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

Question 1:(ii) Find the maximum and minimum values, if any, of the following functions
given by

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

Answer:

Given function is,

$f(x) = 9x^2 + 12x + 2$ 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 n$f(x) = 9x^2 + 12x + 2$ occurs at $x = \frac{-2}{3}$ and the minimum value is

$f\left(\frac{-2}{3}\right) = 9\left(\frac{-2}{3}\right)^2 + 12\left(\frac{-2}{3}\right) + 2 = 4 - 8 + 2 = -2$ and it is clear that there is no maximum value of $f(x) = 9x^2 + 12x + 2$

Question 1: (iii) Find the maximum and minimum values, if any, of the following functions
given by

$f (x) = - (x -1) ^2 + 10$

Answer:

Given function is,
$f (x) = - (x -1) ^2 + 10$
$-(x-1)^2 \leq 0\\ -(x-1)^2+10\leq 10$ for every $x \ \epsilon \ R$
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
$-1 \leq \sin 2x \leq 1$
$-1 + 5 \leq \sin 2x + 5 \leq 1 + 5$
$4 \leq \sin 2x + 5 \leq 6$
Hence, the maximum value of our function $h(x) = \sin(2x) + 5$ is 6 and the minimum value is 4

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

Answer:

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

$-1 \leq \sin 4x \leq 1$

$-1 + 3 \leq \sin 4x + 3 \leq 1 + 3$

$2 \leq \sin 4x + 3 \leq 4$

$2 \leq |\sin 4x + 3| \leq 4$

Hence, the maximum value of our function $f (x) = | \sin 4x + 3|$ is 4 and the minimum value is 2

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

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

$h(x) = \sin x + \cos x$

$h'(x) = \cos x - \sin x$

$h'(x) = 0$

$\cos x - \sin x = 0$

$\cos x = \sin x$

$x = \frac{\pi}{4} \quad \text{as } x \in \left(0, \frac{\pi}{2} \right)$
Now, we use the second derivative test

$h''(x) = -\sin x - \cos x$

$h''\left(\frac{\pi}{4}\right) = -\sin \frac{\pi}{4} - \cos \frac{\pi}{4}$

$h''\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}$

$h''\left(\frac{\pi}{4}\right) = -\frac{2}{\sqrt{2}} = -\sqrt{2} < 0$

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:

$f(x) = x\,\text{sin}\,x - \text{cos}\,x$

Answer:

Given function is

$h(x) = \sin x - \cos x$
$h'(x) = \cos x + \sin x$
$h'(x) = 0$
$\cos x + \sin x = 0$
$\cos x = -\sin x$
$x = \frac{3\pi}{4} \quad \text{as} \quad x \in \left(0, 2\pi\right)$
Now, we use second derivative test

$h''(x) = -\sin x + \cos x$

$h''\left(\frac{3\pi}{4}\right) = -\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right)$

$h''\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}$

$h''\left(\frac{3\pi}{4}\right) = -\frac{2}{\sqrt{2}} = -\sqrt{2} < 0$

Hence, $\frac{\pi}{4}$ is the point of maxima and maximum value is $h\left ( \frac{3\pi}{4} \right )$ which is $\sqrt2$

Question 3: (v) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

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

Answer:

Given function is

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

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

$f'(x) = 0$

$3x^2 - 12x + 9 = 0$

$3(x^2 - 4x + 3) = 0$

$x^2 - 4x + 3 = 0$

$x^2 - x - 3x + 3 = 0$

$x(x - 1) - 3(x - 1) = 0$

$(x - 1)(x - 3) = 0$

$x = 1 \quad \text{and} \quad x = 3$

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. $x = 0 \ \text{and} \ x = \pi$

$f(\frac{\pi}{4}) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4}\\$
$=\frac{1}{\sqrt2}+\frac{1}{\sqrt2} = \frac{2}{\sqrt2} = \sqrt2$
$f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$
$f(\pi) = \sin \pi + \cos \pi = 0 +(-1) = -1$
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

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

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

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

$= 3 \times 81 - 8 \times 27 + 12 \times 9 - 144 + 25$

$= 243 - 216 + 108 - 144 + 25$

$= 16$
$f(0)=3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$
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:

$f(x) = \sin 2x$
$f'(x) = 2\cos 2x$
$f'(x) = 0$
$2\cos 2x = 0$
$\text{As } x \in [0, 2\pi]$
$0 < x < 2\pi$
$0 < 2x < 4\pi$
$\cos 2x = 0 \text{ at } 2x = \frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{5\pi}{2},\ \frac{7\pi}{2}$

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

$f'(x) = 2\cos 2x$

$f'(x) = 0$

$2\cos 2x = 0$

$\text{as } x \in [0, 2\pi]$

$0 < x < 2\pi$

$0 < 2x < 4\pi$

$\cos 2x = 0 \text{ at } 2x = \frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{5\pi}{2},\ \frac{7\pi}{2}$

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

$f(x) = xy = x(24 - x) = 24x - x^2$

$f^{'}(x) = 24 - 2x$

$f^{'}(x) = 0$

$24 - 2x = 0$

$x = 12$

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,

$f^{'}(y) = 180y^2 - 4y^3$

$f^{'}(y) = 0$

$y^2(180 - 4y) = 0$

$y = 0\ \text{and}\ y = 45$
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,

$f^{'}(y) = 6125y^4 - 420y^5 + 7y^6$

$f^{'}(y) = 0$

$y^4(6125 - 420y + 7y^2) = 0$

$y = 0\ \text{and}\ (y - 25)(y - 35) \Rightarrow y = 25,\ y = 35$

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 $\left ( V(x) \right )$ = $x(18-2x)^2$
$V^{'}(x) = (18-2x)^2+(x)2(18-2x)(-2)$

$V'(x) = 0$

$(18 - 2x)^2 - 4x(18 - 2x) = 0$

$324 + 4x^2 - 72x - 72x + 8x^2 = 0$

$12x^2 - 144x + 324 = 0$

$12(x^2 - 12x + 27) = 0$

$x^2 - 9x - 3x + 27 = 0$

$(x - 3)(x - 9) = 0$

$x = 3 \ \text{and} \ x = 9$

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,

$V''(x) = 24x - 144$

$V''(3) = 24 \times 3 - 144$

$\Rightarrow 72 - 144 = -72$

$V''(3) < 0$

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 $V(x) = x(45-2x)(24-2x)$

$V^{'}(x) = (45 - 2x)(24 - 2x) + (-2)(x)(24 - 2x) + (-2)(x)(45 - 2x)$

$= 1080 + 4x^2 - 138x - 48x + 4x^2 - 90x + 4x^2$

$= 12x^2 - 276x + 1080$

$V^{'}(x) = 0$

$12(x^2 - 23x + 90) = 0$

$x^2 - 23x + 90 = 0$

$x^2 - 18x - 5x + 90 = 0$

$(x - 18)(x - 5) = 0$

$x = 18\ \text{and}\ x = 5$

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

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:

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 = $\frac{1}{3}\pi R^2 (r + \sqrt{r^2+R^2})\\ = \frac{1}{3}\pi R^2r + \frac{1}{3}\pi R^2 \sqrt{r^2+R^2}$

$\frac{1}{3}\pi R^2 \left(r + \sqrt{r^2 - R^2}\right)$

$V'(R) = \frac{2}{3}\pi Rr + \frac{2}{3}\pi R \sqrt{r^2 - R^2} + \frac{1}{3}\pi R^2 \cdot \frac{-2R}{2\sqrt{r^2 - R^2}}$
$V'(R) = 0$
$\frac{1}{3}\pi R \left(2r + 2\sqrt{r^2 - R^2} - \frac{R^2}{\sqrt{r^2 - R^2}}\right) = 0$
$\frac{1}{3}\pi R \left( \frac{2r\sqrt{r^2 - R^2} + 2r^2 - 2R^2 - R^2}{\sqrt{r^2 - R^2}} \right) = 0$
$R \neq 0 \Rightarrow 2r\sqrt{r^2 - R^2} = 3R^2 - 2r^2$
$\text{Square both sides:}$
$4r^2(r^2 - R^2) = (3R^2 - 2r^2)^2$
$4r^4 - 4r^2R^2 = 9R^4 + 4r^4 - 12R^2r^2$
$9R^4 - 8R^2r^2 = 0$
$R^2(9R^2 - 8r^2) = 0$
$R \neq 0 \Rightarrow 9R^2 = 8r^2$
$\Rightarrow R = \frac{2\sqrt{2}r}{3}$

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 = $\frac{1}{3}\pi R^2 h = \frac{1}{3}\pi \cdot \frac{8r^2}{9} \cdot \frac{4r}{3} = \frac{8}{27} \cdot \frac{4}{3} \pi r^3 = \frac{8}{27} \times$ 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) = $\pi r l$
$l^2 = r^2 + h^2\\ l = \sqrt{r^2+\frac{9V^2}{\pi^2r^4}}$
$A = \pi r \sqrt{r^2+\frac{9V^2}{\pi^2r^4}} = \pi r^2 \sqrt{1+\frac{9V^2}{\pi^2r^6}}$

$\frac{dA}{dr} = 2\pi r \sqrt{1+\frac{9V^2}{\pi^2r^6}} + \pi r^2 \cdot \frac{1}{2\sqrt{1+\frac{9V^2}{\pi^2r^6}}} \cdot \frac{(-6r^5) \cdot 9V^2}{\pi^2r^7}$

$\frac{dA}{dr} = 0$

$2\pi r \sqrt{1+\frac{9V^2}{\pi^2r^6}} + \pi r^2 \cdot \frac{1}{2\sqrt{1+\frac{9V^2}{\pi^2r^6}}} \cdot \frac{(-6) \cdot 9V^2}{\pi^2r^7} = 0$

$2\pi^2r^6\left(1 + \frac{9V^2}{\pi^2r^6}\right) = 27V^2$

$2\pi^2r^6 \cdot \frac{\pi^2r^6 + 9V^2}{\pi^2r^6} = 27V^2$

$2\pi^2r^6 + 18V^2 = 27V^2$

$2\pi^2r^6 = 9V^2$

$r^6 = \frac{9V^2}{2\pi^2}$

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:

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,

$\frac{dV}{da} = \frac{\pi l^3}{3}\left( 2\sin a \cos^2 a - \sin^3 a \right)$

$\frac{dV}{da} = 0$

$\frac{\pi l^3}{3}\left( 2\sin a \cos^2 a - \sin^3 a \right) = 0$

$2\sin a \cos^2 a = \sin^3 a$

$\tan^2 a = 2$

$a = \tan^{-1} \sqrt{2}$
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:

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

$f(x)= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$= \sqrt{(x - 0)^2 + \left(\frac{x^2}{2} - 5\right)^2}$

$= \sqrt{x^2 + \frac{x^4}{4} - 5x^2 + 25}$

$= \sqrt{ \frac{x^4}{4} - 4x^2 + 25}$

$f'(x) = \frac{x^3 - 8x}{2\sqrt{\frac{x^4}{4} - 4x^2 + 25}}$

$f'(x) = 0$

$\frac{x^3 - 8x}{2\sqrt{\frac{x^4}{4} - 4x^2 + 25}} = 0$

$x(x^2 - 8) = 0$

$x = 0 \quad \text{and} \quad x^2 = 8 \Rightarrow x = 2\sqrt{2}$

$f''(x) = \frac{1}{2} \left( \frac{(3x^2 - 8)\sqrt{\frac{x^4}{4} - 4x^2 + 25} - (x^3 - 8x) \cdot \frac{(x^3 - 8x)}{2\sqrt{\frac{x^4}{4} - 4x^2 + 25}}}{\left(\sqrt{\frac{x^4}{4} - 4x^2 + 25}\right)^2} \right)$

$f''(0) = -8 < 0$

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) $\frac{1}{3}$

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, 1/3 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) $\frac{1}{2}$

(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 so 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

Hence, answer 1 is correct.