Chapter 6: Application of Derivatives

Solution: Let us call the area $z$ of a circle of variable radius $r$.
We know that $z$ (area of circle) $=\pi r^2$
∴ Using the rate of change formula, rate of change of area $z$ with respect to the radius $r$

$
=\frac{d z}{d r}=\pi(2 r)=2 \pi r
$

(a) When $r=3 \mathrm{~cm}$ (given), $\therefore$ From $(i), \frac{d z}{d r}=2 \pi(3)=6 \pi \mathrm{sq}, \mathrm{cm}$
(b) When $r=4 \mathrm{~cm}$ (given), $\therefore$ From (i), $\frac{d z}{d r}=2 \pi(4)=8 \pi \mathrm{sq} . \mathrm{cm}$.

Solution: Let $x \mathrm{~cm}$ be the edge of a cube (for example, a room whose length, breadth and height are equal) at time $t$.
Given: Rate of Increase of volume of cube $=8 \mathrm{~cm}^3 / \mathrm{sec}$.

$
\begin{aligned}
\Rightarrow & \frac{d}{d t}(x . x . x) \text { i.e., } \quad \frac{d}{d t} x^3 \text { is positive and }=8 \\
& \left(8 \mathrm{~cm}^3 / \mathrm{sec} \Rightarrow \text { rate of Increase with respect to time }\right) \\
\Rightarrow & 3 x^2 \frac{d}{d t} x=8 \Rightarrow \frac{d x}{d t}=\frac{8}{3 x^2}
\end{aligned}
$


Let us call the surface area $z$ of the cube. $\therefore z=6 x^2$ (Area of four walls + Area of floor + Area of ceiling)
∴ Rate of change of surface area of cube

$
\begin{aligned}
& =\frac{d z}{d t}=6 \frac{d}{d t} x^2=6\left(2 x \frac{d x}{d t}\right)=12 x\left(\frac{8}{3 x^2}\right) \\
& =4\left(\frac{8}{x}\right)=\frac{32}{x} \mathrm{~cm}^2 / \mathrm{sec} .
\end{aligned}
$


Substituting $x=12 \mathrm{~cm}$ (given), $\frac{d z}{d t}=\frac{32}{12}=\frac{8}{3} \mathrm{~cm}^2 / \mathrm{sec}$
Because $rac{d z}{d t}$ is positive, therefore, surface area is increasing at the rate of $\frac{8}{3} \mathrm{~cm}^2 / \mathrm{sec}$.

Solution: Let $x \mathrm{~cm}$ denote the radius of a circle at time $t$.
Given: Rate of increase of radius of circle $=3 \mathrm{~cm} / \mathrm{sec}$.
$\Rightarrow \frac{d x}{d t}$ is positive and $=3 \mathrm{~cm} / \mathrm{sec}$

Let us call the area $z$ of the circle.
$\therefore \quad z=\pi x^2$.
∴ Rate of change of area of circle $=\frac{d z}{d t}=\pi \frac{d}{d t} x^2$

$
\begin{aligned}
& =\pi \cdot 2 x \frac{d x}{d t}=2 \pi x(3) \\
& =6 \pi x .
\end{aligned}
$


Substituting $x=10 \mathrm{~cm}$ (given), $\frac{d z}{d t}=6 \pi(10)=60 \pi \mathrm{~cm}^2 / \mathrm{sec}$
Because $rac{d z}{d t}$ is positive, therefore area of circle is increasing at the rate of $60 \pi \mathrm{~cm}^2 / \mathrm{sec}$.

Solution: Let $x \mathrm{~cm}$ be the edge of variable cube at time $t$.
Given: Rate of increase of edge $x$ is $3 \mathrm{~cm} / \mathrm{sec}$.

$
\therefore \quad \frac{d x}{d t} \text { is positive and }=3 \mathrm{~cm} / \mathrm{sec}
$


Let us call the volume $z$ of the cube.

$
\therefore \quad z=x^3
$

∴ Rate of change of volume of cube

$
\begin{aligned}
& =\frac{d z}{d t}=\frac{d}{d t} x^3=3 x^2 \frac{d x}{d t}=3 x^2(3) \\
\text { or } \quad \frac{d z}{d t} & =9 x^2 \mathrm{~cm}^3 / \mathrm{sec} .
\end{aligned}
$


Substituting $x=10 \mathrm{~cm}$ (given), $\frac{d z}{d t}=9(10)^2=9(100)=900 \mathrm{~cm}^3 / \mathrm{sec}$.
Because $rac{d z}{d t}$ is positive, therefore volume of the cube is inereasing at the rate of $900 \mathrm{~cm}^3 / \mathrm{sec}$.

Solution: Let $x \mathrm{~cm}$ be radius of circular wave at time $t$.
Given: Waves move in circles at the rate of $5 \mathrm{~cm} / \mathrm{sec}$.
⇒ Radius $x$ of circular wave increases at the rate of $5 \mathrm{~cm} / \mathrm{sec}$.

$
\Rightarrow \frac{d x}{d t} \text { is positive and }=5 \mathrm{~cm} / \mathrm{sec} .
$


Let $z$ denote the enclosed area of the circular wave at time $t$.

$
\therefore \quad z=\pi x^2 .
$

∴ Rate of change of area $=\frac{d z}{d t}=\pi \frac{d}{d t} x^2=\pi \cdot 2 x \frac{d x}{d t}$

$
=2 \pi x(5) \quad[B y(i)]=10 \pi x
$


Substituting $x=8 \mathrm{~cm}$ (given), $\frac{d z}{d t}=10 \pi(8)=80 \pi \mathrm{~cm}^2 / \mathrm{sec}$.
Because $rac{d z}{d t}$ is positive, therefore area of circular wave is increasing at the rate of $80 \pi \mathrm{~cm}^2 / \mathrm{sec}$.

Solution: Let $x$ be the radius of the circle at time $t$.
Given: Rate of increase of radius of circle $=0.7 \mathrm{~cm} / \mathrm{sec}$.

$
\Rightarrow \frac{d x}{d t} \text { is positive and }=0.7 \mathrm{~cm} / \mathrm{sec} .
$
Let us call the circumference $z$ of the circle at time $t$.
$\therefore \quad z=2 \pi x$ (Formula)
∴ Rate of change of circumference of circle

$
\begin{aligned}
& =\frac{d z}{d t}=\frac{d}{d t}(2 \pi x)=2 \pi \frac{d x}{d t}=2 \pi(0.7) \\
& =1.4 \pi \mathrm{~cm} / \mathrm{sec} .
\end{aligned}
$

Solution: Given: Rate of decrease of length $x$ of rectangle is $5 \mathrm{~cm} / \mathrm{minute}$.
$\Rightarrow \frac{d x}{d t}$ is negative and $=-5 \mathrm{~cm} /$ minute

Given: Rate of increase of width $y$ of rectangle is $4 \mathrm{~cm} /$ minute.
$\Rightarrow \frac{d y}{d t}$ is positive and $=4 \mathrm{~cm} /$ minute
(a) Let us call the perimeter $z$ of rectangle.

$
\begin{aligned}
& \therefore \quad z=x+y+x+y=2 x+2 y \\
& \therefore \quad \frac{d z}{d t}=2 \frac{d x}{d t}+2 \frac{d y}{d t}
\end{aligned}
$


Plugging in the values from (i) and (ii),

$
\frac{d z}{d t}=2(-5)+2(4)=-10+8=-2 \text { is negative. }
$

Therefore, the perimeter $z$ of the rectangle is decreasing at the rate of $2 \mathrm{~cm} / \mathrm{sec}$.
(Even when $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$ ).
(b) Let us call the area $z$ of rectangle.

$
\begin{aligned}
& \therefore \quad \mathbf{z}=\boldsymbol{x} \boldsymbol{y} \\
& \therefore \quad \frac{d z}{d t}=x \frac{d y}{d t}+y \frac{d x}{d t}
\end{aligned}
$


I By Product Rule
Substituting $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$ (given) and putting values of

$
\begin{aligned}
& \frac{d x}{d t} \text { and } \frac{d y}{d t} \text { from }(i) \text { and }(i i), \\
& \frac{d z}{d t}=8(4)+6(-5)=32-30=2 \text { is positive. }
\end{aligned}
$

Therefore, the area $z$ of the rectangle is increasing at the rate of $2 \mathrm{sqcm} /$ minute even when $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$.

Solution: Let $x \mathrm{~cm}$ be the radius of the spherical balloon at time $t$.
Given: Rate at which the balloon is being inflated i.e., rate at which the volume of the balloon is increasing $=900 \mathrm{cu} . \mathrm{cm}$ sec.

$
\begin{aligned}
& \Rightarrow \frac{d}{d t}\left(\frac{4 \mathbf{m}}{3} x^3\right)=900 \\
& \Rightarrow \quad \frac{4 \pi}{3} \frac{d}{d t} x^3=900 \Rightarrow \frac{4 \pi}{3} \cdot 3 x^2 \frac{d x}{d t}=900 \\
& \Rightarrow \quad 4 \pi x^2 \frac{d x}{d t}=900 \Rightarrow \frac{d x}{d t}=\frac{900}{4 \pi x^2} \\
& \text { Putting } x=15 \mathrm{~cm}(g \text { iven }), \frac{d x}{d t}=\frac{900}{4 \pi(15)^2}=\frac{900}{4 \pi(225)} \\
& \quad=\frac{900}{900 \pi}=\frac{1}{\pi} \text { is positive. }
\end{aligned}
$

Therefore, the radius of balloon is increasing at the rate of $\frac{1}{\pi} \mathrm{~cm} \mathrm{sec}$.

Solution: We know that the volume $V$ of a balloon with radius $x$ is

$
V=\frac{4}{3} \pi x^3
$

∴ Rate of change of volume with respect to radius $x$ is given by

$
\frac{d V}{d x}=\frac{d}{d x}\left(\frac{4}{3} \pi x^3\right)=\frac{4}{3} \pi+3 x^2=4 \pi x^2
$

∴ When $x=10 \mathrm{~cm}, \frac{d \mathrm{~V}}{d x}=4 \mathrm{~m}(10)^2=400 \mathrm{~s}$
i.e., the volume is increasing at the rate of $4 \pi(10)^2=400 \pi \mathrm{~cm}^3 / \mathrm{cm}^2$.

Solution: Let AB be the ladder and C , the junction of wall and ground, $\mathrm{AB}=5 \mathrm{~m}$
Let $\mathrm{CA}=x$ metres, $\mathrm{CB}=y$ metres. We know that as the end A moves away from C, the end B moves towards C.
[‘: Length of ladder can’t change]
i.e., as $x$ increases, $y$ decreases.

$
\text { Now } \frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{s}
$


-(i) (given)
In $\triangle \mathrm{ABC}$, by Pythagoras Theorem $\mathrm{AC}^2+\mathrm{BC}^2=\mathrm{AB}^2$ or $\quad x^2+y^2=5^2=25$

Differentiating both sides with respect to $t$, we have

$
\begin{array}{rlrl}
& & 2 x \frac{d x}{d t}+2 y \frac{d y}{d t} & =0 \\
& \text { or } & & 2 x(2)+2 y \\
& =0 \text { or } & 2 y \frac{d y}{d t}=-4 x \\
& \therefore & \frac{d y}{d t} & =-\frac{2 x}{y}
\end{array}
$


When $x=4$ (given), from (ii), $16+y^2=25$ or $y^2=9$ or $y=3$
∴ From (tit), $\frac{d y}{d t}=-\frac{2 \times 4}{3}=-\frac{8}{3} \mathrm{~cm} / \mathrm{s}$.
Note. The negative sign indicates that $y$ decreases as $t$ increases.

Solution: Given: Equation of the curve is $6 y=x^3+2$

Let ( $x, y$ ) be the required point on curve ( $i$ ).
Given: $y$-coordinate is changing 8 times as fast as the $x$ coordinate.
⇒ Rate of change of $y$ with respect to $x$ is 8

$
\Rightarrow \quad \frac{d y}{d x}=8
$


Differentiating both sides of ( $i$ ) with respect to $x$, we have $6 \frac{d y}{d x}=3 x^2$
Putting $\frac{d y}{d x}=8$ from $(t t), 48=3 x^2 \Rightarrow x^2=\frac{48}{3}=16 \therefore x= \pm 4$
When $x=4$, from $(i), 6 y=64+2=66 \quad \therefore \quad y=\frac{66}{6}=11$
∴ One required point is (4, 11).
When $x=-4$, from (i) $6 y=-64+2=-62$,

$
\therefore \quad y=\frac{-62}{6}=\frac{-31}{3}
$

∴ Second required point is $\left(-4, \frac{-31}{3}\right)$.
Therefore, the required points on curve $(i)$ are $(4,11)$ and $\left(-4, \frac{-31}{3}\right)$.

Solution: Let $x \mathrm{~cm}$ be the radius of the air bubble at time $t$.
Given: Rate of increase of radius of air bubble (spherical as we all know) $=\frac{1}{2} \mathrm{~cm} / \mathrm{sec}$.
$\Rightarrow \frac{d x}{d t}$ is positive and $=\frac{1}{2} \mathrm{~cm} / \mathrm{sec}$.

Let us call the volume $z$ of the air bubble.
$\therefore \quad z=\frac{4 \pi}{3} x^3$
$\therefore \quad \frac{d z}{d t}=$ Rate of change of volume of air bubble

$
=\frac{4 \pi}{3} \frac{d}{d t} x^3=\frac{4 \pi}{3} \cdot 3 x^2 \frac{d x}{d t}=4 \pi x^2\left(\frac{1}{2}\right)[B y(i)]=2 \pi x^2
$


Substituting $x=1 \mathrm{~cm}$ (given), $\frac{d z}{d t}=2 \pi(1)^2=2 \pi$ this is positive.
Therefore, the required rate of increase of volume of air bubble is $2 \pi \mathrm{~cm}^3 / \mathrm{sec}$.

Solution: Diameter of the balloon $=\frac{3}{2}(2 x+1)$ (given)
Therefore, the radius of balloon $=\frac{1}{2}($ diameter $)=\frac{1}{2}-\frac{3}{2}(2 x+1)=\frac{3}{4}(2 x+1)$
Therefore, the volume of balloon $(\mathrm{V})=\frac{4}{3} \pi(\text { radius })^3$

$
\begin{aligned}
& =\frac{4 \pi}{3}\left(\frac{3}{4}(2 x+1)\right)^3=\frac{4}{3} \pi \cdot \frac{27}{64}(2 x+1)^2 \\
& =\frac{9 \pi}{16}(2 x+1)^3 \mathrm{cu} \text { units }
\end{aligned}
$

∴ Rate of change of volume with respect to $x$,

$
\begin{aligned}
& =\frac{d V}{d x}=\frac{9 \pi}{16} \cdot 3(2 x+1)^2 \cdot \frac{d}{d x}(2 x+1) \\
& =\frac{27 \pi}{16}(2 x+1)^2 \cdot 2=\frac{27 \pi}{8}(2 x+1)^2
\end{aligned}
$

Solution: Let the height and base radius of the sandcone formed at time $t \mathrm{sec}$ be $y \mathrm{~cm}$ and $x \mathrm{~cm}$ respectively. Then $y=\frac{1}{6} x$ (given) or $x=6 y$.
Volume of cone $(\mathbf{V})=\frac{1}{3} \pi x^2 y$

Substituting $x=6 y$,

$
V=\frac{1}{3} \pi(6 y)^2 y=12 \pi y^3
$


$
\therefore \quad \frac{d \mathrm{~V}}{d y}=36 \pi y^2
$
It is given that sand is pouring from a pipe to form a sand-cone at the rate of $12 \mathrm{~cm}^3 / \mathrm{sec}$.

$
\begin{array}{lcl}
\therefore & \frac{d V}{d t}=12 & \Rightarrow \quad \frac{d V}{d y} \times \frac{d y}{d t}=12 \\
\Rightarrow & 36 \pi y^2 \times \frac{d y}{d t}=12(\mathrm{By}(t)) & \Rightarrow \frac{d y}{d t}=\frac{1}{3 \pi y^2}
\end{array}
$


When $y=4 \mathrm{~cm}$, (given); $\frac{d y}{d t}=\frac{1}{3 \pi \times 4^2}=\frac{1}{48 \pi} \mathrm{~cm} / \mathrm{sec}$.

Solution: Marginal cost is defined as the rate of change of total cost with respect to the number of units produced.

$
\begin{aligned}
& \therefore \text { Marginal cost }(\mathrm{MC})=\frac{d \mathrm{C}}{d x} \\
& \quad=\frac{d}{d x}\left(0.007 x^2-0.003 x^2+15 x+4000\right) \\
& \quad=0.021 x^2-0.006 x+15 \\
& \begin{aligned}
\therefore \text { When } x & =17, \mathrm{MC}=0.021 \times(17)^2-0.006 \times(17)+15 \\
& =0.021(289)-0.102+15 \\
& =6.069-0.102+15=20.967
\end{aligned}
\end{aligned}
$


Hence, the required marginal cost $=₹ 20.97$.

Solution: Marginal Revenue is defined as the rate of change of total revenue with respect to the number of units sold.

$
\begin{aligned}
\therefore \text { Marginal revenue }(\mathrm{MR}) & =\frac{d \mathrm{R}}{d x} \\
& =\frac{d}{d x}\left(13 x^2+26 x+15\right)=26 x+26
\end{aligned}
$


When $x=7, \mathrm{MR}=26 \times 7+26=208$
Hence, the required marginal revenue $=₹ 208$.
Choose the correct answer in Exercises 17 and 18.

Solution: Let us call the area $z$ of a circle of radius $r$.

$
\therefore \quad z=\pi r^2
$

∴ Rate of change of area $z$ with respect to the radius $r=\frac{d z}{d r}=2 \pi r$
Substituting $r=6 \mathrm{~cm}$ (given), $\frac{d z}{d r}=2 \pi(6)=12 \pi$
Hence, option (B) is correct.

Solution: Given: Total revenue $R(x)=3 x^2+36 x+5$

$
\begin{aligned}
& \therefore \quad \text { Marginal revenue }=\frac{d}{d x} \mathbf{R}(x)=6 x+36 \\
& \text { Putting } x=15(\text { given }), \begin{array}{l}
\frac{d(R(x))}{d x}=6(15)+36 \\
\therefore \quad 90+36=126
\end{array}
\end{aligned}
$

Hence, option (D) is correct.

Solution: Given: $f(x)=3 x+17$

$
\therefore \quad f^{\prime}(x)=3(1)+0=3>0 \text { i.e., + ve for all } x \in \mathbf{R} .
$

$\therefore f(x)$ is strictly increasing on R .

Solution: Given: $f(x)=e^{2 x}$

$
\therefore \quad f^{\prime}(x)=e^{2 x} \frac{d}{d x} 2 x=e^{2 x}(2)=2 e^{2 x}>0 \text { i.e., } \quad+\text { ve for all } x \in \mathbb{R}
$

[– We know that e is approximately equal to 2.718 and is always positive]
$\therefore f(x)$ is strictly increasing on R .
Remark. $e^{-2}=\frac{1}{\left(e^2\right)}>0$ and $e^0=1>0$.

Solution: Given: $f(x)=\sin x$

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

(a) We know that $f^{\prime}(x)=\cos x>0$ i.e., + ve in first quadrant i.e., in $\left(0, \frac{\pi}{2}\right)$.
$\therefore f(x)$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.
(b) We know that $f^{\prime}(x)=\cos x<0$ i.e., - ve in second quadrant i.e., in $\left(\frac{\pi}{2}, \pi\right)$.
$\therefore f(x)$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$.
(c) Because $f^{\prime}(x)=\cos x>0$ i.e., + ve in $\left(0, \frac{\pi}{2}\right)$ and $f^{\prime}(x)=\cos x<0$ i.e., -ve in $\left(\frac{\pi}{2}, \pi\right)$ and $f^{\prime}\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0$
$\therefore f^{\prime}(x)$ does not keep the same sign in the interval ( $0, \pi$ ). Hence $f(x)$ is neither increasing nor decreasing in ( $0, \pi$ ).

Solution: Given:

$
f(x)=2 x^2-3 x
$


$
\therefore \quad f^{\prime}(x)=4 x-3
$


Step I. Let us put $f^{\prime}(x)=0$ to find the critical points i.e., points on the given curve where tangent is parallel to $x$-axis.
∴ From (i), $4 x-3=0$ i.e., $4 x=3$
or $\quad x=\frac{3}{4}(=0.75)$.

This critical point divides the real line in two disjoint subintervals $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.
Step II.

\begin{tabular}{|l|l|l|}
\hline Interval & sign of $f^{\prime}(x)=4 x-3$ …(i) & Nature of function $\boldsymbol{f}$ \\
\hline $\left(-\infty, \frac{3}{4}\right)$ & Take $x=0.5$ (say) then from (i) $f^{\prime}(x)<0$ & $\therefore f$ is strictly decreasing $\downarrow$ \\
\hline $\left(\frac{3}{4}, \infty\right)$ & Take $x=1$ (say) then from (i), $f^{\prime}(x)>0$ & $\therefore f$ is strictly increasing 1 \\
\hline
\end{tabular}

Thus, (a) $f$ is strictly increasing in $\left(\frac{3}{4}, \infty\right)$.
(b) $f$ is strictly decreasing in $\left(-\infty, \frac{3}{4}\right)$.

Solution: Given: $f(x)=2 x^3-3 x^2-36 x+7$

$
\therefore \quad f^{\prime}(x)=6 x^2-6 x-36
$


Step I. Form factors of $f^{\prime}(x)$

$
f^{\prime}(x)=6\left(x^2-x-6\right)
$

(Caution: Don’t omit 6. It can’t be cancelled only from R.H.S.)
or

$
\begin{aligned}
f^{\prime}(x) & =6\left(x^2-3 x+2 x-6\right)=6(x(x-3)+2(x-3)] \\
& =6(x+2)(x-3)
\end{aligned}
$


Step II. Put $f^{\prime}(x)=0 \Rightarrow 6(x+2)(x-3)=0$
But $6 \neq 0 \quad \therefore$ Either $x+2=0$ or $x-3=0$
i.e., $x=-2, x=3$.

These critical points $x=-2$ and $x=3$ divide the real line into three disjoint sub-intervals $(-\infty,-2),(-2,3)$ and $(3, \infty)$.
Step III.

\begin{tabular}{|l|l|l|}
\hline Interval & sign of $\boldsymbol{f}^{\prime}(\boldsymbol{x})$
$
=6(x+2)(x-3) \ldots(i)
$ & Nature of function $\boldsymbol{f}$ \\
\hline ( $-\infty,-2$ ) & Talke $x=-3$ (say). Then from ( $i$ ),
$
\begin{aligned}
f^{\prime}(x) & =(+)(-)(-) \\
& =(+) \text { i.e. }_{,}>0
\end{aligned}
$ & $\therefore f$ is strictly increasing $\uparrow$ in ( $-\infty,-2$ ) \\
\hline ( $-2,3$ ) & Take $x=2$ (say). Then from ( $i$ ), $f^{\prime}(x)=(+)(+)(-)$
$
=(-) \text { i.e., }<0
$ & $\therefore f$ is strictly decreasing $\downarrow$ in ( $-2,3$ ) \\
\hline (3, $\infty$ ) & Take $x=4$ (say). Then from ( $i)$,
$
\begin{aligned}
f^{\prime}(x) & =(+)(+)(+) \\
& =(+) \text { i.e. }>0
\end{aligned}
$ & $\therefore \quad f$ is strictly increasing $\uparrow$ in $(3, \infty)$ \\
\hline
\end{tabular}

Thus, ( $a$ ) $f$ is strictly increasing in ( $-\infty,-2$ ) and ( $3, \infty$ ).
(b) $f$ is strictly decreasing in ( $-2,3$ ).

Solution: (a) Given: $f(x)=x^2+2 x-5$

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


Step I. Put $\left.f^{\prime}(x)=0 \Rightarrow 2 i x+1\right)=0$
But $2 \neq 0$. Therefore, $x+1=0$ i.e., $x=-1$.
This critical point $x=-1$ divides the real line into two
disjoint sub-intervals ( $-\infty$, -1) and ( $-1, \infty$ ).

\begin{table}
\captionsetup{labelformat=empty}
\caption{Step II.}
\begin{tabular}{|l|l|l|}
\hline Interval & sign of $f^{\prime}(x)$
$
=2(x+1) \ldots(i)
$ & Nature of function $\boldsymbol{f}$ \\
\hline ( $-\infty,-1$ ) & Take $x=-2$ (say). Then from (i), $f^{\prime}(x)=(-)$ i.e., <0 & $\therefore f$ is strictly decreasing $\downarrow$ \\
\hline ( $-1, \infty$ ) & Take $x=0$ (say). Then from (i), $f^{\prime}(x)=(+)$ i.e., > 0 & $\therefore \quad f$ is strictly increasing $\uparrow$ \\
\hline
\end{tabular}
\end{table}

Thus, $f$ is strictly increasing in ( $-1, \infty$ ) (i.e., $x>-1$ ) and strictly decreasing in $(-\infty,-1)$ (i.e., $x<-1$ ).
(b) Given: $f(x)=10-6 x-2 x^2$

$
\therefore \quad f^{\prime}(x)=-6-4 x=-2(3+2 x)
$


Step 1. Put $f^{\prime}(x)=0 \Rightarrow-2(3+2 x)=0$
But $-2 \neq 0$. Therefore, $3+2 x=0$ i.e., $2 x=-3$

$
\text { i.e. } x=-\frac{3}{2}, \quad 0_{-\infty}^0 \quad-\frac{3}{2} \quad 0
$


This critical point $x=-\frac{3}{2}$ divides the real line into two disjoint sub-intervals $\left(-\infty,-\frac{3}{2}\right)$ and $\left(-\frac{3}{2}, \infty\right)$.
Step III.

\begin{tabular}{|l|l|l|}
\hline Interval & sign of $f^{\prime}(x)$
$
=-2(3+2 x) \quad \ldots(i)
$ & Nature of function $\boldsymbol{f}$ \\
\hline $\left(-\infty,-\frac{3}{2}\right)$ & \begin{tabular}{l}
Take $x=-2$ (say). \\
Then from (i),
$
\begin{aligned}
f^{\prime}(x) & =(-)(-) \\
& =(+) \text { i.e. },>0
\end{aligned}
$
\end{tabular} & $\therefore \quad f$ is strictly increasing $\uparrow$ \\
\hline $\left(-\frac{3}{2}, \infty\right)$ & \begin{tabular}{l}
Take $x=-1$ (say). \\
Then from (i),
$
\begin{aligned}
f^{\prime}(x) & =(-)(+) \\
& =(-) \text { i.e. }<0
\end{aligned}
$
\end{tabular} & $\therefore f$ is strictly decreasing $\downarrow$ \\
\hline
\end{tabular}

Thus, $f$ is strictly increasing in $\left(-\infty,-\frac{3}{2}\right)$ (i.e., for $x<-\frac{3}{2}$ ) and
strictly decreasing in $\left(-\frac{3}{2}, \infty\right)$ (i.e., for $x>-\frac{3}{2}$ ).
(e) Let $f(x)=-2 x^3-9 x^2-12 x+1$

$
\therefore \quad f^{\prime}(x)=-6 x^2-18 x-12=-6\left(x^2+3 x+2\right)
$


Step I. Forming factors of $f^{\prime}(x)$

$
=-6\left(x^2+x+2 x+2\right)=-6(x(x+1)+2(x+1))
$

or $f^{\prime}(x)=-6(x+1)(x+2)$

Step II. $f^{\prime}(x)=0$ gives $x=-1$ or $x=-2$
The points $x=-2$ and $x=-1$ (arranged in ascending order) divide the real line into 3 disjoint intervals, namely, ( $-\infty,-2$ ), $(-2,-1)$ and $(-1, \infty)$.

\begin{table}
\captionsetup{labelformat=empty}
\caption{Step III. Nature of $f(x)$}
\begin{tabular}{|l|l|l|}
\hline Interval & sign of $f^{\prime}(x)$
$
=-6(x+1)(x+2)(i)
$ & Nature of function $\boldsymbol{f}$ \\
\hline ( $-\infty,-2$ ) & Take $x=-3$ (say), Then from ( $i$ ), $f^{\prime}(x)=(-)(-)(-) =(-)$ i.e., $<0$ & $\therefore f$ is strictly decreasing in $(-\infty,-2) \downarrow$ \\
\hline ( $-2,-1$ ) & Take $x=-1.5$ (say), Then from ( $i$ ), $f^{\prime}(x)=(-)(-)(+)=+$ i.e., $>0$ & $\therefore \quad f$ is strictly increasing in $(-2,-1) \uparrow$ \\
\hline $(-1, \infty)$ & Take $x=0$ (say), then from ( $i$ ) $f^{\prime}(x)=(-)(+)(+)=(-)$ i.e., < 0 & $\therefore f$ is strictly decreasing in $(-1, \infty) \downarrow$ \\
\hline
\end{tabular}
\end{table}
$\therefore f$ is strictly increasing in $(-2,-1)$ and strictly decreasing in $(-\infty,-2)$ and $(-1, \infty)$
(d) Let $f(x)=6-9 x-x^2 \quad \therefore f^{\prime}(x)=-9-2 x$. $f(x)$ is strictly increasing if $f^{\prime}(x)>0$, i.e., if $-9-2 x>0$
or $\quad-2 x>9 \quad$ or $\quad x<-\frac{9}{2}$
$\therefore f$ is strictly increasing $\uparrow$, in the interval $\left(-\infty,-\frac{9}{2}\right)$.
$f(x)$ is strictly decreasing if $f^{\prime}(x)<0$, i.e, if $-9-2 x<0$
or $\quad-2 x<9 \quad$ or $\quad x>-\frac{9}{2}$
$\therefore f$ is strictly decreasing $\downarrow$ in the interval $\left(-\frac{9}{2}, \infty\right)$.
(e) Let $f(x)=(x+1)^3(x-3)^3$
then $f^{\prime}(x)=(x+1)^3, 3(x-3)^2+(x-3)^3 \cdot 3(x+1)^2$

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


The factors $(x+1)^2$ and $(x-3)^2$ are non-negative for all $x$.
$\therefore f(x)$ is strictly increasing if

$
f^{\prime}(x)>0, \quad \text { i.e., if } x-1>0 \quad \text { or } \quad x>1
$

$f(x)$ is strictly decreasing if $f^{\prime}(x)<0$, i.e., if $x-1<0$
or

$
x<1
$


Thus, $f$ is strictly increasing $\uparrow$ in ( $1, \infty$ ) and strictly decreasing † in ( $-\infty, 1$ ).

Solution: Given: $y=\log (1+x)-\frac{2 x}{2+x}$

$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{1}{1+x} \frac{d}{d x}(1+x)-\left[\frac{(2+x) \frac{d}{d x}(2 x)-2 x \frac{d}{d x}(2+x)}{(2+x)^2}\right] \\
& =\frac{1}{1+x}-\left[\frac{(2+x) 2-2 x}{(2+x)^2}\right]=\frac{1}{1+x}-\frac{(4+2 x-2 x)}{(2+x)^2} \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{1}{1+x}-\frac{4}{(2+x)^2}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2} \\
& =\frac{4+x^2+4 x-4-4 x}{(1+x)(2+x)^2}=\frac{x^2}{(1+x)(2+x)^2}
\end{aligned}
$


Domain of the given function is given to be $x>-1$
$\Rightarrow x+1>0$. Also $(2+x)^2>0$ and $x^2 \geq 0$
∴ From (i), $\frac{d y}{d x} \geq 0$ for all $x$ in the domain $(x>-1)$.
∴ The given function is an increasing function of $x$ (in its domain namely $x>-1$ ).
Note 1. For an increasing function $\frac{d y}{d x}=f^{\prime}(x) \geq 0$ and for a strictly increasing function $\frac{d y}{d x}=f^{\prime}(x)>0$.
Note 2. For a decreasing function $\frac{d y}{d x}=f^{\prime}(x) \leq 0$ and for a strictly decreasing function $\frac{d y}{d x}=f^{\prime}(x)<0$.

Solution: Given: $y(=f(x))=(x(x-2))^2$.
Step I. Find $\frac{d y}{d x}$ and form factors of R.H.S. of value of $\frac{d y}{d x}$.

$
\begin{aligned}
& \therefore \quad \frac{d y}{d x}=2 x(x-2) \frac{d}{d x}[x(x-2)] \\
& \begin{aligned}
\Rightarrow \quad & \left.\quad \frac{d y}{d x}=2 x(x-2)\left[x \frac{d}{d x}(f(x))^n=n(f(x))\right)^{n-1} \frac{d}{d x} f(x)\right] \\
& \left.=2 x(x-2)[x+x-2]=2 x(x-2) \frac{d}{d x} x\right] \quad \text { (Product Rule) }
\end{aligned} \\
& \text { or } \quad \frac{d y}{d x}=4 x(x-2)(x-1)
\end{aligned}
$


Step II. Put $\frac{d y}{d x}=0$.

$
\text { ∴ From (i) } 4 x(x-2)(x-1)=0
$


But $4 \neq 0 \therefore$ Either $x=0$ or $x-2=0$ or $x-1=0$

$
\Rightarrow x=0, x=2, x=1
$


These three critical points $x=0, x=1, x=2$ (arranged in their ascending order divide the real line into three sub-intervals ( $-\infty$, $0],[0,1],[1,2],[2, \infty)$.

\begin{table}
\captionsetup{labelformat=empty}
\caption{Step III}
\begin{tabular}{|l|l|l|}
\hline Interval & sign of $\frac{d y}{d x} =4 x(x-2)(x-1) \ldots(i)$ & Nature of $y=f(x)$ \\
\hline ( $-\infty$, 0 ] & Take $x=-1$ (say). Then from ( $i$ ), $\frac{d y}{d x}=(-)(-)(-) =(-)$ (or $=0$ at $x=0$ ) i.e., $\leq 0$ & $\therefore f(x)$ is decreasing 4 \\
\hline [0, 1] & \begin{tabular}{l}
Take $x=\frac{1}{2}$ (say). \\
Then from $(i)$, $\frac{d y}{d x}=(+)(-)(-) =(+)$ (or $=0 \mathrm{at} x=0, x=1$ ) \\
i.e., $\geq 0$
\end{tabular} & $\therefore f(x)$ is increasing $\uparrow$ \\
\hline [1, 2] & Take $x=1.5$ (say). Then from ( $i$ ), $\frac{d y}{d x}=(+)(-)(+)$ & $\therefore f(x)$ is decreasing + \\
\hline
\end{tabular}
\end{table}
\begin{tabular}{|l|l|l|}
\hline & $
\begin{aligned}
& \begin{array}{r}
=(-)(\mathrm{or}=0 \mathrm{at} \\
x=1, x=2)
\end{array} \\
& \text { i.e., } \leq 0
\end{aligned}
$ & \\
\hline [2, $\infty$ ) & \begin{tabular}{l}
Take $x=3$ (say). \\
Then from ( $i$ ),
$
\frac{d y}{d x}=(+)(+)(+)
$
\\
$=(+)$ (or $=0$ at $x=2$ )
\end{tabular} & $\therefore f(x)$ is increasing $\uparrow$ \\
\hline & & \\
\hline & ie., $\geq 0$ & \\
\hline
\end{tabular}

Therefore, $f(x)$ is an increasing function in the intervals $[0,1]$ (i.e., $0 \leq x \leq 1$ ) and ( $2, \infty$ ) (i.e., $x \geq 2$ ).

Remark. (We have included the critical points in the subintervals because we are to discuss for increasing function and not for strictly increasing function. See Notes 1 and 2 at the end of solution of Q. No. 7).

Solution: Here $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$

$
\begin{aligned}
& \Rightarrow \frac{d y}{d \theta}=\frac{(2+\cos \theta) \cdot 4 \cos \theta-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^2}-1 \\
& =\frac{8 \cos \theta+4 \cos ^2 \theta+4 \sin ^2 \theta}{\left(2+\cos ^2 \theta\right)^2}-1 \\
& =\frac{8 \cos \theta+4\left(\cos ^2 \theta+\sin ^2 \theta\right)-(2+\cos \theta)^2}{(2+\cos \theta)^2} \quad(\text { Taking LC.M.) } \\
& =\frac{8 \cos \theta+4-(2+\cos \theta)^2}{(2+\cos \theta)^2}=\frac{(8 \cos \theta+4)-\left(4+4 \cos \theta+\cos ^2 \theta\right)}{(2+\cos \theta)^2} \\
& \text { or } \frac{d y}{d \theta}=\frac{4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}
\end{aligned}
$


Since $0 \leq \theta \leq \frac{\pi}{2}$,
we have $0 \leq \cos \theta \leq 1$ and, therefore, $4-\cos \theta>0$. Also $(2+\cos \theta)^2>0$
∴ From (i), $\frac{d y}{d \theta} \geq 0$ for $0 \leq \theta \leq \frac{\pi}{2}$.
Hence, $y$ is an increasing function of $\theta$ in $\left[0, \frac{\pi}{2}\right]$.

Solution: Given: $f(x)=\log x$

$
\therefore f^{\prime}(x)=\frac{1}{x}>0 \text { for all } x \text { in }(0, \infty) \quad[\because x \in(0, \infty) \Rightarrow x>0]
$

$\therefore f(x)$ is strictly increasing on ( $0, \infty)$,

Solution: Given: $f(x)=x^2-x+1$

$
\therefore \quad f^{\prime}(x)=2 x-1
$

$f^{\prime}(x)$ is strictly increasing if $f^{\prime}(x)>0$ i.e., if $2 x-1>0$
i.e., if $2 x>1$ or $x>\frac{1}{2}$
$f(x)$ is strictly decreasing if

$
f^{\prime}(x)<0 \text { i.e., if } 2 x-1<0 \text { i.e., } x<\frac{1}{2}
$

$\therefore f(x)$ is strictly increasing for $x>\frac{1}{2}$ i.e., on the interval $\left(\frac{1}{2}, 1\right) [\because$ The given interval is $(-1,1)]$
and $f(x)$ is strictly decreasing for $x<\frac{1}{2}$ i.e., on the interval $\left(-1, \frac{1}{2}\right)$.
[:- The given interval is $(-1,1)]$
$\therefore f(x)$ is neither strictly incressing nor strictly decreasing on the interval ( $-1,1$ ).

Solution: (A) Let $f(x)=\cos x$ then $f^{\prime}(x)=-\sin x$

$
\because \quad 00
$

[Because sin $x$ is positive in both first and second quadrants]

$
\begin{aligned}
& \Rightarrow-\sin x<0 \quad \therefore f^{\prime}(x)=-\sin x<0 \text { on }\left(0, \frac{\pi}{2}\right) \\
& \Rightarrow f(x) \text { is strictly decreasing on }\left(0, \frac{\pi}{2}\right) .
\end{aligned}
$

(B) Let $f(x)=\cos 2 x$ then $f^{\prime}(x)=-2 \sin 2 x$

$
\begin{array}{llc}
\because & 0\Rightarrow & \sin 2 x>0 \Rightarrow & -2 \sin 2 x<0
\end{array}
$
$
\therefore f^{\prime}(x)=-2 \sin 2 x<0 \text { on }\left(0, \frac{\pi}{2}\right)
$

$\Rightarrow f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
(C) Let $f(x)=\cos 3 x$ then $f^{\prime}(x)=-3 \sin 3 x$

$
\because \quad 0$


Now for $0<3 x<\pi$.

$
\left(\text { i.e. } 00
$

( $\because \sin \theta$ is positive in first two quadrants)

$
\Rightarrow f^{\prime}(x)=-3 \sin 3 x<0 \Rightarrow f^{\prime}(x)<0
$

$\Rightarrow f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{3}\right)$
and for $\quad \pi<3 x<\frac{3 \pi}{2}, \quad \sin 3 x<0$
[Because sin $\theta$ is negative in third quadrant]

$
\therefore f^{\prime}(x)=-3 \sin 3 x>0 \Rightarrow f^{\prime}(x)>0
$

$\Rightarrow f(x)$ is strictly increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
$\therefore f(x)$ is neither strictly increasing nor strictly decreasing on

$
\left(0, \frac{\pi}{2}\right)
$

(D) Let $f(x)=\tan x$ then $f^{\prime}(x)=\sec ^2 x>0$
$\Rightarrow f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.
Hence, only the functions in options (A) and (B) are strictly decreasing.

Solution: Given: $f(x)=x^{100}+\sin x-1$

$
\therefore f^{\prime}(x)=100 x^{90}+\cos x
$


Let us test option (A) (0,1)
On (0, 1); $x>0$ and hence $100 x^{90}>0$
For cos $x$; interval $(0,1) \Rightarrow(0,1$ radian $)$
$\Rightarrow\left(0,57^{\circ}\right.$ nearly) $\left(\because \pi\right.$ radians $=180^{\circ}$

$
\begin{aligned}
& \Rightarrow 1 \text { radian }=\frac{180^{\circ}}{\pi} \\
& =\frac{180^{\circ}}{\left(\frac{22}{7}\right)}=180^{\circ} \times \frac{7}{22}=\frac{90^{\circ} \times 7}{11}=\frac{630^{\circ}}{11}=57^{\circ} \text { nearly) }
\end{aligned}
$
$\Rightarrow x$ is in first quadrant and hence cos $x$ is positive.
∴ From $(i), f^{\prime}(x)=100 x^{99}+\cos x>0$ and hence $f(x)$ is strictly increasing on ( 0,1 ).
∴ Option (A) is not the correct option.
Let us test option (B) $\quad\left(\frac{\pi}{2}, \pi\right)$
For $100 x^{50}, x \in\left(\frac{\pi}{2}, \pi\right)$
$\Rightarrow x \in\left(\frac{\left(\frac{22}{7}\right)}{2}, \frac{22}{7}\right)=\left(\frac{11}{7}, \frac{22}{7}\right)=(1.5,3.1)$
$\Rightarrow x>1 \Rightarrow x^{90}>1$ and hence $100 x^{90}>100$.
For $\cos x,\left(\frac{\pi}{2}, \pi\right) \Rightarrow$ Second quadrant and hence cos $x$ is negative and has value between -1 and 0 .
$(\because-1 \leq \cos \theta \leq 1)$
∴ From $(i), f^{\prime}(x)=100 x^{90}+\cos x>100-1=99>0$
$\therefore f(x)$ is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.
∴ Option (B) is not the correct option.
Let us test option (C) $\left(0, \frac{\pi}{2}\right)$
On $\left(0, \frac{\pi}{2}\right)$ i.e., $(0,1.5)$ both terms $100 x^{90}$ and cos $x$ are positive and hence from ( $0, f^{\prime}(x)=100 x^{99}+$ cos $x$ is positive.
$\therefore f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ also.
∴ Option (C) is also not the correct option.
Hence, option (D) is correct.

Solution: Here $f(x)=x^2+a x+1$

Differentiating $(i)$ with respect to $x, f^{\prime}(x)=2 x+a$

Because $f(x)$ is strictly increasing on ( 1,2 ) (given),
$\therefore f^{\prime}(x)=2 x+a>0$ for all $x$ in $(1,2)$

Now on $(1,2), 1Multiplying by $2,2<2 x<4$ for all $x$ in ( 1,2 ).
Adding $a$ to all sides
$2+a<2 x+a<4+a$ for all $x$ in ( 1,2 )
or $\quad 2+a[By (ii)]
∴ Minimum value of $f^{\prime}(x)$ is $2+a$ and maximum value of
$
f^{\prime}(x) \text { is } 4+a .
$


But from (iii), $f^{\prime}(x)>0$ for all $x$ in $(1,2)$
$\therefore 2+a>0$ and $4+a>0$
[By (iv)]
$\therefore a>-2$ and $a>-4$
$\therefore a>-2$
$[\because a>-2 \Rightarrow a>-4$ automatically $]$
∴ Least value of $a$ is -2 .

Solution: Given: $f(x)=x+\frac{1}{x}=x+x^{-1}$

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


Forming factors, $f^{\prime}(x)=\frac{(x-1(x+1)}{x^2}$

Given: $I$ is an interval disjoint from $[-1,1]$.
i.e., $1=(-\infty, \infty)-[-1,1]=(-\infty,-1) \cup(1, \infty)$
∴ For every $x \in 1$, either $x<-1$ or $x>1$
For $x<-1$ (For example, $x=-2$ (say)),
from $(f), f^{\prime}(x)=\frac{(-)(-)}{(+)}=(+)$ i.e., $>0$
For $x>1$ (For example, $x=2$ (say)),
from (i), $f^{\prime}(x)=\frac{(+)(+)}{(+)}=(+)$ i.e., $>0$
$\therefore f^{\prime}(x)>0$ for all $x \in \mathrm{I} \quad \therefore \quad f(x)$ is strictly increasing on I .

Solution: Given: $f(x)=\log \sin x$

$
\therefore r^{\prime}(x)=\frac{1}{\sin x} \frac{d}{d x} \sin x=\frac{1}{\sin x}(\cos x)=\cot x
$


On the interval $\left(0, \frac{\pi}{2}\right)$ i.e., in first quadrant, from (i), $f^{\prime}(x)=\cot x>0$
$\therefore f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.
On the interval $\left(\frac{\pi}{2}, \pi\right)$ i.e., in second quadrant, from $(i), f^{\prime}(x)= \cot x<0$.
$\therefore f(x)$ is strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

Solution: Given: $f(x)=\log \cos x$

$
\therefore f^{\prime}(x)=\frac{1}{\cos x} \frac{d}{d x}(\cos x)=\frac{1}{\cos x}(-\sin x)=-\tan x
$


We know that on the interval $\left(0, \frac{\pi}{2}\right)$ i.e., in first quadrant, $\tan x$ is positive and hence from $(i), f^{\prime}(x)=-\tan x$ is negative i.e., < 0.
$\therefore f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
We know that on the interval $\left(\frac{\pi}{2}, \pi\right)$ i.e., in second quadrant; $\tan x$ is negative and hence from (i),

$
f^{\prime}(x)=-\tan x \text { is positive i.e., }>0 .
$

$\therefore f(x)$ is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.

Solution: Given: $f(x)=x^3-3 x^2+3 x-100$.
Then

$
\begin{aligned}
f^{\prime}(x) & =3 x^2-6 x+3=3\left(x^2-2 x+1\right) \\
& =3(x-1)^2 \geq 0 \text { for all } x \text { in } R
\end{aligned}
$

$\therefore f(x)$ is increasing on R .

Solution: Given: y $(=f(x))=x^2 e^{-x}$

$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =x^2 \frac{d}{d x} e^{-x}+e^{-x} \frac{d}{d x} x^2=x^2 e^{-x}(-1)+e^{-x}(2 x) \\
& =-x^2 e^{-x}+2 x e^{-x}=x e^{-x}(-x+2) \\
\text { or } \quad \frac{d y}{d x} & =\frac{x(2-x)}{e^x}
\end{aligned}
$


Out of the intervals mentioned in the options (A), (B), (C) and (D), $\frac{d y}{d x}>0$ for all $x$ in interval $(0,2)$ of option (D).
$\therefore \quad y(=f(x))$ is strictly increasing and hence increasing in interval $(0,2)$ of option D.
Note. For a subjective solution of this question, proceed as in solution of Q. No. 6 (a), (b), (c).
Remark. Increasing (decreasing) function or monotonically increasing (or monotonically decreasing) function have the same meaning.

Solution: (i) Given: $f(x)=(2 x-1)^2+3$
We know that for all $x \in \mathbf{R},(2 x-1)^2 \geq 0$
⇒ Adding 3 to both sides, $(2 x-1)^2+320+3$
$\Rightarrow \quad f(x) \geq 3$
The minimum value of $f(x)$ is 3 and is obtained when $2 x-1=0, \quad\left[\because\right.$ Minimum value of $(2 x-1)^2$ is 0$]$ i.e., when $x=\frac{1}{2}$. There is no maximum value of $f(x)$.
[ ∵ Maximum value of $f(x)=(2 x-1)^2+3 \rightarrow \infty$ as $x \rightarrow \infty$ and hence does not exist].
(ii) Given: $f(x)=9 x^2+12 x+2$

Making coefficient of $x^2$ unity,

$
=9\left[x^2+\frac{12 x}{9}+\frac{2}{9}\right]=9\left[x^2+\frac{4 x}{3}+\frac{2}{9}\right]
$
Add and subtract $\left(\frac{1}{2} \text { coeff of } x\right)^2$

$
\begin{gathered}
=\left(\frac{1}{2} \times \frac{4}{3}\right)^2=\left(\frac{2}{3}\right)^2 \\
=9\left[x^2+\frac{4 x}{3}+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2+\frac{2}{9}\right]=9\left[\left(x+\frac{2}{3}\right)^2-\frac{4}{9}+\frac{2}{9}\right] \\
\text { or } f(x)=9\left(x+\frac{2}{3}\right)^2-4+2=9\left(x+\frac{2}{3}\right)^2-2
\end{gathered}
$


We know that for all $x \in R, 9\left(x+\frac{2}{3}\right)^2 \geq 0$
Adding -2 to both sides, $9\left(x+\frac{2}{3}\right)^2-2 \geq-2$
⇒ Using $(i), f(x) \geq-2$
∴ Minimum value of $f(x)$ is -2 and is obtained when

$
x+\frac{2}{3}=0 \text { i.e., when } x=-\frac{2}{3}
$


From (i), maximum value of $f(x) \rightarrow \infty$ as $x \rightarrow \infty$ (or as $x \rightarrow-\infty$ ) and hence does not exist.
(iii) Given: $f(x)=-(x-1)^2+10$

We know that for all $x \in \mathbf{R},(x-1)^2 \geq 0$
Multiplying by $-1,-(x-1)^2 \leq 0$
Adding 10 to both sides, $-(x-1)^2+10 \leq 10$
⇒ Using $(i), f(x) \leq 10$
∴ Maximum value of $f(x)$ is 10 and is obtained when $x-1 =0$ i.e., when $x=1$.
From $(i)$, minimum value of $f(x) \rightarrow-\infty$ as $x \rightarrow \infty$ (or as $x \rightarrow-\infty$ ) and hence does not exist.
(iv) Given: $g(x)=x^3+1$

As $x \rightarrow \infty, g(x) \rightarrow \infty$
As $x \rightarrow-\infty, g(x) \rightarrow-\infty$
∴ Maximum value of $g(x)$ does not exist and also minimum value of $g(x)$ does not exist.

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

We know that for all $x \in \mathbf{R},|x+2| \geq 0$
Adding -1 to both sides, $|x+2|-1 \geq-1$
⇒ Using $(i), f(x) \geq-1$
∴ Minimum value of $f(x)$ is -1 and is obtained when $x+2=0$ i.e, when $x=-2$.
From (i), maximum value of $f(x) \rightarrow \infty$ as $x \rightarrow \infty$ (or as $x \rightarrow-\infty$ ) and hence does not exist.
(ii) Given: $g(x)=-|x+1|+3$

We know that for all $x \in \mathrm{R},|x+1| \geq 0$
Multiplying by -1 to both sides

$
\Rightarrow-|x+1| \leq 0
$


Adding 3 to both sides,

$
\Rightarrow-|x+1|+3 \leq 3 \Rightarrow g(x) \leq 3
$

∴ The maximum value of $g(x)$ is 3 and is obtained when $|x+1|=0$, i.e., when $x+1=0$ i.e., when $x=-1$. There is no minimum value of $g(x)$. [Because minimum value of $g(x)=-|x+1|+3 \rightarrow-\infty$ as $x \rightarrow \pm \infty$ and hence does not exist].
(iii) Given: $h(x)=\sin (2 x)+5$

We know that for all $x \in \mathbf{R},-1 \leq \sin 2 x \leq 1$
Adding 5 to all sides, $-1+5 \leq \sin 2 x+5 \leq 1+5$

$
\Rightarrow \quad 4 \leq h(x) \leq 6
$

∴ Minimum value of $h(x)$ is 4 and maximum value is 6 .
(iv) Given: $\quad f(x)=|\sin 4 x+3|$

We know that for all $x \in \mathrm{R},-1 \leq \sin 4 x \leq 1$
Adding 3 throughout,

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

$[\because \sin 4 x+3 \geq 2$ and hence $>0, \therefore|\sin 4 x+3|=\sin 4 x+3]$
∴ The minimum value of $f(x)$ is 2 and the maximum value of $f(x)$ is 4.
(v) Given: $h(x)=x+1, x \in(-1,1)$

Givens $x \in(-1,1) \Rightarrow-1Adding 1 to all sides, $1-1
$
\text { i.e., } 0$

∴ Neither minimum value nor maximum value of $h(x)$ exists.
(U) Equality sign is absent at both ends of inequality (ii). We know from (ii) that minimum value of $h(x)$ is $>0$ and maximum value is < 2 but what exactly they are can't be said).

Solution: (i) Given: $f(x)=x^2$

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


Putting $f^{\prime}(x)=0$ to locate the critical points, we have $2 x=0$

$
\text { or } \quad x=\frac{0}{2}=0
$

(Critical point)
Applying the second derivative test.
When $\quad x=0, f^{\prime \prime}(x)=2$ (positive)

$
\begin{aligned}
& \therefore \quad \begin{aligned}
x & =0 \text { is a local minimum point and local minimum } \\
\text { value }=f(0) & =0^2 \\
& =0
\end{aligned} \\
& \text { [From (i)] }
\end{aligned}
$


Therefore, local minima at $x=0$ and local minimum value $=0$.
(ii)

$
\begin{aligned}
& \text { Given: } g(x)=x^3-3 x \\
& \therefore \quad g^{\prime}(x)=3 x^2-3 \text { and } g^{\prime \prime}(x)=6 x
\end{aligned}
$


Putting $g^{\prime}(x)=0$ to locate the critical points, we have

$
3 x^2-3=0 \text { or } 3\left(x^2-1\right)=0 \text { or } 3(x+1)(x-1)=0
$


But $3=0$. Therefore, either $x+1=0$ or $x-1=0$

$
\text { i.e., } x=-1 \text { or } x=1 \text {. }
$

(Critical points)
Applying the second derivative test.
At $x=-1, g^{\prime \prime}(x)=6 x=6(-1)=-6$
(Negative)
$\therefore x=-1$ is a local maximum point and local maximum
value $=g(-1)=(-1)^3-3(-1) \quad$ [From (i)] $=-1+3=2$.
At $x=1, g^{\prime \prime}(x)=6 x=6(1)=6$ (positive)
$\therefore x=1$ is a local minimum point and local minimum value

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


Therefore, Local maximum at $x=-1$ and local maximum value $=2$. Local minimum at $x=1$ and local minimum value $=-2$.
(iii) Given: $h(x)=\sin x+\cos x\left(0$\therefore \quad h^{\prime}(x)=\cos x-\sin x$ and $h^{\prime \prime}(x)=-\sin x-\cos x$
Putting $h^{\prime}(x)=0$ to locate the critical points,
we have cos $x-\sin x=0 \Rightarrow-\sin x=-\cos x$
Dividing by $-\cos x, \frac{\sin x}{\cos x}=1$ or $\tan x=1$
this is positive. Therefore, $x$ can have values in both Ist and IIIrd quadrants.
Here, $x$ is only in Ist quadrant

$
\left[\because 0$
$
\begin{aligned}
& \therefore \quad \tan x=1=\tan \frac{\pi}{4} \\
& \therefore \quad x=\frac{\pi}{4} \text { (only critical point) } \\
& \text { At } \quad x=\frac{\pi}{4}, h^{\prime \prime}(x)=-\sin x-\cos x \\
& =-\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\
& =-\frac{2}{\sqrt{2}}=-\sqrt{2} \text { is negative. }
\end{aligned}
$

$\therefore x=\frac{\pi}{4}$ is a local maximum point and local maximum value

$
\begin{aligned}
& =h\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}
\end{aligned}
$

(From (i))
∴ Local maximum at $x=\frac{\pi}{4}$, and local maximum value $=\sqrt{2}$.
(iv) Given: $f(x)=\sin x-\cos x$
..(i) $(0$\therefore \quad f^{\prime}(x)=\cos x+\sin x$
and $\quad f^{\prime \prime}(x)=-\sin x+\cos x$
Putting $f^{\prime}(x)=0$ to locate the critical points, we have

$
\cos x+\sin x=0 \Rightarrow \sin x=-\cos x
$


Dividing by $\cos x, \frac{\sin x}{\cos x}=-1$
$\Rightarrow \tan x=-1$ is negative.
Therefore, $x$ is in both second and fourth quadrants.

$
\begin{aligned}
\therefore \quad \tan x & =-1=-\tan \frac{\pi}{4} \\
& =\tan \left(\pi-\frac{\pi}{4}\right) \text { or } \tan \left(2 \pi-\frac{\pi}{4}\right) \\
\Rightarrow \quad \tan x & =\tan \frac{3 \pi}{4} \text { or } \tan \frac{7 \pi}{4} \\
\therefore \quad x & =\frac{3 \pi}{4} \text { and } x=\frac{7 \pi}{4}
\end{aligned}
$


Applying the second derivative test.
At $x=\frac{3 \pi}{4}, f^{\prime \prime}(x)=-\sin x+\cos x=-\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}$
$=-\sin \frac{4 \pi-\pi}{4}+\cos \frac{4 \pi-\pi}{4}=-\sin \left(\pi-\frac{\pi}{4}\right)+\cos \left(\pi-\frac{\pi}{4}\right)$
$
\begin{aligned}
& =-\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\
& =-\frac{2}{\sqrt{2}}=-\sqrt{2} \text { (negative) }
\end{aligned}
$

$\therefore x=\frac{3 \pi}{4}$ is a local maximum point and local maximum value

$
\begin{aligned}
& =f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{4}-\cos \frac{3 \pi}{4} \\
& =\sin \left(\pi-\frac{\pi}{4}\right)-\cos \left(\pi-\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}
\end{aligned}
$


At $x=\frac{7 \pi}{4}$,

$
\begin{aligned}
f^{\prime \prime}(x) & =-\sin x+\cos x=-\sin \frac{7 \pi}{4}+\cos \frac{7 \pi}{4} \\
& =-\sin \left(\frac{8 \pi-\pi}{4}\right)+\cos \left(\frac{8 \pi-\pi}{4}\right) \\
& =-\sin \left(2 \pi-\frac{\pi}{4}\right)+\cos \left(2 \pi-\frac{\pi}{4}\right) \\
& =\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \quad \text { (Ponitive) }
\end{aligned}
$

$\therefore \quad x=\frac{7 \pi}{4}$ is a local minimum point and local minimum value

$
\begin{aligned}
& =f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4} \text { (From (i)) } \\
& =\sin \left(2 \pi-\frac{\pi}{4}\right)-\cos \left(2 \pi-\frac{\pi}{4}\right)=-\sin \frac{\pi}{4}-\cos \frac{\pi}{4} \\
& =\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}
\end{aligned}
$


Therefore, there is a local maximum at $x=\frac{3 \pi}{4}$ and local maximum value $=\sqrt{2}$. Local minima at $x=\frac{7 \pi}{4}$ and local minimum value $=-\sqrt{2}$.
(v) Given: $f(x)=x^3-6 x^2+9 x+15$

$
\therefore \quad f^{\prime}(x)=3 x^2-12 x+9 \quad \text { and } \quad f^{\prime \prime}(x)=6 x-12
$


Putting $f^{\prime}(x)=0$ to locate the critical points, we have

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


Dividing by
or

$
\begin{aligned}
3, x^2-4 x+3 & =0 \\
x^2-x-3 x+3 & =0 \quad \text { or } \quad(x-1)(x-3)=0
\end{aligned}
$
∴ Either $x-1=0$ or $x-3=0$
ie.
$x=1$ or $x=3$. (Critical points)
Applying the second derivative test.
When $x=1, f^{\prime \prime}(x)=6 x-12$

$
=6-12=-6
$

(negative)
$\therefore x=1$ is a local maximum point and local maximum value

$
=f(1)=(1)^3-6(1)^2+9(1)+15=1-6+9+15=19
$


When $x=3 \quad f^{\prime \prime}(x)=6 x-12=6(3)-12=6 \quad$ (positive)
$\therefore x=3$ is a local minimum point and local minimum value

$
\begin{aligned}
& =f(3)=(3)^2-6(3)^2+9(3)+15 \\
& =27-54+27+15=15
\end{aligned}
$


Therefore, there is a local maximum at $x=1$ and local maximum value $=19$.
Local minima at $x=3$ and local minimum value $=15$.
(vi)

$
\begin{aligned}
& \text { Given: } g(x)=\frac{x}{2}+\frac{2}{x}, x>0 \\
& \therefore \quad g^{\prime}(x)=\frac{1}{2}-\frac{2}{x^2}=\frac{x^2-4}{2 x^2}=\frac{(x+2)(x-2)}{2 x^2}
\end{aligned}
$


For critical points, putting $g^{\prime}(x)=0$

$
\begin{aligned}
& \Rightarrow \quad \frac{(x+2)(x-2)}{2 x^2}=0 \\
& \Rightarrow(x+2)(x-2)=0 \Rightarrow x=-2,2
\end{aligned}
$


But $x>0$ (given) $\therefore x=-2$ is rejected.
Hence $x=2$ is the only critical point.
Applying the first derivative test
When $x$ is slightly $<2$, let $x=1.9$
From (i), $\quad g^{\prime}(1.9)=\frac{(1.9+2)(1.9-2)}{21.9)^2}=\frac{(+\mathrm{ve})(-\mathrm{ve})}{(+\mathrm{ve})}=-\mathrm{ve}$
When $x$ is slightly $>2$, let $x=2.1$
$g^{\prime}(2.1)=\frac{(2.1+2)(2.1-2)}{22.1)^2}=\frac{(+\mathrm{ve})(+\mathrm{ve})}{(+\mathrm{ve})}=+\mathrm{ve}$
Thus, $g^{\prime}(x)$ changes sign from negative to positive as $x$ increases through 2.
$\therefore \quad x=2$ is a local minimum point and local minimum value

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


Note. Second derivative test, g “(x)

$
=\frac{d}{d x}\left(\frac{1}{2}-\frac{2}{x^2}\right)=\frac{4}{x^1} \quad \therefore \quad g^{\prime \prime}(2)=\frac{4}{8}=\frac{1}{2}>0
$

$\Rightarrow g(x)$ has local minimum value at $x=2$ and local minimum value $=g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2$.
(vii) Given: $h(x)=\frac{1}{x^2+2}=\left(x^2+2\right)^{-1}$
$
\text { and } \quad \begin{aligned}
h^{\prime}(x) & =(-1)\left(x^2+2\right)^{-2}(2 x)=-\frac{2 x}{\left(x^2+2\right)^2} \\
& =-\left[\frac{\left(x^2+2\right)^2 \cdot 2-2 x \cdot 2\left(x^2+2\right) 2 x}{\left(x^2+2\right)^4}\right] \\
& =\frac{-2\left(x^2+2\right)\left[x^2+2-4 x^2\right)}{\left(x^2+2\right)^4}=\frac{-2\left(2-3 x^2\right)}{\left(x^2+2\right)^3}
\end{aligned}
$


Putting $h^{\prime}(x)=0$ to locate the critical points, we have

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


Applying the second derivative test.
At $x=0, \quad h^{\prime \prime}(x)=\frac{-2\left(2-3 x^2\right)}{\left(x^2+2\right)^3}=\frac{-2(2-0)}{(0+2)^3}=\frac{-4}{8}$

$
=\frac{-1}{2} \quad \text { (Negative) }
$

$\therefore \quad x=0$ is a local maximum point and local maximum value

$
=h(0)=\frac{1}{0+2}=\frac{1}{2} \quad(\text { From }(i))
$


Therefore, there is a local maximum at $x=0$ and local maximum value $=\frac{1}{2}$.
(viii) Given: $f(x)=x \sqrt{1-x}, x \leq 1$

$
\begin{array}{ll}
\therefore & f^{\prime}(x)=x \frac{1}{2}(1-x)^{-12} \frac{d}{d x}(1-x)+\sqrt{1-x} \cdot 1 \\
\therefore & f^{\prime}(x)=x+\frac{1}{2 \sqrt{1-x}}(-1)+\sqrt{1-x} \cdot 1 \\
= & \frac{-x}{2 \sqrt{1-x}}+\sqrt{1-x}=\frac{-x+2(1-x)}{2 \sqrt{1-x}}=\frac{2-3 x}{2 \sqrt{1-x}}
\end{array}
$


For critical points, putting $f^{\prime}(x)=0$

$
\Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0 \Rightarrow 2-3 x=0 \quad \therefore x=\frac{2}{3}
$


Applying the first derivative test.
When $x$ is slightly $<\frac{2}{3}$, let $x=0.6$
From $(i), f^{\prime}(0.6)=\frac{2-3(0.6)}{2 \sqrt{1-0.6}}=\frac{2-1.8}{2 \sqrt{0.4}}=\frac{0.2}{2 \sqrt{0.4}}>0$
When $x$ is slightly $>\frac{2}{3}$, let $x=0.7$
From $(i), f^{\prime}(0.7)=\frac{2-3 .(0.7)}{2 \sqrt{1-0.7}}=\frac{2-2.1}{2 \sqrt{0.3}}=\frac{-0.1}{2 \sqrt{0.3}}<0$
Thus, $f^{\prime}(x)$ changes sign from positive to negative as $x$ increases through $\frac{2}{3}$.
$\therefore x=\frac{2}{3}$ is a local maximum point and local maximum value

$
=f\left(\frac{2}{3}\right)=x \sqrt{1-x}=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \times \frac{1}{\sqrt{3}}=\frac{2 \sqrt{3}}{9} .
$


Note. Apply second derivative test.
Differentiating both sides of ( $i$ ) with respect to $x$,

$
\begin{aligned}
f^{\prime \prime}(x) & =\frac{1}{2} \cdot \frac{\sqrt{1-x} \cdot(-3)-(2-3 x) \cdot \frac{1}{2 \sqrt{1-x}}(-1)}{1-x} \\
f^{\prime \prime}\left(\frac{2}{3}\right) & =\frac{1}{2} \cdot \frac{\left(\frac{1}{\sqrt{3}}\right)(-3)-0}{\frac{1}{3}}=\frac{-9}{2 \sqrt{3}}=-\frac{3 \sqrt{3}}{2}<0
\end{aligned}
$

$\therefore f(x)$ has local maximum value at $x=\frac{2}{3}$.

Solution: (i) Given: $f(x)=e^x$

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


Putting $f^{\prime}(x)=0$ to locate the critical points, we have $e^2=0$. But this gives no real value of $x$.
[ ∵ $e^x>0$ for all real $x$ ]
∴ No critical point.
Hence $f(x)$ does not have maxima or minima.
(ii) Given: $g(x)=\log x$

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


Putting $g^{\prime}(x)=0$ to locate the critical points, we have

$
\frac{1}{x}=0 \Rightarrow 1=0
$


But this is impossible.
∴ No critical point.
Hence $f(x)$ does not have maxima or minima.
(iii) $\quad h(x)=x^3+x^2+x+1$
$h^{\prime}(x)=3 x^2+2 x+1$
Putting $h^{\prime}(x)=0$, we have $3 x^2+2 x+1=0$
$\therefore x=\frac{-2 \pm \sqrt{4-12}}{6}=\frac{-2 \pm \sqrt{-8}}{6}$
$
=\frac{-2 \pm 2 \sqrt{2} i}{6}=\frac{-1 \pm \sqrt{2} i}{3}
$


These values of $x$ are imaginary.
$\therefore h(x)$ does not have maxima or minima.

Solution: (i) Given: $f(x)=x^3, x \in[-2,2]$

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


Putting $f^{\prime}(x)=0$ to locate the critical points, we have

$
3 x^2=0 \Rightarrow x^2=0 \Rightarrow x=0 \in[-2,2]
$


To find absolute maximum and absolute minimum value of the function, we need to find values of $f(x)$ at (a) critical point(s) and (b) at end points of the given closed interval [-2, 2].
Substituting $x=0$ in (i), $f(0)=0$
Substituting $x=-2$ in (i), $f(-2)=(-2)^3=-8$
Substituting $x=2$ in $(i), f(2)=(2)^3=8$
Out of these three values of $f(x)$; absolute minimum value $=-8$ and absolute maximum value is 8 .
Remark. Ahsolute maximum and absolute minimum values of $f(x)$ are also called maximum and minimum values of $f(x)$.
(ii) Given: $f(x)=\sin x+\cos x, x \in[0, \pi]$

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


Putting $f^{\prime}(x)=0$ to locate the critical points, we have

$
\cos x-\sin x=0 \Rightarrow-\sin x=-\cos x
$


Dividing by – cos $x, \tan x=1$ is positive.
$\therefore x$ is in I and III quadrants.
But $x \in[0, \pi]$ (given) can’t be in third quadrant.
$\therefore x$ is in lst quadrant.
Therefore, $\tan x=1=\tan \frac{\pi}{4}$

$
\therefore \quad x=\frac{\pi}{4}
$


Now let us find values of $f(x)$ at critical point $x=\frac{\pi}{4}$ and at end points $x=0$ and $x=\pi$ of given closed interval $[0, \pi]$.

Putting

$
x=\frac{\pi}{4} \text { in }(i), f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}
$
$
=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}
$


Putting

$
\begin{aligned}
x & =0 \text { in }(i), f(0)=\sin 0+\cos 0 \\
& =0+1=1
\end{aligned}
$


Putting

$
\begin{aligned}
x & =\pi \text { in }(i), f(\pi)=\sin \pi+\cos \pi \\
& =0-1=-1
\end{aligned}
$

$\left[\because \sin \pi=\sin 180^{\circ}=\sin \left(180^{\prime \prime}-0^{\prime}\right)=\sin 0^{\prime}=0\right.$
and $\cos \pi=\cos 180^{\circ}=\cos \left(180^{\circ}-0^{\circ}\right)=-\cos 0^{\circ}=-1$ ]
∴ Absolute minimum value $=-1$ and absolute maximum value $=\sqrt{2}$.
(iii) Given: $f(x)=4 x-\frac{1}{2} x^2, x \in\left[-2, \frac{9}{2}\right]$

$
\therefore \quad f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x .
$


Putting $f^{\prime}(x)=0$ to find the critical points,
we have $4-x=0$

$
\text { i.e., }-x=-4 \text { i.e., } x=4 \in\left[-2, \frac{9}{2}\right]
$


Now let us find values of $f(x)$ at critical point $x=4$ and at end points $x=-2$ and $x=\frac{9}{2}$ of the given closed interval $\left[-2, \frac{9}{2}\right]$.
Substituting $x=4$ in $(i), f(4)=16-\frac{1}{2}(16)=16-8=8$
Substituting $x=-2$ in $(i), f(-2)=4(-2)-\frac{1}{2}(4)=-8-2=-10$

$
\begin{aligned}
\text { Putting } x & =\frac{9}{2} \text { in }(i), 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}=\frac{144-81}{8}=\frac{63}{8}
\end{aligned}
$

∴ Absolute minimum value is -10 and absolute maximum value is 8 .
(iv)

Given: $f(x)=(x-1)^2+3, x \in[-3,1]$

$
\therefore \quad f^{\prime}(x)=2(x-1) \frac{d}{d x}(x-1)+0=2(x-1)
$


Putting $f^{\prime}(x)=0$ to find the critical points, we have $2(x-1)=0$

$
\Rightarrow x-1=\frac{0}{2}=0 \Rightarrow x=1 \in[-3,1]
$


Now let us find values of $f(x)$ at critical point $x=1$ and at the end point $x=-3$ of the given closed
interval $[-3,1](\because$ the other end point $x=1$ has already come out to be a critical point)

$
\begin{aligned}
\text { Putting } x & =-3 \text { in }(i), f(-3)=(-3-1)^2+3 \\
& =(-4)^2+3=16+3=19
\end{aligned}
$


Substituting $x=1$ in (i), $f(1)=(1-1)^2+3=0+3=3$
∴ Absolute minimum value is 3 and absolute maximum value is 19.
Note. To find absolute maximum or absolute minimum value of a function $f(x)$ when only one critical point comes out to be there and no closed interval is given, then we get only one value of $f(x)$ at such points and out of one value of $f^{\prime}(x)$ we can’t make a decision about maximum value or minimum value. In such problems, we have to depend upon local minimum value and local maximum value.

Solution: Given: Profit function is $p(x)=41+24 x-18 x^2$

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

(The logic of finding $p^{\prime \prime}(x)$ is given in the note above)
Putting $p^{\prime}(x)=0$ to locate the critical points, we have

$
\begin{array}{ll}
& 24-36 x=0 \Rightarrow-36 x=-24 \\
\Rightarrow & x=\frac{24}{36}=\frac{2}{3}
\end{array}
$


At $x=\frac{2}{3}, p^{\prime \prime}(x)=-36 \quad$ (Negative)
$\therefore p(x)$ has a local maximum value and hence maximum value at $x=\frac{2}{3}$.
Substituting $x=\frac{2}{3}$ in (i),

$
\text { Maximum profit }=41+24\left(\frac{2}{3}\right)-18\left(\frac{4}{9}\right)=41+16-8=49
$


Remark. The original statement in N.C.E.R.T. book $p(x)=41-24 x-18 x^2$ is wrong, because with this $p(x)$, critical point comes out to be $x=-\frac{2}{3}$ which being the number of units produced or sold can’t be negative.

Solution: Let $f(x)=3 x^4-8 x^3+12 x^2-48 x+25$ on $[0,3]$

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


Putting $f^{\prime}(x)=0$ to find the critical points, we have

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


Dividing every term by $12, x^3-2 x^2+2 x-4=0$
$
\begin{array}{llc}
\text { or } \quad x^2(x-2)+2(x-2)=0 & \text { or } \quad(x-2)\left(x^2+2\right)=0 \\
\therefore \quad \text { Either } x-2=0 & \text { or } \quad x^2+2=0 \\
\Rightarrow \quad x=2 & \Rightarrow \quad x^2=-2 \\
& \Rightarrow \quad x= \pm \sqrt{-2}
\end{array}
$


These values of $x$ are imaginary and hence rejected.
Critical point $x=2 \in[0,3]$.
Now let us find values of $f(x)$ at critical point $x=2$ and end points $x=0$ and $x=3$ of closed interval [ 0,3 ].
Substituting $x=2$ in $(i), f(2)=3(16)-8(8)+12(4)-48(2)+25$

$
=48-64+48-96+25=-39
$


Substituting $x=0$ in (i), $f(0)=25$

$
\text { Putting } x=3 \text { in }(i), \begin{aligned}
f(3) & =3(81)-8(27)+12(9)-48(3)+25 \\
& =243-216+108-144+25 \\
& =27-36+25=16
\end{aligned}
$

∴ Minimum (absolute) value is – 39 (at $x=2$ ) and maximum (absolute) value is 25 (at $x=0$ ).

Solution: Let $f(x)=\sin 2 x$, then $f^{\prime}(x)=2 \cos 2 x$
For maxima or minima, $f^{\prime}(x)=0$

$
\Rightarrow \cos 2 x=0 \quad \therefore 2 x=(2 n+1) \frac{\pi}{2} \text { or } x=(2 n+1) \frac{\pi}{4}
$


Putting $n=0,1,2,3 ; x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \in[0,2 \pi]$
Now let us find values of $f^{\prime}(x)$ at these critical points.
Now $f(x)=\sin 2 x$

$
\begin{aligned}
\therefore\left[f(2 n+1) \frac{\pi}{4}\right] & =\sin (2 n+1) \frac{\pi}{2}=\sin \left(n \pi+\frac{\pi}{2}\right)=(-1)^n \sin \frac{\pi}{2} \\
& =(-1)^n \quad\left[\because \sin (n \pi+\alpha)=(-1)^n \sin \alpha ; n \in 1\right]
\end{aligned}
$


Putting $n=0,1,2,3$,

$
\begin{array}{ll}
f\left(\frac{\pi}{4}\right)=(-1)^0=1, & f\left(\frac{3 \pi}{4}\right)=(-1)^1=-1 \\
f\left(\frac{5 \pi}{4}\right)=(-1)^2=1, & f\left(\frac{7 \pi}{4}\right)=(-1)^3=-1
\end{array}
$


Also let us find $f(x)$ at the end-points $x=0$ and $x=2 \pi$ of [0, 2π].
$f(0)=\sin 0=0, f(2 \pi)=\sin 4 \pi=0 \quad[\because \sin n \pi=0$ where $n \in \mathrm{I}]$
$\therefore f(x)$ attains its maximum value 1 at $x=\frac{\pi}{4}$ and $x=\frac{5 \pi}{4}$.
Hence, the required points are $\left(\frac{\pi}{4}, 1\right)$ and $\left(\frac{5 \pi}{4}, 1\right)$.

Solution: Let $f(x)=\sin x+\cos x$

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


Putting $f^{\prime}(x)=0$ to find the critical points, we have

$
\cos x-\sin x=0 \Rightarrow-\sin x=-\cos x
$


Dividing by $-\cos x, \frac{\sin x}{\cos x}=1 \Rightarrow \tan x=1=\tan \frac{\pi}{4}$

$
\therefore \quad x=n \pi+\frac{\pi}{4} \text { where } n \in Z \text { (critical points) }
$

(: If $\tan \theta=\tan \alpha$, then $\theta=n \pi+\alpha$ where $n \in \mathbb{Z}$ )
Substituting $x=n \pi+\frac{\pi}{4}$ in $(i)$,

$
\begin{aligned}
f\left(n \pi+\frac{\pi}{4}\right) & =\sin \left(n \pi+\frac{\pi}{4}\right)+\cos \left(n \pi+\frac{\pi}{4}\right) \\
& =(-1)^n \sin \frac{\pi}{4}+(-1)^n \cos \frac{\pi}{4}
\end{aligned}
$

$\left[\because \sin (n \pi+\alpha)=(-1)^n \sin \alpha\right.$ and cos $\left.(n \pi+\alpha)=(-1)^n \cos \alpha\right]$

$
\begin{aligned}
& =(-1)^n \frac{1}{\sqrt{2}}+(-1)^n \frac{1}{\sqrt{2}}=2(-1)^n \frac{1}{\sqrt{2}} \quad(\because t+t=2 t) \\
& =(-1)^n \sqrt{2}
\end{aligned}
$


If $n$ is even; then $(-1)^n=1$ and then $f\left(n \pi+\frac{\pi}{4}\right)=\sqrt{2}$
If $n$ is odd, then $(-1)^n=-1$; and then $f\left(n \pi+\frac{\pi}{4}\right)=-\sqrt{2}$
∴ Maximum value of $f(x)$ is $\sqrt{2}$.
Note. Minimum value of $f(x)$ is $-\sqrt{2}$.

Solution: Let

$
f(x)=2 x^3-24 x+107
$


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


Let us put $f^{\prime}(x)=0$ to find the critical points.
i.e., $\quad 6 x^2-24=0 \Rightarrow 6 x^2=24$
Dividing by $6, x^2=24 \Rightarrow x= \pm 2$
$\therefore x=-2$ and $x=2$ are two critical points.
(a) Let us find maximum value of $f(x)$ given by (i) in the interval [1, 3].
From (ii), critical point $x=-2$ e $[1,3]$.
So let us find values of $f(x)$ at critical point $x=2$ and at end points $x=1$ and $x=3$ of closed interval [1, 3].

$
\text { Putting } x=2 \text { in }(i), \begin{aligned}
f(2) & =2 / 8)-24(2)+107 \\
& =16-48+107=123-48=75
\end{aligned}
$
Substituting $x=1$ in (i), $f(1)=2-24+107=109-24=85$
Substituting $x=3$ in $(f), f(3)=2(27)-24(3)+107$

$
=54-72+107=161-72=89
$

∴ Maximum value of $f(x)$ given by (i) in [1,3] is 89 (at $x=3$ ).
(b) Let us find maximum value of $f(x)$ given by (i) in the interval [- 3, – 1].
From (ii), critical point $x=2$ ig $[-3,-1]$
So let us find values of $f(x)$ at critical point $x=-2$ and at end points $x=-3$ and $x=-1$ of closed interval [-3, -1]
Putting

$
\begin{aligned}
x=-2 \text { in }(i), f(-2) & =2(-8)-24(-2)+107 \\
& =-16+48+107=139
\end{aligned}
$


Putting

$
\begin{aligned}
x=-3 \mathrm{in}(i), f(-3) & =2(-27)-24(-3)+107 \\
& =-54+72+107=125
\end{aligned}
$


Putting

$
\begin{aligned}
x=-1 \text { in }(i), f(-1) & =2(-1)-24(-1)+107 \\
& =-2+24+107=129
\end{aligned}
$

∴ Maximum value of $f(x)$ is 139 (at $x=-2$ ).

Solution: Let $f(x)=x^4-62 x^2+a x+9$

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


Because $f(x)$ attains its maximum value at $x=1$ in the interval $[0,2]$, therefore, $f^{\prime}(1)=0$.
Substituting $x=1$ in $(i), f^{\prime}(1)=4-124+a=0$
or $a-120=0$ or $a=120$.

Solution: Let $f(x)=x+\sin 2 x$

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


Putting $f^{\prime}(x)=0$ to find the critical points, we have

$
\begin{array}{rlrl}
& & 1+2 \cos 2 x & =0 \Rightarrow \\
\Rightarrow & \cos 2 x & =\frac{-1}{2} & =-\cos \frac{\pi}{3} \quad \\
\therefore & & =\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} \\
& \therefore & 2 x & =2 n \pi \\
& & \frac{2 \pi}{3} \text { where } n \in Z \\
& \because I f \cos \theta & =\cos \alpha, \text { then } \theta=2 n \pi \pm \alpha \text { where } n \in Z]
\end{array}
$


Dividing by $2, x=n \pi \pm \frac{\pi}{3}$ where $n \in Z$
For $n=0, x= \pm \frac{\pi}{3}$. But $x=-\frac{\pi}{3} \in[0,2 \pi] \quad \therefore \quad x=\frac{\pi}{3}$
For $n=1, x=\pi \pm \frac{\pi}{3}=\pi+\frac{\pi}{3}$ and $\pi-\frac{\pi}{3}$
i.e., $\frac{4 \pi}{3}$ and $\frac{2 \pi}{3}$ and both belong to $[0,2 \pi]$
For $n=2, x=2 \pi \pm \frac{\pi}{3}$. But $x=2 \pi+\frac{\pi}{3}>2 \pi$ and hence $E[0,2 \pi]$

$
\therefore \quad x=2 \pi-\frac{\pi}{3}=\frac{5 \pi}{3}
$


It can be easily observed that for all other $n \in \mathrm{Z}$,

$
x=n \pi \pm \frac{\pi}{3} \neq[0,2 \pi]
$

∴ The only critical points of $f(x)$ given by ( $i$ ) which belong to given closed interval [ $0,2 \pi$ ] are

$
x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}
$


Now let us find values of $f(x)$ at these four critical points and at the end points $x=0$ and $x=2 \pi$ of $[0,2 \pi]$.

Substituting $x=\frac{\pi}{3}$ in (i), $f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\sin \frac{2 \pi}{3}$

$
\begin{aligned}
& =\frac{\pi}{3}+\frac{\sqrt{3}}{2}=1.05+0.87 \\
& =1.92 \text { nearly }
\end{aligned}
$


$
\left(\because \frac{\pi}{3}=\frac{\frac{22}{7}}{3}=\frac{22}{21}=1.05 \text { and } \frac{\sqrt{3}}{2}=\frac{1.732}{2}=0.866=0.87\right)
$


$
\text { and }\left(\because \sin \frac{2 \pi}{3}=\sin \frac{3 \pi-\pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right)
$


Substituting $x=\frac{2 \pi}{3}$ in (i), $f\left(\frac{2 \pi}{3}\right)=\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}$

$
\begin{aligned}
& =2 \pi-\frac{\sqrt{3}}{2}=2.10-0.87=1.23 \text { nearly } \\
& \left(\because \sin \frac{4 \pi}{3}=\sin \frac{3 \pi+\pi}{3}=\sin \left(\pi+\frac{\pi}{3}\right)=-\sin \frac{\pi}{3}=\frac{-\sqrt{3}}{2}\right)
\end{aligned}
$


Substituting $x=\frac{4 \pi}{3}$ in $(i), f\left(\frac{4 \pi}{3}\right)=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}$

$
\begin{gathered}
=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}=4(1.05)+0.87=4.20+0.87=5.07 \\
\left(\because \sin \frac{8 \pi}{3}=\sin \left(\frac{6 \pi+2 \pi}{3}\right)=\sin \left(2 \pi+\frac{2 \pi}{3}\right)=\sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}\right)
\end{gathered}
$


$
\begin{aligned}
\text { Putting } x & =\frac{5 \pi}{3} \text { in (i), } f\left(\frac{5 \pi}{3}\right) \\
& =\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}
\end{aligned}
$
$
\begin{aligned}
& =5(1.05)-0.87=5.25-0.87=4.38 \text { nearly } \\
& {\left[\because \sin \frac{10 \pi}{3}=\sin \frac{6 \pi+4 \pi}{3}=\sin \left(2 \pi+\frac{4 \pi}{3}\right)=\sin \frac{4 \pi}{3}=\frac{-\sqrt{3}}{2}\right]}
\end{aligned}
$


Substituting $x=0$ in (i), $f(0)=0+\sin 0=0$
Substituting $x=2 \pi$ in (i), $f(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi$

$
=23.14)=6.28 \text { nearly }(\% \sin n \pi=0 \text { for every integer } n)
$

∴ Maximum value $=2 \pi$ (at $x=2 \pi$ )
and minimum value $=0$ (at $x=0$ ).

Solution: Let the two numbers be $x$ and $y$.
Their sum $=24$ (given) $\Rightarrow x+y=24$

$
\therefore \quad y=24-x
$


Let $z$ denote their product i.e., product of $x$ and $y$
i.e., $\quad z=x y$
Substituting $y=24-x$ from (i),

$
z=x(24-x)=24 x-x^2
$


Now $z$ is a function of $x$ only.

$
\therefore \quad \frac{d z}{d x}=24-2 x \text { and } \frac{d^2 z}{d x^2}=-2
$


Putting $\frac{d z}{d x}=0$ to find the critical points, we have
$24-2 x=0$ i.e., $-2 x=-24$. Therefore, $x=12$.

$
\begin{aligned}
\text { At } x & =12, \frac{d^2 z}{d x^2}=-2 \text { (negative) } \\
\therefore \quad x & =12 \text { is a point of (local) maxima. }
\end{aligned}
$

(See Note at the end of solution of Q. No. 5)
$\therefore z$ is maximum at $x=12$.
Substituting $x=12$ in (i), $y=24-12=12$
∴ The two required numbers are 12 and 12.

Solution: Here $x+y=60, x>0, y>0$
-(i) (Given condition)
Let $\quad \mathbf{P}=x y^3$
…(ii) (To be maximised)
To express $\mathbf{P}$ in terms of one independent variable, (here better $y$, because power of $y$ is larger in the value of $\mathbf{P}$ ), we have from ( $i$ )

$
x=60-y,
$


Substituting $x=60-y$ in (ii), $\mathrm{P}=(60-y) y^2=60 y^2-y^4$

$
\therefore \quad \frac{d \mathrm{P}}{d y}=180 y^2-4 y^3=4 y^2(45-y)
$


To find maxima or minima, Putting $\frac{d \mathrm{P}}{d y}=0$

$
\Rightarrow 4 y^2(45-y)=0 \quad \Rightarrow y=0,45
$
Rejecting $y=0$ because $y>0 \quad \therefore y=45$
When $y$ is slightly $<45$, from (iii), $\frac{d \mathrm{P}}{d y}=(+\mathrm{ve})(+\mathrm{ve})=+\mathrm{ve}$
When $y$ is slightly $>45$, from (iii), $\frac{d P}{d y}=(+v e)(-v e)=-v e$
Thus, $\frac{d \mathrm{P}}{d y}$ changes sign from + ve to -ve as $y$ increases through 45 .
$\therefore P$ is maximum when $y=45$.
Hence, $x y^2$ is maximum when $x=60-45=15$
and

$
y=45 .
$

Solution: Given: $x+y=35 \Rightarrow y=35-x$

Let $\quad z=x^2 y^5$
Substituting $y=35-x$ from $(i), z=x^2(35-x)^5$
Now $z$ is a function of $x$ alone.

$
\begin{aligned}
& \therefore \quad \frac{d z}{d x}=x^2 \cdot 5(35-x)^4(-1)+(35-x)^5 2 x \\
& \text { or } \quad \frac{d z}{d x}=x(35-x)^4[-5 x+(35-x) 2] \\
& \text { or } \quad \frac{d z}{d x}=x(35-x)^4(-5 x+70-2 x)=x(35-x)^4(70-7 x) \\
& \text { or } \quad \frac{d z}{d x}=7 x(35-x)^4(10-x)
\end{aligned}
$


Putting $\frac{d z}{d x}=0$ to find the critical points, we have

$
7 x(35-x)^4(10-x)=0
$


But $7 \neq 0 \quad \therefore$ Either $x=0$ or $35-x=0$ or $10-x=0$
i.e., $x=0$ or $x=35$ or $x=10$.
Now $x=0$ is rejected because $x$ is positive number (given).
Also, $x=35$ is rejected because for $x=35$, from $(i)$
$y=35-35=0$; but $y$ is given to be positive.
$\therefore \quad x=10$ is the only admissible critical point.
Applying the first derivative test because finding $\frac{d^2 z}{d x^2}$ is tedious as you and we think so.
We know that $(35-x)^4$ is never negative because index 4 is even. When $x$ is slightly $<10$ (say $x=9.8)$; from (ii),

$
\frac{d z}{d x}=(+)(+)(+)=(+)
$


When $x$ is slightly $>10$ (say $x=10.1$ ); from (ii),

$
\frac{d z}{d x}=(+)(+)(-)=(-)
$
$\therefore \quad \frac{d z}{d x}$ changes sign from ( + ) to ( – ) as $x$ increases while passing through 10 .
$\therefore \quad x=10$ gives a point of (local) maxima.
(See Note at the end of solution of Q. No. 5)
i.e., $z$ is maximum when $x=10$.
Substituting $x=10$ in (i), $y=35-10=25$.
∴ The two required numbers are 10 and 25 .

Solution: Let the two positive numbers be $x$ and $y$.
Given: $x+y=16 \Rightarrow y=16-x$

Let $z$ denote the sum of their cubes i.e., $z=x^3+y^3$.
Substituting $y=16-x$ from $(i), z=x^3+(16-x)^3$

$
\begin{aligned}
& \Rightarrow \quad z=x^3+(16)^3-x^3-48 x(16-x) \\
& \quad\left[\because(a-b)^3=a^3-b^3-3 a b(a-b)\right] \\
& \Rightarrow \quad z=(16)^3-768 x+48 x^2
\end{aligned}
$


Now $z$ is a function of $x$ alone.

$
\therefore \quad \frac{d z}{d x}=-768+96 x \text { and } \frac{d^2 z}{d x^2}=96
$


Putting $\frac{d z}{d x}=0$ to find the critical points, we have

$
-768+96 x=0 \Rightarrow 96 x=768 \Rightarrow x=\frac{768}{96}=8
$


At $x=8, \frac{d^2 z}{d x^2}=96 \quad$ (+)
$\therefore \quad x=8$ is a point of (local minima.
(See Note at the end of solution of Q. No. 5)
$\therefore \quad z$ is minimum when $x=8$.
Substituting $x=8$ in (i), $y=16-8=8$
∴ The required numbers are 8 and 8 .

Solution: Given: Each side of square piece of tin is 18 cm .

Let $x \mathrm{~cm}$ be the side of each of the four squares cut off from each corner.
Then dimensions of the open box formed by folding the flaps after cutting off squares are $(18-2 x),(18-2 x), x \mathrm{~cm}$.
Let us call the volume $z$ of the open box.

$
\begin{array}{ll}
\therefore & z=\text { length } x \text { breadth } x \text { height } \quad=(18-2 x)(18-2 x) x \\
\text { or } & z=(18-2 x)^2 x=\left(324+4 x^2-72 x\right) x \\
\text { or } & z=4 x^2-72 x^2+324 x \\
\therefore & \frac{d z}{d x}=12 x^2-144 x+324 \text { and } \frac{d^2 z}{d x^2}=24 x-144
\end{array}
$


Putting $\frac{d z}{d x}=0$ to find the critical points, we have

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


Dividing by $12, x^2-12 x+27=0$

$
\begin{aligned}
& \Rightarrow \quad x^2-9 x-3 x+27=0 \Rightarrow x(x-9)-3(x-9)=0 \\
& \Rightarrow \quad(x-9)(x-3)=0 \\
& \therefore \text { Either } x-9=0 \text { or } x-3=0 \\
& \text { i.e., } \begin{aligned}
x & =9 \\
x-9 & \text { or } \quad x=3
\end{aligned}
\end{aligned}
$


But $x=9$ is rejected because for $x=9$,
length of box $=18-2 x=18-18=0$ which is clearly impossible.
$\therefore x=3$ is the only critical point.
At $x=3, \frac{d^2 z}{d x^2}=24 x-144=72-144=-72$ (Negative)
$\therefore z$ is maximum at $x=3$.
i.e., side of (each) square to be cut off from each corner for maximum volume is 3 cm .
Remark. The reader is suggented to take a paper sheet in square shape and cut off four equal squares from four corners and fold the flaps to form a box for himself or herself.

Solution: Dimensions of the rectangular sheet of tin are 45 cm and 24 cm . Let the side of the square cut off from each corner be $x \mathrm{~cm}$. Therefore, dimensions of the box are $45-2 x, 24-2 x$ and $x \mathrm{~cm}$.

The volume V of the box in cubic cm is given by

$
\begin{aligned}
V & =(45-2 x)(24-2 x)(x) \quad \text {

Solution: Let PQRS be the rectangle inscribed in a given circle with centre O and radius $a$.
Let $x$ and $y$ be the length and breadth of the rectangle ( $\therefore x>$ 0 and $y>0$ )
In right angled triangle $\mathrm{PQR}, \mathrm{By}$ Pythagoras Theorem,
(given condition)

\section*{“}
(given condition)

Let A denote the area of the rectangle.

$
\therefore \quad \mathrm{A}=x y
$

-(ii) ( A is to be maximised)
To express A in terms of one independent variable, putting the value of $y$ from ( $i$ ) in ( $i i$ ), we have
or
or

$
\begin{gathered}
\mathrm{PQ}^2+\mathrm{QR}^2=\mathrm{PR}^2 \\
x^2+y^2=(2 a)^2 \\
y^2=4 a^2-x^2 \quad \therefore y=\sqrt{4 a^2-x^2}
\end{gathered}
$

or
“

$
\mathrm{A}=x \sqrt{4 a^2-x^2}
$


Let

$
z=A^2=x^2\left(4 a^2-x^2\right)=4 a^2 x^2-x^4
$


Let us maximise

$
z\left(=\mathrm{A}^2\right)
$


From (iii),

$
\frac{d z}{d x}=8 a^2 x-4 x^3
$

and

$
\frac{d^2 z}{d x^2}=8 a^2-12 x^2
$


To find maxima or minima; put $\frac{d z}{d x}=0$
$
\therefore 8 a^2 x-4 x^3=0 \text { or } 4 x\left(2 a^2-x^2\right)=0
$


But $x$ being side of rectangle cannot be zero.

$
\begin{aligned}
& \therefore 2 a^2-x^2=0 \text { or } x^2=2 a^2 \\
& \therefore \quad x=\sqrt{2} \quad a \\
& \text { At } \quad x=\sqrt{2} a, \quad \frac{d^2 z}{d x^2}=8 a^2-12\left(2 a^2\right)=8 a^2-24 a^2 \\
& \quad=-16 a^2 \text { is negative. } \\
& \quad z\left(=A^2\right) \text { is maximum when } x=\sqrt{2} a . \\
& \begin{aligned}
& \text { Putting } x=\sqrt{2} a \text { in }(i), y=\sqrt{4 a^2-2 a^2}=\sqrt{2 a^2} \quad=\sqrt{2} . a \\
& \therefore x=y=\sqrt{2} a \therefore \text { A is maximum when } x=y=\sqrt{2} a .
\end{aligned}
\end{aligned}
$


Hence, the area of the inscribed rectangle is maximum when it is a square.

Solution: Let $x$ be the radius of the circular base and $y$ be the height of closed right circular cylinder. Total surface area of cylinder is given $(x>0, y>0)$
⇒ It is constant $=S$ (say)
∴ Curved surface area + area of two ends $=S$
$\Rightarrow 2 \pi x y+2 \pi x^2=S$ (Given condition)
Dividing every term by $2 \pi$
[To get a simpler relation in independent variables $x$ and $y$ ]

$
\begin{aligned}
x y+x^2 & =\frac{\mathbf{S}}{2 \pi}=k(\operatorname{say}) \\
\therefore \quad x y & =k-x^2 \text { or } y=\frac{k-x^2}{x}
\end{aligned}
$


Let us call the volume $z$ of cylinder

$
\therefore \quad z=\pi x^2 y
$

[Here $z$ is to be maximised]. Putting the value of $y$ from (i) in (ii) [to express $z$ in terms of one independent variable $x$ ]

$
\begin{aligned}
z & =\pi x^2\left(\frac{k-x^2}{x}\right) \text { or } z=\pi x\left(k-x^2\right)=\pi\left(k x-x^3\right) \\
\therefore \frac{d z}{d x} & =\pi\left(k-3 x^2\right) \text { and } \frac{d^2 z}{d x^2}=\pi(-6 x)=-6 \pi x
\end{aligned}
$


To find maxima or minima put $\frac{d z}{d x}=0 \therefore \pi\left(k-3 x^2\right)=0$
But $\pi \neq 0 \therefore k-3 x^2=0$ or $3 x^2=k$ or $x^2=\frac{k}{3} \quad \therefore x=\sqrt{\frac{k}{3}}$
At $x=\sqrt{\frac{k}{3}}, \frac{d^2 z}{d x^2}=-6 \pi x=-6 \pi \sqrt{\frac{k}{3}}$ is negative
$\therefore \quad z$ is max. at $x=\sqrt{\frac{k}{3}}$

Putting

$
\begin{aligned}
x & =\sqrt{\frac{k}{3}} \text { in }(i), y=\frac{k-\frac{k}{3}}{\sqrt{\frac{k}{3}}}=\frac{2 \frac{k}{3}}{\sqrt{\frac{k}{3}}} \\
& =2 \sqrt{\frac{k}{3}} \quad\left[\because \frac{t}{\sqrt{t}}=\sqrt{t}\right]
\end{aligned}
$

or

$
y=2 \sqrt{\frac{k}{3}}=2 x
$

[By (iii)]
t.e., Height $=$ Diameter

Hence, the volume of cylinder is maximum when its height is equal to the diameter of its base.
Remark 1. Right circular cylinder ⇒ Closed right circular cylinder.
Remark 2. Total surface area of open cylinder $=2 \pi x y+\pi x^2$.

Solution: Let $x \mathrm{~cm}$ be the radius and $y \mathrm{~cm}$ be the height of closed cylinder.
Given: Volume of closed cylinder $=100 \mathrm{cu} \mathrm{cm}$

$
\begin{array}{rlrl}
\Rightarrow & & \pi x^2 y & =100 \\
\Rightarrow & y & =\frac{100}{\pi x^2}
\end{array}
$


Let us call the surface area $z$ of cylinder.

$
\therefore \quad z=2 \pi x y \quad+\quad 2 \pi x^2
$

(Curved surface area) (Area of two ends) or $\quad z=2 \pi\left(x y+x^2\right)$

Substituting $y=\frac{100}{\pi x^2}$ from (i),

$
z=2 \pi\left(x \cdot \frac{100}{\pi x^2}+x^2\right)=2 \pi\left(\frac{100}{\pi x}+x^2\right)=2 \pi\left(\frac{100}{\pi} x^{-1}+x^2\right)
$


Now $z$ is a function of $x$ alone.

$
\therefore \quad \frac{d z}{d x}=2 \pi\left(-\frac{100}{\pi} x^{-2}+2 x\right)
$
$
\text { and } \frac{d^2 z}{d x^2}=2 \pi\left(\frac{200}{\pi} x^{-3}+2\right)
$


Putting $\frac{d z}{d x}=0$ to find the critical points, we have

$
\begin{aligned}
2 \pi\left(\frac{-100}{\pi x^2}+2 x\right) & =0 . \text { But } 2 \pi \neq 0 \\
\frac{-100}{\pi x^2}+2 x & =0 \Rightarrow \frac{-100}{\pi x^2}=-2 x
\end{aligned}
$


Cross-multiplying. $2 \pi x^3=100$

$
\begin{aligned}
& \Rightarrow \quad x^3=\frac{100}{2 \pi}=\frac{50}{\pi} \\
& \therefore \quad x=\left(\frac{50}{\pi}\right)^{1 / 3} \\
& \text { At } \quad \begin{aligned}
x & =\left(\frac{50}{\pi}\right)^{1 / 3}, \frac{d^2 z}{d x^2}=2 \pi\left(\frac{200}{\pi x^3}+2\right) \\
& =2 \pi\left(\frac{200}{\pi\left(\frac{50}{\pi}\right)}+2\right)=2 \pi(4+2)=12 \pi(\text { positive })
\end{aligned}
\end{aligned}
$

$\therefore z$ is minimum (local) when radius $x=\left(\frac{50}{\pi}\right)^{1 / 3} \mathrm{~cm}$
(See Note at the end of solution of Q. No. 5)
Substituting $x=\left(\frac{50}{\pi}\right)^{1 / 3}$ in (i), $y=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{2 / \pi}}$
⇒ Height $y=2 \cdot \frac{\frac{50}{\pi}}{\left(\frac{50}{\pi}\right)^{2 / 3}}=2\left(\frac{50}{\pi}\right)^{1-2 / 3}=2\left(\frac{50}{\pi}\right)^{1 / 3} \mathrm{~cm}$.
Remark. $y=2 x \quad$ (By (ii))

Solution: Let $x$ metres be the side of the square and $y$ metres, the radius of the circle.

$
\begin{aligned}
\text { Length of wire } & =\text { Perimeter of square + Circumference of circle } \\
& =4 x+2 \pi y
\end{aligned}
$
4x
2nty

According to the question, $4 x+2 \pi y=28$ (Given condition)
Dividing by 2 ,

$
2 x+\pi y=14 \quad \therefore y=\frac{14-2 x}{\pi}
$


Area of square $=x^2$ sq. m . Area of circle $=\pi y^2$ sq. m
Let A denote their combined area, then

$
A=x^2+\pi y^2
$

[Here A is to be minimised]
Putting the value of $y$ from eqn. (i),
[To express A in terms of one independent variable x ]

$
A=x^2+\pi\left(\frac{14-2 x}{\pi}\right)^2=x^2+\pi\left(\frac{2(7-x)}{\pi}\right)^2=x^2+\pi \cdot \frac{4}{\pi^2}(7-x)^2
$

or $\quad \mathrm{A}=x^2+\frac{4}{\pi}(7-x)^2$

$
\therefore \quad \frac{d \mathrm{~A}}{d x}=2 x-\frac{8}{\pi}(7-x) \text { and } \frac{d^2 \mathrm{~A}}{d x^2}=2+\frac{8}{\pi}
$


To find maxima or minima, $\frac{d \mathrm{~A}}{d x}=0 \quad \therefore 2 x-\frac{8}{\pi}(7-x)=0$

$
\begin{aligned}
\therefore & & 2 x & =\frac{8}{\pi}(7-x) \\
\text { or } & & 2 \pi x & =56-8 x \text { or }(2 \pi+8) x=56 \\
\therefore & & x & =\frac{56}{2 \pi+8}=\frac{56}{2(\pi+4)}=\frac{28}{\pi+4}
\end{aligned}
$


Also $\frac{d^2 \mathrm{~A}}{d x^2}=2+\frac{8}{\pi}$ is +ve .

$
\therefore A \text { is minimum when } \quad x=\frac{28}{\pi+4} .
$


Hence, the wire should be cut at a distance $4 x=\frac{112}{\pi+4} \mathrm{~m}$ from one end.
Note 1. Length of circle $=2 \pi y$.
Putting the value of $y$ from ( $i$ ),

$
=28-4 x
$


Putting the value of $x$ from (ii),
Length of circle $=28-\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4}$
∴ The length of two parts (square and circle) are respectively $\frac{112}{\pi+4} \mathrm{~m}$ and $\frac{28 \pi}{\pi+4} \mathrm{~m}$.
Note 2. Side of square $=x=\frac{28}{\pi+4}$
From (i), Radius of circle $=y=\frac{14-2 x}{\pi}=\frac{14-2\left(\frac{28}{\pi+4}\right)}{\pi}$

$
=\frac{14 \pi+56-56}{\pi(\pi+4)}=\frac{14 \pi}{\pi(\pi+4)}=\frac{14}{\pi+4}
$


Therefore, diameter of circle $=\frac{28}{\pi+4}$
∴ Side of square $=$ Diameter of circle.

Solution: Let O be the centre and R be the radius of the given sphere.
Let $\mathbf{B M}=x$ and $\mathbf{A M}=y$ be the radius and height of any cone inscribed in the given sphere.
In right angled triangle OMB, By Pythagoras Theorem,

$
\begin{aligned}
& \mathrm{OM}^2+\mathrm{BM}^2=\mathrm{OB}^2 \\
\Rightarrow & (y-\mathrm{R})^2+x^2=\mathrm{R}^2[\because \quad \mathrm{OM}=\mathrm{AM}-\mathrm{OA}=y-\mathrm{R}] \\
\Rightarrow & y^2+\mathrm{R}^2-2 \mathrm{Ry}+x^2=\mathrm{R}^2 \Rightarrow x^2+y^2-2 \mathrm{Ry}=0 \\
\Rightarrow & x^2=2 \mathrm{Ry}-y^2
\end{aligned}
$


Let us call the volume $z$ of any cone inscribed in the given sphere.

$
\therefore \quad z=\frac{1}{3} \pi r^2 y
$


Putting the value of $x^2$ from $(i)$,

$
z=\frac{\pi}{3}\left(2 R y-y^2\right) y=\frac{\pi}{3}\left(2 R y^2-y^3\right)
$


Now $z$ is a function of $y$ alone.

$
\therefore \quad \frac{d z}{d y}=\frac{\pi}{3}\left(4 R y-3 y^2\right) \text { and } \frac{d^2 z}{d y^2}=\frac{\pi}{3}(4 R-6 y)
$


Putting $\frac{d z}{d y}=0$ to find the critical points, we have $\frac{\pi}{3}\left(4 R y-3 y^2\right)=0$
But $\frac{\pi}{3} \neq 0$. Therefore $4 \mathrm{Ry}-3 y^2=0 \Rightarrow-3 y^2=-4 \mathrm{Ry}$
Dividing both sides by $-y \neq 0$,

$
3 y=4 R \Rightarrow y=\frac{4 R}{3}
$


At

$
\begin{aligned}
y & =\frac{4 R}{3}, \frac{d^2 z}{d y^2}=\frac{\pi}{3}(4 R-6 y)=\frac{\pi}{3}(4 R-8 R) \\
& =\frac{\pi}{3}(-4 R)=-\frac{4 R}{3} \quad \text { (Negative) }
\end{aligned}
$

$\therefore \quad z$ is maximum at $y=\frac{4 R}{3}$

Substituting $y=\frac{\mathbf{4 R}}{3}$ from (fit) in (f), we have

$
\begin{aligned}
x^2 & =2 R \cdot \frac{4 R}{3}-\left(\frac{4 R}{3}\right)^2=\frac{8 R^2}{3}-\frac{16 R^2}{9} \\
& =8 R^2\left(\frac{1}{3}-\frac{2}{9}\right)=8 R^2\left(\frac{3-2}{9}\right) \quad \Rightarrow \quad x^2=\frac{8 R^2}{9}
\end{aligned}
$

∴ Maximum volume $z$ of the cone

$
\begin{aligned}
& =\frac{1}{3} \pi x^2 y=\frac{1}{3} \pi \cdot \frac{8 \mathbf{R}^2}{9} \cdot \frac{4 \mathbf{R}}{3}=\frac{8}{27} \cdot \frac{4 \boldsymbol{\pi}}{3} \mathbf{R}^3 \\
& =\frac{8}{27} \text { (Volume of the sphere). }
\end{aligned}
$

Solution: Let $x$ be the base radius and $y$ be the height of cone.
Given Volume ⇒ Volume of the cone is constant

$
\begin{aligned}
& \text { and }=\mathrm{V}(\text { say }) \\
& \therefore \quad \frac{1}{3} \pi x^2 y=\mathrm{V}(\text { Given condition }) \\
& \therefore \quad x^2 y=\frac{3 \mathrm{~V}}{\pi}=k \quad(\text { say }) \ldots(i)
\end{aligned}
$


Let S denote the curved surface of the cone

$
\therefore S=\pi x \sqrt{x^2+y^2} \quad \text { (formula } S=\beta r l \text { ) }
$

( S is to be minimised here)
Let $z=S^2=\pi^2 x^2\left(x^2+y^2\right)$

Putting $x^2=\frac{k}{y}$ from (i) in (ii) to get $z$ as a function of single independent variable $y$.
[Here, we are getting $z$ as a simpler function

of $y$ as compared to $z$ as a function of $x]$
$
\begin{aligned}
& \therefore \quad z=\pi^2 \frac{k}{y}\left(\frac{k}{y}+y^2\right)=\pi^2 k\left(\frac{k}{y^2}+y\right) \\
& \text { or } z=\pi^2 k\left(k y^{-2}+y\right) \\
& \therefore \quad \frac{d z}{d y}=\pi^2 k\left[-2 k y^{-3}+1\right] \text { and } \frac{d^2 z}{d y^2}=\pi^2 k\left[6 k y^{-4}\right]=\frac{6 \pi^2 k^2}{y^4}
\end{aligned}
$


To find maxima or minima, put $\frac{d z}{d y}=0$

$
\begin{array}{rlrlrl}
\therefore & & \pi^2 k\left(-\frac{2 k}{y^3}+1\right) & =0 & \text { But } & \\
\pi^2 k & =0 \\
\therefore & -\frac{2 k}{y^3}+1 & =0 & & \text { or } & \\
\therefore & \frac{2 k}{y^3} & =1 \\
y^3 & =2 k & & & y & =(2 k)^{1 . a}
\end{array}
$


At $\quad y=(2 k)^{1 / 2}, \frac{d^2 z}{d y^2}=\frac{6 \pi^2 k^2}{(2 k)^{4 / 3}}$ this is positive.
$\therefore \quad z$ is least when $y=(2 k)^{1 / 3}$

$
\text { ∴ From }(i), x^2=\frac{k}{y}=\frac{k}{(2 k)^{1 / 5}}
$

or

$
x^2=\frac{2 k}{2(2 k)^{1 / 3}}=\frac{(2 k)^{2 / 3}}{2}=\frac{y^2}{2}
$

[By (iii)]
or $y^2=2 x^2 \quad \therefore y=\sqrt{2} x$
$\therefore \quad z$ or $S$ is least when height $=\sqrt{2}$ (radius of base).

Solution: Let $x$ be the base radius, $y$ the height, $l$ the given slant height and $\theta$, the semi-vertical angle of cone, $(x>0, y>0)$
In AAMB,
By Pythagoras Theorem, $x^2+y^2=l^2$

$
\therefore \quad x^2=l^2-y^2
$


Let V denote the volume of cone, then

$
v=\frac{1}{3} \pi x^2 y
$

-(ii) [V is to
be maximised here] Putting $x^2=l^2-y^2$ from (i) in (ii), to express V as a function of single independent variable $y$.
or

$
\mathrm{v}=\frac{1}{3} \pi\left(t^2-y^2\right) y
$

or $\quad \mathrm{V}=\frac{\pi}{3}\left(l^2 y-y^3\right)$

$
\therefore \quad \frac{d V}{d y}=\frac{\pi}{3}\left(l^2-3 y^2\right)
$
and $\frac{d^2 V}{d y^2}=\frac{\pi}{3}(-6 y)=-2 \pi y$
To find maxima or minima put $\frac{d V}{d y}=0$

$
\begin{aligned}
& \therefore \quad \frac{\pi}{3}\left(l^2-3 y^2\right)=0 \quad \text { But } \quad \frac{\pi}{3} \neq 0 \\
& \therefore \quad l^2-3 y^2=0 \quad \text { or } \quad 3 y^2=l^2 \quad \text { or } y^2=\frac{l^2}{3} \\
& \therefore \quad y=\frac{l}{\sqrt{3}}
\end{aligned} \quad(\because y>0)
$


At $\quad y=\frac{l}{\sqrt{3}}, \frac{d^2 V}{d y^2}=-2 \pi y=-\frac{2 \pi l}{\sqrt{3}}$ this is negative.
$\therefore \mathrm{V}$ is maximum at $y=\frac{l}{\sqrt{3}}$.
Substituting $y=\frac{l}{\sqrt{3}}$ in eqn. (i), $\quad x^2=l^2-\frac{l^2}{3}=\frac{2 l^2}{3}$

$
\therefore \quad x=\sqrt{2} \frac{l}{\sqrt{3}}
$


In right angled $\triangle \mathrm{AMB}, \quad \tan \theta=\frac{\mathrm{MB}}{\mathrm{AM}}=\frac{x}{y}=\frac{\sqrt{2} \frac{l}{\sqrt{3}}}{\frac{l}{\sqrt{3}}}=\sqrt{2}$
∴ Semi-vertical angle

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

Solution: Let $x$ be the radius of base of cone and $y$ be its height. (Total) surface area of cone is given.
$\therefore \pi r l$
$\downarrow$
(Curved $\quad$ (Area of base)
Surface area)

$
\Rightarrow \pi x \sqrt{x^2+y^2}+\pi x^2=S(\operatorname{san})
$


Dividing both sides by $\pi$,

$
\begin{aligned}
& x \sqrt{x^2+y^2}+x^2=\frac{S}{\pi}=k \text { (say) } \\
\Rightarrow & x \sqrt{x^2+y^2}=k-x^2
\end{aligned}
$


Squaring both sides, we have $x^2\left(x^2+y^2\right)=\left(k-x^2\right)^2$

$
\text { or } \quad x^4+x^2 y^2=k^2+x^4-2 k x^2
$
$
\begin{aligned}
& \Rightarrow \quad x^2 y^2+2 k x^2=k^2 \quad \Rightarrow x^2\left(y^2+2 k\right)=k^2 \\
& \Rightarrow \quad x^2=\frac{k^2}{y^2+2 k}
\end{aligned}
$


Let us call the volume $z$ of the cone.

$
\therefore \quad z=\frac{1}{3} \pi x^2 y
$


Putting the value of $x^2$ from $(i)$,

$
\begin{array}{rlrl}
z & =\frac{1}{3} \pi \frac{k^2}{y^2+2 k} y=\frac{1}{3} \pi k^2 \frac{y}{y^2+2 k} \\
\therefore & \frac{d z}{d y} & =\frac{1}{3} \pi k^2 \frac{d}{d y} \frac{y}{y^2+2 k} \\
\text { or } & \frac{d z}{d y} & =\frac{1}{3} \pi k^2\left[\frac{\left(y^2+2 k\right) \cdot 1-y \cdot 2 y}{\left(y^2+2 k\right)^2}\right] \text { (By quotient rule) } \\
\therefore & \frac{d z}{d y} & =\frac{1}{3} \pi k^2 \frac{\left(2 k-y^2\right)}{\left(y^2+2 k\right)^2}
\end{array}
$


Putting $\frac{d z}{d y}=0$ to find the critical points, we have

$
\frac{\pi k^2\left(2 k-y^2\right)}{3\left(y^2+2 k\right)^2}=0 \Rightarrow \pi k^2\left(2 k-y^2\right)=0
$


But $\pi k^2 \neq 0 \quad \therefore \quad 2 k-y^2=0 \Rightarrow y^2=2 k$

$
\therefore \quad y= \pm \sqrt{2 k}
$


Rejecting negative sign because height $y$ of cone can’t be negative.
$\therefore y=\sqrt{2 k}$ is the only critical point.
Applying the first derivative test.
(because finding $\frac{d^2 z}{d x^2}$ looks to be tedious)
Now in R.H.S. of (ii), $\frac{\pi k^2}{3\left(y^2+2 k\right)^2}>0$ clearly.
When $y$ is slightly $<\sqrt{2 k}$; then $y^2<2 k$
$\Rightarrow 0<2 k-y^2 \Rightarrow 2 k-y^2>0$, therefore from (ii), $\frac{d z}{d y}>0$ i.e., (positive)
When $y$ is slightly $>\sqrt{2 k}$, then $y^2>2 k \Rightarrow 0>2 k-y^2$
$\Rightarrow 2 k-y^2<0$; therefore from (ii) $\frac{d z}{d y}<0$ i.e., (negative)
$\therefore \quad \frac{d z}{d y}$ changes sign from $(+)$ to $(-)$ as $y$ increases through $\sqrt{2 k}$
$\therefore \quad$ Volume $z$ is maximum at $y=\sqrt{2 k}$
Substituting $y=\sqrt{2 k}$ in $(i), x^2=\frac{k^2}{2 k+2 k}=\frac{k^2}{4 k}=\frac{k}{4}$

$
\therefore \quad x=\sqrt{\frac{k}{4}}=\frac{\sqrt{k}}{2}
$


Let $\alpha$ be the semi-vertical angle of the cone.
In right angled $\triangle O M B$,

$
\sin \alpha=\frac{\mathrm{MB}}{\mathrm{OB}}=\frac{x}{\sqrt{x^2+y^2}}
$


Putting values of $x$ and $y, \sin \alpha=\frac{\frac{\sqrt{k}}{2}}{\sqrt{\frac{k}{4}+2 k}}=\frac{\frac{\sqrt{k}}{2}}{\sqrt{\frac{9 k}{4}}}=\frac{\frac{\sqrt{k}}{2}}{3 \frac{\sqrt{k}}{2}}=\frac{1}{3}$

$
\therefore \quad \alpha=\sin ^{-1} \frac{1}{3}
$


Choose the correct answer in the Exercises 27 to 29.

Solution: Equation of the curve (upward parabola here) is

$
x^2=2 y
$


The given point is $A(0,5)$.
Let $\mathrm{P}(x, y)$ be any point on curve (i).
∴ Distance $z=$ AP

$
=\sqrt{(x-0)^2+(y-5)^2}
$


1 Distance formula
Let

$
\mathrm{Z}=z^2=x^2+(y-5)^2
$


Putting

$
\begin{array}{ll}
& \mathrm{Z} \\
\text { or } & \mathrm{Z} \\
\therefore & \frac{d \mathrm{Z}}{d y} \\
\therefore & =2 y-8 \text { and } \frac{d^2 \mathrm{Z}}{d y^2}=2
\end{array}
$


Putting $\quad \frac{d Z}{d y}=0$ to get critical point(s), we have

$
\frac{d Z}{d y}=0 \text { i.e., } 2 y-8=0 \Rightarrow 2 y=8 \Rightarrow y=4
$


At

$
y=4, \quad \frac{d^2 \mathrm{Z}}{d y^2}=2 \text { is }(+\mathrm{ve})
$

$\therefore \mathbb{Z}\left(=z^2\right)$ is minimum and hence $z$ is minimum at $y=4$.
Substituting $y=4$ in $(i), x^2=8 \quad \therefore \quad x= \pm \sqrt{8}= \pm 2 \sqrt{2}$.
$\therefore \quad(2 \sqrt{2}, 4)$ and $(-2 \sqrt{2}, 4)$ are two points on curve $(i)$ which are nearest to the given point ( 0,5 ).
∴ Option (A) is correct answer.

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

$
\begin{aligned}
\therefore f^{\prime}(x) & =\frac{\left(1+x+x^2\right) \frac{d}{d x}\left(1-x+x^2\right)-\left(1-x+x^2\right) \frac{d}{d x}\left(1+x+x^2\right)}{\left(1+x+x^2\right)^2} \\
& =\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} \\
\text { or } f^{\prime}(x) & =\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}{\left(1+x+x^2\right)^2} \\
\text { or } f^{\prime}(x) & =\frac{-2+2 x^2}{\left(1+x+x^2\right)^2}=\frac{-2\left(1-x^2\right)}{\left(1+x+x^2\right)^2}
\end{aligned}
$


Let us put $f^{\prime}(x)=0$ to locate the critical points.
Therefore $\frac{-2\left(1-x^2\right)}{\left(1+x+x^2\right)^2}=0 \Rightarrow-2\left(1-x^2\right)=0$
But $-2 \neq 0$. Therefore, $1-x^2=0$ or $-x^2=-1$
$\therefore x^2=1 \Rightarrow x= \pm 1$
$\therefore x=-1$ and $x=1$ are two critical points.
Let us find values of $f(x)$ at these two critical points only because no closed interval is given to be domain of $f(x)$.
Substituting $x=-1$ in $(i), \quad f(-1)=\frac{1+1+1}{1-1+1}=3$
Substituting $x=1$ in $(i), f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
Therefore, minimum value of $f(x)$ is $\frac{1}{3}$.
Hence, option (D) is correct.
Note. Maximum value of $f(x)$ for the above question is 3 .

Solution: Let $f(x)=(x(x-1)+1)^{1 / 3}=\left(x^2-x+1\right)^{1 / 3}, 0 \leq x \leq 1$
$\therefore f^{\prime}(x)=\frac{1}{3}\left(x^2-x+1\right)^{-23} \frac{d}{d x}\left(x^2-x+1\right)$
or $f^{\prime}(x)=\frac{(2 x-1)}{3\left(x^2-x+1\right)^{2 / 3}}$
Putting $f^{\prime}(x)=0$ to locate the critical points, we have

$
\frac{2 x-1}{3\left(x^2-x+1\right)^{2 / 3}}=0
$


Cross-multiplying $2 x-1=0 \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}$
This critical point $x=\frac{1}{2}$ belongs to the given closed interval $0 \leq x \leq 1$ i.e., $[0,1]$
Now let us find values of $f(x)$ at the critical point $x=\frac{1}{2}$ and end points $x=0$ and $x=1$ of given closed interval $[0,1]$.
Substituting $x=\frac{1}{2}$ in $(i)$,

$
f\left(\frac{1}{2}\right)=\left(\frac{1}{4}-\frac{1}{2}+1\right)^{1 / 3}=\left(\frac{1-2+4}{4}\right)^{1 / 3}=\left(\frac{3}{4}\right)^{1 / 3}<1
$


Substituting $x=0$ in (i), $f(0)=(1)^{1 / 2}=1$
Substituting $x=1$ in $(i), f(1)=(1-1+1)^{1 / 3}=(1)^{1 / 3}=1$
∴ Maximum value of $f(x)$ is 1 .
Hence, option (C) is correct.
Note. Minimum value of $f(x)$ for the above question is $\left(\frac{3}{4}\right)^{1 / 3}$.

Solution:
$\begin{array}{l}
\text { Here } f(x)=\frac{\log x}{x}, x>0 \\
\therefore f^{\prime}(x)=\frac{x \cdot \frac{1}{x}-\log x \cdot 1}{x^2}=\frac{1-\log x}{x^2} \\
\text { and } f^{\prime \prime}(x)=\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x) \cdot 2 x}{x^4}=\frac{-x-2 x+2 x \log x}{x^4} \\
\quad=\frac{2 x \log x-3 x}{x^4}=\frac{x(2 \log x-3)}{x^4}=\frac{2 \log x-3}{x^4}
\end{array}$


For local max. or local min., put $f^{\prime}(x)=0$

$
\begin{aligned}
& \Rightarrow \frac{1-\log x}{x^2}=0 \\
& \Rightarrow 1-\log x=0 \Rightarrow \log x=1=\log e \Rightarrow x=e
\end{aligned}
$


Now from (iii), $f^{\prime \prime}(e)=\frac{2 \log e-3}{e^3}=\frac{2-3}{e^3}=-\frac{1}{e^3}<0$
$\Rightarrow x=e$ is a local maximum point and hence $f(x)$ has a maximum (value) at $x=e$.

Solution: Let $\mathrm{BC}=b$ be the fixed base and $\mathrm{AB}=\mathrm{AC}=x$ be the two equal sides of the isosceles triangle ABC.
Given: $\frac{d x}{d t}=-3 \mathrm{~cm} / \mathrm{s}$
Draw $\mathrm{AM} \perp \mathrm{BC}$, then M is mid-point of BC ( $\because \mathrm{AB}=\mathrm{AC}$ (given))

$
\therefore \quad \mathrm{BM}=\mathrm{CM}=\frac{1}{2} b
$


Area of $\triangle \mathrm{ABC}$ is $(\Delta)=\frac{1}{2} \mathrm{BC} \times \mathrm{AM}$

$
\begin{aligned}
& \left\lvert\, \frac{1}{2}\right. \text { Base × Height } \\
= & \frac{b}{2} \sqrt{\mathrm{AC}^2-\mathrm{MC}^2}=\frac{b}{2} \sqrt{x^2-\frac{b^2}{4}} \\
= & \frac{b}{2} \sqrt{\frac{4 x^2-b^2}{4}}=\frac{b}{4} \sqrt{4 x^2-b^2}
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d \Delta}{d t}= & \frac{d}{d t}\left(\frac{b}{4} \sqrt{4 x^2-b^2}\right) \\
= & \frac{b}{4} \times \frac{d}{d x}\left(\sqrt{4 x^2-b^2}\right) \times \frac{d x}{d t}[\text { By Chain Rule }] \\
= & \frac{b}{4} \times \frac{8 x}{2 \sqrt{4 x^2-b^2}} \times(-3)[\text { Using }(i)] \\
& {\left[\because \frac{d}{d x} \sqrt{f(x)}=\frac{f(x)}{2 \sqrt{f(x)}}\right] }
\end{aligned}
$

or $\frac{d \Delta}{d t}=-\frac{3 b x}{\sqrt{4 x^2-b^2}} \mathrm{~cm}^2 / \mathrm{s}$
When the two equal sides are equal to the base (given) i.e., when $x=b$,
We have on putting $x=b, \frac{d \Delta}{d t}=-\frac{3 b \times b}{\sqrt{4 b^2-b^2}}=-\frac{3 b^2}{\sqrt{3} b}$

$
=-\sqrt{3} b \mathrm{~cm}^2 / \mathrm{s}
$


Hence, the area is decreasing ( $\because \frac{d \Delta}{d t}$ is negative) at the rate of $\sqrt{3} b \mathrm{~cm}^2 / \mathrm{s}$.

Solution: $
\begin{aligned}
& \text { Given: } f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x} \\
& =\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} \\
& \Rightarrow \quad f(x)=\frac{4 \sin x}{2+\cos x}-x \\
& \therefore \quad 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 \\
& \text { or } \quad 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}{(2+\cos x)^2}-1=\frac{8 \cos x+4}{(2+\cos x)^2}-1
\end{aligned}
$

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


Now $4-\cos x>0$ for all real $x$ because we know that -1 Scos $x \leq 1$ always. Also $(2+\cos x)^2$ being a square of a real number is $>0$.
$\therefore(i) f(x)$ is increasing if $f^{\prime}(x) \geq 0$
i.e., if cos $x \geq 0$ (From (i))
i.e., if $x$ lies in Ist and 4th quadrants.
i.e., $f(x)$ is increasing for $0 \leq x \leq \frac{\pi}{2}$ and $\frac{3 \pi}{2} \leq x \leq 2 \pi$.
(ii) $f(x)$ is decreasing if $f^{\prime}(x) \leq 0$
i.e., if $\cos x \leq 0$ (From (i))
i.e., if $x$ lies in IInd and III quadrants
$\therefore f(x)$ is decreasing for $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$.

Solution: (i) Given: $f(x)=x^3+\frac{1}{x^3}, x \neq 0$

$
\begin{array}{ll}
\Rightarrow & f(x)=x^3+x^{-3} \\
\therefore & f^{\prime}(x)=3 x^2-3 x^{-4}=3\left(x^2-\frac{1}{x^4}\right)
\end{array}
$


Step I. Forming factors $=3\left(\frac{x^6-1}{x^4}\right)=\frac{3}{x^4}\left[\left(x^2\right)^3-1^3\right]$

$
\begin{aligned}
& f^{\prime}(x)=\frac{3}{x^4}\left(x^2-1\right)\left(\left(x^2\right)^2+x^2 \cdot 1+1^2\right) \\
& \quad\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]
\end{aligned}
$

or

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

(Note, For forming factors, we could also use $a^2-b^2=\left(a-b(a+b)\right.$ but $a^3-b^3$ is much better here)
Step II. Let us find critical points by putting $f^{\prime}(x)=0$
$
\therefore \quad \text { From }(i), \frac{3\left(x^4+x^2+1\right)(x+1)(x-1)}{x^4}=0
$


Cross-multiplying, $3\left(x^4+x^2+1\right)(x+1)(x-1)=0$
But $3\left(x^4+x^2+1\right)$ is positive for all real $x$
(as both terms of $x$ have even powers and all terms are positive) and hence $\neq 0$.

$
\begin{aligned}
& \therefore \text { Either } x+1=0 \text { or } x-1=0 \\
& \text { i.e., } x=-1 \text { or } x=1
\end{aligned}
$


These two critical points $x=-1$ and $x=1$ divide the whole real line into three sub-intervals ( $-\infty$, -1$],[-1,1]$ and $[1, \infty)$,

In R.H.S. of (1), $\frac{3\left(x^4+x^2+1\right)}{x^4}$ is positive for all real $x \neq 0$.
Step III.

\begin{tabular}{|l|l|l|}
\hline Values of $\boldsymbol{x}$ & sign of $f^{\prime}(x)$
$
\begin{aligned}
= & \frac{3\left(x^4+x^2+1\right)}{x^4} \\
& (x+1)(x-1)
\end{aligned}
$ & Nature of $f(x)$ \\
\hline ( $-\infty,-$ 1] i.e., $x \leq-1$ & For example, at $x=-2$,
$
\begin{aligned}
f^{\prime}(x= & (+)(-)(-)=(+) \\
& \text { or } 0 \text { at } x=-1
\end{aligned}
$ & Increasing † \\
\hline [- 1, 1] i.e.,
$
-1 \leq x \leq 1
$ & $
\text { For example, at } x=\frac{1}{2} \text {, }
$
$
\begin{aligned}
f^{\prime} x & =(+)(+)(-) \\
& =(-) \text { or } 0 \text { both at } \\
x & =-1,1
\end{aligned}
$ & Decreasing ↓ \\
\hline \begin{tabular}{l}
$
[1, \infty)
$
\\
i.e., $x \geq 1$
\end{tabular} & For example, at $x=2$,
$
\begin{aligned}
f^{\prime}(x & =(+)(+)(+) \\
& =(+) \text { or } 0 \text { at } x=1
\end{aligned}
$ & Increasing † \\
\hline
\end{tabular}
$\therefore f(x)$ is ( $i$ ) an increasing function for $x \leq-1$ and for $x \geq 1$ and (ii) decreasing function for $-1 \leq x \leq 1$.

Solution: Equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

…(i)
We also know that the extremities of the major axis of the ellipse are $\mathrm{A}(a, 0)$ and $\mathrm{A}^{\prime}(-a, 0)\left(\Rightarrow \mathrm{OA}=a, \mathrm{OA}^{\prime}=a\right)$

Comparing eqn. (i) with

$
\cos ^2 \theta+\sin ^2 \theta=1,
$

we have $\frac{x}{a}=\cos \theta$ and $\frac{y}{b}=\sin \theta$
$\therefore \quad x=a \cos \theta$ and $y=b \sin \theta$
∴ Any point on the ellipse is $\mathrm{P}(a \cos \theta, b \sin \theta)$
Draw PM ⟂ on $x$-axis and produce it to meet the ellipse in the point Q.
$\therefore \mathrm{OM}=a \cos \theta$ and $\mathrm{PM}=b \sin \theta \quad$ (By def. of coordinates of a point)
PM = QM $\quad[\because$ The ellipse ( $i$ ) is symmetrical about major axis (here $x$-axis)]
$\therefore \mathrm{M}$ is the mid-point of PQ .
$\therefore \triangle \mathrm{APQ}$ is isosceles.
Let us call the area $z$ of isosceles triangle $\triangle A P Q$ inscribed in the ellipse with one vertex A coinciding with an extremity of major axis.

$
\begin{aligned}
\therefore \quad z & =\frac{1}{2} \text { Base × Height }=\frac{1}{2} \mathrm{PQ} \cdot \mathrm{AM} \\
& =\frac{1}{2} \cdot 2 \mathrm{PM} \cdot \mathrm{AM}=\mathrm{PM}(\mathrm{OA}-\mathrm{OM})
\end{aligned}
$


Putting values of OA, OM and PM

$
\begin{aligned}
& =b \sin \theta(a-a \cos \theta)=b a \sin \theta(1-\cos \theta) \\
& =a b(\sin \theta-\sin \theta \cos \theta) .
\end{aligned}
$


$
\text { or } z=\frac{a b}{2}(2 \sin \theta-2 \sin \theta \cos \theta)=\frac{a b}{2} \quad(2 \sin \theta-\sin 2 \theta)
$


Differentiating (i) with respect to 0 ,

$
\frac{d z}{d \theta}=\frac{a b}{2}(2 \cos \theta-2 \cos 2 \theta)=a b(\cos \theta-\cos 2 \theta)
$


Again differentiating with respect to $\theta$,

$
\frac{d^2 z}{d \theta^2}=a b(-\sin \theta+2 \sin 2 \theta)
$
Putting $\frac{d z}{d \theta}=0$, we have $a b(\cos \theta-\cos 2 \theta)=0$
But $a b=0 \quad \therefore \cos \theta-\cos 2 \theta=0$
or $\cos \theta=\cos 2 \theta=\cos \left(360^{\circ}-20\right)$
∴ Either $\theta=20$ or $\theta=360^{\circ}-20$
i.e., $0=0$ or $30=360^{\circ} \quad \therefore 0=120^{\circ}$
$\theta=0$ is impossible because otherwise the point $\mathrm{P}(a \cos \theta, b \sin \theta)$
$=(a \cos 0, b \sin 0)=(a, 0)$ will coincide with the point A .
$\therefore \quad 0=120^{\circ}$
At $0=120^{\circ}, \frac{d^2 z}{d \theta^2}=a b\left(-\sin 120^{\circ}+2 \sin 240^{\circ}\right)$

$
=a b\left[-\frac{\sqrt{3}}{2}-\frac{2 \sqrt{3}}{2}\right]=a b\left(-\frac{3 \sqrt{3}}{2}\right)=\text { Negative }
$

$\therefore z$ is maximum at $\theta=120^{\circ}$
Putting $\theta=120^{\circ}$ in (i),

$
\begin{aligned}
\text { Maximum Area } & =\frac{a b}{2}\left[2 \sin 120^{\circ}-\sin 240^{\circ}\right] \\
& =\frac{a b}{2}\left[\frac{2 \sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]=\frac{a b}{2}\left(\frac{3 \sqrt{3}}{2}\right)=\frac{3 \sqrt{3}}{4} a b .
\end{aligned}
$

Solution: Given: A tank with rectangular base and rectangular sides, open at the top.
(The reader is suggested to visualize this tank as a room with four rectangular walls, floor but no ceiling).
Given: Depth of tank $=2 \mathrm{~m}$
Let $x \mathrm{~m}$ be the length and $y \mathrm{~m}$ be the breadth of the base of tank.
Volume of tank $(=I b h)=x y .2=8 \mathrm{~m}^3$ (given)

$
\therefore \quad y=\frac{8}{2 x}=\frac{4}{x}
$


Now cost of building the base of the tank at the given rate of ₹ 70 per square metre is ₹ $70 x y$

Again cost of building the four sides (walls) of the tank at the rate of ₹ 45 per square metre.

$
\begin{aligned}
& =45(x .2+x .2+y .2+y .2)=45(4 x+4 y) \\
& =₹(180 x+180 y)
\end{aligned}
$


Let $z$ denote the total cost of building the tank.
Adding (ii) and (iii), $z=70 x y+180 x+180 y$
$
\begin{aligned}
& \text { Putting } y=\frac{4}{x} \text { from }(i), z=70 x \cdot \frac{4}{x}+180 x+180 \cdot \frac{4}{x} \\
& \text { or } \quad z=280+180 x+\frac{720}{x} \\
& \begin{aligned}
& \therefore \quad \frac{d z}{d x}= 0+180-\frac{720}{x^2} \text { and } \frac{d^2 z}{d x^2}=\frac{1440}{x^3} \\
& {\left[\because \frac{d}{d x}\left(\frac{1}{x}\right)=\frac{d}{d x} x^{-1}=(-1) x^{-2}=\frac{-1}{x^2}\right. \text { and }} \\
&\left.\frac{d}{d x}\left(\frac{1}{x^2}\right)=\frac{d}{d x} x^{-2}=-2 x^{-3}=\frac{-2}{x^3}\right]
\end{aligned}
\end{aligned}
$


Putting $\frac{d z}{d x}=0$ to find the critical points, we have

$
\begin{aligned}
& \quad 180-\frac{720}{x^2}=0 \Rightarrow 180=\frac{720}{x^2} \Rightarrow 180 x^2=720 \\
& \Rightarrow x^2=\frac{720}{180}=4 \Rightarrow x=2(\because x \text { being length can’t be negative) } \\
& \text { At } x=2, \frac{d^2 z}{d x^2}=\frac{1440}{x^4}=\frac{1440}{8}=180(+\mathrm{ve})
\end{aligned}
$

$\therefore z$ is minimum at $x=2$
Substituting $x=2$ in (iv), minimum cost

$
\begin{aligned}
z & =280+180(2)+\frac{720}{2}=280+360+360 \\
& =280+720=₹ 1000
\end{aligned}
$

Solution: Let $x$ be the radius of the circle and $y$ be the side of the square.
Given: Perimeter (circumference) of circle + perimeter of square $=k$

$
\begin{aligned}
\Rightarrow & & 2 \pi x+4 y & =k \\
\Rightarrow & & 4 y & =k-2 \pi x \\
\Rightarrow & & y & =\frac{k-2 \pi x}{4}
\end{aligned}
$


Let $z$ denote the sum of areas of circle and square.

$
\therefore \quad z=\pi x^2+y^2 \quad \text { [Area of square }=(\text { side })^2 \text { ] }
$


Putting the value of $y$ from (i),

$
\begin{aligned}
z & =\pi x^2+\frac{(k-2 \pi x)^2}{16}=\frac{16 \pi x^2+k^2+4 \pi^2 x^2-4 k \pi x}{16} \\
\text { or } \quad z & =\frac{1}{16}\left[\left(16 \pi+4 \pi^2 x^2-4 k \pi x+k^2\right]\right. \\
\therefore \quad \frac{d z}{d x} & =\frac{1}{16}\left[\left(16 \pi+4 \pi^2\right) 2 x-4 k \pi\right] \text { and } \frac{d^2 z}{d x^2}=\frac{1}{16}\left(16 \pi+4 \pi^2\right) 2
\end{aligned}
$
Putting $\frac{d z}{d x}=0$ to find the critical points, we have

$
\begin{aligned}
& \quad \begin{aligned}
\frac{1}{16}\left[\left(16 \pi+4 \pi^2\right) 2 x-4 k \pi\right] & =0 \\
\Rightarrow & \begin{aligned}
\left(16 \pi+4 \pi^2\right) 2 x-4 k \pi & =0 \times 16=0 \\
4 \pi / 4+\pi) 2 x & =4 k \pi
\end{aligned} \\
\Rightarrow & \begin{aligned}
x= & \frac{4 k \pi}{4 \pi(4+\pi) 2}=\frac{k}{2(4+\pi)}
\end{aligned} \\
\Rightarrow & \text { At } x=\frac{k}{2(4+\pi)}, \frac{d^2 z}{d x^2}=\frac{1}{16}\left(16 \pi+4 \pi^2 g \text { is }+\mathrm{ve} .\right.
\end{aligned} \\
& \therefore \quad z \text { is minimum when } x=\frac{k}{2(4+\pi)}
\end{aligned}
$


Putting this value of $x$ in (i),

$
\begin{aligned}
y & =\frac{1}{4}\left[k-2 \pi \frac{k}{2(4+\pi)}\right]=\frac{1}{4}\left[k-\frac{\pi k}{4+\pi}\right] \\
& =\frac{1}{4}\left[\frac{k(4+\pi)-\pi k}{4+\pi}\right]=\frac{4 k+\pi k-\pi k}{4(4+\pi)}
\end{aligned}
$

or

$
y=\frac{4 k}{4(4+\pi)}=\frac{k}{4+\pi}=2 \frac{k}{2(4+\pi)}
$

$\Rightarrow y=2 x$
(By (ii))
$\therefore z$ (sum of areas) is minimum (least) when side $(y)$ of the square is double the radius $(x)$ of the circle.

Solution: Let $x$ metres be the radius of the semi-circular opening of the window. Therefore one side of rectangle part of window is $2 x$. Let $y$ metres be the other side of the rectangle.

$
\begin{aligned}
& \therefore \text { Perimeter of window }=\text { Semi-cire } \\
& \text { are } \mathrm{AB}+\text { Length }(\mathrm{AD}+\mathrm{DC}+\mathrm{BC}) \\
& =10 \mathrm{~m} \text { (given) } \\
& \Rightarrow \quad \frac{1}{2}(2 \pi x)+y+2 x+y=10 \\
& \Rightarrow \quad \begin{array}{l}
\pi x+2 x+2 y=10 \\
\Rightarrow \quad 2 y=10-\pi x-2 x
\end{array} \\
& \Rightarrow \quad y=\frac{10-(\pi+2) x}{2}
\end{aligned}
$

(It may be noted that length (side) AB can’t be a part of the window because if it so, then light after passing through the semi-circle will not enter the rectangle part)
Let $z$ sq. $m$ be the area of the window.
For maximum light to be admitted through the window, area $z$ of window should be maximum.

$
\begin{aligned}
z=\text { Area of window } & =\text { Area of semi-circle + Area of rectangle } \\
& =\frac{1}{2}\left(\pi x^2\right)+(2 x) y
\end{aligned}
$


Putting the value of $y$ from (i),
or

$
\begin{aligned}
& z=\frac{1}{2} \pi x^2+2 x\left[\frac{10-(\pi+2) x}{2}\right]=\frac{1}{2}\left[\pi x^2+20 x-2\left(\pi+2 x^2\right]\right. \\
& z=\frac{1}{2}\left[\pi x^2+20 x-2 \pi x^2-4 x^2\right]=\frac{1}{2}\left[-\pi x^2-4 x^2+20 x\right]
\end{aligned}
$


Now $z$ is a function of $x$ alone.

$
\begin{array}{ll}
\therefore & \frac{d z}{d x}=\frac{1}{2}[-2 \pi x-8 x+20] \\
\text { and } & \frac{d^2 z}{d x^2}=\frac{1}{2}(-2 \pi-8)=\frac{-2}{2}(\pi+4)=-(\pi+4)
\end{array}
$


Putting $\quad \frac{d z}{d x}=0$ to find the critical points, we have

$
\begin{aligned}
& \frac{-2 \pi x-8 x+20}{2}=0 \Rightarrow-2 \pi x-8 x+20=0 \\
& \Rightarrow \quad-2 x(\pi+4)=-20 \Rightarrow x=\frac{20}{2(\pi+4)}=\frac{10}{\pi+4}
\end{aligned}
$


At $x=\frac{10}{\pi+4}, \frac{d^2 z}{d x^2}=-(\pi+4)$ is negative.
$\therefore z$ is maximum at $x=\frac{10}{\pi+4}$.
Substituting $x=\frac{10}{\pi+4}$ in $(i), y=\frac{10-(\pi+2) \frac{10}{\pi+4}}{2}$

$
\begin{aligned}
& =\frac{10(\pi+4)-10(\pi+2)}{2(\pi+4)}=\frac{10 \pi+40-10 \pi-20}{2 \pi+4)} \\
& =\frac{20}{2(\pi+4)}=\frac{10}{\pi+4} \mathrm{~m}
\end{aligned}
$

∴ For maximum light to be admitted through window, dimensions of the window are:
Length of rectangle $=2 x=\frac{20}{\pi+4} \mathrm{~m}$
Width of rectangle $=y=\frac{10}{\pi+4} \mathrm{~m}$
Radius of semi-circle $=x=\frac{10}{\pi+4} \mathrm{~m}$.

Solution: Let P be a point on the hypotenuse AC of a right triangle ABC such that $\mathrm{PL}(\perp \mathrm{AB})=a$ and $\mathrm{PM}(\perp \mathrm{BC})=b$.
Let $\angle \mathrm{BAC}=\angle \mathrm{MPC}=0$,
then in right-angled $\triangle \mathrm{ALP}, \frac{\mathrm{AP}}{\mathrm{PL}}=$ oosec $\theta$
$\therefore \mathrm{AP}=\mathrm{PL}$ cosee $\theta=\alpha$ cosec $\theta$
and in right-angled $\triangle P M C, \frac{P C}{P M}=\sec \theta$
$\therefore \quad \mathrm{PC}=\mathrm{PM} \sec \theta=b \sec 0$.
Let $\mathrm{AC}=z$, then

$
z=A P+P C=a \operatorname{cosec} \theta+b \sec \theta, 0<\theta<\frac{\pi}{2}
$

( $\because 6$ is an angle of right-angled triangle)
$\therefore \quad \frac{d z}{d \theta}=-a \operatorname{cosec} \theta \cot \theta+b \sec \theta \tan \theta$
For maxima or minima, put $\frac{d z}{d \theta}=0$
$\Rightarrow-a \operatorname{cosec} \theta \cot \theta+b \sec \theta \tan \theta=0$

$
\begin{aligned}
& \Rightarrow \frac{b \sin \theta}{\cos ^2 \theta}=\frac{a \cos \theta}{\sin ^2 \theta} \\
& \Rightarrow b \sin ^2 \theta=a \cos ^2 \theta \quad \Rightarrow \frac{\sin ^3 \theta}{\cos ^2 \theta}=\frac{a}{b} \\
& \Rightarrow \quad \tan ^2 \theta=\frac{a}{b} \quad \therefore \tan \theta=\left(\frac{a}{b}\right)^{1 / a}
\end{aligned}
$


$
\begin{array}{r}
\text { Now, } \quad \begin{array}{r}
\frac{d^2 z}{d \theta^2}=-a \quad\left[\operatorname{cosec} \theta\left(-\operatorname{cosec}^2 \theta\right)+\cot \theta(-\operatorname{cosec} \theta \cot \theta)\right. \\
+b\left[\sec \theta \sec ^2 \theta+\tan \theta \sec \theta \tan \theta\right]
\end{array} \\
=a\left(\operatorname{cosec}^3 \theta+\operatorname{cosec} \theta \cot ^2 \theta\right)+b\left(\sec ^3 \theta+\sec \theta \tan ^2 \theta\right)
\end{array}
$


Since $a>0, b>0$ and all $t$-ratios of $\theta$ are positive $(\because 0<\theta<\pi / 2)$

$
\therefore \quad \frac{d^2 z}{d \theta^2}>0
$

$\Rightarrow z$ is least when $\tan \theta=\left(\frac{a}{b}\right)^{1 / 3}$

$
\Rightarrow \quad \sec ^2 \theta=1+\tan ^2 \theta=1+\left(\frac{\boldsymbol{a}}{b}\right)^{2 / 3}=\frac{b^{2 / 3}+a^{2 / 3}}{b^{2 / 3}}
$
$
\Rightarrow \quad \sec \theta=\frac{\left(a^{2 / 3}+b^{2 / 2}\right)^{1 / 2}}{b^{1 / 3}}
$


Also $\operatorname{cosec}^2 \theta=1+\cot ^2 \theta=1+\left(\frac{b}{a}\right)^{2 / 3} \quad[$ By (ii) $]=\frac{a^{2 / 3}+b^{2 / 3}}{a^{2 / 3}}$

$
\Rightarrow \quad \operatorname{cosec} \theta=\frac{\left(a^{2 / 3}+b^{2 / 3}\right)^{1 / 2}}{a^{1 / 3}}
$


Putting these values of $\sec \theta$ and cosec $\theta$ in (i),
∴ Minimum length of hypotenuse z

$
\begin{aligned}
& =a \text { cosec } \theta+b \sec \theta \\
& =a \cdot \frac{\left(a^{2 / 3}+b^{2 / 3}\right)^{1 / 2}}{a^{1 / 3}}+b \cdot \frac{\left(a^{2 / 3}+b^{2 / 3}\right)^{1 / 2}}{b^{1 / 3}} \\
& =\left(a^{23}+b^{23}\right)^{1 / 2}\left(a^{27}+b^{23}\right)=\left(a^{273}+b^{20}\right)^{12}
\end{aligned}
$


Note. Since $\theta$ is a positive acute angle, we may draw a right triangle OMP as shown in the figure, then

$
\begin{aligned}
& \sec \theta=\frac{\mathrm{OP}}{\mathrm{OM}}=\frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{b^{1 / 3}} \\
& \operatorname{cosec} \theta=\frac{\mathrm{OP}}{\mathrm{PM}}=\frac{\sqrt{a^{2 / 1}+b^{2 / 1}}}{a^{1 / 2}} .
\end{aligned}
$

Solution: Given: $f(x)=(x-2)^4(x+1)^3$

$
\begin{aligned}
\therefore \quad f^{\prime}(x) & =(x-2)^4 \frac{d}{d x}(x+1)^3+\frac{d}{d x}(x-2)^4 \cdot(x+1)^2 \\
& =(x-2)^4 3(x+1)^2+4(x-2)^3(x+1)^2
\end{aligned}
$


Forming factors $=(x-2)^3(x+1)^2[3(x-2)+4(x+1)]$

$
\begin{array}{r}
\Rightarrow \quad f^{\prime}(x)=(x-2)^3(x+1)^2(7 x-2) \\
{[\because 3 x-6+4 x+4=7 x-2]}
\end{array}
$


Putting $f^{\prime}(x)=0$ to locate the critical points, we have

$
\begin{aligned}
& (x-2)^3(x+1)^2(7 x-2)=0 \\
& \therefore \quad \text { Either } x-2=0 \text { or } x+1=0 \text { or } 7 x-2=0 \\
& \Rightarrow \quad x=2, \text { or } x=-1, \text { or } x=\frac{2}{7}
\end{aligned}
$


Let us apply First derivative Test
(as we and you think that finding $f^{\prime \prime}(x)$ is tedious)
Clearly the factor $(x+1)^2$ in the value of $f^{\prime}(x)$ being square of a real number is never negative.
At $x=2$
When $x$ is slightly $<2$, (say $x=1.9$ ); from (ii)

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


When $x$ is slightly $>2$, (say $x=2.1$ ), from (ii)

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

$\therefore f^{\prime}(x)$ changes sign from $(-)$ to $(+)$ as $x$ increases through 2.
$\therefore x=2$ gives a point of local minima.
At $x=-1$
When $x$ is slightly $<-1$, (say $x=-1-0.1=-1.1$ ), from (ii)

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


When $x$ is slightly $>-1 \quad($ say $x=-1+0.1=-0.9)$,
from (ii) $f^{\prime}(x)=(-)^3(+)(-)=(+)$
$\therefore f^{\prime}(x)$ does not change sign as $x$ increases through -1 .
$\therefore x=-1$ gives a point of inflexion.
At $x=\frac{2}{7}$
When $x$ is slightly $<\frac{2}{7}$, (say $x=\frac{1}{7}$ ) from (ii)

$
f^{\prime}\left(x=(-)^3(+)(-)=(-)(+)(-)=(+)\right.
$


When $x$ is slightly $>\frac{2}{7}$, (say $x=\frac{3}{7}$ ) from (ii)

$
f^{\prime}(x)=(-)^3(+)(+)=(-)
$

$\therefore f^{\prime}(x)$ changes sign from $(+)$ to $(-)$ as $x$ increases through $\frac{2}{7}$.
$\therefore \quad x=\frac{2}{7}$ gives a point of local maxima.

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

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


Let us put $f^{\prime}(x)=0$ to locate the critical points.
$\therefore \cos x(-2 \sin x+1)=0$
∴ Either cos $x=0$ or $-2 \sin x+1=0$
i.e., $x=\frac{\pi}{2}$ or $-2 \sin x=-1$ i.e., $\sin x=\frac{1}{2}$.
Now $\sin x=\frac{1}{2}$ is positive and hence $x$ lies in Ist quadrant and second quadrant.
$
\begin{aligned}
\therefore \quad & \sin x \\
\therefore \quad & \frac{1}{2}=\sin \frac{\pi}{6} \text { and } \sin \left(\pi-\frac{\pi}{6}\right)=\sin \frac{5 \pi}{6} . \\
\therefore \quad x & =\frac{\pi}{6} \text { and } x=\frac{5 \pi}{6}
\end{aligned}
$

∴ Critical points are $x=\frac{\pi}{2}, x=\frac{\pi}{6}$ and $x=\frac{5 \pi}{6} \in[0, \pi]$
Let us find values of $f(x)$ at these critical points.

$
\begin{aligned}
& \therefore \text { From }(i), f\left(\frac{\pi}{2}\right)=\cos ^2 \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1 \\
& 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{3}{4}+\frac{1}{2}=\frac{5}{4} \\
& f\left(\frac{5 \pi}{6}\right)=\cos ^2 \frac{5 \pi}{6}+\sin \frac{\pi}{6}=\left(\frac{-\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{3}{4}+\frac{1}{2}=\frac{5}{4} \\
& {\left[\because \cos \frac{5 \pi}{6}=\cos \frac{6 \pi-\pi}{6}=\cos \left(\pi-\frac{\pi}{6}\right)=-\cos \frac{\pi}{6}=\frac{-\sqrt{3}}{2}\right]}
\end{aligned}
$


Now let us find values of $f(x)$ at the end points $x=0$ and $x=\pi$ of closed interval $[0, \pi]$
From (i), $f(0)=\cos ^2 0+\sin 0=1+0=1$
and $\quad f(\pi)=\cos ^2 \pi+\sin \pi=(-1)^2+0=1$

$
\begin{aligned}
& \text { I: } \quad \cos \pi=\cos 180^{\circ}=\cos \left(180^{\circ}-0\right)=-\cos 0=-1 \\
& \text { and } \left.\sin \pi=\sin 180^{\circ}=\sin \left(180^{\circ}-0\right)=\sin 0=0\right]
\end{aligned}
$


Therefore absolute maximum is $\frac{5}{4}$ and absolute minimum is 1 .

Solution: Let $x$ be the radius of base of cone and $y$ be the height of the cone inscribed in a sphere of radius $r$.

$
\therefore \quad \mathrm{OD}=\mathrm{AD}-\mathrm{AO}=y-r
$


In right-angled $\triangle \mathrm{OBD}$,

$
\mathrm{OD}^2+\mathrm{BD}^2=\mathrm{OB}^2
$

(By Pythagoras Theorem)

$
(y-r)^2+x^2=r^2
$

or

$
\begin{aligned}
y^2+r^2-2 r y+x^2 & =r^2 \\
x^2 & =2 r y-y^2
\end{aligned}
$

or
Let V denote the volume of the cone.

$
\therefore \quad \mathrm{v}=\frac{1}{3} \pi x^2 y=\frac{1}{3} \pi\left(2 r y-y^2\right) y
$
or

$
V=\frac{\pi}{3}\left(2 r y^2-y^3\right)
$


Diff. with respect to $y, \frac{d V}{d y}=\frac{\pi}{3}\left(4 r y-3 y^2\right)$
and $\quad \frac{d^2 V}{d y^2}=\frac{\pi}{3}(4 r-6 y)$
Put $\quad \frac{d V}{d y}=0 \quad \therefore \frac{\pi}{3}\left(4 \rho y-3 y^2\right)=0$
or $\frac{\pi y}{3}(4 r-3 y)=0$
But $\frac{\pi y}{3} \neq 0 \quad \therefore 4 r-3 y=0$ or $y=\frac{4 r}{3}$
At $\quad y=\frac{4 r}{3}, \quad \frac{d^2 V}{d y^2}=\frac{\pi}{3}(4 r-8 r)=-\frac{4 \pi r}{3}<0$
$\therefore \quad V$ is maximum at $y=\frac{4 r}{3}$.

Solution: Let I denote the interval ( $a, b$ ).
Given: $f^{\prime}(x)>0$ for all $x$ in an interval $L$
Let $x_1, x_2 \in I$ with $x_1Since derivability implies continuity, therefore $f(x)$ is continuous in the closed interval $\left[x_1, x_2\right]$ and derivable in the open interval $\left(x_1, x_2\right)$.
∴ By Lagrange’s Mean Value Theorem, we have

$
\begin{aligned}
& \quad \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=f^{\prime}(c), \text { where } x_1& \quad f^{\prime}\left(x_2\right)-f\left(x_1\right)=\left(x_2-x_1\right) f^{\prime}(c) \text { where } x_1& \text { Now } \quad x_1& \text { Also } \quad f^{\prime}(x)>0 \text { for all } x \text { in } \mathbf{I}(\text { given }) \quad \Rightarrow x_2-x_1>0 \\
& \therefore \quad \text { From }(i), \quad f\left(x_2\right)-f\left(x_1\right)>0 \quad \text { or }\left(f\left(x_1\right)\end{aligned}
$


Thus, for every pair of points $x_1, x_2 \in \mathbf{I}, x_1
$
\Rightarrow \quad f\left(x_1\right)$


Hence, $f(x)$ is strictly increasing in $\mathbf{I}$.
Remark. Geometrically. Let the graph of a strictly increasing function $y=f(x)$ be represented by curve AB. The tangent at any point $P$ on the curve makes an acute angle $\psi$ with the $x$-axis, $\left(\right.$ where $\left.0<\psi<\frac{\pi}{2}\right)$.
$0<\psi<\frac{\pi}{2} \Rightarrow \tan \psi>0$ i.e., slope of the tangent is $>0$
$\Rightarrow f^{\prime \prime}(x)>0$.
Remark. Graph of a strictly increasing function is a rising graph i.e., graph moves up as $x$ moves to the right.

Solution: Let $x$ be the base radius and $y$ be the height of the cylinder inscribed in a sphere having centre $O$ and radius $R$. ( $x>0, y>0$ )
In right-angled $\triangle O A M$, By Pythagoras Theorem, $\mathrm{AM}^2+ O M^2=O A^2$
i.e. $x^2+\left(\frac{y}{2}\right)^2=\mathrm{R}^2$
$\therefore \quad x^2=\mathrm{R}^2-\frac{y^2}{4}$

Let V denote the volume of right circular cylinder,

$
\therefore \quad \mathbf{V}=\pi x^2 y
$


Putting the value of $x^2$ from (i) in (ii),

$
\begin{aligned}
& V & =\pi\left(R^2-\frac{y^2}{4}\right) y=\pi\left(R^2 y-\frac{1}{4} y^3\right) \\
\therefore & \frac{d V}{d y} & =\pi\left(R^2-\frac{3}{4} y^2\right), \\
\text { and } & \frac{d^2 V}{d y^2} & =\pi\left(-\frac{3}{2} y\right)=-\frac{3 \pi y}{2} \\
\text { Put } & \frac{d V}{d y} & =0 \quad \therefore \pi\left(R^2-\frac{3}{4} y^2\right)=0
\end{aligned}
$


But $\quad \pi \neq 0 \quad \therefore R^2-\frac{3}{4} y^2=0$ or $R^2=\frac{3}{4} y^2$
or

$
y^2=\frac{4 R^2}{3} \quad \therefore y=\frac{2 R}{\sqrt{3}}
$


At

$
\begin{aligned}
y & =\frac{2 R}{\sqrt{3}}, \frac{d^2 V}{d y^2}=-\frac{3 \pi}{2} y \\
& =-\frac{3 \pi}{2}\left(\frac{2 R}{\sqrt{3}}\right)=-\pi R \sqrt{3}
\end{aligned}
$

this is negative.
$\therefore V$ is maximum at

$
y=\frac{2 r}{\sqrt{3}}
$


Substituting $y=\frac{2 R}{\sqrt{3}}$ in eqn. (iii),

$
\begin{aligned}
\text { Maximum volume of cylinder } & =\pi\left[R^2 \cdot \frac{2 R}{\sqrt{3}}-\frac{1}{4} \cdot \frac{4 R^2}{3} \cdot \frac{2 R}{\sqrt{3}}\right] \\
& =\pi R^2 \frac{2 R}{\sqrt{3}}\left(1-\frac{1}{3}\right)=\frac{2 \pi R^3}{\sqrt{3}} \cdot \frac{2}{3}=\frac{4 \pi R^3}{3 \sqrt{3}}
\end{aligned}
$

Solution: Let $r$ be the radius of the given right circular cone of given height $h$.
Let the radius of the inscribed cylinder be $x$ and its height be $y$. In similar triangles APQ and ARC, we have

$
\frac{\mathrm{PQ}}{\mathrm{RC}}=\frac{\mathrm{AP}}{\mathrm{AR}}
$

i.e.,

$
\frac{x}{r}=\frac{h-y}{h}
$


$
\mathbf{I} \because \mathbf{A P}=\mathbf{A R}-\mathbf{P R}=h-y]
$


Cross-multiplying, $h x=r h-r y$

$
\begin{array}{rlrl}
\therefore & & r y & =r h-h x=h(r-x) \\
\therefore & & y=\frac{h}{r}(r-x)
\end{array}
$

…(i) (given condition)
Let us call the volume $z$ of the cylinder
∴ $\quad z=\pi x^2 y$
. (ii) [Here $z$ is to be maximised]
Putting the value $y$ from (i) in (ii),

$
\begin{array}{rlr}
z & =\pi x^2 \frac{h}{r}(r-x) & =\frac{\pi h}{r}\left(x^2-x^3\right) \\
\therefore \quad \frac{d z}{d x} & =\frac{\pi h}{r}\left(2 r x-3 x^2\right), \frac{d^2 z}{d x^2} & =\frac{\pi h}{r}(2 r-6 x)
\end{array}
$
To find maxima or minima put $\frac{d z}{d x}=0$

$
\therefore \quad \frac{\pi h}{r}\left(2 r x-3 x^2\right)=0 \quad \text { or } \quad \frac{\pi h}{r} x(2 r-3 x)=0
$


But $x \neq 0, \therefore 2 r-3 x=0$ or $3 x=2 r$ or $x=\frac{2 r}{3}$
At $x=\frac{2 r}{3}, \frac{d^2 z}{d x^2}=\frac{\pi h}{r}\left(2 r-\frac{12 r}{3}\right)=\frac{\pi h}{r}(-2 r)=-2 \pi / h$ this is negative.
$\therefore z$ is maximum at $x=\frac{2 r}{3}$.
Substituting $x=\frac{2 r}{3}$ in $(i), y=\frac{h}{r}\left(r-\frac{2 r}{3}\right)=\frac{h}{r} \cdot \frac{r}{3}=\frac{h}{3}$
Substituting $x=\frac{2 r}{3}$ in eqn. (iii),
Maximum volume of cylinder $=\frac{\pi h}{r}\left[r \cdot \frac{4 r^2}{9}-\frac{8 r^2}{27}\right]$

$
\begin{aligned}
& =\frac{\pi h}{r} r^3\left(\frac{4}{9}-\frac{8}{27}\right)=\pi h r^2\left(\frac{12-8}{27}\right) \\
& =\frac{4}{27} \pi h r^2=\frac{4}{27} \pi h\left(h \tan \omega^2\right. \\
& {\left[\because \operatorname{In} \Delta A R C, \frac{R C}{A R}=\tan \alpha \text { i.e., } \frac{r}{h}=\tan \alpha . \therefore r=h \tan \alpha\right]} \\
& =\frac{4}{27} \pi h^3 \tan ^2 \alpha .
\end{aligned}
$


Choose the correct answer in the Exercises from 19 to 24:

Solution: Let $y \mathrm{~m}$ be the depth of the wheat in the cylindrical tank of radius 10 m at time $t$.
$\therefore \mathrm{V}=$ Volume of wheat in
cylindrical tank at time $t$

$
\begin{aligned}
& =\pi(10)^2 y \\
& =100 \pi y \mathrm{cu} . \mathrm{m}
\end{aligned}
$
Given: rate of increase (:- wheat in being filled in the tank) of volume of wheat $=\frac{d \mathrm{~V}}{d t}=314 \mathrm{cu} . \mathrm{m} / \mathrm{hr}$.

$
\begin{aligned}
& \Rightarrow \quad \frac{d}{d t}(100 \pi y)(\mathrm{By}(i))=314 \\
& \Rightarrow \quad 100 \pi \frac{d y}{d t}=314 \\
& \text { Using } \pi=\frac{22}{7}=3.14 \text { nearly, } \\
& \Rightarrow \quad 100(3.14) \frac{d y}{d t}=314 \quad \Rightarrow \quad 314 y=314 \\
& \Rightarrow \quad y=\frac{314}{314}=1 \mathrm{~m} / \mathrm{h}
\end{aligned}
$

Hence, option (A) is correct.