Solution

(i) The given function is $f(x)=(2 x-1)^{2}+3$.

It can be observed that $(2 x-1)^{2} \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=(2 x-1)^{2}+3 \geq 3$ for every $x \square \mathbf{R}$.

The minimum value of $f$ is attained when $2 x-1=0$.

$2 x-1=0$

$ x=\frac{1}{2} $

$\square$ Minimum value of $f=f(\frac{1}{2})=(2 \cdot \frac{1}{2}-1)^{2}+3=3$

Hence, function $f$ does not have a maximum value.

(ii) The given function is $f(x)=9 x^{2}+12 x+2=(3 x+2)^{2}-2$.

It can be observed that $(3 x+2)^{2} \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=(3 x+2)^{2}-2 \geq-2$ for every $x \square \mathbf{R}$.

The minimum value of $f$ is attained when $3 x+2=0$.

$3 x+2=0 \square^{x=\frac{-2}{3}}$

$\square$ Minimum value of $f=f(-\frac{2}{3})=(3(\frac{-2}{3})+2)^{2}-2=-2$

Hence, function $f$ does not have a maximum value.

(iii) The given function is $f(x)=-(x-1)^{2}+10$.

It can be observed that $(x-1)^{2} \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=-(x-1)^{2}+10 \leq 10$ for every $x \square \mathbf{R}$.

The maximum value of $f$ is attained when $(x-1)=0$.

$(x-1)=0 \square x=0$

$\square$ Maximum value of $f=f(1)=-(1-1)^{2}+10=10$

Hence, function $f$ does not have a minimum value.

(iv) The given function is $g(x)=x^{3}+1$.

Hence, function $g$ neither has a maximum value nor a minimum value.

Solution

(i) $f(x)=|x+2|-1$

We know that $|x+2| \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=|x+2|-1 \geq-1$ for every $x \square \mathbf{R}$.

The minimum value of $f$ is attained when $|x+2|=0$.

$|x+2|=0$

$\Rightarrow x=-2$

$\square$ Minimum value of $f=f(-2)=|-2+2|-1=-1$

Hence, function $f$ does not have a maximum value.

(ii) $g(x)=-|x+1|+3$

We know that $-|x+1| \leq 0$ for every $x \square \mathbf{R}$.

Therefore, $g(x)=-|x+1|+3 \leq 3$ for every $x \square \mathbf{R}$.

The maximum value of $g$ is attained when $|x+1|=0$.

$|x+1|=0$

$\Rightarrow x=-1$

$\square$ Maximum value of $g=g(-1)=-|-1+1|+3=3$

Hence, function $g$ does not have a minimum value. (iii) $h(x)=\sin 2 x+5$

We know that $-1 \leq \sin 2 x \leq 1$.

$\square-1+5 \leq \sin 2 x+5 \leq 1+5$

$\square 4 \leq \sin 2 x+5 \leq 6$

Hence, the maximum and minimum values of $h$ are 6 and 4 respectively.

(iv) $f(x)=|\sin 4 x+3|$

We know that $-1 \leq \sin 4 x \leq 1$.

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

$\square 2 \leq^{|\sin 4 x+3|} \leq 4$

Hence, the maximum and minimum values of $f$ are 4 and 2 respectively.

(v) $h(x)=x+1, x \square(-1,1)$

Here, if a point $x_0$ is closest to -1 , then we find $\frac{x_0}{2}+1<x_0+1$ for all $x_0 \square(-1,1)$.

Also, if $x_1$ is closest to 1 , then $x_1+1<\frac{x_1+1}{2}+1$ for all $x_1 \square(-1,1)$.

Hence, function $h(x)$ has neither maximum nor minimum value in $(-1,1)$.

Solution

(i) $f(x)=x^{2}$

$\therefore f^{\prime}(x)=2 x$

Now,

$f^{\prime}(x)=0 \Rightarrow x=0$

Thus, $x=0$ is the only critical point which could possibly be the point of local maxima or local minima of $f$.

We have $f^{\prime \prime}(0)=2$, which is positive.

Therefore, by second derivative test, $x=0$ is a point of local minima and local minimum value of $f$ at $x=0$ is $f(0)=0$.

(ii) $g(x)=x^{3}-3 x$

$\therefore g^{\prime}(x)=3 x^{2}-3$

Now,

$g^{\prime}(x)=0 \Rightarrow 3 x^{2}=3 \Rightarrow x= \pm 1$

$g^{\prime}(x)=6 x$

$g^{\prime}(1)=6>0$

$g^{\prime}(-1)=-6<0$

By second derivative test, $x=1$ is a point of local minima and local minimum value of $g$ at $x=1$ is $g(1)=1^{3}-3=1-3=-2$. However, $x=-1$ is a point of local maxima and local maximum value of $g$ at $x=-1$ is $g(1)=(-1)^{3}-3(-1)=-1+3=2$.

(iii) $h(x)=\sin x+\cos x, 0<x<\frac{\pi}{2}$

$\therefore h^{\prime}(x)=\cos x-\sin x$

$h^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \in(0, \frac{\pi}{2})$

$h^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)$

$h^{\prime \prime}(\frac{\pi}{4})=-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$

Therefore, by second derivative test, $x=\frac{\pi}{4}$ is a point of local maxima and the local

maximum value of $h$ at $x=\frac{\pi}{4}$ is $h(\frac{\pi}{4})=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$.

(iv) $f(x)=\sin x-\cos x, 0<x<2 \pi$

$\therefore f^{\prime}(x)=\cos x+\sin x$

$f^{\prime}(x)=0 \Rightarrow \cos x=-\sin x \Rightarrow \tan x=-1 \Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \in(0,2 \pi)$

$f^{\prime \prime}(x)=-\sin x+\cos x$

$f^{\prime \prime}(\frac{3 \pi}{4})=-\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}>0$

$f^{\prime \prime}(\frac{7 \pi}{4})=-\sin \frac{7 \pi}{4}+\cos \frac{7 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$

Therefore, by second derivative test, $x=\frac{3 \pi}{4}$ is a point of local maxima and the local

maximum value of $f$ at $x=\frac{3 \pi}{4}$ is

$f(\frac{3 \pi}{4})=\sin \frac{3 \pi}{4}-\cos \frac{3 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$ However, $x=\frac{7 \pi}{4}$ is a point of local minima and

the local minimum value of $f$ at $x=\frac{7 \pi}{4}$ is $f(\frac{7 \pi}{4})=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}$.

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

$\therefore f^{\prime}(x)=3 x^{2}-12 x+9$

$f^{\prime}(x)=0 \Rightarrow 3(x^{2}-4 x+3)=0$

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

$\Rightarrow x=1,3$

Now, $f^{\prime \prime}$

$(x)=6 x-12=6(x-2)$

$f^{\prime \prime}(1)=6(1-2)=-6<0$

$f^{\prime \prime}(3)=6(3-2)=6>0$

Therefore, by second derivative test, $x=1$ is a point of local maxima and the local maximum value of $f$ at $x=1$ is $f(1)=1-6+9+15=19$. However, $x=3$ is a point of local minima and the local minimum value of $f$ at $x=3$ is $f(3)=27-54+27+15=$ 15.

(vi) $g(x)=\frac{x}{2}+\frac{2}{x}, x>0$

$\therefore g^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$

Now,

$g^{\prime}(x)=0$ gives $\frac{2}{x^{2}}=\frac{1}{2} \Rightarrow x^{2}=4 \Rightarrow x= \pm 2$

Since $x>0$, we take $x=2$.

Now,

$g^{\prime \prime}(x)=\frac{4}{x^{3}}$

$g^{\prime \prime}(2)=\frac{4}{2^{3}}=\frac{1}{2}>0$

Therefore, by second derivative test, $x=2$ is a point of local minima and the local

minimum value of $g$ at $x=2$ is $g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2$.

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

$\therefore g^{\prime}(x)=\frac{-(2 x)}{(x^{2}+2)^{2}}$

$g^{\prime}(x)=0 \Rightarrow \frac{-2 x}{(x^{2}+2)^{2}}=0 \Rightarrow x=0$

Now, for values close to $x=0$ and to the left of $0, g^{\prime}(x)>0$. Also, for values close to $x=$ 0 and to the right of $0, g^{\prime}(x)<0$.

Therefore, by first derivative test, $x=0$ is a point of local maxima and the local maximum value of $g(0)$ is $\frac{1}{0+2}=\frac{1}{2}$.

(viii) $f(x)=x \sqrt{1-x}, x>0$

$\therefore f^{\prime}(x)=\sqrt{1-x}+x \cdot \frac{1}{2 \sqrt{1-x}}(-1)=\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}}$

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

$f^{\prime}(x)=0 \Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0 \Rightarrow 2-3 x=0 \Rightarrow x=\frac{2}{3}$

$f^{\prime \prime}(x)=\frac{1}{2}[\frac{\sqrt{1-x}(-3)-(2-3 x)(\frac{-1}{2 \sqrt{1-x}})}{1-x}]$

$=\frac{\sqrt{1-x}(-3)+(2-3 x)(\frac{1}{2 \sqrt{1-x}})}{2(1-x)}$

$=\frac{-6(1-x)+(2-3 x)}{4(1-x)^{\frac{3}{2}}}$

$=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}$

$f^{\prime \prime}(\frac{2}{3})=\frac{3(\frac{2}{3})-4}{4(1-\frac{2}{3})^{\frac{3}{2}}}=\frac{2-4}{4(\frac{1}{3})^{\frac{3}{2}}}=\frac{-1}{2(\frac{1}{3})^{\frac{3}{2}}}<0$

Therefore, by second derivative test, $x=\frac{2}{3}$ is a point of local maxima and the local

maximum value of $f$ at $x=\frac{2}{3}$ is $f(\frac{2}{3})=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9}$.

Solution

i. We have,

$f(x)=e^{x}$

$\therefore f^{\prime}(x)=e^{x}$

Now, if $f^{\prime}(x)=0$, then $e^{x}=0$. But, the exponential function can never assume 0 for any value of $x$.

Therefore, there does not exist $c \square \mathbf{R}$ such that $f^{\prime}(c)=0$.

Hence, function $f$ does not have maxima or minima.

ii. We have,

$g(x)=\log x$

$\therefore g^{\prime}(x)=\frac{1}{x}$

Since $\log x$ is defined for a positive number $x, g^{\prime}(x)>0$ for any $x$.

Therefore, there does not exist $c \square \mathbf{R}$ such that $g^{\prime}(c)=0$.

Hence, function $g$ does not have maxima or minima.

iii. We have,

$h(x)=x^{3}+x^{2}+x+1$

$\therefore h^{\prime}(x)=3 x^{2}+2 x+1$

Now,

$h(x)=0 \square 3 x^{2}+2 x+1=0 \square x=\frac{-2 \pm 2 \sqrt{2} i}{6}=\frac{-1 \pm \sqrt{2} i}{3} \notin \mathbf{R}$

Therefore, there does not exist $c \square \mathbf{R}$ such that $h^{\prime}(c)=0$.

Hence, function $h$ does not have maxima or minima.

Solution

(i) The given function is $f(x)=x^{3}$.

$\therefore f^{\prime}(x)=3 x^{2}$

Now,

$f^{\prime}(x)=0 \Rightarrow x=0$

Then, we evaluate the value of $f$ at critical point $x=0$ and at end points of the interval $[-2,2]$.

$f(0)=0$

$f(-2)=(-2)^{3}=-8$

$f(2)=(2)^{3}=8$

Hence, we can conclude that the absolute maximum value of $f$ on $[-2,2]$ is 8 occurring at $x=2$. Also, the absolute minimum value of $f$ on $[-2,2]$ is -8 occurring at $x=-2$.

(ii) The given function is $f(x)=\sin x+\cos x$.

$\therefore f^{\prime}(x)=\cos x-\sin x$

Now,

$f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}$

Then, we evaluate the value of $f$ at critical point $x=\frac{\pi}{4}$ and at the end points of the interval $[0, \pi]$.

Solution

The profit function is given as $p(x)=41-24 x-18 x^{2}$.

$\therefore p^{\prime}(x)=-24-36 x$

$p^{\prime \prime}(x)=-36$

Now,

$p^{\prime}(x)=0 \Rightarrow x=\frac{-24}{36}=-\frac{2}{3}$

Also,

$p^{\prime \prime}(\frac{-2}{3})=-36<0$

By second derivative test, $x=-\frac{2}{3}$ is the point of local maxima of $p$.

$\therefore$ Maximum profit $=p(-\frac{2}{3})$

$ \begin{aligned} & =41-24(-\frac{2}{3})-18(-\frac{2}{3})^{2} \\ & =41+16-8 \\ & =49 \end{aligned} $

Hence, the maximum profit that the company can make is 49 units.

Solution

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

$f^{\prime}(x)=0$ for finding critical points

$f^{\prime}(x)=12 x^3-24 x^2+24 x-48$

$12 x^3-24 x^2+24 x-48=0$

$12 x^2(x-2)+24(x-2)=0$

$\left(12 x^2+24\right)(x-2)=0$

Since $12 x^2+24 \neq 0$ and $x-2=0 \Longrightarrow x=2$

We need to check at $x=0,2,3$ $f(0)=25$

$f(2)=-39$

$f(3)=16$

$\therefore$ Maximum of $f(x)$ at $x=0$ is 25

Minimum of $f(x)$ at $x=2$ is -39

Solution

Let $f(x)=\sin 2 x$.

$\therefore f^{\prime}(x)=2 \cos 2 x$

Now,

$ \begin{aligned} & f^{\prime}(x)=0 \Rightarrow \cos 2 x=0 \\ & \Rightarrow 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \end{aligned} $

Then, we evaluate the values of $f$ at critical points $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ and at the end points of the interval $[0,2 \pi]$.

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

$f(\frac{5 \pi}{4})=\sin \frac{5 \pi}{2}=1, f(\frac{7 \pi}{4})=\sin \frac{7 \pi}{2}=-1$

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

Hence, we can conclude that the absolute maximum value of $f$ on $[0,2 \pi]$ is occurring

at $x=\frac{\pi}{4}$ and $\quad x=\frac{5 \pi}{4}$.

Solution

Let $f(x)=\sin x+\cos x$.

$\therefore f^{\prime}(x)=\cos x-\sin x$

$f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4} \ldots$,

$f^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)$

Now, $f^{\prime \prime}(x)$ will be negative when $(\sin x+\cos x)$ is positive i.e., when $\sin x$ and $\cos x$ are both positive. Also, we know that $\sin x$ and $\cos x$ both are positive in the first

quadrant. Then, $f^{\prime \prime}(x)$ will be negative when $x \in(0, \frac{\pi}{2})$.

Thus, we consider $x=\frac{\pi}{4}$.

$f^{\prime \prime}(\frac{\pi}{4})=-(\sin \frac{\pi}{4}+\cos \frac{\pi}{4})=-(\frac{2}{\sqrt{2}})=-\sqrt{2}<0$ $\square$ By second derivative test, $f$ will be the maximum at $x=\frac{\pi}{4}$ and the maximum value of $f$ is $f(\frac{\pi}{4})=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$.

Solution

Let $f(x)=2 x^{3}-24 x+107$.

$\therefore f^{\prime}(x)=6 x^{2}-24=6(x^{2}-4)$

Now,

$f^{\prime}(x)=0 \Rightarrow 6(x^{2}-4)=0 \Rightarrow x^{2}=4 \Rightarrow x= \pm 2$

We first consider the interval $[1,3]$.

Then, we evaluate the value of $f$ at the critical point $x=2 \square[1,3]$ and at the end points of the interval $[1,3]$.

$f(2)=2(8)-24(2)+107=16-48+107=75$

$f(1)=2(1)-24(1)+107=2-24+107=85$

$f(3)=2(27)-24(3)+107=54-72+107=89$

Hence, the absolute maximum value of $f(x)$ in the interval $[1,3]$ is 89 occurring at $x=$ 3.

Next, we consider the interval $[-3,-1]$.

Evaluate the value of $f$ at the critical point $x=-2 \square[-3,-1]$ and at the end points of the interval $[1,3]$.

$f(-3)=2(-27)-24(-3)+107=-54+72+107=125$

$f(-1)=2(-1)-24(-1)+107=-2+24+107=129$

$f(-2)=2(-8)-24(-2)+107=-16+48+107=139$

Hence, the absolute maximum value of $f(x)$ in the interval $[-3,-1]$ is 139 occurring at $x$ $=-2$.

Solution

Let $f(x)=x^{4}-62 x^{2}+a x+9$.

$\therefore f^{\prime}(x)=4 x^{3}-124 x+a$

It is given that function $f$ attains its maximum value on the interval $[0,2]$ at $x=1$.

$\therefore f^{\prime}(1)=0$

$\Rightarrow 4-124+a=0$

$\Rightarrow a=120$

Hence, the value of $a$ is 120 .

Solution

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

$\therefore f^{\prime}(x)=1+2 \cos 2 x$

Now, $f^{\prime}(x)=0 \Rightarrow \cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos (\pi-\frac{\pi}{3})=\cos \frac{2 \pi}{3}$

$2 x=2 \pi \pm \frac{2 \pi}{3} \quad n \in \mathbf{Z}$

$\Rightarrow x=n \pi \pm \frac{\pi}{3}, n \in \mathbf{Z}$

$\Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \in[0,2 \pi]$

Then, we evaluate the value of $f$ at critical points $x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}$ and at the end points of the interval $[0,2 \pi]$. $f(\frac{\pi}{3})=\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}$

$f(\frac{2 \pi}{3})=\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$

$f(\frac{4 \pi}{3})=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}$

$f(\frac{5 \pi}{3})=\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}$

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

$f(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi$

Hence, we can conclude that the absolute maximum value of $f(x)$ in the interval $[0,2 \pi]$ is $2 \pi$ occurring at $x=2 \pi$ and the absolute minimum value of $f(x)$ in the interval $[0,2 \pi]$ is 0 occurring at $x=0$.

Solution

Let one number be $x$. Then, the other number is $(24-x)$.

Let $P(x)$ denote the product of the two numbers. Thus, we have:

$ \begin{aligned} & P(x)=x(24-x)=24 x-x^{2} \\ & \therefore P^{\prime}(x)=24-2 x \\ & P^{\prime \prime}(x)=-2 \end{aligned} $

Now,

$P^{\prime}(x)=0 \Rightarrow x=12$

Also,

$P^{\prime \prime}(12)=-2<0$

$\square$ By second derivative test, $x=12$ is the point of local maxima of $P$. Hence, the product of the numbers is the maximum when the numbers are 12 and $24-12=12$.

Solution

The two numbers are $x$ and $y$ such that $x+y=60$.

$\square y=60-x$

Let $f(x)=x y^{3}$.

$\Rightarrow f(x)=x(60-x)^{3}$

$\therefore f^{\prime}(x)=(60-x)^{3}-3 x(60-x)^{2}$

$=(60-x)^{2}[60-x-3 x]$

$=(60-x)^{2}(60-4 x)$

And, $f^{\prime \prime}(x)=-2(60-x)(60-4 x)-4(60-x)^{2}$

$ \begin{aligned} & =-2(60-x)[60-4 x+2(60-x)] \\ & =-2(60-x)(180-6 x) \\ & =-12(60-x)(30-x) \end{aligned} $

Now, $f^{\prime}(x)=0 \Rightarrow x=60$ or $x=15$

When $x=60, f^{\prime \prime}(x)=0$.

When $x=15, f^{\prime \prime}(x)=-12(60-15)(30-15)=-12 \times 45 \times 15<0$.

$\square$ By second derivative test, $x=15$ is a point of local maxima of $f$. Thus, function $x y^{3}$ is maximum when $x=15$ and $y=60-15=45$.

Hence, the required numbers are 15 and 45 .

Solution

Let one number be $x$. Then, the other number is $y=(35-x)$.

Let $P(x)=x^{2} y^{5}$. Then, we have:

$ \begin{aligned} & P(x)=x^{2}(35-x)^{5} \\ & \begin{aligned} \therefore P^{\prime}(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) \end{aligned} \end{aligned} $

And, $P^{\prime \prime}(x)=7(35-x)^{4}(10-x)+7 x[-(35-x)^{4}-4(35-x)^{3}(10-x)]$

$ \begin{aligned} & =7(35-x)^{4}(10-x)-7 x(35-x)^{4}-28 x(35-x)^{3}(10-x) \\ & =7(35-x)^{3}[(35-x)(10-x)-x(35-x)-4 x(10-x)] \\ & =7(35-x)^{3}[350-45 x+x^{2}-35 x+x^{2}-40 x+4 x^{2}] \\ & =7(35-x)^{3}(6 x^{2}-120 x+350) \end{aligned} $

Now, $P^{\prime}(x)=0 \Rightarrow x=0, x=35, x=10$

When $x=35,{ }^{\prime}(x)=f(x)=0$ and $y=35-35=0$. This will make the product $x^{2} y^{5}$ equal to 0 .

When $x=0, y=35-0=35$ and the product $x^{2} y^{2}$ will be 0 .

$\square x=0$ and $x=35$ cannot be the possible values of $x$.

When $x=10$, we have:

$ \begin{aligned} P^{\prime \prime}(x) & =7(35-10)^{3}(6 \times 100-120 \times 10+350) \\ & =7(25)^{3}(-250)<0 \end{aligned} $

$\square$ By second derivative test, $P(x)$ will be the maximum when $x=10$ and $y=35-10=$ 25.

Hence, the required numbers are 10 and 25 .

Solution

Let one number be $x$. Then, the other number is $(16-x)$.

Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,

$S(x)=x^{3}+(16-x)^{3}$

$\therefore S^{\prime}(x)=3 x^{2}-3(16-x)^{2}, S^{\prime \prime}(x)=6 x+6(16-x)$

Now, $S^{\prime}(x)=0 \Rightarrow 3 x^{2}-3(16-x)^{2}=0$

$\Rightarrow x^{2}-(16-x)^{2}=0$

$\Rightarrow x^{2}-256-x^{2}+32 x=0$

$\Rightarrow x=\frac{256}{32}=8$

Now, $S^{\prime \prime}(8)=6(8)+6(16-8)=48+48=96>0$

$\square$ By second derivative test, $x=8$ is the point of local minima of $S$.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16-8=8$.

Solution

Let the side of the square to be cut off be $x cm$. Then, the length and the breadth of the box will be $(18-2 x) cm$ each and the height of the box is $x cm$.

Therefore, the volume $V(x)$ of the box is given by,

$V(x)=x(18-2 x)^{2}$ $\therefore V^{\prime}(x)=(18-2 x)^{2}-4 x(18-2 x)$

$=(18-2 x)[18-2 x-4 x]$

$=(18-2 x)(18-6 x)$

$=6 \times 2(9-x)(3-x)$

$=12(9-x)(3-x)$

And, $V^{\prime \prime}(x)=12[-(9-x)-(3-x)]$

$=-12(9-x+3-x)$

$=-12(12-2 x)$

$=-24(6-x)$

Now, $V^{\prime}(x)=0 \Rightarrow x=9$ or $x=3$

If $x=9$, then the length and the breadth will become 0 .

$\therefore x \neq 9$.

$\Rightarrow x=3$.

Now, $V^{\prime \prime}(3)=-24(6-3)=-72<0$

$\therefore$ By second derivative test, $x=3$ is the point of maxima of $V$.

Hence, 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.

Solution

Let the side of the square to be cut off be $x cm$. Then, the height of the box is $x$, the length is $45-2 x$, and the breadth is $24-2 x$.

Therefore, the volume $V(x)$ of the box is given by,

$ \begin{aligned} V(x) & =x(45-2 x)(24-2 x) \\ & =x(1080-90 x-48 x+4 x^{2}) \\ & =4 x^{3}-138 x^{2}+1080 x \end{aligned} $

$\therefore V^{\prime}(x)=12 x^{2}-276 x+1080$

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

$ =12(x-18)(x-5) $

$V^{\prime \prime}(x)=24 x-276=12(2 x-23)$

Now, ${ }^{V^{\prime}}(x)=0 \Rightarrow x=18$ and $x=5$

It is not possible to cut off a square of side $18 cm$ from each corner of the rectangular sheet. Thus, $x$ cannot be equal to 18 .

$\square x=5$

Now, $V^{\prime \prime}(5)=12(10-23)=12(-13)=-156<0$

$\therefore$ By second derivative test, $x=5$ is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is $5 cm$.

Solution

Let a rectangle of length / and breadth $b$ be inscribed in the given circle of radius $a$. Then, the diagonal passes through the centre and is of length $2 a cm$.

Now, by applying the Pythagoras theorem, we have:

$ \begin{aligned} & (2 a)^{2}=l^{2}+b^{2} \\ & \Rightarrow b^{2}=4 a^{2}-l^{2} \\ & \Rightarrow b=\sqrt{4 a^{2}-l^{2}} \end{aligned} $

$\square$ Area of the rectangle, $A=l \sqrt{4 a^{2}-l^{2}}$

$\therefore \frac{d A}{d l}=\sqrt{4 a^{2}-l^{2}}+l \frac{1}{2 \sqrt{4 a^{2}-l^{2}}}(-2 l)=\sqrt{4 a^{2}-l^{2}}-\frac{l^{2}}{\sqrt{4 a^{2}-l^{2}}}$

$ =\frac{4 a^{2}-2 l^{2}}{\sqrt{4 a^{2}-l^{2}}} $

$\frac{d^{2} A}{d l^{2}}=\frac{\sqrt{4 a^{2}-l^{2}}(-4 l)-(4 a^{2}-2 l^{2}) \frac{(-2 l)}{2 \sqrt{4 a^{2}-l^{2}}}}{(4 a^{2}-l^{2})}$

$ =\frac{(4 a^{2}-l^{2})(-4 l)+l(4 a^{2}-2 l^{2})}{(4 a^{2}-l^{2})^{\frac{3}{2}}} $

$ =\frac{-12 a^{2} l+2 l^{3}}{(4 a^{2}-l^{2})^{\frac{3}{2}}}=\frac{-2 l(6 a^{2}-l^{2})}{(4 a^{2}-l^{2})^{\frac{3}{2}}} $

Now, $\frac{d A}{d l}=0$ gives $4 a^{2}=2 l^{2} \Rightarrow l=\sqrt{2} a$

$ \Rightarrow b=\sqrt{4 a^{2}-2 a^{2}}=\sqrt{2 a^{2}}=\sqrt{2} a $

Now, when $l=\sqrt{2} a$,

$\frac{d^{2} A}{d l^{2}}=\frac{-2(\sqrt{2} a)(6 a^{2}-2 a^{2})}{2 \sqrt{2} a^{3}}=\frac{-8 \sqrt{2} a^{3}}{2 \sqrt{2} a^{3}}=-4<0$

$\therefore$ By the second derivative test, when $l=\sqrt{2} a$, then the area of the rectangle is the maximum.

Since $l=b=\sqrt{2} a$, the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Solution

Let $r$ and $h$ be the radius and height of the cylinder respectively.

Then, the surface area $(S)$ of the cylinder is given by,

$ \begin{aligned} & S=2 \pi r^{2}+2 \pi r h \\ & \Rightarrow h=\frac{S-2 \pi r^{2}}{2 \pi r} \\ &=\frac{S}{2 \pi}(\frac{1}{r})-r \end{aligned} $

Let $V$ be the volume of the cylinder. Then,

$V=\pi r^{2} h=\pi r^{2}[\frac{S}{2 \pi}(\frac{1}{r})-r]=\frac{S r}{2}-\pi r^{3}$

Then, $\frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}, \frac{d^{2} V}{d r^{2}}=-6 \pi r$

Now, $\frac{d V}{d r}=0 \Rightarrow \frac{S}{2}=3 \pi r^{2} \Rightarrow r^{2}=\frac{S}{6 \pi}$

When $r^{2}=\frac{S}{6 \pi}$, then $\frac{d^{2} V}{d r^{2}}=-6 \pi(\sqrt{\frac{S}{6 \pi}})<0$.

By second derivative test, the volume is the maximum when $r^{2}=\frac{S}{6 \pi}$.

Now, when $r^{2}=\frac{S}{6 \pi}$, then $h=\frac{6 \pi r^{2}}{2 \pi}(\frac{1}{r})-r=3 r-r=2 r$.

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

Solution

Let $r$ and $h$ be the radius and height of the cylinder respectively.

Then, volume $(V)$ of the cylinder is given by,

$V=\pi r^{2} h=100 \quad$ (given)

$\therefore h=\frac{100}{\pi r^{2}}$

Surface area $(S$ ) of the cylinder is given by,

$S=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+\frac{200}{r}$

$\therefore \frac{d S}{d r}=4 \pi r-\frac{200}{r^{2}}, \frac{d^{2} S}{d r^{2}}=4 \pi+\frac{400}{r^{3}}$

$\frac{d S}{d r}=0 \Rightarrow 4 \pi r=\frac{200}{r^{2}}$

$\Rightarrow r^{3}=\frac{200}{4 \pi}=\frac{50}{\pi}$

$\Rightarrow r=(\frac{50}{\pi})^{\frac{1}{3}}$

Now, it is observed that when $r=(\frac{50}{\pi})^{\frac{1}{3}}, \frac{d^{2} S}{d r^{2}}>0$.

$\square$ By second derivative test, the surface area is the minimum when the radius of the

cylinder is $(\frac{50}{\pi})^{\frac{1}{3}} cm$.

When $r=(\frac{50}{\pi})^{\frac{1}{3}}, h=\frac{100}{\pi(\frac{50}{\pi})^{\frac{2}{3}}}=\frac{2 \times 50}{(50)^{\frac{2}{3}}()^{1-\frac{2}{3}}}=2(\frac{50}{\pi})^{\frac{1}{3}} cm$.

Hence, the required dimensions of the can which has the minimum surface area is given

by radius $=(\frac{50}{\pi})^{\frac{1}{3}} cm$ and height $=2(\frac{50}{\pi})^{\frac{1}{3}} cm$.

Solution

Let a piece of length / 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 $(28-I) m$.

Now, side of square $=\frac{l}{4}$.

Let $r$ be the radius of the circle. Then,

$ 2 \pi r=28-l \Rightarrow r=\frac{1}{2 \pi}(28-l) \text{. } $

The combined areas of the square and the circle $(A)$ is given by,

$A=(\text{ side of the square })^{2}+r^{2}$

$=\frac{l^{2}}{16}+\pi[\frac{1}{2 \pi}(28-l)]^{2}$

$=\frac{l^{2}}{16}+\frac{1}{4 \pi}(28-l)^{2}$

$\therefore \frac{d A}{d l}=\frac{2 l}{16}+\frac{2}{4 \pi}(28-l)(-1)=\frac{l}{8}-\frac{1}{2 \pi}(28-l)$

$\frac{d^{2} A}{d l^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0$

Now, $\frac{d A}{d l}=0 \Rightarrow \frac{l}{8}-\frac{1}{2 \pi}(28-l)=0$

$\Rightarrow \frac{\pi l-4(28-l)}{8 \pi}=0$

$\Rightarrow(\pi+4) l-112=0$

$\Rightarrow l=\frac{112}{\pi+4}$

Thus, when $l=\frac{112}{\pi+4}, \frac{d^{2} A}{d l^{2}}>0$.

$\therefore$ By second derivative test, the area $(A)$ is the minimum when $l=\frac{112}{\pi+4}$.

Hence, the combined area is the minimum when the length of the wire in making the

square is $\frac{112}{\pi+4} cm$ while the length of the wire in making the circle

is $28-\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4} cm$.

Solution

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 the cone.

Then, $V=\frac{1}{3} \pi r^{2} h$

Height of the cone is given by,

$h=R+AB=R+\sqrt{R^{2}-r^{2}} \quad$ [ $ABC$ is a right triangle]

$ \begin{aligned} & \therefore V=\frac{1}{3} \pi r^{2}(R+\sqrt{R^{2}-r^{2}}) \\ & =\frac{1}{3} \pi r^{2} R+\frac{1}{3} \pi r^{2} \sqrt{R^{2}-r^{2}} \\ & \therefore \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(R^{2}-r^{2})-\pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \\ & =\frac{2}{3} \pi r R+\frac{2 \pi r R^{2}-3 \pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \\ & \frac{d^{2} V}{d r^{2}}=\frac{2 \pi R}{3}+\frac{3 \sqrt{R^{2}-r^{2}}(2 \pi R^{2}-9 \pi r^{2})-(2 \pi r R^{2}-3 \pi r^{3}) \cdot \frac{(-2 r)}{6 \sqrt{R^{2}-r^{2}}}}{9(R^{2}-r^{2})} \\ & =\frac{2}{3} \pi R+\frac{9(R^{2}-r^{2})(2 \pi R^{2}-9 \pi r^{2})+2 \pi r^{2} R^{2}+3 \pi r^{4}}{27(R^{2}-r^{2})^{\frac{3}{2}}} \end{aligned} $

Now, $\frac{d V}{d r}=0 \Rightarrow \frac{2}{3} \quad r R=\frac{3 \pi r^{3}-2 \pi r R^{2}}{3 \sqrt{R^{2}-r^{2}}}$

$\Rightarrow 2 R=\frac{3 r^{2}-2 R^{2}}{\sqrt{R^{2}-r^{2}}} \Rightarrow 2 R \sqrt{R^{2}-r^{2}}=3 r^{2}-2 R^{2}$

$\Rightarrow 4 R^{2}(R^{2}-r^{2})=(3 r^{2}-2 R^{2})^{2}$

$\Rightarrow 4 R^{4}-4 R^{2} r^{2}=9 r^{4}+4 R^{4}-12 r^{2} R^{2}$

$\Rightarrow 9 r^{4}=8 R^{2} r^{2}$

$\Rightarrow r^{2}=\frac{8}{9} R^{2}$

When $r^{2}=\frac{8}{9} R^{2}$, then $\frac{d^{2} V}{d r^{2}}<0$.

By second derivative test, the volume of the cone is the maximum when $r^{2}=\frac{8}{9} R^{2}$.

When $r^{2}=\frac{8}{9} R^{2}, h=R+\sqrt{R^{2}-\frac{8}{9} R^{2}}=R+\sqrt{\frac{1}{9} R^{2}}=R+\frac{R}{3}=\frac{4}{3} R$.

Therefore,

$=\frac{1}{3} \pi(\frac{8}{9} R^{2})(\frac{4}{3} R)$

$=\frac{8}{27}(\frac{4}{3} \pi R^{3})$

$=\frac{8}{27} \times($ Volume of the sphere $)$

Hence, the volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$ the volume of the sphere.

Solution

Let $r$ and $h$ be the radius and the height (altitude) of the cone respectively.

Then, the volume $(V)$ of the cone is given as:

$V=\frac{1}{3} \pi r^{2} h \Rightarrow h=\frac{3 V}{r^{2}}$

The surface area $(S)$ of the cone is given by,

$S=\pi r l$ (where $/$ is the slant height)

$=\pi r \sqrt{r^{2}+h^{2}}$

$=\pi r \sqrt{r^{2}+\frac{9 \hbar^{2}}{\pi^{2} r^{4}}} \stackrel{\pi}{=} \frac{r \sqrt{9^{2} r^{6}+V^{2}}}{\pi r^{2}}$

$=\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{2}}$

$ \begin{aligned} \therefore \frac{d S}{d r} & =\frac{r \cdot \frac{6 \pi^{2} r^{5}}{2 \sqrt{{ }^{2} r^{6} 9 V^{2}}}-\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}}} \\ & =\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \end{aligned} $

Now, $\frac{d S}{d r}=0 \Rightarrow 2 \pi^{2} r^{6}=9 V^{2} \Rightarrow r^{6}=\frac{9 V^{2}}{2 \pi^{2}}$

Thus, it can be easily verified that when $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, \frac{d^{2} S}{d r^{2}}>0$.

$\square$ By second derivative test, the surface area of the cone is the least when $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}$.

When $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, h=\frac{3 V}{\pi r^{2}}=\frac{3}{\pi r^{2}}(\frac{2 \pi^{2} r^{6}}{9})^{\frac{1}{2}}=\frac{3}{\pi r^{2}} \cdot \frac{\sqrt{2} \pi r^{3}}{3}=\sqrt{2} r$.

Hence, 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.

Solution

Let $\theta$ be the semi-vertical angle of the cone.

It is clear that

$ \theta \in[0, \frac{\pi}{2}] $

Let $r, h$, and $/$ be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant.

Now, $r=I \sin \theta$ and $h=I \cos \theta$

The volume $(V)$ of the cone is given by,

$ \begin{aligned} & V=\frac{1}{3} \pi r^{2} h \\ & =\frac{1}{3} \pi(l^{2} \sin ^{2} \theta)(l \cos \theta) \\ & =\frac{1}{3} \pi l^{3} \sin ^{2} \theta \cos \theta \\ & \begin{aligned} \therefore \frac{d V}{d \theta} & =\frac{l^{3} \pi}{3}[\sin ^{2} \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)] \\ & =\frac{l^{3} \pi}{3}[-\sin ^{3}+2 \sin \theta \cos ^{2} \theta] \end{aligned} \\ & \begin{aligned} \frac{d^{2} V}{d \theta^{2}} & =\frac{l^{3} \pi}{3}[-3 \sin ^{2} \theta \cos \theta+2 \cos ^{3} \theta-4 \sin ^{2} \theta \cos \theta] \\ & =\frac{l^{3} \pi}{3}[2 \cos ^{3} \theta-7 \sin ^{2} \theta \cos \theta] \end{aligned} \end{aligned} $

Now, $\frac{d V}{d \theta}=0$

$\Rightarrow \sin ^{3} \theta=2 \sin \theta \cos ^{2} \theta$

$\Rightarrow \tan ^{2} \theta=2$

$\Rightarrow \tan \theta=\sqrt{2}$

$\Rightarrow \theta=\tan ^{-1} \sqrt{2}$

Now, when $\theta=\tan ^{-1} \sqrt{2}$, then $\tan ^{2} \theta=2$ or $\sin ^{2} \theta=2 \cos ^{2} \theta$.

Then, we have:

$\frac{d^{2} V}{d \theta^{2}}=\frac{l^{3} \pi}{3}[2 \cos ^{3} \theta-14 \cos ^{3} \theta]=-4 \pi l^{3} \cos ^{3} \theta<0$ for $\theta \in[0, \frac{\pi}{2}]$

$\square$ By second derivative test, the volume $(V)$ is the maximum when $\theta=\tan ^{-1} \sqrt{2}$.

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is $\tan ^{-1} \sqrt{2}$.

Solution

Let $\mathrm{r}, \mathrm{h}, \mathrm{l}$ be the radius, height and slant height of the right circular cone respectively. Let $S$ be the given surface area of the cone. We have, $l^2=\mathrm{r}^2+\mathrm{h}^2 \ldots .(1)$

$\begin{aligned} & S=\pi r l+\pi r^2 \ & S-\pi r^2=\pi r l \ & \Rightarrow l=\frac{S-\pi r^2}{\pi r} \ldots . . \end{aligned}$

$\begin{aligned} & \mathrm{V}=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\ & \Rightarrow \mathrm{V}=\frac{1}{3} \pi \mathrm{r}^2 \sqrt{1^2-\mathrm{r}^2}(\mathrm{by}(1)) \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \pi^2 \mathrm{r}^4\left(\mathrm{l}^2-\mathrm{r}^2\right) \end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{V}^2=\frac{1}{9} \pi^2 \mathrm{r}^4\left[\left(\frac{\mathrm{S}-\pi \mathrm{r}^2}{\pi \mathrm{r}}\right)^2-\mathrm{r}^2\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \pi^2 \mathrm{r}^4\left[\frac{\left(\mathrm{S}-\pi \mathrm{r}^2\right)^2-\pi^2 \mathrm{r}^4}{\pi^2 \mathrm{r}^2}\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \mathrm{r}^2\left[\left(\mathrm{~S}-\pi \mathrm{r}^2\right)^2-\pi^2 \mathrm{r}^4\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \mathrm{r}^2\left[\mathrm{~S}^2-2 \pi \mathrm{Sr}^2+\pi^2 \mathrm{r}^4-\pi^2 \mathrm{r}^4\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9}\left(\mathrm{r}^2 \mathrm{~S}^2-2 \pi \mathrm{Sr}^4\right) \\ \end{aligned}$

$2 \mathrm{~V} \frac{\mathrm{dV}}{\mathrm{dr}}=\frac{\mathrm{S}^2}{9} 2 \mathrm{r}-\frac{2 \pi \mathrm{S}}{9} 4 \mathrm{r}^3$

$2 \mathrm{~V} \frac{\mathrm{dV}}{\mathrm{dr}}=\frac{2 \mathrm{rS}}{9}\left(\mathrm{~S}-4 \pi \mathrm{r}^2\right)$

For maximum volume, $\frac{\mathrm{dV}}{\mathrm{dr}}=0$

$\begin{aligned} & \Rightarrow \frac{2 \mathrm{rS}}{9}\left(\mathrm{~S}-4 \pi \mathrm{r}^2\right)=0 \\ & \Rightarrow \mathrm{r}=\mathrm{o} \text { or } \mathrm{S}-4 \pi \mathrm{r}^2=0 \end{aligned}$

Since, $r$ cannot be $o$

$\begin{aligned} & \Rightarrow \mathrm{S}=4 \pi \mathrm{r}^2 \ & \Rightarrow \mathrm{r}^2=\frac{\mathrm{S}}{4 \pi} \ & \Rightarrow \mathrm{r}^2=\frac{\pi \mathrm{rl}+\pi \mathrm{r}^2}{4 \pi} \end{aligned}$

$\begin{aligned} & \Rightarrow 4 \pi r^2=\pi r l+\pi r^2 \ & \Rightarrow 3 \pi r^2=\pi r l \ & \Rightarrow l=3 r \end{aligned}$

Let $\alpha$ be the semi-vertical angle.

$\begin{aligned} & \sin \alpha=\frac{r}{l} \\ & \sin \alpha=\frac{r}{3 r} \\ & \Rightarrow \alpha=\sin ^{-1}\left(\frac{1}{3}\right) \end{aligned}$

Solution

The given curve is $x^{2}=2 y$.

For each value of $x$, the position of the point will be $(x, \frac{x^{2}}{2})$.

The distance $d(x)$ between the points $(x, \frac{x^{2}}{2})$ and $(0,5)$ is given by,

$d(x)=\sqrt{(x-0)^{2}+(\frac{x^{2}}{2}-5)^{2}}=\sqrt{x^{2}+\frac{x^{4}}{4}+25-5 x^{2}}=\sqrt{\frac{x^{4}}{4}-4 x^{2}+25}$

$\therefore d^{\prime}(x)=\frac{(x^{3}-8 x)}{2 \sqrt{\frac{x^{4}}{4}-4 x^{2}+25}}=\frac{(x^{3}-8 x)}{\sqrt{x^{4}-16 x^{2}+100}}$

Now, $d^{\prime}(x)=0 \Rightarrow x^{3}-8 x=0$

$\Rightarrow x(x^{2}-8)=0$

$\Rightarrow x=0, \pm 2 \sqrt{2}$

$ \begin{aligned} & \begin{aligned} & \text{ And, } d^{\prime \prime}(x)= \frac{\sqrt{x^{4}-16 x^{2}+100}(3 x^{2}-8)-(x^{3}-8 x) \cdot \frac{4 x^{3}-32 x}{2 \sqrt{x^{4}-16 x^{2}+100}}}{(x^{4}-16 x^{2}+100)} \\ &=\frac{(x^{4}-16 x^{2}+100)(3 x^{2}-8)-2(x^{3}-8 x)(x^{3}-8 x)}{(x^{4}-16 x^{2}+100)^{\frac{3}{2}}} \\ &=\frac{(x^{4}-16 x^{2}+100)(3 x^{2}-8)-2(x^{3}-8 x)^{2}}{(x^{4}-16 x^{2}+100)^{\frac{3}{2}}} \\ & \text{ When, } x=0 \text{, then } d^{\prime \prime}(x)=\frac{36(-8)}{6^{3}}<0 . \end{aligned} \\ & \text{ When, } x= \pm 2 \sqrt{2}, d^{\prime \prime}(x)>0 . \end{aligned} $

$\square$ By second derivative test, $d(x)$ is the minimum at $x= \pm 2 \sqrt{2}$.

When $x= \pm 2 \sqrt{2}, y=\frac{(2 \sqrt{2})^{2}}{2}=4$.

Hence, the point on the curve $x^{2}=2 y$ which is nearest to the point $(0,5)$ is $( \pm 2 \sqrt{2}, 4)$. The correct answer is $A$.

Solution

Let $f(x)=\frac{1-x+x^{2}}{1+x+x^{2}}$.

$ \begin{aligned} \therefore f^{\prime}(x) & =\frac{(1+x+x^{2})(-1+2 x)-(1-x+x^{2})(1+2 x)}{(1+x+x^{2})^{2}} \\ & =\frac{-1+2 x-x+2 x^{2}-x^{2}+2 x^{3}-1-2 x+x+2 x^{2}-x^{2}-2 x^{3}}{(1+x+x^{2})^{2}} \\ & =\frac{2 x^{2}-2}{(1+x+x^{2})^{2}}=\frac{2(x^{2}-1)}{(1+x+x^{2})^{2}} \end{aligned} $

$\therefore f^{\prime}(x)=0 \Rightarrow x^{2}=1 \Rightarrow x= \pm 1$

Now, $f^{\prime \prime}(x)=\frac{2[(1+x+x^{2})^{2}(2 x)-(x^{2}-1)(2)(1+x+x^{2})(1+2 x)]}{(1+x+x^{2})^{4}}$

$ =\frac{4(1+x+x^{2})[(1+x+x^{2}) x-(x^{2}-1)(1+2 x)]}{(1+x+x^{2})^{4}} $

$ \begin{aligned} & =\frac{4[x+x^{2}+x^{3}-x^{2}-2 x^{3}+1+2 x]}{(1+x+x^{2})^{3}} \\ & =\frac{4(1+3 x-x^{3})}{(1+x+x^{2})^{3}} \end{aligned} $

And, $f^{\prime \prime}(1)=\frac{4(1+3-1)}{(1+1+1)^{3}}=\frac{4(3)}{(3)^{3}}=\frac{4}{9}>0$

Also, $f^{\prime \prime}(-1)=\frac{4(1-3+1)}{(1-1+1)^{3}}=4(-1)=-4<0$

$\square$ By second derivative test, $f$ is the minimum at $x=1$ and the minimum value is given

by $f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$.

The correct answer is $D$.

Solution

Let $f(x)=[x(x-1)+1]^{\frac{1}{3}}$.

$\therefore f^{\prime}(x)=\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$

Then, we evaluate the value of $f$ at critical point $x=\frac{1}{2}$ and at the end points of the interval $[0,1]$ {i.e., at $x=0$ and $x=1}$.

$f(0)=[0(0-1)+1]^{\frac{1}{3}}=1$

$f(1)=[1(1-1)+1]^{\frac{1}{3}}=1$

$f(\frac{1}{2})=[\frac{1}{2}(\frac{-1}{2})+1]^{\frac{1}{3}}=(\frac{3}{4})^{\frac{1}{3}}$

Hence, we can conclude that the maximum value of $f$ in the interval $[0,1]$ is 1 .

The correct answer is $C$.