NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6

Q: What are the topics covered in exercise 6.5?

The topic of maxima and minima is covered in Class 12th Maths chapter 6 exercise 6.5.

Q: How many examples are solved before Class 12 Maths exercise 6.5 on the topic maxima and minima?

16 questions are explained before exercise 6.5

Q: What is the total number of solved examples in the NCERT book till Exercise 6.5 Class 12 Maths?

A sum of 41 questions are solved till the Class 12th Maths chapter 6 exercise 6.5

Q: Give some applications of maxima and minima?

The problems of maxima and minima are used in business, science and mathematics etc. Examples are problems related to maximising profit, minimising the distance between the two points etc

Q: What is a monotonic function in a given interval?

In the given interval, a monotonic function is either increasing or decreasing.

Q: For a function f, if k is a point of local maxima, then what is the local maximum value of the function is?

The local maximum value is f(k)

Q: For a function f, if u is a point of local minima, then what is the local minimum value of the function is?

f(u) is the local minimum value of function f

Q: Is the topic maxima and minima important for board exams?

Yes, maxima and minima are one of the important topics of the chapter from where 1 question can be expected for CBSE board exam

NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6 - Application of Derivatives

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:53 PM IST | #CBSE Class 12th

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NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5

NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6 Application of Derivatives are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Throughout the NCERT solutions for exercise 6.5 Class 12 Maths chapter 6 the topic maxima and minima is discussed. NCERT solutions for Class 12 Maths chapter 6 exercise 6.5 uses the concept of derivatives to find the maximum and minimum of different functions. Exercise 6.5 Class 12 Maths also give ideas about absolute minimum and maximum. In the NCERT Class 12 Mathematics Book, some real-life examples of finding maximum and minimum values are given. And certain definitions are discussed after the examples in the NCERT book. Such as definitions of maximum and minimum values, extreme point, monotonic functions, local maxima and minima and certain theorems etc. After these the Class 12 Maths chapter 6 exercise 6.5 is given for practice.

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Application of Derivatives Exercise 6.5

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,
$f (x) = 9x^ 2 + 12x + 2$
add and subtract 2 in given equation
$f (x) = 9x^ 2 + 12x + 2 + 2- 2\\ f(x)= 9x^2 +12x+4-2\\ f(x)= (3x+2)^2 - 2$
Now,
$(3x+2)^2 \geq 0\\ (3x+2)^2-2\geq -2$ for every $x \ \epsilon \ R$
Hence, 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(ii) Find the maximum and minimum values, if any, of the following functions
given by

Answer:

Given function is,
f (x) = 9x^ 2 + 12x + 2 add and subtract 2 in given equation

$f (x) = 9x^ 2 + 12x + 2 + 2- 2\\ f(x)= 9x^2 +12x+4-2\\ f(x)= (3x+2)^2 - 2$

Now, $(3x+2)^2 \geq 0\\ (3x+2)^2-2\geq -2$ for every $x \ \epsilon \ R$

Hence, 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

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

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

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 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:

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

Answer:

Given function is
$h(x) = x + 1$
It is given that the value of $x \ \epsilon (-1,1)$
So, we can not comment about either maximum or minimum value
Hence, function $h(x) = x + 1$ has neither has a maximum or 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
$f ( x) = x^2\\ f^{'}(x) = 2x\\ f^{'}(x) = 0 \Rightarrow 2x = 0 \Rightarrow x = 0$
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 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
$h(x) = \sin x + \cos x\\ h^{'}(x)= \cos x - \sin x\\ h^{'}(x)= 0\\ \cos x - \sin x = 0\\ \cos x = \sin x\\ x = \frac{\pi}{4} \ \ \ \ \ \ as \ x \ \epsilon \ \left ( 0,\frac{\pi}{2} \right )$
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
$h(x) = \sin x - \cos x\\ h^{'}(x)= \cos x + \sin x\\ h^{'}(x)= 0\\ \cos x + \sin x = 0\\ \cos x = -\sin x\\ x = \frac{3\pi}{4} \ \ \ \ \ \ as \ x \ \epsilon \ \left ( 0,2\pi \right )$
Now, we use second derivative test
$h^{''}(x)= -\sin x + \cos x\\ h^{''}(\frac{3\pi}{4}) = -\sin \frac{3\pi}{4} + \cos \frac{3\pi}{4}\\ h^{''}(3\frac{\pi}{4}) = -(\frac{1}{\sqrt2})-\frac{1}{\sqrt2}\\ h^{''}(\frac{\pi}{4})=- \frac{2}{\sqrt2} = -\sqrt2 < 0$
Hence, $\frac{\pi}{4}$ is the point of maxima and 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:

Givrn function is
$f (x) = x^3 - 6x^2 + 9x + 15\\ f^{'}(x) = 3x^2 - 12x + 9\\ f^{'}(x)= 0\\ 3x^2 - 12x + 9 = 0\\ 3(x^2-4x+3)=0\\ x^2-4x+3 = 0\\ x^2 - x -3x + 3=0\\ x(x-1)-3(x-1) = 0\\ (x-1)(x-3) = 0\\ x=1 \ \ \ \ \ \ and \ \ \ \ \ \ \ x = 3$
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:

Gien 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 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 $c \ \epsilon \ R$ 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 $c \ \epsilon \ R$ 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 end points of 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
$f(1) = (1-1)^2+3 = 0^2+3 = 3$

$f(-3) = (-3-1)^2+3= (-4)^2+3 = 16+3= 19$
$f(1) = (1-1)^2+3 = 0^2+3 = 3$
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 second derivative test
$p^{''}(x)= -36< 0$
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 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
1628071745575 $=3\times16 - 8\times 8 + 12\times 4 - 96 + 25 = 48-64+48-96+25 = -39$

$f(3)=3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25\\ =3\times81-8\times27+12\times9-144+25 \\ =243-216+108-144+25 = 16$

$f(0)=3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$
Hence, maximum value of function $f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}$ occurs at x = 0 and vale 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, 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 $x = \pi$

$f(x) = \sin 2x\\ f(\frac{\pi}{4}) = \sin 2\left ( \frac{\pi}{4} \right ) = \sin \frac{\pi}{2} = 1$

$f(x) = \sin 2x\\ f(\frac{3\pi}{4}) = \sin 2\left ( \frac{3\pi}{4} \right ) = \sin \frac{3\pi}{2} = -1$

$f(x) = \sin 2x\\ f(\frac{5\pi}{4}) = \sin 2\left ( \frac{5\pi}{4} \right ) = \sin \frac{5\pi}{2} = 1$

$f(x) = \sin 2x\\ f(\frac{7\pi}{4}) = \sin 2\left ( \frac{7\pi}{4} \right ) = \sin \frac{7\pi}{2} = -1$

$f(x) = \sin 2x\\ f(\pi) = \sin 2(\pi)= \sin 2\pi = 0$

$f(x) = \sin 2x\\ f(0) = \sin 2(0)= \sin 0 = 0$

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
$f(x) = \sin x + \cos x$
$f^{'}(x) = \cos x - \sin x\\ f^{'}(x)= 0\\ \cos x- \sin x= 0\\ \cos = \sin x\\ x = 2n\pi+\frac{\pi}{4} \ where \ n \ \epsilon \ I$
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 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
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 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 vale 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
$f^{'}(1)=0$
$f^{'}(x) = 4x^3-124x+a\\ f^{'}(1) = 4(1)^3-124(1)+a\\ f^{'}(1)=4-124+a = a - 120\\$
Now,
$f^{'}(1)=0\\ a - 120=0\\ a=120$
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, 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 $x = 2\pi$

$f(x) =x+ \sin 2x\\ f(\frac{\pi}{3}) = \frac{\pi}{3}+\sin 2\left ( \frac{\pi}{3} \right ) = \frac{\pi}{3}+\sin \frac{2\pi}{3} = \frac{\pi}{3}+\frac{\sqrt3}{2}$

$f(x) =x+ \sin 2x\\ f(\frac{2\pi}{3}) = \frac{2\pi}{3}+\sin 2\left ( \frac{2\pi}{3} \right ) = \frac{2\pi}{3}+\sin \frac{4\pi}{3} = \frac{2\pi}{3}-\frac{\sqrt3}{2}$

$f(x) =x+ \sin 2x\\ f(\frac{4\pi}{3}) = \frac{4\pi}{3}+\sin 2\left ( \frac{4\pi}{3} \right ) = \frac{4\pi}{3}+\sin \frac{8\pi}{3} = \frac{4\pi}{3}+\frac{\sqrt3}{2}$

$f(x) =x+ \sin 2x\\ f(\frac{5\pi}{3}) = \frac{5\pi}{3}+\sin 2\left ( \frac{5\pi}{3} \right ) = \frac{5\pi}{3}+\sin \frac{10\pi}{3} = \frac{5\pi}{3}-\frac{\sqrt3}{2}$

$f(x) = x+\sin 2x\\ f(2\pi) = 2\pi+\sin 2(2\pi)= 2\pi+\sin 4\pi = 2\pi$

$f(x) = x+\sin 2x\\ f(0) = 0+\sin 2(0)= 0+\sin 0 = 0$

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 are 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
Noe, y = 24 - x
= 24 - 12 = 12
Hence, the value 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,
$f^{'}(y) = 180y^2-4y^3\\ f^{'}(y) = 0\\ y^2(180-4y)=0\\ y= 0 \ and \ y = 45$

Now,
$f^{''}(y) = 360y-12y^2\\ f^{''}(0) = 0\\$
hence, 0 is neither point of minima or maxima
$f^{''}(y) = 360y-12y^2\\ f^{''}(45) = 360(45)-12(45)^2 = -8100 < 0$
Hence, y = 45 is point of maxima
x = 60 - y
= 60 - 45 = 15
Hence, 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,
$f^{'}(y) = 6125y^4-420y^5+7y^6\\ f^{'}(y)=0\\ y^4(6125-420y+7y^2) = 0 \\y =0 \ and \ (y-25)(y-35)\Rightarrow y = 25 , y=35$
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 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 value 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 an d y are positive two 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, 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 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 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 cube is (18-2x) cm and 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 can not be 9 because then the value of breath become 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 top, by cutting off 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 assume the side of the square to be cut off is x cm
Then,
Volume of cube $V(x) = x(45-2x)(24-2x)$
$V^{'}(x) = (45-2x)(24-2x) + (-2)(x)(24-2x)+(-2)(x)(45-2x)\\$
$1080 + 4x^2 - 138x - 48x + 4x^2 - 90x +4x^2\\ 12x^2 - 276x + 1080$
$V^{'}(x) = 0\\ 12(x^2 - 23x+90)=0\\ x^2-23x+90 = 0\\ x^2-18x-5x+23=0\\ (x-18)(x-5)=0\\ x =18 \ and \ x = 5$
But x cannot be equal to 18 because then side (24 - 2x) become 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 assume that length and breadth of rectangle inscribed in a circle is l and b respectively
and the radius of the circle is r
1628071829107 Now, by Pythagoras theorem
$a = \sqrt{l^2+b^2}\\$
a = 2r
$4r^2 = l^2+b^2\\ l = \sqrt{4r^2 - b^2}$
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 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,
$V^{''}(r) = -6\pi r\\ V^{''}(\sqrt{\frac{A}{6\pi}}) = - 6\pi . \sqrt{\frac{A}{6\pi}} = - \sqrt{A6\pi} < 0$
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) = $a^2$
Area of the circle(S) = $\pi r^2$
Given the length of wire = 28 m
Let the length of one of the piece is 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 is = 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 8/27 of the volume of the sphere.

Answer: 1654609173546 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 ca 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 = $\frac{1}{3}\pi R^2 (r + \sqrt{r^2+R^2})\\ = \frac{1}{3}\pi R^2r + \frac{1}{3}\pi R^2 \sqrt{r^2+R^2}$
$\frac{1}{3}\pi R^2 (r + \sqrt{r^2-R^2})\\ V^{'}(R)= \frac{2}{3}\pi Rr + \frac{2}{3}\pi R \sqrt{r^2-R^2}+\frac{1}{3}\pi R^2.\frac{-2R}{2\sqrt{r^2-R^2}}\\ V^{'}(R) = 0\\ \frac{1}{3}\pi R \left ( 2r + 2\sqrt{r^2-R^2} - \frac{R^2}{\sqrt{r^2-R^2}} \right ) = 0\\ \frac{1}{3}\pi R \left ( \frac{2r\sqrt{r^2-R^2} + 2r^2-2R^2 -R^2}{\sqrt{r^2-R^2}} \right ) = 0\\ R \neq 0 \ So, \\ 2r\sqrt{r^2-R^2} = 3R^2 - 2r^2\\ Square \ both \ sides\\ 4r^4-4r^2R^2 = 9R^4 + 4r^4 - 12R^2r^2\\ 9R^4-8R^2r^2 = 0\\ R^2(9R^2-8r^2) = 0\\ R \neq 0 \ So, 9R^2 = 8r^2\\ R = \frac{2\sqrt2r}{3}$
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 = $= \frac{1}{3}\pi R^2h = \frac{1}{3}\pi \frac{8r^2}{9}.\frac{4r}{3} = \frac{8}{27}.\frac{4}{3}\pi r^3 = \frac{8}{27}\times \ volume \ of \ sphere$
Hence 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 varify 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 $\sqrt 2$ time 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:

1628071922992 Let a be the semi-vertical angle of cone
Let r , h , l are the radius , height , slent height of 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 proved

Question:26 Show that semi-vertical angle of the right circular cone of given surface area and maximum volume is $\sin ^{-1} (1/3)$

Answer:

1628071965473 Let r, l, and h are the radius, slant height and height of 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 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
$f(x)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$= \sqrt{(x-0)^2+(\frac{x^2}{2}-5)^2} = \sqrt{x^2+ \frac{x^4}{4}-5x^2+25} = \sqrt{ \frac{x^4}{4}-4x^2+25}$
$f^{'}(x) = \frac{x^3-8x}{2\sqrt{\frac{x^4}{4}-4x^2+25}}\\ f^{'}(x)= 0\\ \frac{x^3-8x}{2\sqrt{\frac{x^4}{4}-4x^2+25}} =0\\ x(x^2 - 8)=0\\x=0 \ and \ x^2 = 8\Rightarrow x = 2\sqrt2$
$f^{''}(x) = \frac{1}{2}\left (\frac{(3x^2-8)(\sqrt{\frac{x^4}{4}-4x^2+25} - (x^3-8x).\frac{(x^3-8x)}{2\sqrt{\frac{x^4}{4}-4x^2+25}}}{(\sqrt{\frac{x^4}{4}-4x^2+25})^2}) \right )$
$f^{''}(0) = -8 < 0$
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 (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) 1/3

Answer:

Given function is
$f(x)= \frac{1- x + x^2 }{1+ x +x^2}$
$f^{'}(x)= \frac{(-1+2x)(1+x+x^2)-(1-x+x^2)(1+2x)}{(1+ x +x^2)^2}$
$= \frac{-1-x-x^2+2x+2x^2+2x^3-1-2x+x+2x^2-x^2-2x^3}{(1+ x +x^2)^2} = \frac{-2+2x^2}{(1+ x +x^2)^2}$
$f^{'}(x)=0\\ \frac{-2+2x^2}{(1+ x +x^2)^2} = 0\\ x^2 = 1\\ x= \pm 1$
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, (D) is the correct answer

Question:29 The maximum value of $[ x ( x-1)+ 1 ] ^{1/3 } , 0\leq x \leq 1$
$(A) \left ( \frac{1}{3} \right ) ^{1/3}\: \: (B) 1 /2\: \: (C) 1\: \: (D) 0$

Answer:

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

option c is correct

Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

The exercise 6.5 Class 12 Maths is solved by expert Mathematics faculties and students can rely on Class 12th Maths chapter 6 exercise 6.5 for exam preparation.
Not only for board exams but also for competitive exams like JEE main the NCERT solutions for Class 12 Maths chapter 6 exercise 6.5 will be helpful.
Students can access these solutions for free.

Key Features Of NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6

Comprehensive Coverage: The solutions encompass all the topics covered in ex 6.5 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 6.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 6.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 6.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 6.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 6.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Questions (FAQs)

1. What are the topics covered in exercise 6.5?

The topic of maxima and minima is covered in Class 12th Maths chapter 6 exercise 6.5.

2. How many examples are solved before Class 12 Maths exercise 6.5 on the topic maxima and minima?

16 questions are explained before exercise 6.5

3. What is the total number of solved examples in the NCERT book till Exercise 6.5 Class 12 Maths?

A sum of 41 questions are solved till the Class 12th Maths chapter 6 exercise 6.5

4. Give some applications of maxima and minima?

The problems of maxima and minima are used in business, science and mathematics etc. Examples are problems related to maximising profit, minimising the distance between the two points etc

5. What is a monotonic function in a given interval?

In the given interval, a monotonic function is either increasing or decreasing.

6. For a function f, if k is a point of local maxima, then what is the local maximum value of the function is?

The local maximum value is f(k)

7. For a function f, if u is a point of local minima, then what is the local minimum value of the function is?

f(u) is the local minimum value of function f

8. Is the topic maxima and minima important for board exams?

Yes, maxima and minima are one of the important topics of the chapter from where 1 question can be expected for CBSE board exam

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
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Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

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Type B: For candidates scoring between 50% and 60%.

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Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6 - Application of Derivatives

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5

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