Ex 6.5 class 12 maths ncert solutions

It is given that \( \mathrm{f}(x)=(2x-1)^{2}+3 \)
Now, we can see that \( (2 x-1)^{2} \geq 0 \) for every \( x \in R \)
\( \Rightarrow \mathrm{f}(x)=(2 x-1)^{2}+3 \geq 3 \) for every \( x \in \mathrm{R} \)
The minimum value of f is attained when \( 2 x-1=0 \)
\(2 x-1=0\)
\(\Rightarrow x=\frac{1}{2}\)
Then, Minimum value of
\(\mathrm{f}=\mathrm{f}\left(\frac{1}{2}\right)=\left(2 \cdot \frac{1}{2}-1\right)^{2}+3=3\)
Therefore, function f does not have a maximum value.

It is given that \( f(x)=9 x^{2}+12 x+2=(3 x+2)^{2}-2 \)
Now, we can see that \( (3 x+2)^{2} \geq 0 \) for every \( x \in R \)
\( \Rightarrow \mathrm{f}(x)=(3 x+2)^{2}-2 \geq-2 \) for every \( x \in \mathrm{R} \)
The minimum value of \( f \) is attained when \( 3 x+2=0 \)
\(3 x+2=0\)
\(\Rightarrow \frac{-2}{3}\)
Then, Minimum value of
\(f=f\left(\frac{-2}{3}\right)=\left(3\left(\frac{-2}{3}\right)+2\right)^{2}-2=-2\)
Therefore, function \( f \) does not have a maximum value.

It is given that \( \mathrm{f}(x)=-(x-1)^{2}+10 \)
Now, we can see that \( (x-1)^{2} \geq 0 \) for every \( x \in R \)
\(\Rightarrow \mathrm{f}(x)=-(x-1)^{2}+10 \leq 10 \text { for every } x \in \mathrm{R}\)
The minimum value of \( f \) is attained when \( x-1=0 \)
\(x-1=0\)
\(\Rightarrow x=1\)
Then, Maximum value of \( \mathrm{f}=\mathrm{f}(1)=-(1-1)^{2}+10=10 \)
Therefore, function f does not have a minimum value.

It is given that \( g(x)=x^{3}+1 \)
Now,
\(x \in \mathbb{R}\)
\(\Rightarrow-\infty \leq x \leq \infty\)
\(\Rightarrow-\infty \leq x^{3} \leq \infty\)
\(\Rightarrow-\infty \leq x^{3}+1 \leq \infty\)
The function g neither has a maximum value nor a minimum value.

It is given that \( \mathrm{f}(x)=|x+2|-1 \)
Now, we can see that \( |x+2| \geq 0 \) for every \( x \in R \)
\(\Rightarrow f(x)=|x+2|-1 \geq-1 \text { for every } x \in R\)
The minimum value of \( f \) is attained when \( |x+2|=0 \)
\(|x+2|=0\)
\(\Rightarrow x=-2\)
Then, Minimum value of \( f=f(-2)=|-2+2|-1=-1 \)
Therefore, function f does not have a maximum value.

It is given that \( \mathrm{g}(x)=-|x+1|+3 \)
Now, we can see that \( -|x+1| \leq 0 \) for every \( x \in R \)
\(\Rightarrow \mathrm{g}(x)=-|x+1|+3 \leq 3 \text { for every } x \in \mathrm{R}\)
The maximum value of f is attained when \( |x+1|=0 \)
\(|x+1|=0\)
\(\Rightarrow x=-1\)
Then, Maximum value of \( g=g(-1)=-|-1+1|+3=3 \)
Therefore, function f does not have a minimum value.

It is given that \( h(x)=\sin (2 x)+5 \)
Now, we can see that \( -1 \leq \sin 2 x \leq 1 \)
\(\Rightarrow-1+5 \leq \sin 2 x+5 \leq 1+5\)
\(\Rightarrow 4 \leq \sin 2 x+5 \leq 6\)
Therefore, the maximum and minimum value of function \( h \) are 6 and 4 respectively.

It is given that \( f(x)=|\sin 4 x+3| \)
Now, we can see that \( -1 \leq \sin 4 x \leq 1 \)
\(\Rightarrow 2 \leq \sin 4 x+3 \leq 4\)
\(\Rightarrow 2 \leq|\sin 4 x+3| \leq 4\)
Therefore, the maximum and minimum value of function h are 4 and 2 respectively.

It is given that \( \mathrm{h}(x)=x+1, x \in(-1,1) \)
Now, if a point \( x_{0} \) is closest to \(-1 \), then,
We find \( \frac{x_{0}}{2}+1 < x_{0}+1 \) for all \( x_{0} \in(-1,1) \)
Also, if a point \( x_{1} \) is closest to \(1 \), then,
We find \( x_{1}+1 < \frac{x_{1}+1}{2}+1 \) for all \( x_{1} \in(-1,1) \)
Therefore, the function \( \mathrm{h}(x )\) has neither maximum nor minimum value in \( (-1,1) \).

\(f(x)=x^{2}\)
\(\Rightarrow f^{\prime}(x)=2 x\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow x=0\)
\( \Rightarrow x=0 \) is the only critical point which could possibly be the point of local maxima or local minima of f .
\( \Rightarrow f^{\prime}(0)=2 \), which is positive.
Then, by second derivative test,
\( \Rightarrow x=0 \) is point of local maxima and local minima of f at \( x=0 \) is \( \mathrm{f}(0)= \) 0 .

\(g(x)=x^{3}-3 x\)
\(\Rightarrow g^{\prime}(x)=3 x^{2}-3\)
Now, \( g^{\prime}(x)=0 \)
\(\Rightarrow 3 x^{2}-3=0\)
\(\Rightarrow 3 x^{2}=3\)
\(\Rightarrow x=1\)
\(g^{\prime \prime}(x)=6 x\)
Now, \( g^{\prime}(1)=6 > 0 \)
and \( g^{\prime}(-1)=-6 < 0 \)
Then, by second derivative test,
\( \Rightarrow x=1 \) is point of local maxima and local minima of g at \( x=1 \) is \( g(1)=1^{3}-3=1-3=-2 \)
And, \( x=-1 \) is point of local maxima and local maximum value of \( g \) at \( x=-1 \) is
\(g(-1)=(-1)^{3}-3(-1)=-1+3=2\)

\(\mathrm{h}(x)=\sin x+\cos x, 0 < x < \frac{\pi}{2}\)
\(\mathrm{~h}^{\prime}(x)=\cos x-\sin x\)
Now, \( h^{\prime}(x)=0 \)
\(\Rightarrow \cos x-\sin x=0\)
\(\Rightarrow \cos x=\sin x\)
\(\Rightarrow \tan x=1\)
\(=x=\frac{\pi}{4}\)
\(h^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)\)
\(h^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2} < 0\)
Then, by second derivative test,
\( \Rightarrow x=\frac{\pi}{4} \) is point of local maxima and local minima of h at \( x=\frac{\pi}{4} \) is \( \mathrm{h}\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} \)

\(\mathrm{f}(x)=\sin x-\cos x, 0 < x < 2 \pi\)
\(\mathrm{f}^{\prime}(x)=\cos x+\sin x\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow \cos x+\sin x=0\)
\(\Rightarrow \cos x=-\sin x\)
\(\Rightarrow \tan x=-1\)
\(\Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4}, \in 2 \pi\)
\((x)=-\sin x+\cos x\)
\(f^{\prime \prime}\left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2} < 0\)
\(f^{\prime \prime}\left(\frac{7 \pi}{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} > 0\)
Then, by second derivative test,
\( \Rightarrow x=\frac{3 \pi}{4} \) is point of local maxima and the local maximum value of f at \( x=\frac{3 \pi}{4} \) is
\(f\left(\frac{3 \pi}{4}\right)=\sin \left(\frac{3 \pi}{4}\right)-\cos \left(\frac{3 \pi}{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
And, \( \Rightarrow x=\frac{7 \pi}{4} \) is point of local minima and the local minimum value of f at \( x=\frac{7 \pi}{4} \) is
\(f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}\)

\(\mathrm{f}(x)=x^{3}-6 x^{2}+9 x+15\)
\(\Rightarrow \mathrm{f}^{\prime}(x)=3 x^{2}-12 x+9\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow 3 x^{2}-12 x+9=0\)
\(\Rightarrow 3(x-1)(x-3)=0\)
\(\Rightarrow x=1,3\)
\(\mathrm{~g}^{\prime \prime}(x)=6 x-12=6(x-2)\)
Now, \( f^{\prime}(1)=6(1-2)=-6 < 0 \)
and \( f^{\prime}(3)=6(3-2)=6 > 0 \)
Then, by second derivative test,
\( \Rightarrow x=1 \) is point of local maxima and local maximum of f at \( x=1 \) is \(f(1)=1^{3}-6+9+15=19\)
And,
\( x=3 \) is point of local minima and local minimum value of \( f \) at \( x=3 \) is \(f(3)=27-54+27+15=15\)

\(\mathrm{g}(x)=\frac{x}{2}+\frac{2}{x}, x > 0\)
\(\Rightarrow g^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}\)
\(\text { Now, } \mathrm{g}^{\prime}(x)=0\)
\(\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0\)
\(\Rightarrow \frac{2}{x^{2}}=\frac{1}{2}\)
\(\Rightarrow x^{2}=4\)
\(\Rightarrow x= 2\)
Since \( x > 0 \), we take \( x=2 \)
\(g^{\prime}(x)=\frac{4}{x^{3}}\)
\(g^{\prime \prime}(2)=\frac{4}{2^{3}}=\frac{1}{2} > 0\)
Then, by second derivative test,
\( \Rightarrow x=2 \) is point of local minima and local minimum of \( g \) at \( x=2 \) is
\(g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2\)

\(\mathrm{g}(x)=\frac{1}{x^{2}+2}\)
\(\Rightarrow g^{\prime}(x)=\frac{-2 x}{\left(x^{2}+2\right)^{2}}=0\)
\(\Rightarrow x=0\)
Now, for values close to \( x=0 \) and to the left of 0 ,
\( \mathrm{g}^{\prime}(x) > 0 \).
Also, for values close to \( x=0 \) and to the right of 0 ,
\(g^{\prime}(x) < 0\)
Then, by first derivative test,
\( \Rightarrow x=0 \) is point of local maxima and local maximum of g at \( x=0 \) is
\(g(0)=\frac{1}{0+2}=\frac{1}{2}\)

\(\mathrm{f}(x)=x \sqrt{1-x} ; 0 < x < 1\)
\(\Rightarrow \mathrm{f}^{\prime}(x)=\sqrt{1-x}+x \frac{1}{2 \sqrt{1-x}}(-1)\)
\(\Rightarrow \sqrt{1-x}-\frac{1}{2 \sqrt{1-x}}\)
\(=\frac{2(1-x)-x}{2 \sqrt{1-x}}\)
\(=\frac{2-3 x}{2 \sqrt{1-x}}\)
\(\mathrm{f}^{\prime}(x)=0\)
\(\Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0\)
\(\Rightarrow 2-3 x=0\)
\(\Rightarrow x=\frac{2}{3}\)
\(f^{\prime \prime}(x)=\frac{1}{2}\left[\frac{\sqrt{1-x}(-3)-(2-3 x)\left(\frac{-1}{2 \sqrt{1-x}}\right)}{1-x}\right]\)
\(=\left[\frac{\sqrt{1-x}(-3)+(2-3 x)\left(\frac{-1}{2 \sqrt{1-x}}\right)}{2(1-x)}\right]\)
\(=\left[\frac{-6(1-x)+(2-3 x)}{2(1-x)^{\frac{3}{2}}}\right]\)
\(=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}\)
\(f^{\prime \prime}\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{\sqrt[3]{3}}=\frac{2 \sqrt{3}}{9}\)

\(f(x)=e^{x}\)
\(\Rightarrow f^{\prime}(x)=e^{x}\)
Now, if \( \mathrm{f}^{\prime}(x)=0 \), then \( \mathrm{e}^{x}=0 \).
But, the exponential function can never assume 0 for any value of \(x \). Therefore, there does not exist \( c \in R \) such that \( f^{\prime}(c)=0 \)
Hence, function f does not have maxima or minima.

\(\mathrm{g}(x)=\log x\)
\(\Rightarrow g^{\prime}(x)=\frac{1}{x}\)
Since, \( \log x \) is defined for a positive number \(x \),
\( \mathrm{g}^{\prime}(x) > 0 \) for any \(x \).
Therefore, there does not exist \( c \in R \) such that \( f^{\prime}(c)=0 \)
Hence, function f does not have maxima or minima.

\(\mathrm{h}(x)=x^{3}+x^{2}+x+1\)
\(\Rightarrow \mathrm{h}^{\prime}(x)=3 x^{2}+2 x+1\)
\(\mathrm{~h}(x)=0\)
\(\Rightarrow 3 x^{2}+2 x+1=0\)
\(\Rightarrow x=\frac{-2 \pm 2 \sqrt{2 i}}{6}\)
\(\Rightarrow \frac{-1 \pm \sqrt{2 i}}{3}, \notin R\)
Therefore, there does not exist \( \mathrm{c} \in \mathrm{R} \) such that \( \mathrm{h}^{\prime}(\mathrm{c})=0 \)
Hence, function \( h \) does not have maxima or minima.

It is given that \( f(x)=x^3, x \in[-2,2] \)
\(\Rightarrow \mathrm{f}^{\prime}(x)=3 x^{2}\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow x=0\)
Now, we evaluate the value of \( f \) at critical point \( x=0 \) and at end points of the interval \( [-2,2] \).
\(\mathrm{f}(0)=0\)
\(\mathrm{f}(-2)=(-2)^{3}=-8\)
\(\mathrm{f}(2)=(2)^{3}=8\)
Therefore, we have the absolute maximum value of \( f \) on \( [-2,2] \) is 8 occurring at \( x=2 \).
And, the absolute minimum value of f on \( [-2,2] \) is \(-8\) occurring at \( x=-2 \).

It is given that \( f(x)=\sin x+\cos x, x \in[0, \pi] \)
\(f^{\prime}(x)=\cos x-\sin x\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow \cos x-\sin x=0\)
\(\Rightarrow \cos x=\sin x\)
\(\Rightarrow \tan x=1\)
\(\Rightarrow x=\frac{\pi}{4}\)
Now, we evaluate the value of \( f \) at critical point \( x=\frac{\pi}{4} \) and at end points of the interval \( [0, \pi] \)
\(f^{\prime}\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \cos \frac{\pi}{4}=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}\)
\(f(0)=\sin 0+\cos 0=0+1=1\)
\(f(\pi)=\sin \pi+\cos \pi=0-1=-1\)
Therefore, we have the absolute maximum value of f on \( [0, \pi] \) is \( \sqrt{2} \) occurring at \( x=\frac{\pi}{4} \).
And, the absolute minimum value of \( f \) on \( [0, \pi] \) is \(-1\) occurring at \( x=\pi \).

It is given that
\(\mathrm{f}(x)=4 x-\frac{1}{2} x^{2}, x \in\left[-2, \frac{9}{2}\right]\)
\(\Rightarrow f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow x=4\)
Now, we evaluate the value of \( f \) at critical point \( x=0 \) and at end points of the interval \( \left[-2, \frac{9}{2}\right] \)
\(f(4)=16-\frac{1}{2}(16)=16-8=8\)
\(f(-2)=-8-\frac{1}{2}(4)=-8-2=-10\)
\(f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^{2}=18-\frac{81}{8}=18-10.125=7.875\)
Therefore, we have the absolute maximum value of f on \( \left[-2, \frac{9}{2}\right] \) is 8 occurring at \( x=4 \).
And, the absolute minimum value of \( f \) on \( \left[-2, \frac{9}{2}\right] \) is -10 occurring at \( x= \) -2 .

It is given that \( f(x)=(x-1)^{2}+3, x \in[-3,1] \)
\(\Rightarrow \mathrm{f}^{\prime}(x)=2(x-1)\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow 2(x-1)\)
\(\Rightarrow x=1\)
Now, we evaluate the value of \( f \) at critical point \( x=1 \) and at end points of the interval \( [-3,1] \).
\(\mathrm{f}(1)=(1-1)^{2}+3=0+3=3\)
\(\mathrm{f}(-3)=(-3-1)^{2}+3=16+3=19\)
Therefore, we have the absolute maximum value of f on \( [-3,1] \) is 19 occurring at \( x=-3 \).
And, the absolute minimum value of \( f \) on \( [-3,1] \) is 3 occurring at \( x=1 \).

It is given that the profit function \( p(x)=41-72 x-18 x^{2} \)
\(\Rightarrow \mathrm{p}^{\prime}(x)=-72-36 x\)
and \( p^{\prime \prime}(x)=-36 \)
Now, \( g^{\prime}(x)=0 \)
\(72=-36 x\)
\(\Rightarrow x=\frac{-72}{36}\)
\(\Rightarrow x=-2\)
\(P^{\prime \prime}(-2)=-36 < 0\)
Then, by second derivative test,
\( x=-2 \) is point of local maxima of \( p \).
Therefore, Maximum Profit \( = \)
\(\mathrm{P}(-2)=41-72(-2)-18(-2)^{2}\)
\(=41+144-72\)
\(=113\)
Therefore, the maximum profit that the company can make is 113 units.

Let \( \mathrm{f}(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25, x \in[0,3] \)
\(\Rightarrow \mathrm{f}^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48\)
\(=12\left(x^{3}-2 x^{2}+2 x-4\right)\)
\(=12\left[x^{2}(x-2)+2(x-2)\right]\)
\(=12(x-2)\left(x^{2}+2\right)\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\( \Rightarrow x=2 \) or \( \left(x^{2}+2\right)=0 \) for which there are no real roots.
Therefore, we will only consider \( x=2 \)
Now, we evaluate the value of \( f \) at critical point \( x=2 \) and at end points of the interval \( [0,3] \).
\(\mathrm{f}(2)=3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25\)
\(=3(16)-8(8)+12(4)+25\)
\(=48-64+48-96+25\)
\(=-39\)
\(\mathrm{f}(0)=3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25\)
\(=0+0+0+25\)
\(=25\)
\(\mathrm{f}(3)=3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25\)
\(=3(81)-8(27)+12(9)+25\)
\(=243-216+108-144+25\)
\(=16\)
Therefore, we have the absolute maximum value of \( f \) on \( [0,3] \) is \(25\) occurring at \( x=0 \).
And, the absolute minimum value of \( f \) on \( [0,3] \) is \(-39\) occurring at \( x=2 \).

It is given that \( f(x)=\sin 2 x, x \in[0,2 \pi] \)
\(\mathrm{f}^{\prime}(x)=2 \cos 2 x\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow \cos 2 x=0\)
\(\Rightarrow 2 x=0\)
\(\Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\)
Now, we evaluate the value of f at critical point \( x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \) and at end points of the interval \( [0,2 \pi] \)
\(f^{\prime}\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{2}-1\)
\(f^{\prime}\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2}=-1\)
\(f^{\prime}\left(\frac{5 \pi}{4}\right)=\sin \frac{5 \pi}{2}=1\)
\(f^{\prime}\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{2}=-1\)
\(f(0)=\sin 0, f(2 \pi)=\sin 2 \pi=0\)
Therefore, we have the absolute maximum value of f on \( [0,2 \pi] \) is 1 occurring at
\(x=\frac{\pi}{4} \text { and } x=\frac{5 \pi}{4}\)

Let \( f(x)=\sin x+\cos x \)
\(\Rightarrow \mathrm{f}^{\prime}(x)=\cos x-\sin x\)
Now, \( f^{\prime}(x)=0 \)
\(\Rightarrow \cos x-\sin x=0\)
\(\Rightarrow \cos x=\sin x\)
\(\Rightarrow \tan x=1\)
\(\Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4} \ldots \ldots \ldots\)
Now,
If \( f^{\prime}(x) \) will be negative when \( (\sin x+\cos x) > 0 \), means both \( \sin x \) and \(\cos x\) are positive.
And, we know that \( \sin x \) and \( \cos x \) both are positive in the first quadrant.
Then, \( \mathrm{f}^{\prime \prime} ( x ) \) will be negative when
\(X \in\left(0, \frac{\pi}{2}\right)\)
\(f^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)\)
Now, let us take \( x=\frac{\pi}{4} \)
\(f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2} < 0\)
Then, by second derivative test,
\(f\) will be maximum at \( x=\frac{\pi}{4} \)
And, the maximum value of f is
\(f^{\prime}\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)

Let \( f(x)=2 x^{3}-24 x+107, x \in[1,3] \)
\(\Rightarrow f^{\prime}(x)=6 x^{2}-24\)
\(=6\left(x^{2}-4\right)\)
Now, \( f^{\prime}(x)=0 \)
\(\Rightarrow 6\left(x^{2}-4\right)=0\)
\(\Rightarrow x^{2}=4\)
\(\Rightarrow x=2\)
Therefore, we will only consider the interval \( [1,3] \)
Now, we evaluate the value of \( f \) at critical point \( x=2 \epsilon[1,3] \) and at end points of the interval \( [1,3] \).
\(\mathrm{f}(2)=2(2)^{3}-24(2)+107\)
\(=2(8)-24(2)+107\)
\(=75\)
\(f(1) =2(1)^{3}-24(1)+107\)
\( =2-24+107\)
\( =85\)
\(f(3)=2(3)^{3}-24(3)+107\)
\(=2(27)-24(3)+107\)
\(=89\)
Therefore, we have the absolute maximum value of \( f \) on \( [1,3] \) is 89 occurring at \( x=3 \).
Now, we will only consider the interval \( [-3,-1] \)
Now, we evaluate the value of \( f \) at critical point \( x=-2 \epsilon[-3,-1] \) and at end points of the interval \( [1,3] \).
\(\mathrm{f}(-3)=2(-3)^{3}-24(-3)+107\)
\(= 2(-27)-24(-3)+107\)
\(= 125\)
\( f(-1)=2(-1)^{3}-24(-1)+107\)
\(= -2+24+107\)
\(= 129\)
\(f(-2)=2(-2)^{3}-24(-2)+107\)
\(= 2(-8)-24(-2)+107\)
\(=139\)
Therefore, we have the absolute maximum value of \( f \) on \( [-3,-1] \) is 89 occurring at \( x=-2 \).

It is given that \( f(x)=x^{4}-62 x^{2}+a x+9 \)
Then, \( \mathrm{f}^{\prime}(x)=4 x^{3}-124 x+\mathrm{a} \)
It is given that function \( f \) attains its maximum value on the interval \( [0,2] \) at \( x=1 \).
\(\Rightarrow \mathrm{f}^{\prime}(1)=0\)
\(\Rightarrow 4-124+\mathrm{a}=0\)
\(\Rightarrow \mathrm{a}=120\)
Therefore, the value of a is 120 .

It is given that \( f(x)=x+\sin 2 x, x \in[0,2 \pi] \)
\( f^{\prime}(x)=1+2 \cos 2 x \)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\( \Rightarrow \cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} \)
\( 2 x=2 \pi \frac{2 \pi}{3}, \mathrm{n} \in \mathrm{Z} \)
\( \Rightarrow x=\mathrm{n} \pi \frac{\pi}{3}, \mathrm{n} \in \mathrm{Z} \)
\( \Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \in[0,2 \pi] \)
Now, we evaluate the value of f at critical point \( x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \) and at end points of the interval \( [0,2 \pi] \)
\(f^{\prime}\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}\)
\(f^{\prime \prime}\left(\frac{\pi}{3}\right)=\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\)
\(f^{\prime}\left(\frac{4 \pi}{3}\right)=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}\)
\(f^{\prime}\left(\frac{5 \pi}{4}\right)=\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}+\frac{\sqrt{3}}{2}\)
\(f^{\prime}(0)=0+\sin 0=0\)
\(f^{\prime}(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi\)
Therefore, we have the absolute maximum value of \( f \) on \( [0,2 \pi] \) is \( 2 \pi \) occurring at
\( x=2 \pi \) and absolute minimum value of \( f(x) \) in the interval \( [0,2 \pi] \) is 0 occuring at \( x=0 \).

Let one number be \(x \). Then, the other number is \( (24-x) \).
Let \( \mathrm{P}(x) \) denote the product of the two numbers.
Then, we get,
\(P(x)=x(24-x)=24 x-x^2\)
\(\Rightarrow P^{\prime}(x)=24-2 x\)
\(\Rightarrow P^{\prime \prime}(x)=-2\)
Now, \( P^{\prime}(x)=0 \)
\(\Rightarrow x=12\)
And
\(P^{\prime \prime}(12)=-2 < 0\)
Then, by second derivative test,
\( x=12 \) is the point of local maxima of P .
Therefore, the product of the numbers is the maximum when the numbers are 12 and \( 24-12=12 \).

Let the two numbers are \(x\) and \(y\) such that \( x+y=60 \)
\(\Rightarrow y=60-x\)
Let \( f(x)=xy^{3} \)
\(\Rightarrow \mathrm{f}(x)=x(60-x)^{3}\)
\(\Rightarrow f^{\prime}(x)=(60-x)^{3}-3 x(60-x)^{2}\)
\(=(60-x)^{2}[60-x-3 x]\)
\(=(60-x)^{2}[60-4 x]\)
And \( f^{\prime}(x)=-2(60-x)(60-4 x)-4(60-x)^{2} \)
\(=-2(60-x)[60-4 x+2(60-x)]\)
\(=-2(60-x)(180-6 x)\)
\(=-12(60-x)(30-x)\)
Now, \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow x=60 \text { or } x=15\)
When \( x=60, f^{\prime}(x)=0 \).
When \( x=15, f^{\prime}(x)=-12(60-15)(30-15)=-12 \times 45 \times 15 < 0 \)
Then, by second derivative test, \( x=15 \) is a point of local maxima of f . Then, function \( \mathrm{xy}^{3} \) is maximum when \( x=15 \) and \( y=60-15=45 \).
Therefore, required numbers are 15 and 45 .

Let one number be \( x \). Then, the other number is \( y=(35-x) \)
Let \( P(x)=x^{2} y^{5} \)
Then, we get, \( P(x) \)
\(=2 x(35-x)^{5}-5 x^{2}(35-x)^{4}\)
\(=x(35-x)^{4}[2(35-x)-5 x]\)
\(=x(35-x)^{4}[70-7 x]\)
\(=7 x(35-x)^{4}(10-x)\)
Now, \( \mathrm{P}^{\prime\prime} (x)=7(35-x)^{4}(10-x)+7 x\left[-(35-x)^{4}-4(35-x)^{3}(10-x)\right] \)
\(\left.=7(35-x)^{4}(10-x)-7 x(35-x)^{4}-28 x(35-x)^{3}(10-x)\right]\)
\(=7(35-x)^{3}[(35-x)(10-x)-x(35-x)-4 x(10-x)]\)
\(=7(35-x)^{3}\left[350-45 x+x^{2}-35 x+x^{2}-40 x+4 x^{2}\right]\)
\(=7(35-x)^{3}\left[6 x^{2}-120 x+350\right]\)
Now, \( P^{\prime}(x)=0 \)
\(\Rightarrow x=0,35,10\)
When \( x=0,35 \) This will make the product \(x^2 y^5\) equal to 0 .
Therefore, \( x=0,35 \) cannot be possible values of \( x \).
And when \( x=10 \)
Then, we have,
\(\mathrm{P}^{\prime \prime}(x)=7(35-10)^{3}\left[6(10)^{2}-120(10)+350\right]\)
\(=7(25)^{3}[-250] < 0\)
Then, by second derivative test,
\( x=10 \) and \( y=35-10=25 \) is the point of local maxima of \( P \).
Therefore, the required number are 10 and 25 .

Let one number be \(x \). Then, the other number is \( (16-x) \).
Let \( S(x) \) be the sum of these number. Then,
\(S(x)=x^{3}+(16-x)^{3}\)
\(\Rightarrow S^{\prime}(x)=3 x^{2}-3(16-x)^{2}\)
\(\Rightarrow S^{\prime \prime}(x)=6 x+6(16-x)\)
Now, \( S^{\prime}(x)=0 \)
\(\Rightarrow 3 x^{2}-3(16-x)^{2}=0\)
\(\Rightarrow x^{2}-(16-x)^{2}=0\)
\(\Rightarrow x^{2}-256-x^{2}+32 x=0\)
\(\Rightarrow x=8\)
Now, \( S^{\prime \prime}(8)=6(8)+6(16-8) \)
\(=48+48=96 > 0\)
Then, by second derivative test, \( x=8 \) is the point of local minima of \(S \).
Therefore, the sum of the cubes of the numbers is the minimum when the numbers are \(8\) and \( 16-8=8 \).

Let the side of the square to be cut off be \( x \), then, the length and the breadth of the box will be \( (18-x) \mathrm{cm} \) each and the height of the box is \(x\) cm .
Then, the volume \( \{\mathrm{V}(x)\} \) of the box is given by:
\(V(x)=x(18-x)^{2}\)
\(\Rightarrow V^{\prime}(x)=(18-x)^{2}-2 x(18-x)\)
\(=(18-x)[18-x-2 x]\)
\(=(18-x)(18-3 x)\)
Now, \(V^{\prime\prime} (x)=(18-x)(-3)+(18-3 x)(-1) \)
\(=-3(18-x)-(18-3 x)=-54+3 x-18+3 x=6 x-72\)
Now, \( V^{\prime}(x)=0 \)
\(\Rightarrow x=18 \text { or } 3\)
If \( x=18 \) then breadth becomes 0 which is not possible Therefore, \( x=3 \)
\(V^{\prime \prime}(3)=6.3-72=-\mathrm{ve}\)
Then, by second derivative test, \( x=3 \) is the point of maxima of \( V \).
Therefore, If we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Let the side of the square to be cut off be \( x \), then, the height of the box is \(x\) and the length is \( 45-2 x \) and the breadth is \( 24-2 x \).
Then, the volume \( \{\mathrm{V}(x)\} \) of the box is given by:
\(V(x)=x(45-2 x)(24-x)\)
\(=x\left(1080-90 x-48 x+4 x^{2}\right)\)
\(=4 x^{3}-138 x^{2}+1080 x\)
\(\Rightarrow V^{\prime}(x)=12 x^{2}-276 x+1080\)
\(=12\left(x^{2}-23 x+90\right)\)
\(=12(x-18)(x-5)\)
Now, \( \mathrm{V}^{\prime\prime} \) \( (x)=24 x-276=12(2 x-23) \)
Now, \( V^{\prime}(x)=0 \)
\(\Rightarrow x=18 \text { or } 5\)
It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. So, \( x \) cannot be equal to 18 .
Therefore, \( x=5 \)
\(V^{\prime \prime}(5)=12(10-23)=-156 < 0\)
Then, by second derivative test, \( x=5 \) is the point of maxima of V .
Therefore, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm .

The figure is given below:

Let a rectangle of length 1 and breadth \( b \) be inscribed in the circle of radius a.
Then, the diagonal passes through the centre and is of length 2 acm .
Now, by Pythagoras theorem, we get,
\((2 a)^{2}=1^{2}+b^{2}\)
\(\Rightarrow b^{2}=4 a^{2}-1^{2}\)
\(\Rightarrow b=\sqrt{4 a^{2}-1^{2}}\)
Therefore, Area of rectangle, \( \mathrm{A}=1 \sqrt{4 a^{2}-1^{2}} \)
\(\Rightarrow \frac{d A}{d l}=\sqrt{4 a^{2}-1^{2}}+1 \frac{1}{2 \sqrt{4 a^{2}-1^{2}}}(-21)\)
\(=\sqrt{4 a^{2}-1^{2}}-\frac{1^{2}}{\sqrt{4 a^{2}-1^{2}}}\)
\(=\frac{4 a^{2}-1^{2}}{\sqrt{4 a^{2}-1^{2}}}\)
\( \frac{d^{2} A}{d 1^{2}}=\frac{\sqrt{4 a^{2}-1^{2}}(-4 I)-\left(4 a^{2}-21^{2}\right) \frac{(-21)}{2 \sqrt{4 a^{2}-1^{2}}}}{\left(\sqrt{4 a^{2}-1^{2}}\right)} \)
\(=\frac{\left(4 a^{2}-1^{2}\right)(-4 I)+1\left(4 a^{2}-21^{2}\right)}{\left(4 a^{2}-1^{2}\right)^{\frac{3}{2}}}\)
\(=\frac{-12 a^{2} 1+21^{3}}{\left(4 a^{2}-1^{2}\right)^{\frac{3}{2}}}\)
\(=\frac{-2 i\left(6 a^{2}-1^{2}\right)}{\left(4 a^{2}-1^{2}\right)^{\frac{3}{2}}}\)
\(\frac{d A}{d l}=0\)
Gives \( 4 a^{2}=21^{2} \)
\( \Rightarrow l=\sqrt{2} \mathrm{a} \)
\( \Rightarrow \mathrm{b}=\sqrt{4 a^{2}-2 a^{2}}=\sqrt{2 a^{2}}=\sqrt{2 a} \)
when \( l=\sqrt{2 a} \)
Then, \( =\frac{d^{2} A}{d I^{2}}=\frac{-2(\sqrt{2} a)\left(6 a^{2}-2 a^{2}\right)}{2 \sqrt{2} a^{3}}=\frac{-8 \sqrt{2} a^{3}}{2 \sqrt{2} a^{3}}=-4=0 \)
Then, by second derivative test, when \( l=\sqrt{2 a} \), then the area of the rectangle is the maximum.
Since, \( l=b=\sqrt{2 a} \),
Therefore, the rectangle is square.
Hence proved.

Let r be the radius and \( h \) be the height of the cylinder.
Let V be the volume of the cylinder. Then
\(\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}=100 \text { (given) }\)
\(\Rightarrow \mathrm{h}=\frac{100}{\pi \mathrm{r} 2}\)
hence, the surface area ( S ) of the cylinder is given by:
\(\mathrm{S}=2 \pi \mathrm{r}^{2}+2 \pi \mathrm{rh}\)
\(=2 \pi \mathrm{r}^{2}+\frac{200}{r}\)
Now,
\(\frac{d s}{d r}=4 \pi r-\frac{200}{r^{2}}, \frac{d^{2} s}{d r^{2}}=4 \pi+\frac{400}{r^{3}} < 0\)
If \( \frac{d s}{d r}=0 \Rightarrow \frac{200}{r^{2}} \Rightarrow r^{3}=\frac{200}{4 \pi}=\frac{50}{\pi} \Rightarrow \mathrm{r}=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \)
So, when \( \mathrm{r}=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \) then \( \frac{d^{2} S}{d r^{2}} > 0 \)
Then, by second derivative test, the surface area is the minimum when \( r \) \( =\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \)
Now, when \( \mathrm{r}=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \) then \( \mathrm{h}=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{1}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm} \).
Therefore, the dimensions of the can which has the minimum surface area are \( \mathrm{r}=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \) and \( \mathrm{h}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm} \).

Let a piece of length 1 be cut from the given wire to make a square. Then, the other piece of wire to be made into a circle is of length \( (28-1) \mathrm{m} \).
Now, side of square \( =\frac{1}{4} \)
Let \( r \) be the radius of the circle,
Then, \( 2 \pi r=28-1 \)
\(\Rightarrow r=\frac{1}{2 \pi}(28-1)\)
Therefore, the required area (a) is given by
\( \mathrm{A}=( \) side of the square \( ) 2+\pi \mathrm{r}^{2} \)
\(=\frac{1^{2}}{16}+\pi\left[\frac{1}{2 \pi}(28-1)\right]^{2}\)
\(=\frac{1^{2}}{16}+\frac{1}{4 \pi}\left[(28-1)^{2}\right]\)
Then, \( \frac{d A}{d l}=\frac{21}{16}+\frac{2}{4 \pi}(28-1)(-1) \)
\(=\frac{1}{8}-\frac{2}{4 \pi}(28-1)\)
\(\frac{d^{2} A}{d l^{2}}=\frac{1}{8}-\frac{2}{2 \pi}(28-1)=0\)
Now, if \( \frac{d A}{d l}=0 \)
Then, \( \frac{1}{8}-\frac{2}{2 \pi}(28-1)=0 \)
\(\begin{array}{l}
\Rightarrow \frac{\pi 1-4(28-1)}{8 \pi}=0 \\
\Rightarrow(\pi+4) l-112=0 \\
\Rightarrow 1=\frac{112}{\pi+4}
\end{array}\)
So, when \( 1=\frac{112}{\pi+4} \)
Then, \( \frac{d^{2} A}{d l^{2}} > 0 \)
Then, by second derivative test, the area \( (\mathrm{A}) \) is the minimum when \( 1= \) \( \frac{112}{\pi+4} \).
Therefore, the combined area is the minimum when the length of the wire in making the square is \( \frac{112}{\pi+4} \mathrm{~cm} \) while the length of the wire in making the circle is \( 28-\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4} \mathrm{~cm} \).

Let \( r \) and \( h \) be the radius and height of the cone respectively inscribed in a sphere of radius \( R \).
Let V be the volume of cone.
Then, \( \mathrm{V}=\frac{1}{3} \pi r^{2} \mathrm{~h} \)
And height of cone \( \mathrm{h}=\mathrm{R}+\sqrt{R^{2}-r^{2}} \)
\(\therefore \mathrm{V}=\frac{1}{3} \pi r^{2}\left(\mathrm{R}+\sqrt{R^{2}-r^{2}}\right)\)
Now, \( \frac{d V}{d r}=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}+\frac{1}{3} \pi r^{2} \cdot \frac{(-2 r)}{2 \sqrt{R^{2}-r^{2}}} \)
\( =\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}+\frac{1}{3} \pi \frac{r^{3}}{\sqrt{R^{2}-r^{2}}} \)
\( =\frac{2}{3} \pi r R+\frac{2 \pi r\left(R^{2}-r^{2}\right)-\pi r^{3}}{\sqrt[3]{R^{2}-r^{2}}} \)
\( =\frac{2}{3} \pi r R+\frac{2 \pi r R^{2}-3 \pi r^{3}}{\sqrt[3]{R^{2}-r^{2}}} \)
\( \frac{d^{2} V}{d r^{2}}=\frac{2}{3} \pi r R+\frac{\sqrt[3]{R^{2}-r^{2}}\left(2 \pi R^{2}-9 \pi r^{2}\right)-\left(2 \pi r R^{2}-3 \pi r^{3}\right) \cdot \frac{(-2 r)}{\sqrt[6]{R^{2-r^{2}}}}}{9\left(R^{2}-r^{2}\right)} \)
\( =\frac{2}{3} \pi r R+\frac{9\left(R^{2}-r^{2}\right)\left(2 \pi R^{2}-9 \pi r^{2}\right)+\left(2 \pi r R^{2}-3 \pi r^{3}\right)}{27\left(R^{2}-r^{2}\right)} \)
Now, if \( \frac{d V}{d r}=0 \Rightarrow \frac{2}{3} \pi r R=-\left(\frac{2 \pi r R^{2}-3 \pi r^{3}}{\sqrt[3]{R^{2}-r^{2}}}\right) \)
After solving this we get, \( r^{2}=\frac{8}{9} R^{2} \)
So, when, \( r^{2}=\frac{8}{9} R^{2} \), then \( \frac{d^{2} V}{d r^{2}} < 0 \)
Then, by second derivative test, the volume of the cone is the maximum when \( r^{2}=\frac{8}{9} R^{2} \)
So, when,
\( r^{2}=\frac{8}{9} R^{2}, \mathrm{~h}=\mathrm{R}+\sqrt{R^{2}-\frac{8}{9} R^{2}}=\mathrm{R}+\sqrt{\frac{1}{9} R^{2}}=\mathrm{R}+\frac{R}{3}=\frac{4}{3} \mathrm{R} \)
Therefore, \( \mathrm{V}=\frac{1}{3} \pi\left(\frac{8}{9} R^{2}\right)\left(\frac{4}{3} \mathrm{R}\right) \)
\( =\frac{8}{27}\left(\frac{4}{3} \pi R^{2}\right)=\frac{8}{27} \times( \) volume of sphere \( ) \)
Therefore, the volume of the largest cone that can be inscribed in the sphere is \( \frac{8}{27} \) the volume of the sphere.

Let \( r \) and \( h \) be the radius and height of the cone respectively.
Then, the volume \( (\mathrm{V}) \) of the cone is given by:
\(\mathrm{V}=\frac{1}{3} \pi r^{2} \mathrm{~h} \Rightarrow \mathrm{h}=\frac{3 V}{\pi r^{2}}\)
The surface area \( (\mathrm{S}) \) of the cone
\(\mathrm{S}=\pi \mathrm{rl}\)
\(=\pi \mathrm{r} \sqrt{r^{2}+h^{2}}\)
\(=\pi \mathrm{r} \sqrt{r^{2}+\frac{9 V}{\pi^{2} r^{4}}}\)
\(=\pi \mathrm{r} \sqrt{\frac{\pi^{2} r^{6}+9 V^{2}}{\pi^{2} r^{4}}}\)
\(=\frac{\pi r}{\pi r^{2}} \sqrt{\pi^{2} r^{6}+9 V^{2}}\)
\(\text { Then, } \frac{d S}{d r}=\frac{r \cdot \frac{1}{2 \sqrt{\pi^{2} r^{6}+9 V^{2}}} \cdot 6 \pi^{2} r^{5}-\sqrt{\pi^{2} r^{6}+9 V^{2}}}{r^{2}}\)
\(=\frac{3 \pi^{2} r^{6}-\pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}}\)
\(=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}}\)
Now, if \( \frac{d S}{d r}=0 \Rightarrow 2 \pi^{2} r^{6}=9 V^{2} \Rightarrow r^{6}=\frac{9 V^{2}}{2 \pi^{2}} \)
So, when \( r^{6}=\frac{9 V^{2}}{2 \pi^{2}} \) then \( \frac{d^{2} s}{d r^{2}} > 0 \)
Then, by second derivative test, the surface area of the cone is the least when \( r^{6}=\frac{9 V^{2}}{2 \pi^{2}} \)
So when, \( r^{6}=\frac{9 V^{2}}{2 \pi^{2}} \),
then \( =\mathrm{h}=\frac{3 V}{\pi r^{2}}=\frac{3}{\pi r^{2}}\left(\frac{2 \pi^{2} r^{6}}{9}\right)^{\frac{1}{2}}=\frac{3}{\pi r^{2}} \cdot \frac{\sqrt{2} \pi r^{3}}{3}=\sqrt{2 r} \).
Therefore, for a given volume, the right circular cone of the least curved surface has an altitude equal to \( \sqrt{2} \) times the radius of the base.

Let \( \theta\) be the semi-vertical angle of the cone.
Let \( \mathrm{r}, \mathrm{h} \) and \(l\) be the radius, height and the slant height of the cone respectively.
It is given that slant height is constant.
Now, \( \mathrm{r}=l \sin \theta\) and \( \mathrm{h}=l \cos \theta\)
Then, the volume of the cone \( (\mathrm{V}) \)
\(\mathrm{V}=\frac{1}{3} \pi r^{2} \mathrm{~h}\)
\(=\frac{1}{3} \pi\left(1^{2} \sin ^{2} \theta\right)(l \cos \theta)\)
\(=\frac{1}{3} \pi 1^{3} \sin ^{2} \theta \cos \theta\)
\(\therefore \frac{d V}{d \theta}=\frac{1^{3} \pi}{3}\left[\sin n^{2} \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)\right]\)
\(=\frac{1^{3} \pi}{3}\left[-\sin ^{3} \theta+2 \sin \theta \cos ^{2} \theta\right]\)
\(\frac{d V^{2}}{d \theta^{2}}=\frac{1^{3} \pi}{3}\left[\sin ^{2} \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)\right]\)
\(=\frac{1^{3} \pi}{3}\left[-\sin ^{3} \theta+2 \sin \theta \cos ^{2} \theta\right]\)
\(\frac{d V^{2}}{d \theta^{2}}=\frac{1^{3} \pi}{3}\left[-3 \sin ^{2} \theta \cos \theta+2 \cos ^{3} \theta-4 \sin ^{2} \theta \cos \theta\right]\)
\(=\frac{1^{3} \pi}{3}\left[2 \cos ^{3} \theta-7 \sin ^{2} \theta \cos \theta\right]\)
Now, if \( \frac{d V}{d \theta}=0 \)
\(\sin ^{3} \theta=2 \sin \theta \cos ^{2} \theta\)
\(\Rightarrow \tan ^{2} \theta=2\)
\(\Rightarrow \tan \theta=\sqrt{2}\)
\(\Rightarrow \theta=\tan ^{-1} \sqrt{2}\)
Now, when, \( \theta=\tan ^{-1} \sqrt{2} \tan ^{2} \theta=2 \) or \( \sin ^{2} \theta=2 \cos ^{2} \theta\).
Then, we get
\(\frac{d V^{2}}{d \theta^{2}}=\frac{1^{3} \pi}{3}\left[2 \cos ^{3} \theta-14 \cos ^{3} \theta\right]\)
\(=-4 \pi l^{3} \cos ^{3} \theta < 0 \text { for } \theta\epsilon\left[0, \frac{\pi}{2}\right]\)
Then, by second derivative test, the volume \( (\mathrm{V}) \) is the maximum when \( \theta=\tan ^{-1} \sqrt{2} \)
Therefore, the semi-vertical angle of the cone of the maximum volume and of given slant height is \( \tan ^{-1} \sqrt{2} \).
Hence Proved.

We know that total surface area of the cone \( =S=\pi r(1+r) \ldots(1)\)
and Volume of the cone \( (\mathrm{V})=\frac{1}{3} \pi r^{2} \mathrm{~h} \)
\(V^{2}=\frac{1}{9} \pi^{2} r^{4}\left(1^{2}-r^{2}\right)\)
Then by (1), we get,
\(V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\left(\frac{s}{\pi r}-r\right)^{2}-r^{2}\right]\)
\(=\frac{1}{9}\left(S\left(S r^{2}-2 \pi r^{4}\right)\right)\)
\(\mathrm{P}=\mathrm{V}^{2}\)
Now, differentiating P with respect to r , we get,
\(\frac{d P}{d r}=\frac{1}{9}\left(S\left(2 S r-8 \pi r^{3}\right)\right.\)
Now, if \( \frac{d P}{d r}=0 \), then
\(\mathrm{S}=4 \pi r^{2}\)
Now again differentiating with respect to r , we get \( \frac{d^{2} P}{d r^{2}} < 0 \)
Therefore, P is maximum when \( \mathrm{S}=4 \pi \mathrm{r}^{2} \)
And V is maximum when \( \mathrm{S}=4 \pi \mathrm{r}^{2} \)
\(\Rightarrow \pi r(1+r)=4 \pi r^{2}\)
\(\Rightarrow 1=3 r\)
\( \operatorname{Sin} \theta=\frac{r}{1}=\frac{1}{3} \)
\(\Rightarrow \theta=\operatorname{Sin}^{-1}\left(\frac{1}{3}\right)\)
Therefore, semi-vertical angle of right circular cone of given surface area and maximum volume is \( \operatorname{Sin}^{-1}\left(\frac{1}{3}\right) \).
Hence Proved.

It is given that \( x^{2}=2 y \)
For each value of \(x \), the position of the point will be \( \left(x, \frac{x^{2}}{2}\right) \)
The distance \( \mathrm{d}(x) \) between the points \( \left(x, \frac{x^{2}}{2}\right) \) and \( (0,5) \) is given by:
\(\mathrm{d}(x)=\sqrt{(x-0)^{2}+\left(\frac{x^{2}}{2}-5\right)^{2}}\)
\(=\sqrt{x^{2}+\frac{x^{2}}{4}+25-5 x^{2}}\)
\(=\sqrt{\frac{x^{2}}{4}-4 x^{2}+25}\)
\(\therefore d^{\prime}(x)=\frac{\left(x^{3}-8 x\right)}{2 \sqrt{\frac{x^{4}}{4}-4 x^{2}+25}}\)
\(=\frac{\left(x^{3}-8 x\right)}{\sqrt{x^{4}-16 x^{2}+100}}\)
\(d^{\prime}(x)=0\)
\(\Rightarrow x^{3}-8 x=0\)
\(\Rightarrow x\left(x^{2}-8\right)=0\)
\(\Rightarrow x=0, \pm 2 \sqrt{2}\)
And,
\(d^{\prime \prime}(x)=\frac{\sqrt{x^{4}-16 x^{2}+100}\left(3 x^{2}-8\right)-\left(x^{3}-8 x\right) \cdot \frac{4 x^{3}-32 x}{2 \sqrt{x^{4}-16 x^{2}+100}}}{x^{4}-16 x^{2}+100}\)
\(=\frac{\left(x^{4}-16 x^{2}+100\right)\left(3 x^{2}-8\right)-2\left(x^{3}-8 x\right)\left(x^{3}-8 x\right)}{\left(x^{4}-16 x^{2}+100\right)^{\frac{3}{2}}}\)
So, now when \( x=0 \), then \( d^{\prime \prime}(x)=\frac{36(-8)}{6^{3}} < 0 \)
And when, \( x= \pm \sqrt[2]{2}, d^{\prime \prime}(x) > 0 \)
Then, by second derivative test, \( d(x) \) is minimum at \( x= \pm \sqrt[2]{2} \)
So when, \( x= \pm \sqrt[2]{2} \cdot y=\frac{(\sqrt[2]{2})^{2}}{2}=4 \)
Therefore, the point on the curve \( x^{2}=2 y \) which is nearest to the point \( (0,5) \) is \( ( \pm \sqrt[2]{2}, 4) \).

Let \( \mathrm{f}(x)=\frac{1-x+x^{2}}{1+x+x^{2}} \)
\(\therefore f^{\prime}(x)=\frac{\left(1+x+x^{2}\right)(-1+2 x)-\left(1-x+x^{2}\right)(1+2 x)}{\left(1+x+x^{2}\right)^{2}}\)
\( =\frac{2 x^{2}-2}{\left(1+x+x^{2}\right)^{2}} \)
\(=\frac{2\left(x^{2}-1\right)}{\left(1+x+x^{2}\right)^{2}}\)
Then, \( \mathrm{f}^{\prime}(x)=0 \)
\( \Rightarrow x^{2}=1 \)
\( \Rightarrow x=\pm 21 \)
Now, \( f^{\prime \prime}(x)=\frac{4\left(1+x+x^{2}\right)\left[\left(1+x+x^{2}\right) x-\left(x^{2}-1\right)(1+2 x)\right]}{\left(1+x+x^{2}\right)^{4}} \)
\( =\frac{4\left(x+x^{2}-x^{3}-x^{2}-2 x^{3}+1+2 x\right)}{\left(1+x+x^{2}\right)^{3}} \)
\( =\frac{4\left[1+2 x-x^{3}\right]}{\left(1+x+x^{2}\right)^{3}} \)
And, \(f {\prime \prime}(1)=\frac{4\left[1+2(1)-(1)^{3}\right]}{\left(1+1+1^{2}\right)^{3}}=\frac{4[3]}{(3)^{3}}=\frac{4}{9} > 0 \)
Also, \( f^{\prime\prime}(-1)=-4 < 0 \)
Then, by second derivative test, \( f \) is minimum at \( x=1 \) and the minimum value is given by
\( f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3} \)

Let \( \mathrm{f}(x)=[x(x-1)+1]^{\frac{1}{3}} \)
\(\therefore f^{\prime}(x)=\frac{2 x-1}{3[[x(x-1)+1]]^{\frac{2}{3}}}\)
Now, if \( \mathrm{f}^{\prime}(x)=0 \)
\(\Rightarrow x=\frac{1}{2}\)
Then, we evaluate the value of \( f \) at critical point \( x=\frac{1}{2} \) and at the end points of the interval \( [0,1] \).
\(f(0)=[0(0-1)+1]^{\frac{1}{3}}=1\)
\( f(1)=[1(1-1)+1]^{\frac{1}{3}}=1 \)
\(f\left(\frac{1}{2}\right)=\left[\frac{1}{2}\left(\frac{1}{2}-1\right)+1\right]^{\frac{1}{3}}=\left(\frac{3}{4}\right)^{\frac{1}{3}}\)
Therefore, we can conclude that the maximum value of in the interval \( [0,1] \) is \(1\).