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

# 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

## 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.

12th class Maths exercise 6.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Application of Derivatives Exercise 6.5

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 $f (x) = 9x^2+12x+2$

Given function is,
add and subtract 2 in given equation

Now, for every

Hence, minimum value occurs when

Hence, the minimum value of function occurs at and the minimum value isand it is clear that there is no maximum value of

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 $f (x) = 9x^2+12x+2$

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

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 $x \ \epsilon \ R$

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 $x \ \epsilon \ R$

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

Given function is
$f (x) = | \sin 4x + 3|$
We know that 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

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

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
$f(0) = (0)^2 = 0$

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
$g(1) = (-1)^3 - 3(-1) = -1 + 3 = 2$

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 $\sqrt2$

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 $\sqrt2$

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
$f (3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27-54+27+15 = 15$

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
$g ( x) = \frac{x}{2} + \frac{2}{x} \\ g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$

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
$g (0) = \frac{1}{0^2 + 2} = \frac{1}{2}$

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
$f (x) = x \sqrt{ 1-x }\\ f(\frac{2}{3}) = \frac{2}{3}\sqrt{1-\frac{2}{3}}\\ f(\frac{2}{3}) = \frac{2}{3}\sqrt{\frac{1}{3}}\\ f(\frac{2}{3}) = \frac{2}{3\sqrt3}\\ f(\frac{2}{3}) = \frac{2\sqrt3}{9}$

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

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

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

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

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

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

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

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

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
$=3\times16 - 8\times 8 + 12\times 4 - 96 + 25 = 48-64+48-96+25 = -39$

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

$f(0)=3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$
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

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 $x \ \epsilon \ [0,2\pi]$

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 $\sqrt2$

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 $x \ \epsilon \ [-3,-1]$

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

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

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

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

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

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

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

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

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
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

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

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

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 $\frac{112}{\pi+4}$

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 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

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

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

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

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)

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

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

## More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

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• 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.

## Subject Wise NCERT Exemplar Solutions

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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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

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.

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.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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 help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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