Miscellaneous Exercise (Revised) - Chapter 6 - Applications Of Derivatives - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 6 - Applications Of Derivatives

Miscellaneous Exercise Question 1:

Show that the function given by $f(x)=\frac{\log x}{x}$ has maximum at $x=e$.

Answer:
The given function is $f(x)=\frac{\log x}{x}$.
$
f^{\prime}(x)=\frac{x\left(\frac{1}{x}\right)-\log x}{x^2}=\frac{1-\log x}{x^2}
$

Now, $f^{\prime}(x)=0$
$
\begin{aligned}
& \Rightarrow 1-\log x=0 \\
& \Rightarrow \log x=1 \\
& \Rightarrow \log x=\log e \\
& \Rightarrow x=e
\end{aligned}
$

Now, $f^{\prime \prime}(x)=\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^4}$
$
\begin{aligned}
& =\frac{-x-2 x(1-\log x)}{x^4} \\
& =\frac{-3+2 \log x}{x^3}
\end{aligned}
$

Now, $f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^3}=\frac{-3+2}{e^3}=\frac{-1}{e^3}<0$
Therefore, by second derivative test, $f$ is the maximum at $x=e$.

Miscellaneous Exercise Question 2:

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 $\mathrm{cm}$ per second. How fast is the area decreasing when the two equal sides are equal to the base?

Answer :
Let $\triangle A B C$ be isosceles where $B C$ is the base of fixed length $b$.
Let the length of the two equal sides of $\triangle A B C$ be $a$.
Draw $A D \perp B C$.

Now, in $\triangle A D C$, by applying the Pythagoras theorem, we have:
$
\mathrm{AD}=\sqrt{a^2-\frac{b^2}{4}}
$
$\therefore$ Area of triangle $(A)=\frac{1}{2} b \sqrt{a^2-\frac{b^2}{4}}$
The rate of change of the area with respect to time $(t)$ is given by,
$
\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^2-\frac{b^2}{4}}} \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^2-b^2}} \frac{d a}{d t}
$

It is given that the two equal sides of the triangle are decreasing at the rate of $3 \mathrm{~cm}$ per second.
$
\begin{aligned}
& \therefore \frac{d a}{d t}=-3 \mathrm{~cm} / \mathrm{s} \\
& \therefore \frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^2-b^2}}
\end{aligned}
$

Then, when $a=b$, we have:
$
\frac{d A}{d t}=\frac{-3 b^2}{\sqrt{4 b^2-b^2}}=\frac{-3 b^2}{\sqrt{3 b^2}}=-\sqrt{3} b
$

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $\sqrt{3} b \mathrm{~cm}^2 / \mathrm{s}$.

Miscellaneous Exercise Question 3:

Find the intervals in which the function $f$ given by
$
f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}
$
is (i) increasing (ii) decreasing

Answer:

Given: $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$
$
\begin{aligned}
& =\frac{4 \sin x-x(2+\cos x)}{2+\cos x} \\
& =\frac{4 \sin x}{2+\cos x}-\frac{x(2+\cos x)}{2+\cos x} \\
& =\frac{4 \sin x}{2+\cos x}-x \\
& \therefore f^{\prime}(x)=\frac{(2+\cos x) \frac{d}{d x}(4 \sin x)-4 \sin x \frac{d}{d x}(2+\cos x)}{(2+\cos x)^2}-1 \\
& \Rightarrow f^{\prime}(x)=\frac{(2+\cos x)(4 \cos x)-4 \sin x(-\sin x)}{(2+\cos x)^2}-1 \\
& =\frac{8 \cos x+4 \cos { }^2 x+4 \sin ^2 x}{\left(2+\cos ^2 x\right)^2}-1
\end{aligned}
$

$\begin{aligned}
& \Rightarrow f^{\prime}(x)=\frac{8 \cos x+4}{(2+\cos x)^2}-1 \\
& =\frac{8 \cos x+4-(2+\cos x)^2}{(2+\cos x)^2} \\
& =\frac{8 \cos x+4--4-\cos ^2 x-4 \cos x}{(2+\cos x)^2}
\end{aligned}$

$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\frac{4 \cos x-\cos ^2 x}{(2+\cos x)^2} \\
& =\cos x \frac{(4-\cos x)}{(2+\cos x)^2} \ldots \ldots \ldots .(\text { i) }
\end{aligned}
$

Now $4-\cos x>0$ for all real $x$ as $-1 \leq \cos x \leq 1$. Also $(2+\cos x)^2>0$
(i) $f(x)$ is increasing if $f^{\prime}(x) \geq 0$, i.e., from eq. (i), $\cos x \geq 0$
$\Rightarrow x$ lies in I and IV quadrants, i.e., $f(x)$ is increasing for $0 \leq x \leq \frac{\pi}{2}$
and $\frac{3 \pi}{2} \leq x \leq 2 \pi$
and (ii) $f(x)$ is decreasing if $f^{\prime}(x) \leq 0$, i.e., from eq. (i), $\cos x \leq 0$
$\Rightarrow x$ lies in II and III quadrants, i.e., $f(x)$ is decreasing for $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$

Miscellaneous Exercise Question 4:

Find the intervals in which the function $f$ given by $f(x)=x^3+\frac{1}{x^3}, x \neq 0$ is
(i) increasing (ii) decreasing

Answer :
$
\begin{aligned}
& f(x)=x^3+\frac{1}{x^3} \\
& \therefore f^{\prime}(x)=3 x^2-\frac{3}{x^4}=\frac{3 x^6-3}{x^4}
\end{aligned}
$

Then, $f^{\prime}(x)=0 \Rightarrow 3 x^6-3=0 \Rightarrow x^6=1 \Rightarrow x= \pm 1$
Now, the points $x=1$ and $x=-1$ divide the real line into three disjoint intervals i.e., $(-\infty,-1),(-1,1)$, and $(1, \infty)$.
In intervals $(-\infty,-1)$ and $(1, \infty)$ i.e., when $x<-1$ and $x>1, f^{\prime}(x)>0$.
Thus, when $x<-1$ and $x>1$, $f$ is increasing.
In interval ( $-1,1$ ) i.e., when $-1 Thus, when $-1

Miscellaneous Exercise Question 5

Find the maximum area of an isosceles triangle inscribed in the ellipse $\overline{a^2}+\overline{b^2}=1$ with its vertex at one end of the major axis.

Answer:

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $A B C$, be the triangle inscribed in the ellipse where vertex $C$ is at $(a, 0)$. Since the ellipse is symmetrical with $x$-axis and $y-a x i s y_1= \pm \frac{b}{a} \sqrt{a^2-x_1^2}$.

Coordinates of $A$ are $\left(-x_1, \frac{b}{a} \sqrt{a^2-x^2 1}\right)$ and coordinates of $B$ are $\left(x_1,-\frac{b}{a} \sqrt{a^2-x_1^2}\right)$ As the point $\left(-x_1, y_1\right)$ lies on the ellipse, the area of triangle $A B C$ is $A=\frac{1}{2} \left\lvert\, a\left(\frac{2 b}{a} \sqrt{a^2-x_1^2}\right)+\left(-x_1\right)\left(-\frac{b}{a} \sqrt{a^2-x_1^2}\right)+\left(-x_1\right)\left(-\frac{b}{a} \sqrt{a^2-x^2 1}\right)\right.$

$\begin{aligned}
& \Rightarrow A=b a \sqrt{a^2-x^2 1}+x_1 \frac{b}{a} \sqrt{a^2-x_1^2} \\
& \therefore \frac{d A}{d x_1}=\frac{-2 x b}{2 \sqrt{a^2-x_1^2}}+\frac{b}{a} \sqrt{a^2-x_1^2}-\frac{2 b x_1^2}{a^2 \sqrt{a^2-x_1^2}} \\
& =\frac{b}{2 \sqrt{a^2-x_1^2}}\left[-x_1 a+\left(a^2-x_1^2\right)-x_1^2\right] \\
& =\frac{b\left(-2 x_1^2-x_1^2+a^2\right)}{a \sqrt{a^2-x_1^2}}
\end{aligned}$

$\begin{aligned}
& \frac{d A}{d x_1}=0 \\
& \Rightarrow-2 x_1^2-x_1 a+a^2=0 \\
& \Rightarrow x_1=\frac{a \pm \sqrt{a^2-4(-2)\left(a^2\right)}}{2(-2)} \\
& =\frac{a \pm \sqrt{9 a^2}}{-4} \\
& =\frac{a \pm 3 a}{-4}
\end{aligned}$

$
\Rightarrow x_1=-a, \frac{a}{2}
$
$x_1$ cannot be equal to $a . \therefore x_1=\frac{a}{2} \Rightarrow y_1=\frac{b}{a} \sqrt{a^2-\frac{a^2}{4}}=\frac{b a}{2 a} \sqrt{3}=\frac{\sqrt{3 b}}{2}$

$\text { Now } \frac{d^2 A}{d x^2 1}=\left\{\frac{b \sqrt{a^2-x_1^2}\left(-4 x_1-a\right)-\left(-2 x_1^2-x_1 a+a^2\right) \frac{\left(-2 x_1\right)}{2 \sqrt{a^2-x_1^2}}}{a}\right.$

Area is the maximum when $x_1=\frac{a}{2}$. Maximum area of the triangle is $A=b \sqrt{a^2-\frac{a^2}{4}}+\left(\frac{a}{2}\right)^b \sqrt{a^2-\frac{a^2}{4}}$
$
\begin{aligned}
& =a b \frac{\sqrt{3}}{2}+\left(\frac{a}{2}\right)^b \frac{b}{a} \times \frac{a \sqrt{3}}{2} \\
& =\frac{a b \sqrt{3}}{2}+\frac{a b \sqrt{3}}{4}=\frac{3 \sqrt{3}}{4} a b
\end{aligned}
$

Miscellaneous Exercise Question 6.

A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is
$
2 m
$
and volume is $8 \mathrm{~m}^3$. If the building of tank costs Rs 70 per sq meters for the base and Rs 45 per sq meters for sides. What is the cost of the least expensive tank?

Answer:

Let
$
l, b
$
and $h$ represent the length, breadth, and height of the tank respectively.
height $(h)=2 m$
Volume of the tank $=8 \mathrm{~m}^3$ Volume of the tank $=l \times b \times \mathrm{h} 8=l \times b \times 2$
$
\Rightarrow l b=4 \Rightarrow b=\frac{4}{l}
$

Area of the base $=l b=4$

Area of 4 walls $(A)=2 h(l+b)$
$
\begin{aligned}
& \therefore A=4\left(l+\frac{4}{l}\right) \\
& \Rightarrow \frac{d A}{d l}=4\left(1-\frac{4}{l^2}\right)
\end{aligned}
$

Now, $\frac{d A}{d l}=0$
$
\begin{aligned}
& \Rightarrow I-\frac{4}{l^2}=0 \\
& \Rightarrow I^2=4 \\
& \Rightarrow I= \pm 2
\end{aligned}
$

Therefore, we have $/=4$.
$
\begin{aligned}
& \therefore b=\frac{4}{l}=\frac{4}{2}=2 \\
& \frac{d^2 A}{d l^2}=\frac{32}{l^3} \\
& I=2, \frac{d^2 A}{d l^2}=\frac{32}{8}=4>0 .
\end{aligned}
$

Area is the minimum when $/=2$.
We have $I=b=h=2$.
Cost of building base = $\operatorname{Rs} 70 \times(/ b)=\operatorname{Rs} 70(4)=\operatorname{Rs} 280$
Cost of building walls $=R s 2 h(l+h) \times 45=R s 90(2)(2+2)=$
Rs $8(90)=$ Rs 720
Required total cost $=R s(280+720)=\operatorname{Rs} 1000$

Miscellaneous Exercise Question 7

The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their area is least when the side of square is double the radius of the circle.

Answer:

$2 \pi r+4 a=k$ (where $k$ is constant) $\Rightarrow a=\frac{k-2 \pi r}{4}$
sum of the areas of the circle and the square $(A)$ is given by,
$
\begin{aligned}
A & =\pi r^2+a^2=\pi r^2+\frac{(k=2 \pi r)^2}{16} \\
& \therefore \frac{d A}{d r}=2 \pi r+\frac{2(k-2 \pi r)(2 \pi)}{16}=2 \pi r \\
& =-\frac{\pi(k-2 \pi r)}{4}
\end{aligned}
$

Now, $\frac{d A}{d r}=0$
$
\begin{aligned}
& \Rightarrow 2 \pi r=\frac{\pi(k-2 \pi r)}{4} \\
& 8 r=k-2 \pi r \\
& \Rightarrow(8+2 \pi) r=k \\
& \Rightarrow r=\frac{k}{8+2 \pi}=\frac{k}{2(4+\pi)}
\end{aligned}
$

Now, $\frac{d^2 A}{d r^2}=2 \pi+\frac{\pi^2}{2}>0$
$\therefore$ where $r=\frac{k}{2(4+\pi)}, \frac{d^2 A}{d r^2}>0$.

area is least when $r=\frac{k}{2(4+\pi)}$ where $r=\frac{k}{2(4+\pi)}$,
$
a=\frac{k-2 \pi\left[\frac{k}{2(4+\pi)}\right]}{4}=\frac{8 k+2 \pi k-2 \pi k}{2(4+\pi) \times 4}=\frac{k}{4+\pi}=2 r
$

Miscellaneous Exercise Question 8:

A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is $10 \mathrm{~m}$. Find the dimensions of the window to admit maximum light through the whole opening.

Answer :
Let $x$ and $y$ be the length and breadth of the rectangular window.
Radius of the semicircular opening $=\frac{x}{2}$

It is given that the perimeter of the window is $10 \mathrm{~m}$.
$
\begin{aligned}
& \therefore x+2 y+\frac{\pi x}{2}=10 \\
& \Rightarrow x\left(1+\frac{\pi}{2}\right)+2 y=10 \\
& \Rightarrow 2 y=10-x\left(1+\frac{\pi}{2}\right) \\
& \Rightarrow y=5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)
\end{aligned}
$
$\therefore$ Area of the window $(A)$ is given by,
$
A=x y+\frac{\pi}{2}\left(\frac{x}{2}\right)^2
$

$\begin{aligned} & \quad=x\left[5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)\right]+\frac{\pi}{8} x^2 \\ & =5 x-x^2\left(\frac{1}{2}+\frac{\pi}{4}\right)+\frac{\pi}{8} x^2 \\ & \therefore \frac{d A}{d x}=5-2 x\left(\frac{1}{2}+\frac{\pi}{4}\right)+\frac{\pi}{4} x \\ & =5-x\left(1+\frac{\pi}{2}\right)+\frac{\pi}{4} x \\ & \therefore \frac{d^2 A}{d x^2}=-\left(1+\frac{\pi}{2}\right)+\frac{\pi}{4}=-1-\frac{\pi}{4} \\ & \text { Now, } \frac{d A}{d x}=0 \\ & \Rightarrow 5-x\left(1+\frac{\pi}{2}\right)+\frac{\pi}{4} x=0\end{aligned}$

$
\begin{aligned}
& \Rightarrow 5-x-\frac{\pi}{4} x=0 \\
& \Rightarrow x\left(1+\frac{\pi}{4}\right)=5 \\
& \Rightarrow x=\frac{5}{\left(1+\frac{\pi}{4}\right)}=\frac{20}{\pi+4}
\end{aligned}
$

Thus, when $x=\frac{20}{\pi+4}$ then $\frac{d^2 A}{d x^2}<0$.
Therefore, by second derivative test, the area is the maximum when length $x=\frac{20}{\pi+4} \mathrm{~m}$.
Now,
$
y=5-\frac{20}{\pi+4}\left(\frac{2+\pi}{4}\right)=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4} \mathrm{~m}
$

Hence, the required dimensions of the window to admit maximum light is given by length $=\frac{20}{\pi+4} \mathrm{~m}$ and breadth $=\frac{10}{\pi+4} \mathrm{~m}$.

Miscellaneous Exercise Question 9

A point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$

Let $\mathbf{P}$ be the point on the hypotenuse $A C$

Given point $P$ is at distance a \& $b$ from sides of triangle

Let's construct $P L \perp A B \& P M \perp B C$
$
\therefore \mathrm{PL}=\mathrm{a} \& \mathbf{P M}=\mathrm{b}
$

Let $\angle A C B=\theta$
Thus, $\angle \mathrm{APL}=\theta$ (Corresponding angles)

We need to find maximum length of the hypotenuse

Let $l$ be the length of the hypotenuse,
$
\therefore \quad l=\mathbf{A P}+\mathbf{P C}
$

$\ln \triangle A P L$
$
\begin{aligned}
& \cos \theta=\frac{\text { Side adjacent to } \theta}{\text { Hypotenuse }} \\
& \cos \theta=\frac{P L}{A P} \\
& \cos \theta=\frac{a}{A P} \\
& \mathrm{AP}=\frac{a}{\cos \theta} \\
& \mathrm{AP}=\boldsymbol{a} \boldsymbol{\operatorname { s e c } \theta}
\end{aligned}
$

In $\triangle P M C$
$
\begin{aligned}
& \sin \theta=\frac{\text { Side opposite to } \theta}{\text { Hypotenuse }} \\
& \sin \theta=\frac{P M}{P C} \\
& \sin \theta=\frac{b}{P C} \\
& P C=\frac{b}{\sin \theta} \\
& \mathbf{P C}=\boldsymbol{b} \operatorname{cosec} \boldsymbol{\theta}
\end{aligned}
$

Now,
$
\begin{aligned}
& l=A P+P C \\
& l=a \sec \theta+b \operatorname{cosec} \theta
\end{aligned}
$

Differentiating w.r.t $\theta$
$
\begin{aligned}
& \frac{d l}{d \theta}=\frac{d(a \sec \theta+b \operatorname{cosec} \theta)}{d \theta} \\
& \frac{d l}{d \theta}=\mathrm{a} \sec \theta \tan \theta-\mathrm{b} \operatorname{cosec} \theta \cot \theta
\end{aligned}
$

Putting $\frac{d l}{d \theta}=0$
$\mathrm{a} \sec \theta \tan \theta-\mathrm{b} \operatorname{cosec} \theta \cot \theta=0$

$a \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta}-b \cdot \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta}=0$

$
\begin{gathered}
\frac{a \sin \theta}{\cos ^2 \theta}-\frac{b \cos \theta}{\sin ^2 \theta}=0 \\
\frac{a \sin \theta}{\cos ^2 \theta}=\frac{b \cos \theta}{\sin ^2 \theta} \\
\mathrm{a} \sin \theta \times \sin ^2 \theta=\mathrm{b} \cos \theta \cos ^2 \theta \\
\mathrm{a} \sin ^3 \theta=\mathrm{b} \cos ^3 \theta \\
\frac{\sin ^3 \theta}{\cos ^3 \theta}=\frac{b}{a} \\
\tan ^3 \theta=\frac{b}{a} \\
\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}
\end{gathered}
$

$
\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}
$

Finding $\frac{d^2 l}{d^2 \theta}$
$
\frac{d l}{d \theta}=a \sec \theta \tan \theta-b \operatorname{cosec} \theta \cot \theta
$

Differentiating w.rt $\theta$
$
\begin{aligned}
& \frac{d^2 l}{d^2 \theta}=\frac{d(a \sec \theta \tan \theta-b \operatorname{cosec} \theta \cot \theta)}{d \theta} \\
& =a \frac{d(\sec \theta \tan \theta)}{d \theta}-b \frac{d(\operatorname{cosec} \theta \cot \theta)}{d \theta} \\
& \text { Using Product Rule } \\
& \text { as }(u v)^{\prime}=u^{\prime} v+v^{\prime} u
\end{aligned}
$

$
\begin{aligned}
& =a\left((\sec \theta)^{\prime} \tan \theta+(\tan \theta)^{\prime} \sec \theta\right)-b\left((\operatorname{cosec} \theta)^{\prime} \cot \theta+(\cot \theta)^{\prime} \operatorname{cosec} \theta\right) \\
& =a\left(\sec \theta \cdot \tan \theta \cdot \tan \theta+\sec ^2 \theta \cdot \sec \theta\right)-b\left((-\operatorname{cosec} \theta \cdot \cot \theta) \cot \theta+\left(-\operatorname{cosec}{ }^2 \theta\right) \cdot \operatorname{cosec} \theta\right) \\
& =a\left(\sec \theta \tan ^2 \theta+\sec ^3 \theta\right)-b\left(-\operatorname{cosec} \theta \cot ^2 \theta-\operatorname{cosec}^3 \theta\right)
\end{aligned}
$

$=a \sec \theta\left(\boldsymbol{\operatorname { t a n }}^2 \boldsymbol{\theta}+\boldsymbol{\operatorname { s e c }}^2 \boldsymbol{\theta}\right)+b \operatorname{cosec} \theta\left(\boldsymbol{\operatorname { c o t }}^2 \boldsymbol{\theta}+\boldsymbol{\operatorname { c o s e c }}^2 \boldsymbol{\theta}\right)$

are always positive 

And
Since $\theta$ is acute, i.e. $0<\theta<\frac{\pi}{2}$
$\therefore \theta$ lies in $1^{\text {st }}$ quadrant

So, $\sec \theta \& \operatorname{cosec} \theta$ will be positive

Thus, $\frac{\boldsymbol{d}^2 \boldsymbol{l}}{\boldsymbol{d}^2 \boldsymbol{\theta}}>0$ at $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$
$\therefore l$ is least when $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$

Now,
$
\begin{gathered}
\tan \theta=\frac{b^{\frac{1}{3}}}{a^{\frac{1}{3}}} \\
\tan \theta=\frac{\text { Height }}{\text { Base }} \\
\text { Height }=b^{\frac{1}{3}} \& \text { base } \boldsymbol{a}^{\frac{1}{3}}
\end{gathered}
$

Using Pythagoras theorem
$
\begin{aligned}
& \text { Hypotenuse }^2=\text { Height }^2+\text { Base }^2 \\
& \text { Hypotenuse }{ }^2=\left(b^{\frac{1}{3}}\right)^2+\left(a^{\frac{1}{3}}\right)^2 \\
& \text { Hypotenuse }=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}} \\
&
\end{aligned}
$

Least value of $l$
Least value of $l=\mathrm{a} \sec \theta+\mathrm{b} \operatorname{cosec} \theta$
$
\begin{gathered}
=a \times \frac{\text { Hypotenuse }}{\text { Base }}+\boldsymbol{b} \times \frac{\text { Hypotenuse }}{\text { Height }} \\
=\mathrm{a} \times \frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}+\mathrm{b} \times \frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}} \\
l=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(a^{1-\frac{1}{3}}+b^{1-\frac{1}{3}}\right) \\
l=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right) \\
l=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{1}{2}+1}
\end{gathered}
$

$
\begin{aligned}
& l=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{1}{2}+1} \\
& l=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}
\end{aligned}
$

Hence $l=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$

Hence proved..

Miscellaneous Exercise Question 10:

Find the points at which the function $f$ given by $f(x)=(x-2)^4(x+1)^3$ has
(i) local maxima
(ii) local minima
(iii) point of inflexion
$
f(x)=(x-2)^4(x+1)^3
$

Finding $f^{\prime}(x)$
$
f^{\prime}(x)=\frac{d\left((x-2)^4(x+1)^3\right)}{d x}
$

Using product rule as $(u v)^{\prime}=u^{\prime} v+v^{\prime} u$
$
=\left((x-2)^4\right)^{\prime}(x+1)^3+\left((x+1)^3\right)^{\prime}(x-2)^4
$

$
\begin{aligned}
& =4(x-2)^3(x+1)^3+3(x+1)^2(x-2)^4 \\
& =(x-2)^3(x+1)^2[4(x+1)+3(x-2)] \\
& =(x-2)^3(x+1)^2[4 x+4+3 x-6] \\
& =(x-2)^3(x+1)^2[7 x-2]
\end{aligned}
$

Putting $f^{\prime}(\boldsymbol{x})=\mathbf{0}$
$
(x-2)^3(x+1)^2(7 x-2)=0
$

$
\begin{aligned}
& (x-2)^3=0 \\
& x-2=0 \\
& x=2
\end{aligned}
$

$
\begin{aligned}
& (x+1)^2=0 \\
& (x+1)=0 \\
& \boldsymbol{x}=-\mathbf{1}
\end{aligned}
$

$
\begin{aligned}
&\begin{aligned}
& 7 x-2=0 \\
& 7 x=2
\end{aligned}\\
&x=\frac{2}{7}
\end{aligned}
$

$
\text { Hence, } x=2 \& x=-1 \& x=\frac{2}{7}=0.28
$

Thus,
- $x=-\mathbf{1}$ is a point of Inflexion
- $x=\frac{2}{7}$ is point of maxima
- $x=2$ is point of minima

Miscellaneous Exercise Question 11

Find the absolute maximum and minimum values of the function $f$ given by

$
f(x)=\cos ^2 x+\sin x, x \in[0, \pi]
$

Answer:

$f(x)=\cos ^2 x+\sin x$
$
\begin{aligned}
& f^{\prime}(x)=2 \cos x(-\sin x)+\cos x \\
& =-2 \sin x \cos x+\cos x \\
& f^{\prime}(x)=0 \\
& \Rightarrow 2 \sin x \cos x=\cos x \Rightarrow \cos x(2 \sin x-1)=0 \\
& \Rightarrow \sin x=\frac{1}{2} \text { or } \cos x=0 \\
& \Rightarrow x=\frac{\pi}{6} \text {, or } \frac{\pi}{2} \text { as } x \in[0, \pi] \\
& f\left(\frac{\pi}{6}\right)=\cos ^2 \frac{\pi}{6}+\sin \frac{\pi}{6}=\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{5}{4}
\end{aligned}
$

$
\begin{aligned}
& f(0)=\cos ^2 0+\sin 0=1+0=1 \\
& f(\pi)=\cos ^2 \pi+\sin \pi=(-1)^2+0=1
\end{aligned}
$
$
f\left(\frac{\pi}{2}\right)=\cos ^2 \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1
$
absolute maximum value of $f$ is $\frac{5}{4}$ at $x=\frac{\pi}{6}$
The absolute minimum value of $f$ is 1 at $x=0, x=\frac{\pi}{2}$, and $\pi$.

Miscellaneous Exercise Question 12

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$.

Answer:

$V=\frac{1}{3} \pi R^2 h$
$
\begin{aligned}
& B C=\sqrt{r^2-R^2} \\
& h=r+\sqrt{r^2-R^2} \\
& \therefore V=\frac{1}{3} \pi R^2\left(r+\sqrt{r^2-R^2}\right)=\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 \pi}{3} \pi R \sqrt{r^2-R^2}+\frac{R^2}{3} \cdot \frac{(-2 R)}{2 \sqrt{r^2-R^2}}
\end{aligned}
$

$
\begin{aligned}
& =\frac{2}{3} \pi R r+\frac{2 \pi}{3} \pi R \sqrt{r^2-R^2}-\frac{R^3}{3 \sqrt{r^2-R^2}} \\
& =\frac{2}{3} \pi R r+\frac{2 \pi R r\left(r^2-R^2\right)-\pi R^3}{3 \sqrt{r^2-R^2}} \\
& =\frac{2}{3} \pi R r+\frac{2 \pi R r^2-3 \pi R r^3}{3 \sqrt{r^2-R^2}} \\
& \frac{d V}{d R^2}=0 \\
& \Rightarrow \frac{2 \pi r R}{3}=\frac{3 \pi R^2-2 \pi R r^2}{3 \sqrt{r^2-R^2}}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 2 r \sqrt{r^2-R^2}=3 R^2-2 r^2 \\
& \Rightarrow 4 r^2\left(r^2-R^2\right)=\left(3 R^2-2 r^2\right)^2 \\
& \Rightarrow 14 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=0 \\
& \Rightarrow 9 R^2=8 r^2 \\
& \Rightarrow R^2=\frac{8 r^2}{9}
\end{aligned}
$

$
\frac{d^2 V}{d R^2}=\frac{2 \pi r}{3}+\frac{3 \sqrt{r^2-R^2}\left(2 \pi r^2-9 \pi R^2\right)-\left(2 \pi R^3-3 \pi R^3\right)(-6 R) \frac{1}{2 \sqrt{r^2-R^2}}}{9\left(r^2-R^2\right)}
$

$
=\frac{2 \pi r}{3}+\frac{3 \sqrt{r^2-R^2}\left(2 \pi r^2-9 \pi R^2\right)-\left(2 \pi R^3-3 \pi R^3\right)(3 R) \frac{1}{2 \sqrt{r^2-R^2}}}{9\left(r^2-R^2\right)}
$
when $R^2=\frac{8 r^2}{9}, \frac{d^2 V}{d R^2}<0$
volume is the maximum when $R^2=\frac{8 r^2}{9} \cdot R^2=\frac{8 r^2}{9}$,
height of the cone $=r+\sqrt{r^2-\frac{8 R^2}{9}}=r+\sqrt{\frac{r^2}{9}}=r+\frac{r}{3}=\frac{4 r}{3}$

Miscellaneous Exercise Question 13

Let $f$ be a function defined on $[a, b]$ such that $f^{\prime}(x)>0$, for all $x$ $\in(a, b)$. Then prove that $f$ is an increasing function on $(a, b)$.

We have to prove that function is always increasing i.e. $f\left(x_1\right) where $x_1, x_2 \in[a, b]$

Proof
Let $\boldsymbol{x}_1, \boldsymbol{x}_2$ be two numbers in the interval $[a, b]$ i.e. $x_1, x_2 \in[a, b]$

And,

$x_1

In Interval $\left[x_1, x_2\right]$
As $f$ is defined everywhere, $f$ is continuous \& differentiable in $\left[x_1, x_2\right]$

By Mean value of theorem, There exists $\mathrm{c}$ in $\left(x_1, x_2\right)$ i.e. $c \in\left(x_1, x_2\right)$ such that
$
f^{\prime}(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}
$

Given that $\mathrm{f}^{\prime}(x)>0$ for all $x \in(a, b)$
So, $\boldsymbol{f}^{\prime}(\boldsymbol{c})>\mathbf{0}$ for all $\mathrm{c} \in\left(\boldsymbol{x}_1, \boldsymbol{x}_2\right)$

$\begin{aligned}
& \frac{f\left(\boldsymbol{x}_2\right)-\boldsymbol{f}\left(\boldsymbol{x}_1\right)}{\boldsymbol{x}_2-\boldsymbol{x}_1}>\mathbf{0} \\
& f\left(x_2\right)-f\left(x_1\right)>0
\end{aligned}$

So, we can write that
For any two points $x_1, x_2$ in interval $[a, b]$
Where $\boldsymbol{x}_2>\boldsymbol{x}_1$
$
f\left(x_2\right)>\boldsymbol{f}\left(x_1\right)
$

Thus, $f$ increasing in the interval $[\boldsymbol{a}, \boldsymbol{b}]$
Hence proved

Miscellaneous Exercise Question 14:

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}}$, also find the maximum volume.

Answer:

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

$\begin{aligned}
& =\frac{4 \pi r\left(R^2-r^2\right)-2 \pi r^3}{\sqrt{R^2-r^2}} \\
& =\frac{4 \pi r R^2-6 \pi r^3}{\sqrt{R^2-r^2}} \\
& \text { Now, } \frac{d V}{d r}=0 \Rightarrow 4 \pi r R^2-6 \pi r^3=0 \\
& \Rightarrow r^2=\frac{2 R^2}{3}
\end{aligned}$

$
r^2=\frac{2 R^2}{3}, \frac{d^2 V}{d r^2}<0
$
volume is maximum when $r^2=\frac{2 R^2}{3} \cdot r^2=\frac{2 R^2}{3}$.
height of the cylinder is $2 \sqrt{R^2-\frac{2 R^2}{3}}=2 \sqrt{\frac{R^2}{3}}=\frac{2 R}{\sqrt{3}}$.
volume of the cylinder is maximum when height of cylinder is $\frac{2 R}{\sqrt{3}}$

Miscellaneous Exercise Question 15

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi veritical angle $a$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27} \pi h^2 \tan ^2 a$.

Answer:

$
r=h \tan a
$
since $\triangle A O G$ is similar to $\triangle C E G$,
$
\begin{aligned}
\frac{A O}{O G} & =\frac{C E}{E G} \\
\Rightarrow \frac{h}{r} & =\frac{H}{r-R} \\
\Rightarrow H & =\frac{h}{r}(r-R)=\frac{h}{h \tan a}(h \tan a-R)=\frac{1}{\tan a}(h \tan a-R)
\end{aligned}
$
volume of the cylinder is $V=\pi R^2 \mathrm{H}=\frac{\pi R^2}{\tan a}(h \tan a-R)$
$
=\pi R^2 h-\frac{\pi R^3}{\tan a}
$

$\begin{aligned}
& \therefore \frac{d V}{d R}=2 \pi R h \frac{3 \pi R^2}{\tan a} \\
& \frac{d V}{d R}=0 \\
& \Rightarrow 2 \pi R h=\frac{3 \pi R^2}{\tan a} \\
& \Rightarrow 2 h \tan a=3 R \\
& \Rightarrow R=\frac{2 h}{3} \tan a \\
& \frac{d^2 V}{d R^2}=2 \pi R h-\frac{6 \pi R}{\tan a}
\end{aligned}$

And, for $R=\frac{2 h}{3} \tan a$, we have:
$
\begin{aligned}
& \frac{d^2 V}{d R^2}=2 \pi h-\frac{6 \pi}{\tan a}\left(\frac{2 h}{3} \tan a\right)=2 \pi h-4 \pi h=-2 \pi h<0 \\
& \text { volume of the cylinder is greatest when } R=\frac{2 h}{3} \tan a .
\end{aligned}
$
$
R=\frac{2 h}{3} \tan a, H=\frac{1}{\tan a}\left(h \tan a-\frac{2 h}{3} \tan a\right)=\frac{1}{\tan a}\left(\frac{h \tan a}{3}\right)=\frac{h}{3} .
$
the maximum volume of cylinder can be obtained as

$
\pi\left(\frac{2 h}{3} \tan a\right)^2\left(\frac{h}{3}\right)=\pi\left(\frac{4 h^2}{9} \tan ^2 a\right)\left(\frac{h}{3}\right)=\frac{4}{27} \pi h^3 \tan ^2 a
$
Miscellaneous Exercise Question 16

A cylindrical tank of radius $10 \mathrm{~m}$ is being filled with wheat at the rate of 314 cubic mere per hour. Then the depth of the wheat is increasing at the rate of
(A) $1 \mathrm{~m} / \mathrm{h}$ (B) $0.1 \mathrm{~m} / \mathrm{h}$ (C) $1.1 \mathrm{~m} / \mathrm{h}$
(D) $0.5 \mathrm{~m} / \mathrm{h}$

Answer:

: $V=\pi(\text { radius })^2 x$ height
$
\begin{aligned}
& =\pi(10)^2 h \quad(\text { radius }=10 \mathrm{~m}) \\
& =100 \pi h
\end{aligned}
$
$
\frac{d V}{d t}=100 \pi \frac{d h}{d t}
$

Tank is being filled with wheat at rate of 314 cubic meters per hour.
$
\begin{aligned}
\frac{d V}{d t} & =314 \mathrm{~m}^3 / \mathrm{h} \\
314 & =100 \pi \frac{d h}{d t} \\
\Rightarrow & \frac{d h}{d t}=\frac{314}{100(3.14)}=\frac{314}{314}=1
\end{aligned}
$

The depth of wheat is increasing at $1 \mathrm{~m} / \mathrm{h}$.
The correct answer is $\mathbf{A}$.