Exercise 6.3 (Revised) - Chapter 6 - Applications Of Derivatives - Ncert Solutions class 12 - Maths
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NCERT Class 12 Maths Solutions: Chapter 6 - Applications Of Derivatives
Ex 6.3 Question 1:
Find the maximum and minimum values, if any, of the following functions given by
(i) $f(x)=(2 x-1)^2+3$
(ii) $f(x)=9 x^2+12 x+2$
(iii) $f(x)=-(x-1)^2+10$
(iv) $g(x)=x^3+1$
Answer :
(i) The given function is $f(x)=(2 x-1)^2+3$.
It can be observed that $(2 x-1)^2 \geq 0$ for every $x \in \mathbf{R}$.
Therefore, $f(x)=(2 x-1)^2+3 \geq 3$ for every $x \in R$.
The minimum value of $f$ is attained when $2 x-1=0$.
$
2 x-1=0 \Rightarrow x=\frac{1}{2}
$
$\therefore$ Minimum value of $f=f\left(\frac{1}{2}\right)=\left(2 \cdot \frac{1}{2}-1\right)^2+3=3$
Hence, function $f$ does not have a maximum value.
(ii) The given function is $f(x)=9 x^2+12 x+2=(3 x+2)^2-2$.
It can be observed that $(3 x+2)^2 \geq 0$ for every $x \in \mathbf{R}$.
Therefore, $f(x)=(3 x+2)^2-2 \geq-2$ for every $x \in R$.
The minimum value of $f$ is attained when $3 x+2=0$.
$
3 x+2=0 \Rightarrow x=\frac{-2}{3}
$
$\therefore$ Minimum value of $f=f\left(-\frac{2}{3}\right)=\left(3\left(\frac{-2}{3}\right)+2\right)^2-2=-2$
Hence, function $f$ does not have a maximum value.
(iii) The given function is $f(x)=-(x-1)^2+10$.
It can be observed that $(x-1)^2 \geq 0$ for every $x \in \mathbf{R}$.
Therefore, $f(x)=-(x-1)^2+10 \leq 10$ for every $x \in \mathbf{R}$.
The maximum value of $f$ is attained when $(x-1)=0$.
$
(x-1)=0 \Rightarrow x=1
$
$\therefore$ Maximum value of $f=f(1)=-(1-1)^2+10=10$
Hence, function $f$ does not have a minimum value.
(iv) The given function is $g(x)=x^3+1$.
Hence, function $g$ neither has a maximum value nor a minimum value.
Ex 6.3 Question 2 :
Find the maximum and minimum values, if any, of the following functions given by
(i) $f(x)=|x+2|-1$
(ii) $g(x)=-|x+1|+3$
(iii) $h(x)=\sin (2 x)+5$ (iv) $f(x)=|\sin 4 x+3|$
(v) $h(x)=x+4, x \in(-1,1)$
Answer :
(i) $f(x)=|x+2|-1$
We know that $|x+2| \geq 0$ for every $x \in \mathbf{R}$.
Therefore, $f(x)=|x+2|-1 \geq-1$ for every $x \in \mathbf{R}$.
The minimum value of $f$ is attained when $|x+2|=0$.
$
\begin{aligned}
& |x+2|=0 \\
& \Rightarrow x=-2
\end{aligned}
$
$\therefore$ Minimum value of $f=f(-2)==|-2+2|-1=-1$
Hence, function $f$ does not have a maximum value.
(ii) $g(x)=-|x+1|+3$
We know that $-|x+1| \leq 0$ for every $x \in \mathbf{R}$.
Therefore, $g(x)=-|x+1|+3 \leq 3$ for every $x \in \mathbf{R}$.
The maximum value of $g$ is attained when $|x+1|=0$.
$
\begin{aligned}
& |x+1|=0 \\
& \Rightarrow x=-1
\end{aligned}
$
$\therefore$ Maximum value of $g=g(-1)=-|-1+1|+3=3$
Hence, function $g$ does not have a minimum value.
(iii) $h(x)=\sin 2 x+5$
We know that $-1 \leq \sin 2 x \leq 1$.
$
\begin{aligned}
& \Rightarrow-1+5 \leq \sin 2 x+5 \leq 1+5 \\
& \Rightarrow 4 \leq \sin 2 x+5 \leq 6
\end{aligned}
$
Hence, the maximum and minimum values of $h$ are 6 and 4 respectively.
(iv) $f(x)=|\sin 4 x+3|$
We know that $-1 \leq \sin 4 x \leq 1$.
$
\begin{aligned}
& \Rightarrow 2 \leq \sin 4 x+3 \leq 4 \\
& \Rightarrow 2 \leq|\sin 4 x+3| \leq 4
\end{aligned}
$
Hence, the maximum and minimum values of $f$ are 4 and 2 respectively.
(v) Given: $h(x)=x+1, x \in(-1,1)$
Since $-1 Adding 1 to both sides, $-1+1 $
\Rightarrow 0 $
Therefore, neither minimum value not maximum value of $h(x)$ exists.
Ex 6.3 Question 3 :
Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i). $f(x)=x^2$
(ii). $g(x)=x^3-3 x$
(iii). $h(x)=\sin x+\cos x, 0 (iv). $f(x)=\sin x-\cos x, 0 (v). $f(x)=x^3-6 x^2+9 x+15$
(vi). $g(x)=\frac{x}{2}+\frac{2}{x}, x>0$
(vii). $g(x)=\frac{1}{x^2+2}$
(viii). $f(x)=x \sqrt{1-x}, x>0$
Answer:
(i) $f(x)=x^2$
$
\therefore f^{\prime}(x)=2 x
$
Now,
$
f^{\prime}(x)=0 \Rightarrow x=0
$
Thus, $x=0$ is the only critical point which could possibly be the point of local maxima or local minima of $f$.
We have $f^{\prime \prime}(0)=2$, which is positive.
Therefore, by second derivative test, $x=0$ is a point of local minima and local minimum value of $f$ at $x=0$ is $f(0)=0$.
$
\begin{aligned}
& \text { (ii) } g(x)=x^3-3 x \\
& \therefore g^{\prime}(x)=3 x^2-3
\end{aligned}
$
Now,
$
\begin{aligned}
& g^{\prime}(x)=0 \Rightarrow 3 x^2=3 \Rightarrow x= \pm 1 \\
& g^{\prime}(x)=6 x \\
& g^{\prime}(1)=6>0 \\
& g^{\prime}(-1)=-6<0
\end{aligned}
$
By second derivative test, $x=1$ is a point of local minima and local minimum value of $g$ at $x=1$ is $g(1)=1^3-3=1-3=-2$. However,
$x=-1$ is a point of local maxima and local maximum value of $g$ at
$
x=-1 \text { is } g(1)=(-1)^3-3(-1)=-1+3=2 \text {. }
$
$
\begin{aligned}
& \text { (iii) } h(x)=\sin x+\cos x, 0 & \therefore h^{\prime}(x)=\cos x-\sin x \\
& h^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right) \\
& h^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x) \\
& h^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0
\end{aligned}
$
Therefore, by second derivative test, $x=\frac{\pi}{4}$ is a point of local maxima and the local maximum value
of $h$ at $x=\frac{\pi}{4}$ is $h\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$.
(iv) $f(x)=\sin x-\cos x, 0$
$
f(x)=\sin x-\cos x, \quad 0 $
Finding $f^{\prime}(x)$
$
\begin{aligned}
& \mathrm{f}^{\prime}(x)=\cos x-(-\sin x) \\
& \mathrm{f}^{\prime}(x)=\cos x+\sin x
\end{aligned}
$
Putting $\mathrm{f}^{\prime}(x)=0$
$
\begin{aligned}
& \cos x+\sin x=0 \\
& \cos x=-\sin x
\end{aligned}
$
$
\begin{gathered}
1=\frac{-\sin x}{\cos x} \\
\frac{-\sin x}{\cos x}=1 \\
-\tan x=1 \\
\tan x=-1
\end{gathered}
$
Since $0
So, value of $x$ is
$
x=\frac{3 \pi}{4} \text { or } \frac{7 \pi}{4}
$
Now finding $\mathrm{f}^{\prime \prime}(x)$
$
\begin{aligned}
& f^{\prime \prime}(x)=\frac{d(\cos x+\sin x)}{d x} \\
& f^{\prime \prime}(x)=-\sin x+\cos x
\end{aligned}
$
Putting $x=\frac{3 \pi}{4}$
$
\begin{aligned}
f^{\prime \prime}\left(\frac{3 \pi}{4}\right) & =-\sin \left(\frac{3 \pi}{4}\right)+\cos \left(\frac{3 \pi}{4}\right) \\
& =-\sin \left(\pi-\frac{\pi}{4}\right)+\cos \left(\pi-\frac{\pi}{4}\right) \\
& =-\sin \left(\frac{\pi}{4}\right)+\left(-\cos \frac{\pi}{4}\right) \\
& =\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\
& =\frac{-2}{\sqrt{2}}
\end{aligned}
$
$
\begin{array}{r}
=-\sqrt{2} \\
<0
\end{array}
$
Hence $\mathrm{f}^{\prime \prime}(x)<0$ when $x=\frac{3 \pi}{4}$
$\begin{aligned}
& \text { As } \sin (180-\theta)=\sin \theta \\
& \& \cos (180-\theta)=-\cos \theta
\end{aligned}$
Thus $x=\frac{3 \pi}{4}$ is point of local maxima
$\therefore \mathrm{f}(x)$ is maximum value at $x=\frac{3 \pi}{4}$
The local maximum value is
$
\begin{aligned}
f(x) & =\sin x-\cos x \\
f\left(\frac{3 \pi}{4}\right) & =\sin \left(\frac{3 \pi}{4}\right)-\cos \left(\frac{3 \pi}{4}\right) \\
& =\sin \left(\pi-\frac{\pi}{4}\right)-\cos \left(\pi-\frac{\pi}{4}\right) \\
& =\sin \left(\frac{\pi}{4}\right)-\left(-\cos \frac{\pi}{4}\right)
\end{aligned}
$
$
\begin{aligned}
& =\sin \frac{\pi}{4}+\cos \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\
& =\frac{2}{\sqrt{2}} \\
& =\sqrt{2}
\end{aligned}
$
Now, for $x=\frac{7 \pi}{4}$
$
f^{\prime \prime}(x)=-\sin x+\cos x
$
Putting $x=\frac{7 \pi}{4}$
$
f^{\prime \prime}\left(\frac{7 \pi}{4}\right)=-\sin \left(\frac{7 \pi}{4}\right)+\cos \left(\frac{7 \pi}{4}\right)
$
Putting $x=\frac{7 \pi}{4}$
$
\begin{aligned}
& f^{\prime \prime}\left(\frac{7 \pi}{4}\right)=-\sin \left(\frac{7 \pi}{4}\right)+\cos \left(\frac{7 \pi}{4}\right) \\
& f^{\prime \prime}\left(\frac{7 \pi}{4}\right)=-\sin \left(2 \pi-\frac{\pi}{4}\right)+\cos \left(2 \pi-\frac{\pi}{4}\right)
\end{aligned}
$
$\begin{aligned}
& \text { As } \sin (2 \pi-\theta)=-\sin \theta \\
& \& \cos (2 \pi-\theta)=\cos \theta
\end{aligned}$
$
\begin{aligned}
& =-\left(-\sin \left(\frac{\pi}{4}\right)\right)+\cos \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} \\
& >0
\end{aligned}
$
$\mathrm{f}^{\prime \prime}(x)>0$ when $x=\frac{7 \pi}{4}$
Thus $x=\frac{7 \pi}{4}$ is point of local minima $\mathrm{f}(x)$ has minimum value at $x=\frac{7 \pi}{4}$
Local minimum value is
$
\begin{aligned}
f(x) & =\sin x-\cos x \\
f\left(\frac{7 \pi}{4}\right) & =\sin \left(\frac{7 \pi}{4}\right)-\cos \left(\frac{7 \pi}{4}\right) \\
& =\sin \left(2 \pi-\frac{\pi}{4}\right)-\cos \left(2 \pi-\frac{\pi}{4}\right) \\
& =-\sin \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{4}\right) \\
& =\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\
& =\frac{-2}{\sqrt{2}} \\
& =-\sqrt{2}
\end{aligned}
$
Thus, $\mathrm{f}(x)$ is maximum at $\mathrm{x}=\frac{3 \pi}{4}$ and maximum value is $\sqrt{2}$ $\& \mathrm{f}(x)$ is minimum at $\mathrm{x}=\frac{7 \pi}{4}$ and maximum value is $-\sqrt{2}$
(v) $f(x)=x^3-6 x^2+9 x+15$
$
f(x)=x^3-6 x^2+9 x+15
$
Finding $\mathbf{f}^{\prime}(x)$
$
\begin{aligned}
& f^{\prime}(x)=\frac{d\left(x^3-6 x^2+9 x+15\right)}{d x} \\
& f^{\prime}(x)=3 x^2-12 x+9 \\
& f^{\prime}(x)=3\left(x^2-4 x+3\right)
\end{aligned}
$
Putting $\mathrm{f}^{\prime}(\boldsymbol{x})=\mathbf{0}$
$
\begin{aligned}
& 3\left(x^2-4 x+3\right)=0 \\
& x^2-4 x+3=0 \\
& x^2-3 x-x+3=0 \\
& x(x-3)-1(x-3)=0 \\
& (x-1)(x-3)=0
\end{aligned}
$
So, $x=1 \quad \& x=3$
Finding $\mathrm{f}^{\prime \prime}(x)$
$
f^{\prime}(x)=3\left(x^2-4 x+3\right)
$
$
\begin{aligned}
f^{\prime \prime}(x) & =\frac{d\left(3\left(x^2-4 x+3\right)\right)}{d x} \\
& =3(2 x-4+0) \\
& =6 x-12
\end{aligned}
$
Putting $x=1$ in $\mathrm{f}^{\prime \prime}(x)$
$
\begin{aligned}
f^{\prime \prime}(1) & =6(1)-12 \\
& =6-12 \\
& =-6 \\
& <0
\end{aligned}
$
Since $\mathrm{f}^{\prime \prime}(x)<0$ when $x=1$
$\Rightarrow x=1$ is point of local maxima
$\therefore \mathrm{f}(x)$ is maximum at $\boldsymbol{x}=\mathbf{1}$
Maximum value of $\mathrm{f}(x)$ at $x=1$
$
\begin{aligned}
f(x) & =x^3-6 x^2+9 x+15 \\
f(1) & =(1)^3-6(1)^2+9(1)+15 \\
& =1-6+9+15=19
\end{aligned}
$
Putting $x=3$ in $\mathrm{f}^{\prime \prime}(\mathrm{x})$
$
\begin{aligned}
f^{\prime \prime}(x) & =6 x-12 \\
f^{\prime \prime}(3) & =6(3)-12 \\
& =18-12=6 \\
& >0
\end{aligned}
$
Since $\mathrm{f}^{\prime \prime}(x)>0$ when $x=3$
$\Rightarrow x=3$ is point of local minima
$\therefore \mathrm{f}(x)$ is minimum at $\boldsymbol{x}=\mathbf{3}$
Minimum value of $\mathrm{f}(x)$ at $x=3$
$
\begin{aligned}
f(x) & =x^3-6 x^2+9 x+15 \\
f(3) & =(3)^3-6(3)^2+9(3)+15 \\
& =27-54+27+15=15
\end{aligned}
$
(vi) $g(x)=\frac{x}{2}+\frac{2}{x}, \quad x>0$
$
\mathrm{g}(x)=\frac{x}{2}+\frac{2}{x}, \quad x>0
$
Finding $\mathrm{g}^{\prime}(x)$
$
\begin{aligned}
g^{\prime}(x) & =\frac{d}{d x}\left(\frac{x}{2}+\frac{2}{x}\right) \\
& =\frac{d}{d x}\left(\frac{x}{2}\right)+\frac{d}{d x}\left(2 x^{-1}\right) \\
& =\frac{1}{2}-2 x^{-2} \\
& =\frac{1}{2}-\frac{2}{x^2}
\end{aligned}
$
Putting $\mathrm{g}^{\prime}(\boldsymbol{x})=\mathbf{0}$
$
\begin{aligned}
& \frac{1}{2}-\frac{2}{x^2}=0 \\
& \frac{x^2-4}{2 x^2}=0 \\
& x^2-4=0 \times 2 x^2 \\
& (x-2)(x+2)=0
\end{aligned}
$
So, $x=2$ \& $x=-2$
Since $x>0$ is given, we consider only $x=2$
Finding $\mathrm{g}^{\prime \prime}(\boldsymbol{x})$
$
\begin{aligned}
g^{\prime}(x) & =\frac{1}{2}-\frac{2}{x^2} \\
g^{\prime \prime}(x) & =\frac{d}{d x}\left(\frac{1}{2}-\frac{2}{x^2}\right) \\
& =0-2 \cdot(-2) x^{-2-1} \\
& =4 x^{-3} \\
& =\frac{4}{x^3}
\end{aligned}
$
Putting value $x=2$ in $\mathrm{g}^{\prime \prime}(x)$
$
g^{\prime \prime}(x)=\frac{4}{(2)^3}=\frac{4}{8}=\frac{1}{2}
$
Since $g^{\prime \prime}(x)>0$, for $x=2$
$g(x)$ is minimum at $x=2$
Minimum value of $\mathrm{g}(x)$ is
$
g(x)=\frac{x}{2}+\frac{2}{x}
$
Putting $x=2$
$
\begin{aligned}
g(2) & =\frac{2}{2}+\frac{2}{2} \\
& =1+1 \\
& =2
\end{aligned}
$
(vii) $g(x)=\frac{1}{x^2+2}$
Finding $\mathrm{g}^{\prime}(x)$
$
\begin{aligned}
& \mathrm{g}^{\prime}(x)=\frac{d}{d x}\left(\frac{1}{x^2+2}\right) \\
& \mathrm{g}^{\prime}(x)=\frac{d\left(x^2+2\right)^{-1}}{d x} \\
& \mathrm{~g}^{\prime}(x)=-1\left(x^2+2\right)^{-1-1} \times(2 x+0) \\
& \mathrm{g}^{\prime}(x)=-2 x\left(x^2+2\right)^{-2}
\end{aligned}
$
$
\mathrm{g}^{\prime}(x)=\frac{-2 x}{\left(x^2+2\right)^2}
$
Putting $\mathrm{g}^{\prime}(\boldsymbol{x})=\mathbf{0}$
$
\begin{aligned}
& \frac{-2 x}{\left(x^2+2\right)^2}=0 \\
& -2 x=0 \times\left(x^2+2\right)^2 \\
& -2 x=0 \\
& x=0
\end{aligned}
$
Finding $\mathrm{g}^{\prime \prime}(x)$
$
\mathrm{g}^{\prime}(x)=\frac{-2 x}{\left(x^2+2\right)^2}
$
Using quotient rule
$
\text { as }\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-v^{\prime} u}{v^2}
$
$g^{\prime \prime}(x)=\frac{\frac{d(-2 x)}{d x} \cdot\left(x^2+2\right)^2-\frac{d\left(x^2+2\right)^2}{d x} \cdot(-2 x)}{\left(\left(x^2+2\right)^2\right)^2}$
$\begin{aligned}
& =\frac{-2\left(x^2+2\right)^2-2\left(x^2+2\right)^{2-1} \cdot \frac{d\left(x^2+2\right)}{d x} \cdot(-2 x)}{\left(\left(x^2+2\right)^2\right)^2} \\
& =\frac{-2\left(x^2+2\right)^2-2\left(x^2+2\right)(2 x+0)(-2 x)}{\left(x^2+2\right)^4} \\
& =\frac{-2\left(x^2+2\right)^2-2\left(x^2+2\right)(2 x)(-2 x)}{\left(x^2+2\right)^4} \\
& =\frac{-2\left(x^2+2\right)^2+8 x^2\left(x^2+2\right)}{\left(x^2+2\right)^4} \\
& =\frac{-2\left(x^2+2\right)\left[\left(x^2+2\right)-4 x^2\right]}{\left(x^2+2\right)^4} \\
& =\frac{-2\left(x^2+2\right)\left(-3 x^2+2\right)}{\left(x^2+2\right)^4} \\
& =\frac{-2\left(-3 x^2+2\right)}{\left(x^2+2\right)^3}
\end{aligned}$
Putting $x=0$ in $g^{\prime \prime}(x)$
$
g^{\prime \prime}(0)=\frac{-2(-3(0)+2)}{\left(0^2+2\right)^3}=\frac{-2(0+2)}{(2)^3}=\frac{-4}{8}=\frac{-1}{2}
$
Hence $\mathrm{g}^{\prime \prime}(x)<0$ when $x=0$
$\therefore \quad x=0$ is point of local maxima
Thus, $\mathrm{g}(x)$ is maximum at $x=0$
Maximum value of $g(x)$ at $x=0$
$
\begin{aligned}
& g(x)=\frac{1}{x^2+2} \\
& g(0)=\frac{1}{0^2+2}=\frac{1}{2}
\end{aligned}
$
Maximum value is $\frac{1}{2}$
(viii) $\mathrm{f}(x)=x \sqrt{1-x}, x>0$
Finding $f^{\prime}(x)$
$
f^{\prime}(x)=\frac{d(x \sqrt{1-x})}{d x}
$
Using product rule as $(u v)^{\prime}=u^{\prime} v+v^{\prime} u$
$
\begin{aligned}
f^{\prime}(x) & =\frac{d(x)}{d x} \cdot \sqrt{1-x}+\frac{d(\sqrt{1-x})}{d x} \cdot x \\
& =1 \cdot \sqrt{1-x}+\frac{1}{2 \sqrt{1-x}} \cdot \frac{d(1-x)}{d x} \cdot x
\end{aligned}
$
$
\begin{aligned}
& =\sqrt{1-x}+\frac{1}{2 \sqrt{1-x}}(0-1) \cdot x \\
& =\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}} \\
& =\frac{2(\sqrt{1-x})^2-x}{2 \sqrt{1-x}} \\
& =\frac{2(1-x)-x}{2 \sqrt{1-x}} \\
& =\frac{2-2 x-x}{2 \sqrt{1-x}} \\
& =\frac{2-3 x}{2 \sqrt{1-x}}
\end{aligned}
$
Putting $f^{\prime}(x)=0$
$
\begin{aligned}
& \frac{2-3 x}{2 \sqrt{1-x}}=0 \\
& 2-3 x=0 \times 2 \sqrt{1-x}
\end{aligned}
$
$
\begin{aligned}
& 2-3 x=0 \\
& -3 x=-2 \\
& x=\frac{2}{3}
\end{aligned}
$
Finding $\mathrm{f}^{\prime \prime}(x)$
$
\begin{aligned}
& f^{\prime}(x)=\frac{2-3 x}{2 \sqrt{1-x}} \\
& f^{\prime \prime}(x)=\frac{d}{d x}\left(\frac{2-3 x}{2 \sqrt{1-x}}\right)
\end{aligned}
$
Using quotient rule
$
\text { as }\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-v^{\prime} u}{v^2}
$
$\begin{aligned}
& =\frac{1}{2}\left[\frac{\frac{d(2-3 x)}{d x} \cdot \sqrt{1-x}-\frac{d(\sqrt{1-x})}{d x} \cdot(2-3 x)}{(\sqrt{1-x})^2}\right] \\
& =\frac{1}{2}\left[\frac{(0-3) \sqrt{1-x}-\frac{1}{2 \sqrt{1-x}} \cdot \frac{d(1-x)}{d x} \cdot(2-3 x)}{(1-x)}\right]
\end{aligned}$
$\begin{aligned}
& =\frac{1}{2}\left[\frac{-3 \sqrt{1-x}-\frac{1}{2 \sqrt{1-x}}(0-1) \cdot(2-3 x)}{(1-x)}\right] \\
& =\frac{1}{2}\left[\frac{-3 \sqrt{1-x}+\frac{2-3 x}{2 \sqrt{1-x}}}{1-x}\right] \\
& =\frac{1}{2}\left[\frac{(-3 \sqrt{1-x})(2 \sqrt{1-x})+2-3 x}{2(1-x) \sqrt{1-x}}\right] \\
& =\frac{1}{2}\left[\frac{-6(1-x)+2-3 x}{2(1-x) \sqrt{1-x}}\right] \\
& =\frac{1}{2}\left[\frac{-6+6 x+2-3 x}{2(1-x) \sqrt{1-x}}\right] \\
& =\frac{1}{4}\left[\frac{-4+3 x}{(1+x)^{\frac{3}{2}}}\right]
\end{aligned}$
Hence, $\mathrm{f}^{\prime \prime}(x)=\frac{1}{4}\left[\frac{-4+3 x}{(1+x)^{\frac{3}{2}}}\right]$
Putting $x=\frac{2}{3}$
$
\begin{aligned}
f^{\prime \prime}\left(\frac{2}{3}\right) & =\frac{1}{4}\left[\frac{-4+3\left(\frac{2}{3}\right)}{\left(1+\frac{2}{3}\right)^{\frac{3}{2}}}\right] \\
& =\frac{1}{4}\left[\frac{-4+2}{\left(\frac{5}{3}\right)^{\frac{3}{2}}}\right] \\
& =\frac{1}{4}\left[\frac{-2}{\left(\frac{5}{3}\right)^{\frac{3}{2}}}\right] \\
& =\frac{-1}{2}\left(\frac{3}{5}\right)^{\frac{3}{2}} \\
& <0
\end{aligned}
$
Since $\mathrm{f}^{\prime \prime}(x)<0$ when $x=\frac{2}{3}$
Hence, $x=\frac{2}{3}$ is the maxima
Finding Maximum value of $\mathrm{f}(x)=x \sqrt{1-x}$
Putting $x=\frac{2}{3}$
$
\begin{aligned}
f\left(\frac{2}{3}\right) & =\frac{2}{3} \sqrt{1-\frac{2}{3}} \\
& =\frac{2}{3} \sqrt{\frac{3-2}{3}}
\end{aligned}
$
Finding Maximum value of $f(x)=x \sqrt{1-x}$
Putting $x=\frac{2}{3}$
$
\begin{aligned}
& f(x)=x \sqrt{1-x} \\
& f\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}} \\
& =\frac{2}{3} \sqrt{\frac{3-2}{3}} \\
& =\frac{2}{3} \sqrt{\frac{1}{3}} \\
& =\frac{2}{3 \sqrt{3}} \\
& =\frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\
&
\end{aligned}
$
$
=\frac{2 \sqrt{3}}{9}
$
Maximum value of $f(x)$ is $\frac{2 \sqrt{3}}{9}$ at $x=\frac{2}{3}$
Ex 6.3 Question 4 :
Prove that the following functions do not have maxima or minima:
(i) $f(x)=e^x$ (ii) $g(x)=\log x$
(iii) $h(x)=x^3+x^2+x+1$
Answer :
i. We have,
$
\begin{aligned}
& f(x)=\mathrm{e}^{\mathrm{x}} \\
& \therefore f^{\prime}(x)=e^x
\end{aligned}
$
Now, if $f^{\prime}(x)=0$, then $e^x=0$. But, the exponential function can never assume 0 for any value of $x$.
Therefore, there does not exist $c \in \mathrm{R}$ such that $f^{\prime}(c)=0$.
Hence, function $f$ does not have maxima or minima.
ii. We have,
$
\begin{aligned}
& g(x)=\log x \\
& \therefore g^{\prime}(x)=\frac{1}{x}
\end{aligned}
$
Since $\log x$ is defined for a positive number $x, g^{\prime}(x)>0$ for any $x$.
Therefore, there does not exist $c \in \mathrm{R}$ such that $g^{\prime}(c)=0$.
Hence, function $g$ does not have maxima or minima.
iii. We have,
$
h(x)=x^3+x^2+x+1
$
$
\therefore h^{\prime}(x)=3 x^2+2 x+1
$
Now,
$
h(x)=0 \Rightarrow 3 x^2+2 x+1=0 \Rightarrow x=\frac{-2 \pm 2 \sqrt{2} i}{6}=\frac{-1 \pm \sqrt{2} i}{3} \notin \mathbf{R}
$
Therefore, there does not exist $c \in \mathbf{R}$ such that $h^{\prime}(c)=0$.
Hence, function $h$ does not have maxima or minima.
Ex 6.3 Question 5 :
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) $f(x)=x^3, x \in[-2,2]$
(ii) $f(x)=\sin x+\cos x, x \in[0, \pi]$
(iii) $f(x)=4 x-\frac{1}{2} x^2, x \in\left[-2, \frac{9}{2}\right]$
(iv) $f(x)=(x-1)^2+3, x \in[-3,1]$
Answer:
(i) The given function is $f(x)=x^3$.
$
\therefore f^{\prime}(x)=3 x^2
$
Now,
$
f^{\prime}(x)=0 \Rightarrow x=0
$
Then, we evaluate the value of $f$ at critical point $x=0$ and at end points of the interval [ - 2, 2].
$
\begin{aligned}
& f(0)=0 \\
& f(-2)=(-2)^3=-8 \\
& f(2)=(2)^3=8
\end{aligned}
$
Hence, we can conclude that the absolute maximum value of $f$ on $[-2,2]$ is 8 occurring at $x=2$. Also, the absolute minimum value of $f$ on $[-2,2]$ is -8 occurring at $x=-2$.
(ii) The given function is $f(x)=\sin x+\cos x$.
$
\therefore f^{\prime}(x)=\cos x-\sin x
$
Now,
$
f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}
$
Then, we evaluate the value of $f$ at critical point $x=\frac{\pi}{4}$ and at the end points of the interval $[0, \pi]$.
$
\begin{aligned}
& 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} \\
& f(0)=\sin 0+\cos 0=0+1=1 \\
& f(\pi)=\sin \pi+\cos \pi=0-1=-1
\end{aligned}
$
Hence, we can conclude that the absolute maximum value of $f$ on $[0, \pi]$ is $\sqrt{2}$ occurring at $x=\frac{\pi}{4}$ and the absolute minimum value of $f$ on $[0, \pi]$ is - 1 occurring at $x=\pi$.
(iii) The given function is $f(x)=4 x-\frac{1}{2} x^2$.
$
\therefore f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x
$
Now,
$
f^{\prime}(x)=0 \Rightarrow x=4
$
Then, we evaluate the value of $f$ at critical point $x=4$ and at the end points of the interval $\left[-2, \frac{9}{2}\right]$.
$
\begin{aligned}
& f(4)=16-\frac{1}{2}(16)=16-8=8 \\
& f(-2)=-8-\frac{1}{2}(4)=-8-2=-10
\end{aligned}
$
$
f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2=18-\frac{81}{8}=18-10.125=7.875
$
Hence, we can conclude that the absolute maximum value of $f \circ n\left[-2, \frac{9}{2}\right]$ is 8 occurring at $x=4$ and the absolute minimum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is - 10 occurring at $x=-2$.
(iv) The given function is $f(x)=(x-1)^2+3$.
$
\therefore f^{\prime}(x)=2(x-1)
$
Now,
$
f^{\prime}(x)=0 \Rightarrow 2(x-1)=0 \Rightarrow x=1
$
Then, we evaluate the value of $f$ at critical point $x=1$ and at the end points of the interval $[-3,1]$.
$
\begin{aligned}
& f(1)=(1-1)^2+3=0+3=3 \\
& f(-3)=(-3-1)^2+3=16+3=19
\end{aligned}
$
Hence, we can conclude that the absolute maximum value of $f$ on $[-3,1]$ is 19 occurring at $x=-3$ and the minimum value of $f$ on $[-3,1]$ is 3 occurring at $x=1$.
Ex 6.3 Question 6
Find the maximum profit that a company can make, if the profit function is given by $p(x)=41-72 x-18 x^2$
The profit function is given by
$
p(x)=41-72 x-18 x^2
$
Finding $p^{\prime}(x)$
$
p^{\prime}(x)=-72-36 x
$
$
\begin{aligned}
& \text { Putting } p^{\prime}(x)=0 \\
& -72-36 x=0 \\
& -36 x=72 \\
& x=\frac{-72}{36} \\
& x=-2
\end{aligned}
$
Finding $p^{\prime \prime}(x)$
Since $p^{\prime}(x)=-72-36 x$
$
\therefore \mathrm{p}^{\prime \prime}(\mathrm{x})=-36
$
Since $\mathrm{p}^{\prime \prime}(\mathrm{x})<0$
$x=-2$ is the maxima
$
\begin{aligned}
\text { Maximum profit } & =\mathrm{p}(-2) \\
& =41-72 x-18 x^2 \\
& =\mathbf{4 1}-\mathbf{7 2}(-\mathbf{2})-\mathbf{1 8}(-\mathbf{2})^{\mathbf{2}} \\
& =41+144-18(4) \\
& =41+144-72 \\
& =\mathbf{1 1 3}
\end{aligned}
$
Hence, the maximum profit is $\mathbf{1 1 3}$ unit.
Ex 6.3 Question 7:
Find both the maximum value and the minimum value of
$
3 x^4-8 x^3+12 x^2-48 x+25 \text { on the interval }[0,3]
$
Answer :
Let $f(x)=3 x^4-8 x^3+12 x^2-48 x+25$.
$
\begin{aligned}
\therefore f^{\prime}(x) & =12 x^3-24 x^2+24 x-48 \\
& =12\left(x^3-2 x^2+2 x-4\right) \\
& =12\left[x^2(x-2)+2(x-2)\right] \\
& =12(x-2)\left(x^2+2\right)
\end{aligned}
$
Now, $f^{\prime}(x)=0$ gives $x=2$ or $x^2+2=0$ for which there are no real roots.
Therefore, we consider only $x=2 \in[0,3]$.
Now, we evaluate the value of $f$ at critical point $x=2$ and at the end points of the interval $[0,3]$.
$
\begin{aligned}
f(2) & =3(16)-8(8)+12(4)-48(2)+25 \\
& =48-64+48-96+25 \\
& =-39 \\
f(0) & =3(0)-8(0)+12(0)-48(0)+25 \\
& =25 \\
f(3) & =3(81)-8(27)+12(9)-48(3)+25 \\
& =243-216+108-144+25=16
\end{aligned}
$
Hence, we can conclude that the absolute maximum value of $f$ on $[0,3]$ is 25 occurring at $x=0$ and the absolute minimum value of $f$ at $[0,3]$ is - 39 occurring at $x=2$.
Ex 6.3 Question 8 :
At what points in the interval $[0,2 \pi]$, does the function $\sin 2 x$ attain its maximum value?
Answer:
Let $f(x)=\sin 2 x$.
$
\therefore f^{\prime}(x)=2 \cos 2 x
$
Now,
$
\begin{aligned}
& f^{\prime}(x)=0 \Rightarrow \cos 2 x=0 \\
& \Rightarrow 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\
& \Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}
\end{aligned}
$
Then, we evaluate the values of $f$ at critical points $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ and at the end points of the interval $[0,2 \pi]$.
$
\begin{aligned}
& f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{2}=1, f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2}=-1 \\
& f\left(\frac{5 \pi}{4}\right)=\sin \frac{5 \pi}{2}=1, f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{2}=-1 \\
& f(0)=\sin 0=0, f(2 \pi)=\sin 2 \pi=0
\end{aligned}
$
Hence, we can conclude that the absolute maximum value of $f$ on $[0,2 \pi]$ is occurring at $x=\frac{\pi}{4}$ and $x=\frac{5 \pi}{4}$
Ex 6.3 Question 9:
What is the maximum value of the function $\sin x+\cos x$ ?
Answer :
Let $f(x)=\sin x+\cos x$.
$
\begin{aligned}
& \therefore f^{\prime}(x)=\cos x-\sin x \\
& f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4} \ldots, \\
& f^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)
\end{aligned}
$
Now, $f^{\prime \prime}(x)$ will be negative when $(\sin x+\cos x)$ is positive i.e., when $\sin x$ and $\cos x$ are both positive. Also, we know that $\sin x$ and $\cos x$ both are positive in the first quadrant. Then, $f^{\prime \prime}(x) \mathrm{w}$ be negative when $x \in\left(0, \frac{\pi}{2}\right)$.
Thus, we consider $x=\frac{\pi}{4}$.
$
f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)=-\left(\frac{2}{\sqrt{2}}\right)=-\sqrt{2}<0
$
$\therefore$ By second derivative test, $f$ will be the maximum at $x=\frac{\pi}{4}$ and the maximum value of $f$ is $f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$.
Ex 6.3 Question 10:
Find the maximum value of $2 x^3-24 x+107$ in the interval $[1,3]$. Find the maximum value of the same function in $[-3,-1]$.
Answer :
Let $f(x)=2 x^3-24 x+107$.
$
\therefore f^{\prime}(x)=6 x^2-24=6\left(x^2-4\right)
$
Now,
$
f^{\prime}(x)=0 \Rightarrow 6\left(x^2-4\right)=0 \Rightarrow x^2=4 \Rightarrow x= \pm 2
$
We first consider the interval $[1,3]$.
Then, we evaluate the value of $f$ at the critical point $x=2 \in[1,3]$ and at the end points of the interval $[1,3]$.
$
\begin{aligned}
& f(2)=2(8)-24(2)+107=16-48+107=75 \\
& f(1)=2(1)-24(1)+107=2-24+107=85 \\
& f(3)=2(27)-24(3)+107=54-72+107=89
\end{aligned}
$
Hence, the absolute maximum value of $f(x)$ in the interval $[1,3]$ is 89 occurring at $x=3$.
Next, we consider the interval $[-3,-1]$.
Evaluate the value of $f$ at the critical point $x=-2 \in[-3,-1]$ and at the end points of the interval [1, 3].
$
\begin{aligned}
& f(-3)=2(-27)-24(-3)+107=-54+72+107=125 \\
& f(-1)=2(-1)-24(-1)+107=-2+24+107=129 \\
& f(-2)=2(-8)-24(-2)+107=-16+48+107=139
\end{aligned}
$
Hence, the absolute maximum value of $f(x)$ in the interval $[-3,-1]$ is 139 occurring at $x=-2$.
Ex 6.3 Question 11 :
It is given that at $x=1$, the function $x^4-62 x^2+a x+9$ attains its maximum value, on the interval $[0,2]$. Find the value of $a$.
Answer:
Let $f(x)=x^4-62 x^2+a x+9$.
$
\therefore f^{\prime}(x)=4 x^3-124 x+a
$
It is given that function $f$ attains its maximum value on the interval $[0,2]$ at $x=1$.
$
\begin{aligned}
& \therefore f^{\prime}(1)=0 \\
& \Rightarrow 4-124+a=0 \\
& \Rightarrow a=120
\end{aligned}
$
Hence, the value of $a$ is 120 .
Ex 6.3 Quetion 12:
Find the maximum and minimum values of $x+\sin 2 x$ on $[0,2 \pi]$.
Answer :
Let $f(x)=x+\sin 2 x$.
$
\therefore f^{\prime}(x)=1+2 \cos 2 x
$
Now, $f^{\prime}(x)=0 \Rightarrow \cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}$
$
\begin{aligned}
& 2 x=2 n \pi \pm \frac{2 \pi}{3}, n \in \mathrm{Z} \\
& \Rightarrow x=n \pi \pm \frac{\pi}{3}, n \in \mathrm{Z} \\
& \Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \in[0,2 \pi]
\end{aligned}
$
Then, we evaluate the value of $f$ at critical points $x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}$ and at the end points of the interval $[0,2 \pi]$.
$
\begin{aligned}
& f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2} \\
& f\left(\frac{2 \pi}{3}\right)=\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2} \\
& f\left(\frac{4 \pi}{3}\right)=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2} \\
& f\left(\frac{5 \pi}{3}\right)=\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2} \\
& f(0)=0+\sin 0=0 \\
& f(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi
\end{aligned}
$
Hence, we can conclude that the absolute maximum value of $f(x)$ in the interval $[0,2 \pi]$ is $2 \pi$ occurring at $x=2 \pi$ and the absolute minimum value of $f(x)$ in the interval $[0,2 \pi]$ is 0 occurring at $x=$ 0.
Ex 6.3 Question 13:
Find two numbers whose sum is 24 and whose product is as large as possible.
Answer :
Let one number be $x$. Then, the other number is $(24-x)$.
Let $P(x)$ denote the product of the two numbers. Thus, we have:
$
\begin{aligned}
& P(x)=x(24-x)=24 x-x^2 \\
& \therefore P^{\prime}(x)=24-2 x \\
& P^{\prime \prime}(x)=-2
\end{aligned}
$
Now,
$
P^{\prime}(x)=0 \Rightarrow x=12
$
Also,
$
P^{\prime \prime}(12)=-2<0
$
$\therefore$ By second derivative test, $x=12$ is the point of local maxima of $P$. Hence, the product of the numbers is the maximum when the numbers are 12 and $24-12=12$.
Ex 6.3 Question 14:
Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^3$ is maximum.
Answer:
The two numbers are $x$ and $y$ such that $x+y=60$.
$
\begin{aligned}
& \Rightarrow y=60-x \\
& \text { Let } f(x)=x y^3 . \\
& \Rightarrow f(x)=x(60-x)^3 \\
& \begin{aligned}
\therefore f^{\prime}(x) & =(60-x)^3-3 x(60-x)^2 \\
& =(60-x)^2[60-x-3 x] \\
& =(60-x)^2(60-4 x)
\end{aligned}
\end{aligned}
$
And, $f^{\prime \prime}(x)=-2(60-x)(60-4 x)-4(60-x)^2$
$
\begin{aligned}
& =-2(60-x)[60-4 x+2(60-x)] \\
& =-2(60-x)(180-6 x) \\
& =-12(60-x)(30-x)
\end{aligned}
$
Now, $f^{\prime}(x)=0 \Rightarrow x=60$ or $x=15$
When $x=60, f^{\prime \prime}(x)=0$.
When $x=15, f^{\prime \prime}(x)=-12(60-15)(30-15)=-12 \times 45 \times 15<0$.
$\therefore$ By second derivative test, $x=15$ is a point of local maxima of $f$. Thus, function $x y^3$ is maximum when $x=15$ and $y=60-15=45$.
Hence, the required numbers are 15 and 45 .
Ex 6.3 Question 15 :
Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^2 y^5$ is a maximum
Answer :
Let one number be $x$. Then, the other number is $y=(35-x)$.
Let $P(x)=x^2 y^5$. Then, we have:
$\begin{aligned}
& P(x)=x^2(35-x)^5 \\
& \begin{aligned}
\therefore P^{\prime}(x) & =2 x(35-x)^5-5 x^2(35-x)^4 \\
& =x(35-x)^4[2(35-x)-5 x] \\
& =x(35-x)^4(70-7 x) \\
& =7 x(35-x)^4(10-x)
\end{aligned}
\end{aligned}$
$\text { And, } \begin{aligned}
P^{\prime \prime}(x) & =7(35-x)^4(10-x)+7 x\left[-(35-x)^4-4(35-x)^3(10-x)\right] \\
= & 7(35-x)^4(10-x)-7 x(35-x)^4-28 x(35-x)^3(10-x) \\
= & 7(35-x)^3[(35-x)(10-x)-x(35-x)-4 x(10-x)] \\
& =7(35-x)^3\left[350-45 x+x^2-35 x+x^2-40 x+4 x^2\right] \\
& =7(35-x)^3\left(6 x^2-120 x+350\right)
\end{aligned}$
Now, $P^{\prime}(x)=0 \Rightarrow x=0, x=35, x=10$
When $x=35, f^{\prime}(x)=f(x)=0$ and $y=35-35=0$. This will make the product $x^2 y^5$ equal to 0 .
When $x=0, y=35-0=35$ and the product $x^2 y^2$ will be 0 .
$\therefore x=0$ and $x=35$ cannot be the possible values of $x$.
When $x=10$, we have:
$
\begin{aligned}
P^{\prime \prime}(x) & =7(35-10)^3(6 \times 100-120 \times 10+350) \\
& =7(25)^3(-250)<0
\end{aligned}
$
$\therefore$ By second derivative test, $P(x)$ will be the maximum when $x=10$ and $y=35-10=25$.
Hence, the required numbers are 10 and 25 .
Ex 6.3 Question 16:
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Answer :
Let one number be $x$. Then, the other number is $(16-x)$.
Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,
$
\begin{aligned}
& S(x)=x^3+(16-x)^3 \\
& \therefore S^{\prime}(x)=3 x^2-3(16-x)^2, S^{\prime \prime}(x)=6 x+6(16-x)
\end{aligned}
$
Now, $S^{\prime}(x)=0 \Rightarrow 3 x^2-3(16-x)^2=0$
$
\begin{aligned}
& \Rightarrow x^2-(16-x)^2=0 \\
& \Rightarrow x^2-256-x^2+32 x=0 \\
& \Rightarrow x=\frac{256}{32}=8
\end{aligned}
$
Now, $S^{\prime \prime}(8)=6(8)+6(16-8)=48+48=96>0$
$\therefore$ By second derivative test, $x=8$ is the point of local minima of $S$.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16-8=$ 8.
Ex 6.3 Question 17:
A square piece of tin of side $18 \mathrm{~cm}$ is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Answer :
Let the side of the square to be cut off be $x \mathrm{~cm}$. Then, the length and the breadth of the box will be $(18-2 x) \mathrm{cm}$ each and the height of the box is $x \mathrm{~cm}$.
Therefore, the volume $V(x)$ of the box is given by,
$
V(x)=x(18-2 x)^2
$
$
\begin{aligned}
\therefore V^{\prime}(x) & =(18-2 x)^2-4 x(18-2 x) \\
& =(18-2 x)[18-2 x-4 x] \\
& =(18-2 x)(18-6 x) \\
& =6 \times 2(9-x)(3-x) \\
& =12(9-x)(3-x)
\end{aligned}
$
And, $V^{\prime \prime}(x)=12[-(9-x)-(3-x)]$
$
\begin{aligned}
& =-12(9-x+3-x) \\
& =-12(12-2 x) \\
& =-24(6-x)
\end{aligned}
$
Now, $V^{\prime}(x)=0 \Rightarrow x=9$ or $x=3$
If $x=9$, then the length and the breadth will become 0 .
$
\begin{aligned}
& \therefore x \neq 9 . \\
& \Rightarrow x=3 .
\end{aligned}
$
Now, $V^{\prime \prime}(3)=-24(6-3)=-72<0$
$\therefore$ By second derivative test, $x=3$ is the point of maxima of $V$.
Hence, if we remove a square of side $3 \mathrm{~cm}$ from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.
Ex 6.3 Question 18:
A rectangular sheet of tin $45 \mathrm{~cm}$ by $24 \mathrm{~cm}$ is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Answer :
Let the side of the square to be cut off be $x \mathrm{~cm}$. Then, the height of the box is $x$, the length is $45-2 x$, and the breadth is $24-2 x$.
Therefore, the volume $V(x)$ of the box is given by,
$
\begin{aligned}
& V(x)=x(45-2 x)(24-2 x) \\
& =x\left(1080-90 x-48 x+4 x^2\right) \\
& =4 x^3-138 x^2+1080 x \\
& \therefore V^{\prime}(x)=12 x^2-276 x+1080 \\
& =12\left(x^2-23 x+90\right) \\
& =12(x-18)(x-5) \\
& V^{\prime \prime}(x)=24 x-276=12(2 x-23) \\
&
\end{aligned}
$
Now, $V^{\prime}(x)=0 \Rightarrow x=18$ and $x=5$
It is not possible to cut off a square of side $18 \mathrm{~cm}$ from each corner of the rectangular sheet. Thus, $x$ cannot be equal to 18 .
$
\therefore x=5
$
Now, $V^{\prime \prime}(5)=12(10-23)=12(-13)=-156<0$
$\therefore$ By second derivative test, $x=5$ is the point of maxima.
Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 $\mathrm{cm}$.
Ex 6.3 Question 19:
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Answer:
Let a rectangle of length / and breadth $b$ be inscribed in the given circle of radius $a$. Then, the diagonal passes through the centre and is of length $2 \mathrm{acm}$.
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Now, by applying the Pythagoras theorem, we have:
$
\begin{aligned}
& (2 a)^2=l^2+b^2 \\
& \Rightarrow b^2=4 a^2-l^2 \\
& \Rightarrow b=\sqrt{4 a^2-l^2}
\end{aligned}
$
$\therefore$ Area of the rectangle, $A=l \sqrt{4 a^2-l^2}$
$
\begin{aligned}
& \therefore \frac{d A}{d l}=\sqrt{4 a^2-l^2}+l \frac{1}{2 \sqrt{4 a^2-l^2}}(-2 l)=\sqrt{4 a^2-l^2}-\frac{l^2}{\sqrt{4 a^2-l^2}} \\
& =\frac{4 a^2-2 l^2}{\sqrt{4 a^2-l^2}} \\
& \frac{d^2 A}{d l^2}=\frac{\sqrt{4 a^2-l^2}(-4 l)-\left(4 a^2-2 l^2\right) \frac{(-2 l)}{2 \sqrt{4 a^2-l^2}}}{\left(4 a^2-l^2\right)} \\
& =\frac{\left(4 a^2-l^2\right)(-4 l)+l\left(4 a^2-2 l^2\right)}{\left(4 a^2-l^2\right)^{\frac{3}{2}}} \\
& =\frac{-12 a^2 l+2 l^3}{\left(4 a^2-l^2\right)^{\frac{3}{2}}}=\frac{-2 l\left(6 a^2-l^2\right)}{\left(4 a^2-l^2\right)^{\frac{3}{2}}} \\
&
\end{aligned}
$
Now, $\frac{d A}{d l}=0$ gives $4 a^2=2 l^2 \Rightarrow l=\sqrt{2} a$
$
\Rightarrow b=\sqrt{4 a^2-2 a^2}=\sqrt{2 a^2}=\sqrt{2} a
$
Now, when $l=\sqrt{2} a$,
$
\frac{d^2 \mathrm{~A}}{d l^2}=\frac{-2(\sqrt{2} a)\left(6 a^2-2 a^2\right)}{2 \sqrt{2} a^3}=\frac{-8 \sqrt{2} a^3}{2 \sqrt{2} a^3}=-4<0
$
$\therefore$ By the second derivative test, when $l=\sqrt{2} a$, then the area of the rectangle is the maximum.
Since $l=b=\sqrt{2} a$, the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.
Ex 6.3 Question 20 :
Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.
Answer :
Let $r$ and $h$ be the radius and height of the cylinder respectively.
Then, the surface area ( $S$ ) of the cylinder is given by,
$
\begin{aligned}
S= & 2 \pi r^2+2 \pi r h \\
\Rightarrow h & =\frac{S-2 \pi r^2}{2 \pi r} \\
& =\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r
\end{aligned}
$
Let $V$ be the volume of the cylinder. Then,
$
V=\pi r^2 h=\pi r^2\left[\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\right]=\frac{S r}{2}-\pi r^3
$
Then, $\frac{d V}{d r}=\frac{S}{2}-3 \pi r^2, \frac{d^2 V}{d r^2}=-6 \pi r$
Now, $\frac{d V}{d r}=0 \Rightarrow \frac{S}{2}=3 \pi r^2 \Rightarrow r^2=\frac{S}{6 \pi}$
When $r^2=\frac{S}{6 \pi}$, then $\frac{d^2 V}{d r^2}=-6 \pi\left(\sqrt{\frac{S}{6 \pi}}\right)<0$.
$\therefore$ By second derivative test, the volume is the maximum when $r^2=\frac{S}{6 \pi}$.
Now, when $r^2=\frac{S}{6 \pi}$, then $h=\frac{6 \pi r^2}{2 \pi}\left(\frac{1}{r}\right)-r=3 r-r=2 r$.
Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.
Ex 6.3 Question 21:
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Answer :
Let $r$ and $h$ be the radius and height of the cylinder respectively.
Then, volume $(V)$ of the cylinder is given by,
$
\begin{aligned}
& V=\pi r^2 h=100 \quad \text { (given) } \\
& \therefore h=\frac{100}{\pi r^2}
\end{aligned}
$
Surface area ( $S$ ) of the cylinder is given by,
$
\begin{aligned}
& S=2 \pi r^2+2 \pi r h=2 \pi r^2+\frac{200}{r} \\
& \therefore \frac{d S}{d r}=4 \pi r-\frac{200}{r^2}, \frac{d^2 S}{d r^2}=4 \pi+\frac{400}{r^3} \\
& \frac{d S}{d r}=0 \Rightarrow 4 \pi r=\frac{200}{r^2} \\
& \Rightarrow r^3=\frac{200}{4 \pi}=\frac{50}{\pi} \\
& \Rightarrow r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}
\end{aligned}
$
Now, it is observed that when $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}, \frac{d^2 \mathrm{~S}}{d r^2}>0$.
$\therefore$ By second derivative test, the surface area is the minimum when the radius of the cylinder is
$
\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm} \text {. }
$
When $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}, h=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=\frac{2 \times 50}{(50)^{\frac{2}{3}}(\pi)^{1-\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$.
Hence, the required dimensions of the can which has the minimum surface area is given by radius $=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$ and height $=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$.
Ex 6.3 Question 22:
A wire of length $28 \mathrm{~m}$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Answer :
Let a piece of length / be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length $(28-1 \mathrm{~m}$.
Now, side of square $=\frac{l}{4}$.
Let $r$ be the radius of the circle. Then, $2 \pi r=28-l \Rightarrow r=\frac{1}{2 \pi}(28-l)$.
The combined areas of the square and the circle $(A)$ is given by,
$
\begin{aligned}
A & =(\text { side of the square })^2+\pi r^2 \\
& =\frac{l^2}{16}+\pi\left[\frac{1}{2 \pi}(28-l)\right]^2 \\
& =\frac{l^2}{16}+\frac{1}{4 \pi}(28-l)^2 \\
\therefore & \frac{d A}{d l}=\frac{2 l}{16}+\frac{2}{4 \pi}(28-l)(-1)=\frac{l}{8}-\frac{1}{2 \pi}(28-l) \\
\frac{d^2 A}{d l^2} & =\frac{1}{8}+\frac{1}{2 \pi}>0
\end{aligned}
$
$
\begin{aligned}
& \text { Now, } \frac{d A}{d l}=0 \Rightarrow \frac{l}{8}-\frac{1}{2 \pi}(28-l)=0 \\
& \Rightarrow \frac{\pi l-4(28-l)}{8 \pi}=0 \\
& \Rightarrow(\pi+4) l-112=0 \\
& \Rightarrow l=\frac{112}{\pi+4}
\end{aligned}
$
Thus, when $l=\frac{112}{\pi+4}, \frac{d^2 \mathrm{~A}}{d l^2}>0$.
$\therefore$ By second derivative test, the area $(A)$ is the minimum when $l=\frac{112}{\pi+4}$.
Hence, the combined area is the minimum when the length of the wire in making the square is $\frac{112}{\pi+4}$ $\mathrm{cm}$ while the length of the wire in making the circle is $28-\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4} \mathrm{~cm}$.
Ex 6.3 Question 23:
Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.
Answer :
Let $r$ and $h$ be the radius and height of the cone respectively inscribed in a sphere of radius $R$.
.png)
Let $V$ be the volume of the cone.
Then, $V=\frac{1}{3} \pi r^2 h$
Height of the cone is given by,
$h=R+\mathrm{AB}=R+\sqrt{R^2-r^2}$
[ $\mathrm{ABC}$ is a right triangle]
$
\begin{aligned}
\therefore V & =\frac{1}{3} \pi r^2\left(R+\sqrt{R^2-r^2}\right) \\
& =\frac{1}{3} \pi r^2 R+\frac{1}{3} \pi r^2 \sqrt{R^2-r^2}
\end{aligned}
$
$\therefore \frac{d V}{d r}=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^2-r^2}+\frac{1}{3} \pi r^2 \cdot \frac{(-2 r)}{2 \sqrt{R^2-r^2}}$
$
\begin{aligned}
& =\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^2-r^2}-\frac{1}{3} \pi \frac{r^3}{\sqrt{R^2-r^2}} \\
& =\frac{2}{3} \pi r R+\frac{2 \pi r\left(R^2-r^2\right)-\pi r^3}{3 \sqrt{R^2-r^2}} \\
& =\frac{2}{3} \pi r \cdot R+\frac{2 \pi r R^2-3 \pi r^3}{3 \sqrt{R^2-r^2}} \\
\frac{d^2 V}{d r^2} & =\frac{2 \pi R}{3}+\frac{3 \sqrt{R^2-r^2}\left(2 \pi R^2-9 \pi r^2\right)-\left(2 \pi r R^2-3 \pi r^3\right) \cdot \frac{(-2 r)}{6 \sqrt{R^2-r^2}}}{9\left(R^2-r^2\right)}
\end{aligned}
$
$
\begin{aligned}
& \quad=\frac{2}{3} \pi R+\frac{9\left(R^2-r^2\right)\left(2 \pi R^2-9 \pi r^2\right)+2 \pi r^2 R^2+3 \pi r^4}{27\left(R^2-r^2\right)^{\frac{3}{2}}} \\
& \text { Now, } \frac{d V}{d r}=0 \Rightarrow \frac{2}{3} r R=\frac{3 \pi r^3-2 \pi r R^2}{3 \sqrt{R^2-r^2}} \\
& \Rightarrow 2 R=\frac{3 r^2-2 R^2}{\sqrt{R^2-r^2}} \Rightarrow 2 R \sqrt{R^2-r^2}=3 r^2-2 R^2 \\
& \Rightarrow 4 R^2\left(R^2-r^2\right)=\left(3 r^2-2 R^2\right)^2 \\
& \Rightarrow 4 R^4-4 R^2 r^2=9 r^4+4 R^4-12 r^2 R^2 \\
& \Rightarrow 9 r^4=8 R^2 r^2 \\
& \Rightarrow r^2=\frac{8}{9} R^2
\end{aligned}
$
When $r^2=\frac{8}{9} R^2$, then $\frac{d^2 V}{d r^2}<0$.
$\therefore$ By second derivative test, the volume of the cone is the maximum when $r^2=\frac{8}{9} R^2$.
When $r^2=\frac{8}{9} R^2, h=R+\sqrt{R^2-\frac{8}{9} R^2}=R+\sqrt{\frac{1}{9} R^2}=R+\frac{R}{3}=\frac{4}{3} R$.
Therefore,
$
\begin{aligned}
& =\frac{1}{3} \pi\left(\frac{8}{9} R^2\right)\left(\frac{4}{3} R\right) \\
& =\frac{8}{27}\left(\frac{4}{3} \pi R^3\right) \\
& =\frac{8}{27} \times(\text { Volume of the sphere })
\end{aligned}
$
Hence, the volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$
Ex 6.3 Question 24.
Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base.
Answer:
Let $x$ be the base radius and $y$ be the height of cone.
Given Volume $\Rightarrow$ Volume of the cone is constant and $=\mathrm{V}$ (say)
.png)
$
\begin{aligned}
& \therefore \quad \frac{1}{3} \pi x^2 y=\mathrm{V} \text { (Given condition) } \\
& \therefore x^2 y=\frac{3 \mathrm{~V}}{\pi}=k \quad \text { (say) ...(i) }
\end{aligned}
$
Let $\mathrm{S}$ denote the curved surface of the cone
$\therefore \mathrm{S}=\pi x \sqrt{x^2+y^2} \quad($ formula $\mathrm{S}=\pi r l$ )
( $\mathrm{S}$ is to be minimised here)
Let $z=\mathrm{S}^2=\pi^2 x^2\left(x^2+y^2\right)$
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 $x]$
$
\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)
$
or $\quad z=\pi^2 k\left(k y^{-2}+y\right)$
$\therefore \frac{d z}{d y}=\pi^2 k\left[-2 k y^{-3}+1\right]$ and $\frac{d^2 z}{d y^2}=\pi^2 k\left[6 k y^{-4}\right]=\frac{6 \pi^2 k^2}{y^4}$
For max. or min., put $\frac{d z}{d y}=0$
$\therefore \quad \pi^2 k\left(-\frac{2 k}{y^3}+1\right)=0 \quad$ But $\quad \pi^2 k \neq 0$
$\therefore \quad-\frac{2 k}{y^3}+1=0 \quad$ or $\quad \frac{2 k}{y^3}=1$
$\therefore \quad y^3=2 k \quad \therefore \quad y=(2 k)^{1 / 3}$
At $\quad y=(2 k)^{1 / 3}, \frac{d^2 z}{d y^2}=\frac{6 \pi^2 k^2}{(2 k)^{4 / 3}}$ which is positive.
$\therefore z$ is least when $y=(2 k)^{1 / 3}$
$\therefore$ From (i), $x^2=\frac{k}{y}=\frac{k}{(2 k)^{1 / 3}}$
[Using (iii)]
or $\quad x^2=\frac{2 k}{2(2 k)^{1 / 3}}=\frac{(2 k)^{2 / 3}}{2}=\frac{y^2}{2}$
[By (iii)]
or $\quad y^2=2 x^2 \quad \therefore y=\sqrt{2} x$
$\therefore \quad z$ or $\mathrm{S}$ is least when height $=\sqrt{2}$ (radius of base).
Ex 6.3 Question 25 :
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$.
Answer :
Let $\theta$ be the semi-vertical angle of the cone.
It is clear that $\theta \in\left[0, \frac{\pi}{2}\right]$.
Let $r, h$, and $/$ be the radius, height, and the slant height of the cone respectively.
The slant height of the cone is given as constant.
.png)
Now, $r=/ \sin \theta$ and $h=/ \cos \theta$
The volume ( $V$ ) of the cone is given by,
$
\begin{aligned}
& V=\frac{1}{3} \pi r^2 h \\
& =\frac{1}{3} \pi\left(l^2 \sin ^2 \theta\right)(l \cos \theta) \\
& =\frac{1}{3} \pi l^3 \sin ^2 \theta \cos \theta \\
& \therefore \frac{d V}{d \theta}=\frac{l^3 \pi}{3}\left[\sin ^2 \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)\right] \\
& =\frac{l^3 \pi}{3}\left[-\sin ^3+2 \sin \theta \cos ^2 \theta\right] \\
& \frac{d^2 V}{d \theta^2}=\frac{l^3 \pi}{3}\left[-3 \sin ^2 \theta \cos \theta+2 \cos ^3 \theta-4 \sin ^2 \theta \cos \theta\right] \\
& =\frac{l^3 \pi}{3}\left[2 \cos ^3 \theta-7 \sin ^2 \theta \cos \theta\right] \\
&
\end{aligned}
$
Now, $\frac{d V}{d \theta}=0$
$
\begin{aligned}
& \Rightarrow \sin ^3 \theta=2 \sin \theta \cos ^2 \theta \\
& \Rightarrow \tan ^2 \theta=2 \\
& \Rightarrow \tan \theta=\sqrt{2} \\
& \Rightarrow \theta=\tan ^{-1} \sqrt{2}
\end{aligned}
$
Now, when $\theta=\tan ^{-1} \sqrt{2}$, then $\tan ^2 \theta=2$ or $\sin ^2 \theta=2 \cos ^2 \theta$.
Then, we have:
$
\frac{d^2 V}{d \theta^2}=\frac{l^3 \pi}{3}\left[2 \cos ^3 \theta-14 \cos ^3 \theta\right]=-4 \pi l^3 \cos ^3 \theta<0 \text { for } \theta \in\left[0, \frac{\pi}{2}\right]
$
$\therefore$ By second derivative test, the volume $(V)$ is the maximum when $\theta=\tan ^{-1} \sqrt{2}$.
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is $\tan ^{-1} \sqrt{2}$.
Ex 6.3 Question 26:
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is
$
\operatorname{Sin}^{-1}\left(\frac{1}{3}\right) \text {. }
$
Answer:
Let $r$ be the radius, $l$ be the slant height and $h$ be the height of the cone of given surface area, $S$.
Also, let $\alpha$ be the semi-vertical angle of the cone.
.png)
.png)
$
\Rightarrow l=\frac{S-\pi r^2}{\pi r}
$
Let $V$ be the volume of the cone.
$
\begin{aligned}
& \text { Then } V=\frac{1}{3} \pi r^2 h \\
& \begin{aligned}
\Rightarrow V^2 & =\frac{1}{9} \pi^2 r^4 h^2 \\
& =\frac{1}{9} \pi^2 r^4\left(l^2-r^2\right)\left[\text { Asl } l^2=r^2+h^2\right] \\
& =\frac{1}{9} \pi^2 r^4\left[\left(\frac{S-\pi r^2}{\pi r}\right)^2-r^2\right] \\
& =\frac{1}{9} \pi^2 r^4\left[\frac{\left(S-\pi r^2\right)^2-\pi^2 r^4}{\pi^2 r^2}\right] \\
& =\frac{1}{9} r^2\left(S^2-2 S \pi r^2\right) \\
\Rightarrow V^2 & =\frac{1}{9} S r^2\left(S-2 \pi r^2\right) \ldots . .(2)
\end{aligned}
\end{aligned}
$
Differentiating (2) with respect to $r$, we get
$
2 V \frac{d V}{d r}=\frac{1}{9} S\left(2 S r-8 \pi r^3\right)
$
For maximum or minimum, put $\frac{d V}{d r}=0$
$
\begin{aligned}
& \Rightarrow \frac{1}{9} S\left(2 S r-8 \pi r^3\right)=0 \\
& \Rightarrow 2 S r-8 \pi r^3=0 \quad(\text { As } S \neq 0) \\
& \Rightarrow S=4 \pi r^2 \quad(\text { As } r \neq 0)
\end{aligned}
$
$
\Rightarrow r^2=\frac{S}{4 \pi}
$
Differentiating again with respect to $r$, we get
$
\begin{aligned}
& 2 V \frac{d^2 V}{d r^2}+2\left(\frac{d V}{d r}\right)^2=\frac{1}{9} S\left(2 S-24 \pi r^2\right) \\
& \begin{aligned}
2 V \frac{d^2 V}{d r^2} & =\frac{1}{9} S\left(2 S-24 \pi \times \frac{S}{4 \pi}\right) \quad\left(\text { As } \frac{d V}{d r}=0 \text { and } r^2=\frac{S}{4 \pi}\right) \\
& =\frac{1}{9} S(2 S-6 S) \\
& =-\frac{4}{9} S^2<0
\end{aligned}
\end{aligned}
$
Thus, $V$ is maximum when $\mathrm{S}=4 \tilde{\mathrm{A}} \hat{a}, \neg r^2$
Ex 6.3 Question 27.
The point on the curve $x^2=2 y$ which is nearest to the point $(0,5)$ is
(A) $(2 \sqrt{2}, 4)$
(B) $(2 \sqrt{2}, 0)$
(C) $(\mathbf{0}, \mathbf{0})$
(D) (2, 2).
Answer:
Equation of the curve (upward parabola here) is
$
x^2=2 y
$
The given point is $\mathrm{A}(0,5)$.
Let $\mathrm{P}(x, y)$ be any point on curve (i).
$\therefore$ Distance $z=\mathrm{AP}$
$
=\sqrt{(x-0)^2+(y-5)^2}
$
.png)
I Distance formula
Let $\quad \mathrm{Z}=z^2=x^2+(y-5)^2$
Putting
or
$
\begin{aligned}
& x^2=2 y \text { from }(i), \\
& \mathrm{Z}=2 y+(y-5)^2 \\
& \mathrm{Z}=y^2-8 y+25
\end{aligned}=2 y+y^2+25-10 y
$
$\therefore \quad \frac{d Z}{d y}=2 y-8$ and $\frac{d^2 Z}{d y^2}=2$
Putting $\quad \frac{d Z}{d y}=0$ to get turning 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 $\quad y=4, \frac{d^2 \mathrm{Z}}{d y^2}=2$ is (+ ve)
$\therefore \mathrm{Z}\left(=z^2\right)$ is minimum and hence $z$ is minimum at $y=4$.
Putting $y=4$ in $(i), x^2=8 \quad \therefore \quad x= \pm \sqrt{8}= \pm 2 \sqrt{2}$.
$\therefore(2 \sqrt{2}, 4)$ and $(-2 \sqrt{2}, 4)$ are two points on curve (i) which are nearest to the given point $(0,5)$.
$\therefore$ Option (A) is correct answer.
Ex 6.3 Question 28 :
For all real values of $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is
(A) 0 (B) 1
(C) 3 (D) $\frac{1}{3}$
Answer :
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)(-1+2 x)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)^2} \\
& =\frac{-1+2 x-x+2 x^2-x^2+2 x^3-1-2 x+x+2 x^2-x^2-2 x^3}{\left(1+x+x^2\right)^2} \\
& =\frac{2 x^2-2}{\left(1+x+x^2\right)^2}=\frac{2\left(x^2-1\right)}{\left(1+x+x^2\right)^2}
\end{aligned}
$
$
\therefore f^{\prime}(x)=0 \Rightarrow x^2=1 \Rightarrow x= \pm 1
$
Now, $f^{\prime \prime}(x)=\frac{2\left[\left(1+x+x^2\right)^2(2 x)-\left(x^2-1\right)(2)\left(1+x+x^2\right)(1+2 x)\right]}{\left(1+x+x^2\right)^4}$
$
=\frac{4\left(1+x+x^2\right)\left[\left(1+x+x^2\right) x-\left(x^2-1\right)(1+2 x)\right]}{\left(1+x+x^2\right)^4}
$
$
\begin{aligned}
& =\frac{4\left[x+x^2+x^3-x^2-2 x^3+1+2 x\right]}{\left(1+x+x^2\right)^3} \\
& =\frac{4\left(1+3 x-x^3\right)}{\left(1+x+x^2\right)^3}
\end{aligned}
$
And, $f^{\prime \prime}(1)=\frac{4(1+3-1)}{(1+1+1)^3}=\frac{4(3)}{(3)^3}=\frac{4}{9}>0$
Also, $f^{\prime \prime}(-1)=\frac{4(1-3+1)}{(1-1+1)^3}=4(-1)=-4<0$
$\therefore$ By second derivative test, $f$ is the minimum at $x=1$ and the minimum value is given by
$
f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3} \text {. }
$
The correct answer is D.
Ex 6.3 Question 29:
The maximum value of $[x(x-1)+1]^{\frac{1}{3}}, 0 \leq x \leq 1$ is
(A) $\left(\frac{1}{3}\right)^{\frac{1}{3}}$
(B) $\frac{1}{2}$
(C) 1 (D) 0
Answer :
$
\begin{aligned}
& \text { Let } f(x)=[x(x-1)+1]^{\frac{1}{3}}, \\
& \therefore f^{\prime}(x)=\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}
\end{aligned}
$
Let $f(x)=[x(x-1)+1]^{\frac{1}{3}}$.
Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$
Then, we evaluate the value of $f$ at critical point $x=\frac{1}{2}$ and at the end points of the interval $[0,1]\{$ i.e., at $x=0$ and $x=1\}$.
$
\begin{aligned}
& f(0)=[0(0-1)+1]^{\frac{1}{3}}=1 \\
& f(1)=[1(1-1)+1]^{\frac{1}{3}}=1 \\
& f\left(\frac{1}{2}\right)=\left[\frac{1}{2}\left(\frac{-1}{2}\right)+1\right]^{\frac{1}{3}}=\left(\frac{3}{4}\right)^{\frac{1}{3}}
\end{aligned}
$
Hence, we can conclude that the maximum value of $f$ in the interval $[0,1]$ is 1 .
The correct answer is C.
