f(x) =4x - \frac{1}{2}x^2$
Now, we need to check the value of function $f(x) =4x - \frac{1}{2}x^2$ at x = 4 and at the end points of given range i.e. at x = -2 and x = 9/2
$f(4) =4(4) - \frac{1}{2}(4)^2$
$=16-\frac{1}{2}.16 = 16-8 = 8$
$f(-2) = 4(-2)-\frac{1}{2}.(-2)^2 = -8-2 = -10$
$f(\frac{9}{2}) =4(\frac{9}{2})-\frac{1}{2}.\left ( \frac{9}{2} \right )^2 = 18-\frac{81}{8} = \frac{63}{8}$
Hence, absolute maximum value of function $f(x) =4x - \frac{1}{2}x^2$ occures at x = 4 and value is 8 and absolute minimum value of function $f(x) =4x - \frac{1}{2}x^2$ occures at x = -2 and value is -10.

Question:5(iv) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $f (x) = ( x-1) ^2 + 3, x \epsilon [ -3, 1 ]$

Answer:
Given function is
$f(x) = (x-1)^2+3$
$f^{'}(x) =2(x-1)$
$f^{'}(x)= 0$
$⇒ 2(x-1)= 0$
$⇒ x=1$
Hence, x = 1 is the critical point of function $f(x) = (x-1)^2+3$
Now, we need to check the value of function $f(x) = (x-1)^2+3$ at x = 1 and at the end points of given range i.e. at x = -3 and x = 1
$f(1) = (1-1)^2+3 = 0^2+3 = 3$
$f(-3) = (-3-1)^2+3= (-4)^2+3 = 16+3= 19$
Hence, absolute maximum value of function $f(x) = (x-1)^2+3$ occurs at x = -3 and value is 19
and absolute minimum value of function $f(x) = (x-1)^2+3$ occurs at x = 1 and value is 3.

Question:6 Find the maximum profit that a company can make if the profit function is
given by $p(x) = 41 - 72x - 18x ^2$

Answer:
Profit of the company is given by the function
$p(x) = 41 - 72x - 18x ^2$
$p^{'}(x)= -72-36x$
$ p^{'}(x) = 0$
$⇒-72-36x= 0$
$⇒ x = -2$
x = -2 is the only critical point of the function $p(x) = 41 - 72x - 18x ^2$
Now, by the second derivative test, we get,
$p^{''}(x)= -36< 0$
At x = -2, $p^{''}(x)< 0$
Hence, maxima of function $p(x) = 41 - 72x - 18x ^2$ occurs at x = -2 and maximum value is
$p(-2) = 41 - 72(-2) - 18(-2) ^2=41+144-72 = 113$
Hence, the maximum profit the company can make is 113 units

Question:7 Find both the maximum value and the minimum value of $3x^4 - 8x^3 + 12x^2 - 48x + 25$ on the interval [0, 3].

Answer:
Given function is
$f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}$
$f^{'}(x)=12x^3 - 24x^2 +24x - 48$
$ f^{'}(x)=0$
$⇒12(x^3-2x^2+2x-4) = 0$
$⇒ x^3-2x^2+2x-4=0\\$
Now, by hit and trial, let's first assume x = 2.
$(2)^3-2(2)^2+2(2)-4$
$= 8-8+4-4=0$
Hence, x = 2 is one value.
Now,
$\frac{x^3-2x^2+2x-4}{x-2} = \frac{(x^2+2)(x-2)}{(x-2)} = (x^2+2)$
$x^2 = - 2$ which is not possible
Hence, x = 2 is the only critical value of function $f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}$
Now, we need to check the value at x = 2 and at the end points of given range, i.e. x = 0 and x = 3
$f(2)=3(2)^4-8(2)^3+12(2)^2-48(2)+25$
$=3\times16 - 8\times 8 + 12\times 4 - 96 + 25 = 48-64+48-96+25 = -39$
$f(3)=3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25$
$=3\times81-8\times27+12\times9-144+25 $
$ =243-216+108-144+25 $
$= 16$
$f(0)=3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$
Hence, maximum value of function $f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}$ occurs at x = 0 and value is 25
and minimum value of function $f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}$ occurs at x = 2 and value is -39.

Question:8: At what points in the interval $[ 0, 2 \pi ]$ does the function $\sin 2x$ attain its maximum value?

Answer:
Given function is
$f(x) = \sin 2x$
$f^{'}(x) = 2\cos 2x$
$ f^{'}(x) = 0$
$⇒ 2\cos 2x = 0$ as $ x \ \epsilon [0,2\pi]$
$0 < x < 2\pi$
$⇒ 0< 2x < 4\pi$
$⇒ \cos 2x = 0$ at $ 2x = \frac{\pi}{2},2x = \frac{3\pi}{2},2x=\frac{5\pi}{2}$ and $2x= \frac{7\pi}{2}\\$
So, the values of x are
$x = \frac{\pi}{4},x = \frac{3\pi}{4},x=\frac{5\pi}{4}$ and $x= \frac{7\pi}{4}\\$
These are the critical points of the function $f(x) = \sin 2x$
Now, we need to find the value of the function $f(x) = \sin 2x$ at $x = \frac{\pi}{4},x = \frac{3\pi}{4},x=\frac{5\pi}{4} \ and \ x= \frac{7\pi}{4}\\$ and at the end points of given range i.e. at x = 0 and $x = \pi$
$f(x) = \sin 2x$
$ f(\frac{\pi}{4}) = \sin 2\left ( \frac{\pi}{4} \right ) = \sin \frac{\pi}{2} = 1$
$f(\frac{3\pi}{4}) = \sin 2\left ( \frac{3\pi}{4} \right ) = \sin \frac{3\pi}{2} = -1$
$ f(\frac{5\pi}{4}) = \sin 2\left ( \frac{5\pi}{4} \right ) = \sin \frac{5\pi}{2} = 1$
$f(\frac{7\pi}{4}) = \sin 2\left ( \frac{7\pi}{4} \right ) = \sin \frac{7\pi}{2} = -1$
$f(\pi) = \sin 2(\pi)= \sin 2\pi = 0$
$f(0) = \sin 2(0)= \sin 0 = 0$
Hence, at $x =\frac{\pi}{4} \ and \ x = \frac{5\pi}{4}$ function $f(x) = \sin 2x$ attains its maximum value i.e. in 1 in the given range of $x \ \epsilon \ [0,2\pi]$.

Question:9 What is the maximum value of the function $\sin x + \cos x$?

Answer:
Given function is
$f(x) = \sin x + \cos x$
$f^{'}(x) = \cos x - \sin x$
$ f^{'}(x)= 0$
$⇒\cos x- \sin x= 0$
$⇒ \cos = \sin x$
$⇒x = 2n\pi+\frac{\pi}{4}$ where $n \in \ I$
Hence, $x = 2n\pi+\frac{\pi}{4}$ is the critical point of the function $f(x) = \sin x + \cos x$
Now, we need to check the value of the function $f(x) = \sin x + \cos x$ at $x = 2n\pi+\frac{\pi}{4}$
Value is the same for all cases, so let assume that n = 0
Now
$f(\frac{\pi}{4}) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4}\\$
$=\frac{1}{\sqrt2}+\frac{1}{\sqrt2} = \frac{2}{\sqrt2} = \sqrt2$
Hence, the maximum value of the function $f(x) = \sin x + \cos x$ is $\sqrt2$

Question:10. Find the maximum value of $2 x^3 - 24 x + 107$ in the interval [1, 3]. Find the maximum value of the same function in [–3, –1].

Answer:
Given function is
$f(x) = 2x^3-24x+107$
$f^{'}(x)=6x^2 - 24 $
$ f^{'}(x)=0$
$⇒ 6(x^2-4) = 0$
$⇒ x^2-4=0$
$⇒ x^{2} = 4$
$⇒ x = \pm2$
We neglect the value x = - 2 because $x \ \epsilon \ [1,3]$
Hence, x = 2 is the only critical value of function $f(x) = 2x^3-24x+107$
Now, we need to check the value at x = 2 and at the end points of given range, i.e. x = 1 and x = 3
$f(2) = 2(2)^3-24(2)+107\\ = 2\times 8 - 48+107\\ =16-48+107 = 75$
$f(3) = 2(3)^3-24(3)+107\\ = 2\times 27 - 72+107\\ =54-72+107 = 89$
$f(1) = 2(1)^3-24(1)+107\\ = 2\times 1 - 24+107\\ =2-24+107 = 85$
Hence, maximum value of function $f(x) = 2x^3-24x+107$ occurs at x = 3 and vale is 89 when $x \ \epsilon \ [1,3]$
Now, when $x \ \epsilon \ [-3,-1]$
We neglect the value x = 2
Hence, x = -2 is the only critical value of function $f(x) = 2x^3-24x+107$
Now, we need to check the value at x = -2 and at the end points of the given range, i.e. x = -1 and x = -3
$f(-1) = 2(-1)^3-24(-1)+107\\ = 2\times (-1) + 24+107\\ =-2+24+107 = 129$
$f(-2) = 2(-2)^3-24(-2)+107\\ = 2\times (-8) + 48+107\\ =-16+48+107 = 139$
$f(-3) = 2(-3)^3-24(-3)+107\\ = 2\times (-27) + 72+107\\ =-54+72+107 = 125$
Hence, the maximum value of function $f(x) = 2x^3-24x+107$ occurs at x = -2 and value is 139 when $x \ \epsilon \ [-3,-1]$.

Question:11: It is given that at x = 1, the function $x ^4 - 62x^2 + ax + 9$ attains its maximum value on the interval [0, 2]. Find the value of a.

Answer:
Given function is
$f(x) =x ^4 - 62x^2 + ax + 9$
Function $f(x) =x ^4 - 62x^2 + ax + 9$ attains maximum value at x = 1 then x must one of the critical point of the given function that means
$f^{'}(1)=0$
$f^{'}(x) = 4x^3-124x+a$
$ f^{'}(1) = 4(1)^3-124(1)+a=a-120$
Now,
$f^{'}(1)=0$
$⇒ a - 120=0$
$\therefore a=120$
Hence, the value of a is 120.

Question:12 Find the maximum and minimum values of $x + \sin 2x$ on $ [ 0 , 2 \pi ]$

Answer:
Given function is
$f(x) =x+ \sin 2x$
$f^{'}(x) =1+ 2\cos 2x$
$ f^{'}(x) = 0$
$ 1+2\cos 2x = 0$ as $ x \ \epsilon \ [0,2\pi]$
$ 0 < x < 2\pi$
$⇒ 0< 2x < 4\pi$
$ \cos 2x = \frac{-1}{2}$ at $2x = 2n\pi \pm \frac{2\pi}{3}$ where $n \ \epsilon \ Z\\ x = n\pi \pm \frac{\pi}{3}$
$ x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3} \ as \ x \ \epsilon \ [0,2\pi]$
So, the values of x are:
$x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
These are the critical points of the function $f(x) = x+\sin 2x$
Now, we need to find the value of the function $f(x) = x+\sin 2x$ at $x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$ and at the end points of given range, i.e. at x = 0 and $x = 2\pi$
$f(x) =x+ \sin 2x$
$ f(\frac{\pi}{3}) = \frac{\pi}{3}+\sin 2\left ( \frac{\pi}{3} \right ) = \frac{\pi}{3}+\sin \frac{2\pi}{3} = \frac{\pi}{3}+\frac{\sqrt3}{2}$
$ f(\frac{2\pi}{3}) = \frac{2\pi}{3}+\sin 2\left ( \frac{2\pi}{3} \right ) = \frac{2\pi}{3}+\sin \frac{4\pi}{3} = \frac{2\pi}{3}-\frac{\sqrt3}{2}$
$ f(\frac{4\pi}{3}) = \frac{4\pi}{3}+\sin 2\left ( \frac{4\pi}{3} \right ) = \frac{4\pi}{3}+\sin \frac{8\pi}{3} = \frac{4\pi}{3}+\frac{\sqrt3}{2}$
$ f(\frac{5\pi}{3}) = \frac{5\pi}{3}+\sin 2\left ( \frac{5\pi}{3} \right ) = \frac{5\pi}{3}+\sin \frac{10\pi}{3} = \frac{5\pi}{3}-\frac{\sqrt3}{2}$
$ f(2\pi) = 2\pi+\sin 2(2\pi)= 2\pi+\sin 4\pi = 2\pi$
$ f(0) = 0+\sin 2(0)= 0+\sin 0 = 0$
Hence, at $x = 2\pi$ function $f(x) = x+\sin 2x$ attains its maximum value and value is $2\pi$ in the given range of $x \ \epsilon \ [0,2\pi]$
and at x = 0 function $f(x) = x+\sin 2x$ attains its minimum value and value is 0.

Question: 13 Find two numbers whose sum is 24 and whose product is as large as possible.

Answer:
Let x and y be the two numbers.
It is given that
x + y = 24
⇒ y = 24 - x
and product of xy is maximum.
let $f(x) = xy=x(24-x)=24x-x^2$
$ f^{'}(x) = 24-2x$
$ f^{'}(x)=0$
$⇒24-2x=0$
$⇒ x=12$
Hence, x = 12 is the only critical value.
Now,
$f^{''}(x) = -2< 0$
At x= 12, $f^{''}(x) < 0$
Hence, x = 12 is the point of maxima.
Now, y = 24 - x = 24 - 12 = 12
Hence, the values of x and y are 12 and 12, respectively.

Question:14: Find two positive numbers x and y such that x + y = 60 and $xy^3$ is maximum.

Answer:
It is given that
x + y = 60
⇒ x = 60 - y
and $xy^3$ is maximum
Let $f(y) = (60-y)y^3 = 60y^3-y^4$
Now,
$f^{'}(y) = 180y^2-4y^3$
$ f^{'}(y) = 0$
So, $ y^2(180-4y)=0$
$⇒ y= 0$ and $ y = 45$
Now,
$f^{''}(y) = 360y-12y^2$
$ f^{''}(0) = 0\\$
Hence, 0 is neither a point of minima or maxima.
$f^{''}(y) = 360y-12y^2$
$ f^{''}(45) = 360(45)-12(45)^2 = -8100 < 0$
Hence, y = 45 is a point of maxima.
x = 60 - y = 60 - 45 = 15
Hence, the values of x and y are 15 and 45, respectively.

Question:15 Find two positive numbers x and y such that their sum is 35 and the product $x^2 y^5$ is a maximum.

Answer:
It is given that
x + y = 35
⇒ x = 35 - y
and $x^2 y^5$ is maximum
Therefore,
Let $ f (y )= (35-y)^2y^5\\ = (1225-70y+y^2)y^5$
$ f(y)=1225y^5-70y^6+y^7$
Now,
$f^{'}(y) = 6125y^4-420y^5+7y^6$
$ f^{'}(y)=0$
So, $ y^4(6125-420y+7y^2) = 0 $
$⇒y =0 $ and $ (y-25)(y-35)$
$\Rightarrow y = 25 , y=35$
Now,
$f^{''}(y)= 24500y^3-2100y^4+42y^5$
$f^{''}(35)= 24500(35)^3-2100(35)^4+42(35)^5 = 105043750 > 0$
Hence, y = 35 is the point of minima.
$f^{''}(0)= 0\\$
Hence, y = 0 is neither a point of maxima or minima.
$f^{''}(25)= 24500(25)^3-2100(25)^4+42(25)^5 = -27343750 < 0$
Hence, y = 25 is the point of maxima
x = 35 - y = 35 - 25 = 10
Hence, the values of x and y are 10 and 25, respectively.

Question:16 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Answer:
Let x and y be two positive numbers.
It is given that
x + y = 16
⇒ y = 16 - x
and $x^3 + y^3$ is minimum.
$f(x) = x^3 + (16-x)^3$
Now,
$f^{'}(x) = 3x^2 + 3(16-x)^2(-1)$
$f^{'}(x) = 0$
$⇒ 3x^2 - 3(16-x)^2 =0$
$ ⇒3x^2-3(256+x^2-32x) = 0$
$⇒ 3x^2 -3x^2+96x-768= 0$
$⇒ 96x = 768$
$⇒ x = 8$
Hence, x = 8 is the only critical point.
Now,
$f^{''}(x) = 6x - 6(16-x)(-1) = 6x + 96 - 6x = 96$
$⇒ f^{''}(x) = 96$
$f^{''}(8) = 96 > 0$
Hence, x = 8 is the point of minima
y = 16 - x = 16 - 8 = 8
Hence, the values of x and y are 8 and 8, respectively.

Question:17 A square piece of tin of side 18 cm is to be made into a box without a top by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Answer:
It is given that the side of the square is 18 cm
Let's assume that the length of the side of the square to be cut off is x cm
So, by this, we can say that the breath of the cube is (18 - 2x) cm and the height is x cm.
Then,
Volume of cube $\left ( V(x) \right )$ = $x(18-2x)^2$
$V^{'}(x) = (18-2x)^2+(x)2(18-2x)(-2)$
$V^{'}(x) = 0$
$ (18-2x)^2-4x(18-2x)=0$
$⇒ 324 + 4x^2 - 72x - 72x + 8x^2 = 0$
$⇒12x^2-144x+324 = 0$
$⇒12(x^2-12x+27) = 0$
$⇒ x^2-9x-3x+27=0$
$⇒ (x-3)(x-9)=0$
$x = 3$ and $x = 9$.
But the value of x cannot be 9 because then the value of breath becomes 0.
So, we neglect value x = 9
Hence, x = 3 is the critical point
Now,
$V^{''}(x) = 24x -144$
$ V^{''}(3) = 24\times 3 - 144 = 72 - 144 = -72$
$ V^{''}(3) < 0$
Hence, x = 3 is the point of maxima
Hence, the length of the side of the square to be cut off is 3 cm so that the volume of the box is the maximum possible.

Question:18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without a top by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Answer:
It is given that the sides of the rectangle are 45 cm and 24 cm.
Let us assume the side of the square to be cut off is x cm.
Then,
Volume of cube $V(x) = x(45-2x)(24-2x)$
$V^{'}(x) = (45-2x)(24-2x) + (-2)(x)(24-2x)+(-2)(x)(45-2x)\\$
$=1080 + 4x^2 - 138x - 48x + 4x^2 - 90x +4x^2$
$= 12x^2 - 276x + 1080$
$V^{'}(x) = 0$
$ ⇒12(x^2 - 23x+90)=0$
$ ⇒x^2-23x+90 = 0$
$ ⇒x^2-18x-5x+23=0$
$⇒ (x-18)(x-5)=0$
So, $ x =18$ and $ x = 5$
But x cannot be equal to 18 because then side (24 - 2x) becomes negative, which is not possible, so we neglect value x = 18
Hence, x = 5 is the critical value.
Now,
$V^{''}(x)=24x-276$
$ ⇒V^{''}(5)=24\times5 - 276$
$⇒ V^{''}(5)= -156 < 0$
Hence, x = 5 is the point of maxima.
Hence, the side of the square to be cut off is 5 cm so that the volume of the box is maximum.

Question:19: Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Answer:
Let us assume that the length and breadth of the rectangle inscribed in a circle are l and b, respectively, and the radius of the circle is r.

Now, by Pythagoras' theorem,
$a = \sqrt{l^2+b^2}\\$
a = 2r [Given]
$⇒4r^2 = l^2+b^2$
$⇒ l = \sqrt{4r^2 - b^2}$
Now, area of reactangle(A) = l $\times$ b
$A(b) = b(\sqrt{4r^2-b^2})$
$A^{'}(b) = \sqrt{4r^2-b^2}+b.\frac{(-2b)}{2\sqrt{4r^2-b^2}}\\ = \frac{4r^2-b^2-b^2}{\sqrt{4r^2-b^2}} = \frac{4r^2-2b^2}{\sqrt{4r^2-b^2}}$
$A^{'}(b) = 0 $
$⇒ \frac{4r^2-2b^2}{\sqrt{4r^2-b^2}} = 0$
$⇒ 4r^2 = 2b^2$
$⇒b = \sqrt2r$
Now,
$A^{''}(b) = \frac{-4b(\sqrt{4r^2-b^2})-(4r^2-2b^2).\left ( \frac{-1}{2(4r^2-b^2)^\frac{3}{2}}.(-2b) \right )}{(\sqrt{4r^2-b^2})^2}$
$⇒ A^{''}(\sqrt2r) = \frac{(-4b)\times\sqrt2r}{(\sqrt2r)^2} = \frac{-2\sqrt2b}{r}< 0$
Hence, $b = \sqrt2r$ is the point of maxima.
$l = \sqrt{4r^2-b^2}=\sqrt{4r^2-2r^2}= \sqrt2r$
Since l = b, we can say that the given rectangle is a square.
Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Question:20: Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Answer:
Let r be the radius of the base of the cylinder and h be the height of the cylinder.
we know that the surface area of the cylinder $(A) = 2\pi r(r+h)$
$h = \frac{A-2\pi r^2}{2\pi r}$
Volume of cylinder
$(V) = \pi r^2 h\\ = \pi r^2 \left ( \frac{A-2\pi r^2}{2\pi r} \right ) = r \left ( \frac{A-2\pi r^2}{2 } \right )$
$V^{'}(r)= \left ( \frac{A-2\pi r^2}{2} \right )+(r).(-2\pi r)\\ = \frac{A-2\pi r^2 -4\pi r^2}{2} = \frac{A-6\pi r^2}{2}$
$V^{'}(r)= 0 $
$⇒ \frac{A-6\pi r^2}{2} = 0$
$⇒ r = \sqrt{\frac{A}{6\pi}}$
Hence, $r = \sqrt{\frac{A}{6\pi}}$ is the critical point.
Now,
$V^{''}(r) = -6\pi r$
$⇒ V^{''}(\sqrt{\frac{A}{6\pi}}) = - 6\pi . \sqrt{\frac{A}{6\pi}} = - \sqrt{6\pi A} < 0$
Hence, $r = \sqrt{\frac{A}{6\pi}}$ is the point of maxima.
$h = \frac{A-2\pi r^2}{2\pi r} = \frac{2-2\pi \frac{A}{6\pi}}{2\pi \sqrt \frac{A} {6\pi}} = \frac{4\pi \frac{A}{6\pi}}{2\pi \sqrt \frac{A} {6\pi}} = 2\pi \sqrt \frac{A} {6\pi} = 2r$
Hence, the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base.

Question:21 Of all the closed cylindrical cans (right circular) of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Answer:
Let r be the radius of base and h be the height of the cylinder
The volume of the cube (V) = $\pi r^2 h$
It is given that the volume of cylinder = 100 $cm^3$
$\pi r^2 h = 100$
$\Rightarrow h = \frac{100}{\pi r^2}$
Surface area of cube(A) = $2\pi r(r+h)$
$A(r)= 2\pi r(r+\frac{100}{\pi r^2})$
$= 2\pi r ( \frac{\pi r^3+100}{\pi r^2}) = \frac{2\pi r^3+200}{ r} = 2\pi r^2+\frac{200}{r}$
$A^{'}(r) = 4\pi r + \frac{(-200)}{r^2} $
$A^{'}(r)= 0$
$⇒ 4\pi r^3 = 200$
$⇒ r^3 = \frac{50}{\pi}$
$⇒ r = \left ( \frac{50}{\pi} \right )^{\frac{1}{3}}$
Hence, $r = (\frac{50}{\pi})^\frac{1}{3}$ is the critical point.
$A^{''}(r) = 4\pi + \frac{400r}{r^3}\\ A^{''}\left ( (\frac{50}{\pi})^\frac{1}{3} \right )= 4\pi + \frac{400}{\left ( (\frac{50}{\pi})^\frac{1}{3} \right )^2} > 0$
Hence, $r = (\frac{50}{\pi})^\frac{1}{3}$ is the point of minima.
$h = \frac{100}{\pi r^2} = \frac{100}{\pi \left ( (\frac{50}{\pi})^\frac{1}{3} \right )^2} = 2.(\frac{50}{\pi})^\frac{1}{3}$
Hence, $r = (\frac{50}{\pi})^\frac{1}{3}$ and $h = 2.(\frac{50}{\pi})^\frac{1}{3}$ are the dimensions of the can which has the minimum surface area.

Question:22: A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Answer:
Area of the square (A) = $a^2$, $a$ = side length
Area of the circle(S) = $\pi r^2$, $r$ = radius
Given the length of wire = 28 m
Let the length of one of the pieces be x m.
Then the length of the other piece is (28 - x) m
Now,
$4a = x$
$\Rightarrow a = \frac{x}{4}$
and
$2 \pi r = (28-x)$
$ \Rightarrow r= \frac{28-x}{2\pi}$
Area of the combined circle and square $f(x)$ = A + S
$=a^2 + \pi r^2 = (\frac{x}{4})^2+\pi (\frac{28-x}{2\pi})^2$
$f^{'}(x) = \frac{2x}{16}+\frac{(28-x)(-1)}{2\pi} $
$⇒ f^{'}(x) = \frac{x\pi+4x-112}{8\pi}$
$ f^{'}(x) = 0$
$⇒ \frac{x\pi+4x-112}{8\pi} = 0$
$⇒x(\pi+4) = 112$
$⇒x = \frac{112}{\pi + 4}$
Now,
$f^{''}(x) = \frac{1}{8}+ \frac{1}{2\pi}$
$ f^{''}(\frac{112}{\pi+4}) = \frac{1}{8}+ \frac{1}{2\pi} > 0$
Hence, $x = \frac{112}{\pi+4}$ is the point of minima.
Other length = 28 - x
= $28 - \frac{112}{\pi+4} = \frac{28\pi+112-112}{\pi+4} = \frac{28\pi}{\pi+4}$
Hence, two lengths are $\frac{28\pi}{\pi+4}$ and $\frac{112}{\pi+4}$.

Question:23: Prove that the volume of the largest cone that can be inscribed in a sphere of radius r is $\frac{8}{27}$ of the volume of the sphere.

Answer:

Volume of cone (V) = $\frac{1}{3}\pi R^2h$
Volume of sphere with radius r = $\frac{4}{3}\pi r^3$
By Pythagoras theorem in $\Delta ADC$, we can say that,
$OD^2 = r^2 - R^2 $
$⇒OD = \sqrt{r^2 - R^2}$
$⇒ h = AD = r + OD = r + \sqrt{r^2 - R^2}$
V = $\frac{1}{3}\pi R^2 (r + \sqrt{r^2+R^2})\\ = \frac{1}{3}\pi R^2r + \frac{1}{3}\pi R^2 \sqrt{r^2+R^2}$
$ V^{'}(R)= \frac{2}{3}\pi Rr + \frac{2}{3}\pi R \sqrt{r^2-R^2}+\frac{1}{3}\pi R^2.\frac{-2R}{2\sqrt{r^2-R^2}}$
$ V^{'}(R) = 0$
$⇒ \frac{1}{3}\pi R \left ( 2r + 2\sqrt{r^2-R^2} - \frac{R^2}{\sqrt{r^2-R^2}} \right ) = 0$
$⇒ \frac{1}{3}\pi R \left ( \frac{2r\sqrt{r^2-R^2} + 2r^2-2R^2 -R^2}{\sqrt{r^2-R^2}} \right ) = 0$
$⇒R \neq 0$ So, $2r\sqrt{r^2-R^2} = 3R^2 - 2r^2$
Square both sides, we get,
$4r^4-4r^2R^2 = 9R^4 + 4r^4 - 12R^2r^2$
$⇒9R^4-8R^2r^2 = 0$
$⇒R^2(9R^2-8r^2) = 0$
$⇒ R \neq 0$
So, $9R^2 = 8r^2$
$⇒R = \frac{2\sqrt2r}{3}$
Now,
$V^{''}(R)= \frac{2}{3}\pi r + \frac{2}{3}\pi \sqrt{r^2-R^2}+\frac{2}{3}\pi R.\frac{-2R}{2\sqrt{r^2-R^2}}-\frac{3\pi R^2}{\sqrt{r^2-R^2}} - \frac{(-1)(-2R)}{(r^2+R^2)\frac{3}{2}}$
$ V^{''}(\frac{2\sqrt2r}{3}) < 0$
Hence, point $R = \frac{2\sqrt2r}{3}$ is the point of maxima.
$h = r + \sqrt{r^2-R^2} = r + \sqrt{r^2-\frac{8r^2}{9}} = r + \frac{r}{3} = \frac{4r}{3}$
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\frac{4r}{3}$.
Volume = $= \frac{1}{3}\pi R^2h = \frac{1}{3}\pi \frac{8r^2}{9}.\frac{4r}{3} = \frac{8}{27}.\frac{4}{3}\pi r^3 = \frac{8}{27}\times$ Volume of the sphere
Hence, it is proved.

Question:24 Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt 2$ time the radius of the base.

Answer:

Volume of cone(V) $=\frac{1}{3}\pi r^2h$
$ \Rightarrow h = \frac{3V}{\pi r^2}$
curved surface area(A) = $\pi r l$
$l^2 = r^2 + h^2\\ l = \sqrt{r^2+\frac{9V^2}{\pi^2r^4}}$
$A = \pi r \sqrt{r^2+\frac{9V^2}{\pi^2r^4}} = \pi r^2 \sqrt{1+\frac{9V^2}{\pi^2r^6}}$
$\frac{dA}{dr} = 2\pi r \sqrt{1+\frac{9V^2}{\pi^2r^6}}+ \pi r^2.\frac{1}{2\sqrt{1+\frac{9V^2}{\pi^2r^6}}}.\frac{(-6r^5)9V^2}{\pi^2r^7}$
$ \frac{dA}{dr} = 0$
$ ⇒2\pi r \sqrt{1+\frac{9V^2}{\pi^2r^6}}+ \pi r^2.\frac{1}{2\sqrt{1+\frac{9V^2}{\pi^2r^6}}}.\frac{(-6)9V^2}{\pi^2r^7} = 0 $
$⇒ 2\pi^2r^6\left ( 1+\frac{9V^2}{\pi^2r^6} \right ) = {27V^2}$
$⇒ 2\pi^2r^6\left ( \frac{\pi^2r^6+9V^2}{\pi^2r^6} \right ) = {27V^2}$
$⇒ 2\pi^2r^6 + 18V^2 = 27V^2$
$ ⇒2\pi^2r^6 = 9V^2$
$ ⇒r^6 = \frac{9V^2}{2\pi^2}$
Now, we can clearly verify that,
$\frac{d^2A}{dr^2} > 0$
when $r^6 =\frac{9V^2}{2\pi^2}$
Hence, $r^6 =\frac{9V^2}{2\pi^2}$ is the point of minima.
$V = \frac{\sqrt2\pi r^3}{3}$
$h = \frac{3V}{\pi r^2} = \frac{3.\frac{\sqrt2\pi r^3}{3}}{\pi r^2} = \sqrt2 r$
Hence, proved that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt 2$ times the radius of the base.

Question:25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt 2$

Answer:

Let a be the semi-vertical angle of the cone.
Let r, h, and l be the radius, height, and slant height of the cone.
Now,
$r = l\sin a$ and $h=l\cos a$
We know that
Volume of cone (V) = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l\sin a)^2(l\cos a) = \frac{\pi l^3\sin^2 a\cos a}{3}$
Now,
$\frac{dV}{da}= \frac{\pi l^3}{3}\left ( 2\sin a\cos a.\cos a+\sin^2a.(-\sin a)\right )= \frac{\pi l^3}{3}\left ( 2\sin a\cos^2a-\sin^3a \right )$
$\frac{dV}{da}=0$
$⇒ \frac{\pi l^3}{3}\left ( 2\sin a\cos^2a-\sin^3a \right ) = 0$
$⇒ 2\sin a\cos^2a-\sin^3a= 0$
$⇒ 2\sin a\cos^2a=\sin^3a$
$⇒ \tan^2 a = 2$
$⇒ a = \tan^{-1}\sqrt 2$
Now,
$\frac{d^2V}{da^2}= \frac{\pi l^3}{3}\left ( 2\cos a\cos^2a+2\cos a(-2\cos a\sin a+3\sin^2a\cos a) \right )$
Now, at $a= \tan ^{-1}\sqrt 2$
$\frac{d^2V}{dx^2}< 0$
Therefore, $a= \tan ^{-1}\sqrt 2$ is the point of maxima
Hence, it is proved.

Question:26 Show that the semi-vertical angle of the right circular cone of given surface area and maximum volume is $\sin ^{-1} (\frac13)$

Answer:

Let r, l, and h be the radius, slant height and height of the cone, respectively.
Now,
$r = l\sin a $ and $ h =l\cos a$
Now,
We know that
The surface area of the cone (A) = $\pi r (r+l)$
$A= \pi l\sin a l(\sin a+1)$
$⇒l^2 = \frac{A}{\pi \sin a(\sin a+1)}$
$⇒ l = \sqrt{\frac{A}{\pi \sin a(\sin a+1)}}$
Now,
Volume of cone(V) =

$\frac{1}{3}\pi r^2h = \frac{1}{3}\pi l^3 \sin^2 a\cos a= \frac{\pi}{3}.\left ( \frac{A}{\pi\sin a(\sin a+1)} \right )^\frac{3}{2}.\sin^2 a\cos a$
On differentiate it w.r.t to a and after that
$\frac{dV}{da}= 0$
We will get
$a = \sin^{-1}\frac{1}{3}$
Now, at $a = \sin^{-1}\frac{1}{3}$
$\frac{d^2V}{da^2}<0$
Hence, we can say that $a = \sin^{-1}\frac{1}{3}$ is the point if maxima
Hence, it is proved.

Question:27 The point on the curve $x^2 = 2y$ which is nearest to the point (0, 5) is

$(A) (2 \sqrt 2,4) \: \: (B) (2 \sqrt 2,0)\: \: (C) (0, 0)\: \: (D) (2, 2)$

Answer:
Given curve is $x^2 = 2y$
Let the points on curve be $\left ( x, \frac{x^2}{2} \right )$
Distance between two points is given by
$f(x)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$= \sqrt{(x-0)^2+(\frac{x^2}{2}-5)^2} = \sqrt{x^2+ \frac{x^4}{4}-5x^2+25} = \sqrt{ \frac{x^4}{4}-4x^2+25}$
$f^{'}(x) = \frac{x^3-8x}{2\sqrt{\frac{x^4}{4}-4x^2+25}}$
$ f^{'}(x)= 0$
$⇒ \frac{x^3-8x}{2\sqrt{\frac{x^4}{4}-4x^2+25}} =0$
$⇒x(x^2 - 8)=0$
$⇒x=0 $ and
$x^2 = 8$
$\Rightarrow x = 2\sqrt2$
$\begin{aligned} & f^{\prime \prime}(x)=\frac{1}{2}\left(\frac{\left(3 x^2-8\right)\left(\sqrt{\frac{x^4}{4}-4 x^2+25}-\left(x^3-8 x\right) \cdot \frac{\left(x^3-8 x\right)}{\left(2 \sqrt{\frac{x^4}{4}-4 x^2+25}\right.}\right)}{\left(\sqrt{\frac{x^4}{4}-4 x^2+25}\right)^2}\right)\end{aligned}$
$f^{''}(0) = -8 < 0$
Hence, x = 0 is the point of maxima.
$f^{''}(2\sqrt2) > 0$
Hence, the point $x = 2\sqrt2$ is the point of minima
$x^2 = 2y$
$\Rightarrow y = \frac{x^2}{2} = \frac{8}{2}=4$
Hence, the point $(2\sqrt2,4)$ is the point on the curve $x^2 = 2y$ which is nearest to the point (0, 5)
Hence, the correct answer is option (A).

Question:28 For all real values of x, the minimum value of $\frac{1- x + x^2 }{1+ x +x^2}$
is: (A) 0 (B) 1 (C) 3 (D) $\frac13$

Answer:
Given function is
$f(x)= \frac{1- x + x^2 }{1+ x +x^2}$
$f^{'}(x)= \frac{(-1+2x)(1+x+x^2)-(1-x+x^2)(1+2x)}{(1+ x +x^2)^2}$
$⇒f^{'}(x)= \frac{-1-x-x^2+2x+2x^2+2x^3-1-2x+x+2x^2-x^2-2x^3}{(1+ x +x^2)^2} = \frac{-2+2x^2}{(1+ x +x^2)^2}$
$f^{'}(x)=0$
$⇒ \frac{-2+2x^2}{(1+ x +x^2)^2} = 0$
$ ⇒x^2 = 1$
$⇒ x= \pm 1$
Hence, x = 1 and x = -1 are the critical points.
Now,
$f^{''}(x)= \frac{4x(1+ x +x^2)^2-(-2+2x^2)2(1+x+x^2)(2x+1)}{(1+ x +x^2)^4}$
$f^{''}(1) = \frac{4\times(3)^2}{3^4} = \frac{4}{9} > 0$
Hence, x = 1 is the point of minima, and the minimum value is
$f(1)= \frac{1- 1 + 1^2 }{1+ 1 +1^2} = \frac{1}{3}$
$f^{''}(-1) =-4 < 0$
Hence, x = -1 is the point of maxima
Hence, the minimum value of
$\frac{1- x + x^2 }{1+ x +x^2}$ is $\frac{1}{3}$
Hence, the correct answer is option (D).

Question:29 The maximum value of $[ x ( x-1)+ 1 ] ^{1/3 } , 0\leq x \leq 1$
$(A) \left ( \frac{1}{3} \right ) ^{\frac13}\: \: (B) \frac12\: \: (C) 1\: \: (D) 0$

Answer:
Given function is
$f(x) = [ x ( x-1)+ 1 ] ^{\frac13 }$
$f^{'}(x) = \frac{1}{3}.[(x-1)+x].\frac{1}{[x(x-1)+1]^\frac{2}{3}} = \frac{2x-1}{3[x(x-1)+1]^\frac{2}{3}}$
$f^{'}(x) = 0$
$ ⇒\frac{2x-1}{3[x(x-1)+1]^\frac{2}{3}} = 0$
$⇒ x =\frac{1}{2}$
Hence, $x = \frac12$ is the critical point s0 we need to check the value at x = $\frac12$ and at the end points of given range, i.e. at x = 1 and x = 0
$f(\frac{1}{2}) = [ \frac{1}{2} ( \frac{1}{2}-1)+ 1 ] ^{\frac13 } = \left ( \frac{3}{4} \right )^\frac{1}{3}$
$f(0) = [ 0 ( 0-1)+ 1 ] ^{\frac13} = \left ( 1 \right )^\frac{1}{3} = 1$
$f(1) = [ 1 ( 1-1)+ 1 ] ^{\frac13 } = \left ( 1 \right )^\frac{1}{3} = 1$
Hence, by this, we can say that the maximum value of the given function is 1 at x = 0 and x = 1

Hence, the correct answer is option (C).

Question:1. Show that the function given by $f ( x ) = \frac{\log x}{x}$ has a maximum at x = e.

Answer:
Given function is
$f ( x ) = \frac{\log x}{x}$
$f^{'}(x) = \frac{1}{x}.\frac{1}{x} + log x\frac{-1}{x^2} = \frac{1}{x^2}(1-\log x)$
$f^{'}(x) =0$
$⇒ \frac{1}{x^2}(1-\log x) = 0$
$⇒ \frac{1}{x^2} \neq 0 $
So, $\log x = 1$
$\Rightarrow x = e$
Hence, x = e is the critical point
Now,
$f^{''}(x) = \frac{-2x}{x^3}(1-\log x)+\frac{1}{x^2}(-\frac{1}{x}) = \frac{1}{x^3}(-2x+2x \log x-1)$
$f^{''}(e) = \frac{-1}{e^3} < 0$
Hence, x = e is the point of maxima.

Question:2 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Answer:
It is given that the base of the triangle is b.
and let the side of the triangle be x cm , $\frac{dx}{dt} = -3 $ cm/s
We know that the area of the triangle(A) = $\frac{1}{2}bh$
now, $h = \sqrt{x^2-(\frac{b}{2})^2}$
$A= \frac{1}{2}b \sqrt{x^2-(\frac{b}{2})^2}$
$\frac{dA}{dt}=\frac{dA}{dx}.\frac{dx}{dt}= \frac{1}{2}b\frac{2x}{2\sqrt{x^2-(\frac{b}{2})^2}}.(-3)$
Now, at x = b
$\frac{dA}{dx} = \frac{1}{2}b\frac{2b}{\frac{\sqrt3b}{2}}.(-3)=-\sqrt3b$
Hence, the area decreasing when the two equal sides are equal to the base is $\sqrt3b$ cm2/s.

Question:3(i) Find the intervals in which the function f given by $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is increasing

Answer:
Given function is
$f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$
$f^{'}(x) = \frac{(4\cos x - 2-\cos x+x\sin x)(2+\cos x)-(4\sin x - 2x - x\cos x )(-\sin x)}{(2+\cos x)^2}$
$=\frac{4\cos x - \cos^2x}{2+\cos x}$
$f^{'}(x)=0$
$⇒ \frac{4\cos x - \cos^2x}{2+\cos x} =0$
$ ⇒\cos x(4-\cos x) = 0$
$ ⇒\cos x = 0$ and $ \cos x =4$
But $\cos x \neq 4$
So,
$\cos x = 0$
$ x = \frac{\pi}{2}$ and $ \frac{3\pi}{2}$
Now three ranges are there $\left ( 0,\frac{\pi}{2} \right ),\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ) $ and $ \left ( \frac{3\pi}{2},2\pi \right )$
In interval $\left ( 0,\frac{\pi}{2} \right ) $ and $ \left ( \frac{3\pi}{2},2\pi \right )$ , $f^{'}(x) > 0$

Hence, the given function $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is increasing in the interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
in interval $,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ), f^{'}(x) < 0$ so function is decreasing in this interval.

Question:3(ii) Find the intervals in which the function f given by f x is equal to $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is decreasing

Answer:
Given function is
$f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$
$f^{'}(x) = \frac{(4\cos x - 2-\cos x+x\sin x)(2+\cos x)-(4\sin x - 2x - x\cos x )(-\sin x)}{(2+\cos x)^2}$
$=\frac{4\cos x - \cos^2x}{2+\cos x}$
$f^{'}(x)=0$
$ \frac{4\cos x - \cos^2x}{2+\cos x} =0$
$ \cos x(4-\cos x) = 0$
$ \cos x = 0 $ and $ \cos x =4$
But $\cos x \neq 4$
So,
$\cos x = 0 \\ x = \frac{\pi}{2}$ and $ \frac{3\pi}{2}$
Now three ranges are there $\left ( 0,\frac{\pi}{2} \right ),\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
In interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$ , $f^{'}(x) > 0$

Hence, given function $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is increasing in interval $\left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )$
in interval $,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ), f^{'}(x) < 0$
Hence, given function $f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }$ is decreasing in interval $,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right )$.

Question:4(i) Find the intervals in which the function f given by $f (x) = x ^3 + \frac{1}{x^3}, x \neq 0$ Increasing

Answer:

Given function is
$f (x) = x ^3 + \frac{1}{x^3}$
$f^{'} (x) = 3x ^2 + \frac{-3x^2}{x^4}\\ f^{'}(x) = 0\\ 3x ^2 + \frac{-3x^2}{x^4} = 0\\ x^4 = 1\\ x = \pm1$
Hence, three intervals are their $(-\infty,-1),(-1,1) \ and (1,\infty)$
In interval $(-\infty,-1) \ and \ (1,\infty) , f^{'})x > 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is increasing in interval $(-\infty,-1) \ and \ (1,\infty)$
In interval (-1,1) , $f^{'}(x)< 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is decreasing in interval (-1,1)

Question:4(ii) Find the intervals in which the function f given by $f ( x) = x ^3 + \frac{1}{x^3} , x \neq 0$ decreasing

Answer:
Given function is
$f (x) = x ^3 + \frac{1}{x^3}$
$f^{'} (x) = 3x ^2 + \frac{-3x^2}{x^4}$
$f^{'}(x) = 0$
$ ⇒3x ^2 + \frac{-3x^2}{x^4} = 0$
$ ⇒x^4 = 1$
$⇒x = \pm1$

Hence, three intervals are their $(-\infty,-1),(-1,1) $ and $(1,\infty)$
In interval $(-\infty,-1)$ and $(1,\infty) , f^{'})x > 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is increasing in interval $(-\infty,-1)$ and $ (1,\infty)$
In interval (-1, 1), $f^{'}(x)< 0$
Hence, given function $f (x) = x ^3 + \frac{1}{x^3}$ is decreasing in interval (-1, 1).

Question:5 Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1$ with its vertex at one end of the major axis.

Answer:

Given the equation of the ellipse
$\frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1$
Now, we know that an ellipse is symmetrical about the x and y-axis.
Therefore, let's assume coordinates of A = (-n, m) then,
Now,
Put(-n,m) in equation of ellipse
We will get
$m = \pm \frac{b}{a}.\sqrt{a^2-n^2}$
Therefore, Now
Coordinates of A = $\left ( -n,\frac{b}{a}.\sqrt{a^2-n^2} \right )$
Coordinates of B = $\left ( -n,-\frac{b}{a}.\sqrt{a^2-n^2} \right )$
Now,
Length AB(base) = $2\frac{b}{a}.\sqrt{a^2-n^2}$
And height of triangle ABC = (a+n)
Now,
Area of triangle = $\frac{1}{2}bh$
$A = \frac{1}{2}.\frac{2b}{a}.\sqrt{a^2-n^2}.(a+n)= ab\sqrt{a^2-n^2}+bn\sqrt{a^2-n^2}$
Now,
$\frac{dA}{dn} = \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{bn^2}{\sqrt{a^2-n^2}}$
Now,
$\frac{dA}{dn} =0$
$\frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{bn^2}{\sqrt{a^2-n^2}} =0$
$-abn + n(a^2-n^2)-bn^2 = 0$
$ \Rightarrow n = -a,\frac{a}{2}$
but n cannot be zero,
therefore, $n = \frac{a}{2}$
Now, at $n = \frac{a}{2}$
$\frac{d^2A}{dn^2}< 0$
Therefore, $n = \frac{a}{2}$ is the point of maxima
Now,
$b = 2\frac{b}{a}.\sqrt{a^2- (\frac{a}{2})^2}= \sqrt3b$
$h = (a+ n ) = a+ \frac{ a}{2} = \frac{ 3a}{2}$
Now,
Therefore, Area (A) $= \frac{1}{2}bh = \frac{1}{2}\sqrt3 b \frac{3a}{2} = \frac{3\sqrt3ab}{4}$

Question:6 A tank with a rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of the least expensive tank?

Answer:

Let l, b, and h be the length, breath and height of the tank.
Then, volume of tank = l X b X h = 8 $m^3$
h = 2m (given)
lb = 4 = $l = \frac{4}{b}$
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
$A(b) = 4 + 2h(\frac{4}{b}+b)$
$A^{'}(b) = 2h(\frac{-4}{b^2}+1)$
$A^{'}(b)=0$
$ ⇒2h(\frac{-4}{b^2}+1) = 0$
$⇒ b^2= 4$
$⇒ b = 2$
Now,
$A^{''}(b) = 2h(\frac{-4\times-2b}{b^3})$
$A^{''}(2) = 8 > 0$
Hence, b = 2 is the point of minima
$l = \frac{4}{b} = \frac{4}{2} = 2$
So, l = 2 , b = 2 and h = 2 m
Area of base = l × B = 2 × 2 = 4 m2
The building of the tank costs Rs 70 per square meter for the base
Therefore, for 4 m2, Rs = 4 × 70 = Rs. 280
Area of 4 side walls = 2h(l + b)
= 2 × 2(2 + 2) = 16 m2
The building of the tank costs Rs 45 per square metre for sides
Therefore, for 16 m2, Rs = 16 × 45 = Rs. 720
Therefore, total cost for making the tank is = 720 + 280 = Rs. 1000

Question:7 The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

Answer:
It is given that the sum of the perimeters of a circle and a square is
$k = 2\pi r + 4a = k$
$\Rightarrow a = \frac{k - 2\pi r}{4}$
Let the sum of the area of a circle and square(A) = $\pi r^2 + a^2$
$A = \pi r^2 + (\frac{k-2\pi r}{4})^2$
$A^{'}(r) = 2\pi r + 2(\frac{k-2\pi r}{16})(- 2\pi)$
$ A^{'}(r) = 0$
$⇒ 2\pi (\frac{8r-k-2\pi r}{8}) = 0$
$⇒ r = \frac{k}{8-2\pi}$
Now,
$A^{''}(r) = 2\pi (\frac{8-2\pi }{8}) = 0$
$⇒A^{''}(\frac{k}{8-2\pi}) > 0$
Hence, $r= \frac{k}{8-2\pi}$ is the point of minima
$a = \frac{k-2\pi r}{4} = \frac{k-2\pi \frac{k}{8-2\pi}}{4}=2 \frac{k}{8-2\pi} = 2r$
Hence, it is proved that the sum of their areas is the least when the side of the square is double the radius of the circle.

Question:8: A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Answer:
Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle
$(r = \frac{l}{2})$
The total perimeter of the window = perimeter of the rectangle + perimeter of the semicircle
= $l+2b + \pi \frac{l}{2}$

$l+2b + \pi \frac{l}{2} = 10$
$⇒ l = \frac{2(10-2b)}{2+\pi}$
Area of window id given by (A) = $lb + \frac{\pi}{2}\left ( \frac{l}{2} \right )^2$
$= \frac{2(10-2b)}{2+\pi}b + \frac{\pi}{2}\left ( \frac{10-2b}{2+\pi} \right )^2\\$
$A^{'}(b) = \frac{20-8b}{2+\pi}+\frac{\pi}{2}.2(\frac{10-2b}{2+\pi}).\frac{(-2)}{2+\pi}$
$= \frac{20-8b}{2+\pi}-2\pi(\frac{10-2b}{(2+\pi)^2})$
$ A^{'}(b) = 0$
$⇒ \frac{20-8b}{2+\pi}=2\pi(\frac{10-2b}{(2+\pi)^2})$
$⇒ 40 + 20\pi -16b -8\pi b = 20\pi - 4\pi b$
$⇒40 = 4b(\pi+4)$
$⇒b = \frac{10}{\pi+4}$
Now,
$A^{''}(b) = \frac{-8}{2+\pi}+\frac{4\pi}{(2+\pi)^2} = \frac{-16-8\pi+4\pi}{(2+\pi)^2} = \frac{-16-4\pi}{(2+\pi)^2} $
$⇒A^{''}(\frac{10}{\pi+4}) < 0$
Hence, $b=\frac52$ is the point of maxima.
$l = \frac{2(10-2b)}{2+\pi} = \frac{2(10-2.\frac{10}{4+\pi})}{2+\pi} = \frac{20}{4+\pi}$
$r= \frac{l}{2}= \frac{20}{2(4+\pi)}=\frac{10}{4+\pi}$
Hence, these are the dimensions of the window to admit maximum light through the whole opening

Question:9: A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is $( a ^{\frac{2}{3}}+ b ^\frac{2}{3}) ^ \frac{3}{2}$

Answer:
It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle.

Let the angle between AC and BC be $\theta$.
So, the angle between AD and ED is also $\theta$
Now,
CD = $b \operatorname{ cosec}\theta$
And
AD = $a \sec\theta$
AC = H = AD + CD = $a \sec\theta$ + $b \operatorname{ cosec}\theta$
$\frac{dH}{d\theta} = a \sec\theta\tan\theta - b\cot\theta \operatorname{ cosec} \theta$
$ \frac{dH}{d\theta} = 0$
$a \sec\theta\tan\theta - b\cot\theta \operatorname{ cosec} \theta =0$
$⇒ a \sec\theta\tan\theta = b\cot\theta \operatorname{ cosec} \theta$
$⇒ a\sin^3\theta = b\cos^3\theta$
$\tan^3\theta = \frac{b}{a} \\ \tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$
Now,
$\frac{d^2H}{d\theta^2} > 0$
When $\tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$
Hence, $\tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}$ is the point of minima.
$\sec \theta = \frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}}$ and $\operatorname{ cosec} \theta = \frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}$

AC = $\frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}} +$ $\frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}$ = $(a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}$
Hence, it is proved.

Question:10 Find the points at which the function f given by $f(x) = (x-2)^4(x+1)^3$ has
(i) local maxima (ii) local minima (iii) point of inflexion

Answer:
Given function is
$f(x) = (x-2)^4(x+1)^3$
$f^{'}(x) = 4(x-2)^3(x+1)^3 + 3(x+1)^2(x-2)^4$
$ f^{'}(x)= 0$
$ 4(x-2)^3(x+1)^3 + 3(x+1)^2(x-2)^4=0$
$⇒ (x-2)^3(x+1)^2(4(x+1) + 3(x-2))=0$
$ x = 2 , x = -1 $ and $ x = \frac{2}{7}$
Now, for value x close to $\frac{2}{7}$ and to the left of $\frac{2}{7}$,
$f^{'}(x) > 0$, and for value close to $\frac{2}{7}$ and to the right of $\frac{2}{7}$ $f^{'}(x) < 0$
Thus, point x = $\frac{2}{7}$ is the point of maxima.
Now, for value x close to 2 and to the right of 2, $f^{'}(x) > 0$, and for value close to 2 and to the left of 2, $f^{'}(x) < 0$.
Thus, point x = 2 is the point of minima.
There is no change in the sign when the value of x is -1.
Thus, x = -1 is the point of inflexion.

Question:11: Find the absolute maximum and minimum values of the function f given by
$f (x) = \cos ^2 x + \sin x , x \epsilon [ 0 , \pi ]$

Answer:
Given function is
$f (x) = \cos ^2 x + \sin x$
$f^{'} (x) = 2\cos x(-\sin x) + \cos x$
$ f^{'}(x) = 0$
$ ⇒ -2\cos x\sin x + \cos x=0$
$⇒ \cos x(1-2\sin x) = 0$ either $ \cos x = 0$ and $\sin x = \frac{1}{2}$
$⇒ x = \frac{\pi}{2}$ and $ x = \frac{\pi}{6}$ as $\ x \ \epsilon [0,\pi]$
Now,
$f^{''} (x) = -2(-\sin x)\sin x - 2\cos x\cos x + (-\sin x) $
$⇒ f^{''}(x)= 2\sin^2x - 2\cos^2x - \sin x$
$⇒ f^{''}(\frac{\pi}{6}) = \frac{-3}{2} < 0$
Hence, the point $x = \frac{\pi}{6}$ is the point of maxima and the maximum value is
$f (\frac{\pi}{6}) = \cos ^2 \frac{\pi}{6} + \sin \frac{\pi}{6} = \frac{3}{4}+\frac{1}{2} = \frac{5}{4}$
And
$f^{''}(\frac{\pi}{2}) = 1 > 0$
Hence, the point $x = \frac{\pi}{2}$ is the point of minima and the minimum value is
$f (\frac{\pi}{2}) = \cos ^2 \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1$

Question:12 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\frac{4r}3$

Answer:

The volume of a cone (V) = $\frac{1}{3}\pi R^2h$
The volume of the sphere with radius r = $\frac{4}{3}\pi r^3$
By Pythagoras' theorem in $\Delta ADC$, we can say that
$OD^2 = r^2 - R^2 $
$ OD = \sqrt{r^2 - R^2}$
$ h = AD = r + OD = r + \sqrt{r^2 - R^2}$
$V = \frac{1}{3}\pi R^2 (r + \sqrt{r^2+R^2})$
$⇒ V = \frac{1}{3}\pi R^2r + \frac{1}{3}\pi R^2 \sqrt{r^2+R^2}$
$⇒V=\frac{1}{3}\pi R^2 (r + \sqrt{r^2-R^2})$
$V^{'}(R)= \frac{2}{3}\pi Rr + \frac{2}{3}\pi R \sqrt{r^2-R^2}+\frac{1}{3}\pi R^2.\frac{-2R}{2\sqrt{r^2-R^2}}$
Also, $ V^{'}(R) = 0$
$ ⇒\frac{1}{3}\pi R \left ( 2r + 2\sqrt{r^2-R^2} - \frac{R^2}{\sqrt{r^2-R^2}} \right ) = 0$
$ ⇒\frac{1}{3}\pi R \left ( \frac{2r\sqrt{r^2-R^2} + 2r^2-2R^2 -R^2}{\sqrt{r^2-R^2}} \right ) = 0$
$R \neq 0$
So, $2r\sqrt{r^2-R^2} = 3R^2 - 2r^2$
Squaring both sides, we get,
$⇒4r^4-4r^2R^2 = 9R^4 + 4r^4 - 12R^2r^2$
$⇒ 9R^4-8R^2r^2 = 0$
$⇒ R^2(9R^2-8r^2) = 0$
$ R \neq 0 $
So, $9R^2 = 8r^2$
$⇒R = \frac{2\sqrt2r}{3}$
Now,
$V^{''}(R)= \frac{2}{3}\pi r + \frac{2}{3}\pi \sqrt{r^2-R^2}+\frac{2}{3}\pi R.\frac{-2R}{2\sqrt{r^2-R^2}}-\frac{3\pi R^2}{\sqrt{r^2-R^2}} - \frac{(-1)(-2R)}{(r^2+R^2)\frac{3}{2}}$
$ ⇒V^{''}(\frac{2\sqrt2r}{3}) < 0$
Hence, the point $R = \frac{2\sqrt2r}{3}$ is the point of maxima.
$h = r + \sqrt{r^2-R^2} = r + \sqrt{r^2-\frac{8r^2}{9}} = r + \frac{r}{3} = \frac{4r}{3}$
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\frac{4r}{3}$.

Question:13 Let f be a function defined on [a, b] such that $f (x) > 0$, for all $x \: \: \epsilon \: \: ( a,b)$. Then, prove that f is an increasing function on (a, b).

Answer:
Let's do this question by taking an example.
Suppose
$f(x)= x^3 > 0 , (a.b)$
Now, also
$f{'}(x)= 3x^2 > 0 , (a,b)$
Hence, by this, we can say that f is an increasing function on (a, b).

Question:14: Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2 R }{\sqrt 3 }$. Also, find the maximum volume.

Answer:

The volume of the cylinder (V) = $\pi r^2 h$
By Pythagoras' theorem in $\Delta OAB$
$OA = \sqrt{R^2-r^2}$
h = 2OA
$h = 2\sqrt{R^2-r^2}$
$V = 2\pi r^2\sqrt{R^2-r^2}$
$V^{'}(r) = 4\pi r\sqrt{R^2-r^2}+2\pi r^2 . \frac{-2r}{2\sqrt{R^2-r^2}}$
$ V^{'}(r) = 0$
$⇒ 4\pi r\sqrt{R^2-r^2}- \frac{2\pi r^3}{\sqrt{R^2-r^2}} = 0$
$ ⇒4\pi r (R^2-r^2 ) - 2\pi r^3 = 0$
$⇒ 6\pi r^3 = 4\pi rR^2$
$⇒r =\frac{\sqrt6R}{3}$
Now,
$V^{''}(r) = 4\pi \sqrt{R^2-r^2}+4\pi r.\frac{-2r}{2\sqrt{R^2-r^2}}- \frac{6\pi r^2}{\sqrt{R^2-r^2}}.\frac{(-1)-2r}{2(R^2-r^2)\frac{3}{2}}$
$⇒V^{''}(\frac{\sqrt6R}{3}) < 0$
Hence, the point $r = \frac{\sqrt6R}{3}$ is the point of maxima
$h = 2\sqrt{R^2-r^2} = 2\sqrt{R^2 - \frac{2R^2}{3}} =\frac{2R}{\sqrt3}$
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2 R }{\sqrt 3 }$
and maximum volume is
$V = \pi r^2 h = \pi \frac{2R^2}{3}.\frac{2R}{\sqrt3} = \frac{4\pi R^3}{3\sqrt3}$

Question:15 Show that the height of the cylinder of the greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle a is one-third that of the cone, and the greatest volume of the cylinder is $\frac{4}{27}\pi h ^3 \tan ^2 \alpha$

Answer:

Let's take the radius and height of cylinder = r and h ' respectively
Let's take the radius and height of the cone = R and h respectively
Volume of cylinder = $\pi r^2 h'$
Volume of cone = $\frac{1}{3}\pi R^2 h$
Now, we have
$R = h\tan a$
Now, since $\Delta AOG$ and $ \Delta CEG$ are similar.
$\frac{OA}{OG} = \frac{CE}{EG}$
$\frac{h}{R} = \frac{h'}{R-r}$
$h'=\frac{h(R-r)}{R}$
$h'=\frac{h(h\tan a-r)}{h\tan a} = \frac{h\tan a- r}{\tan a}$
Now,
$V = \pi r^ 2 h' = \pi r^2 .\frac{h\tan a-r}{\tan a} = \pi r^2 h - \frac{\pi r^3}{\tan a}$
Now,
$\frac{dV}{dr}= 2\pi rh- \frac{3\pi r^2}{\tan a} $
$ \frac{dV}{dr}=0$
$ ⇒2\pi rh- \frac{3\pi r^2}{\tan a} = 0$
$⇒ 2\pi rh = \frac{3\pi r^2}{\tan a}$
$⇒r = \frac{2h\tan a}{3}$
Now,
$\frac{d^2V}{dr^2}= 2\pi h- \frac{6\pi r}{\tan a}$
at $r = \frac{2h\tan a}{3}$
$\frac{d^2V}{dr^2} = 2\pi h- 4\pi h < 0$
Hence, $r = \frac{2h\tan a}{3}$ is the point of maxima.
$h' = \frac{h\tan a-r}{\tan a} = \frac{h\tan a- \frac{2h\tan a}{3}}{\tan a}= \frac{1}{3}h$
Hence, it is proved.
Now, Volume (V) at $h' = \frac{1}{3}h$ and $r = \frac{2h\tan a}{3}$ is
$V = \pi r^2 h' = \pi \left ( \frac{2h\tan a}{3} \right )^2.\frac{h}{3}= \frac{4}{27}.\pi h^3\tan^2 a$
Hence, it is proved.

Question:16 A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meters per hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

Answer:

It is given that
$\frac{dV}{dt} = 314$ m3/h
Volume of cylinder (V) = $\pi r^2 h = 100\pi h \ \ \ \ \ \ \ \ \ \ \ (\because r = 10 m)$
$\frac{dV}{dt} = 100\pi \frac{dh}{dt}$
$⇒314 = 100\pi \frac{dh}{dt}$
$⇒ \frac{dh}{dt} = \frac{3.14}{\pi} = 1$ m/h
Hence, the correct answer is option (A).