Exercise 7.6 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 7.6$
Question 1.
Find the absolute extrema of the following functions on the given closed interval.
(i) $f(x)=x^3-12 x+10 ;[1,2]$
(ii) $f(x)=3 x^4-4 x^3 ;[-1,2]$
(iii) $f(x)=6 x^{\frac{4}{3}}-3 x^{\frac{1}{3}} ;[-1,1]$
(iv) $f(x)=2 \cos x+\sin 2 x ;\left[0, \frac{\pi}{2}\right]$

Solution:
(i)
$
\begin{aligned}
& f(x)=y=x^3-12 x+10 \\
& \frac{d y}{d x}=3 x^2-12 \\
& \frac{d y}{d x}=0 \Rightarrow 3 x^2=12 \\
& x^2=4 \\
& x=\pm 2 \\
& \text { Here } x=2 \in[1,2] \\
& \text { Now } f(1)=1-12+10=-1 \\
& f(2)=8-24+10=-6 \\
&
\end{aligned}
$
$\therefore$ Absolute maximum is $-1$ and absolute minimum is $-6$
(ii) $f(x)=3 x^4-4 x^3$
$
\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=12 \mathrm{x}^3-12 \mathrm{x}^2 \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 12 \mathrm{x}^2(\mathrm{x}-1)=0 \\
& \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=1
\end{aligned}
$
$[$ Here $x=0,1 \in[-1,2]]$
Now $\mathrm{f}(-1)=4$
$\mathrm{f}(0)=0$
$f(1)=-1$
$f(2)=16$
so absolute maximum $=16$ and absolute minimum $=-1$

(iii)
$
\begin{aligned}
f(x)= & 6 x^{\frac{4}{3}}-3 x^{\frac{1}{3}} \\
f^{\prime}(x) & =6 \times \frac{4}{3} x^{\frac{1}{3}}-3 \frac{1}{3} x^{\frac{-2}{3}} \\
& =8 x^{\frac{1}{3}}-x^{\frac{-2}{3}} \\
f^{\prime}(x) & =0 \Rightarrow 8 x^{\frac{1}{3}}-x^{\frac{-2}{3}}=0 \\
& \Rightarrow 8 x^{\frac{1}{3}}=x^{\frac{-2}{3}}=\frac{1}{\frac{2}{3}} \\
& \Rightarrow x^{\frac{1}{3}} \cdot x^{\frac{2}{3}}=\frac{1}{8}(\text { i.e., }) x=\frac{1}{8} \\
\text { Now } f(-1) & =9 \\
f(1) & =3 \\
f\left(\frac{1}{8}\right) & =\frac{-9}{8}
\end{aligned}
$
so absolute maximum $=9$ and absolute minimum $=\frac{-9}{8}$
(iv) $f(x)=2 \cos x+\sin 2 x$
$f^{\prime}(x)=-2 \sin x+2 \cos 2 x$

$
\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \cos 2 \mathrm{x}=\sin \mathrm{x} \\
& \text { (i.e.,) } 1-2 \sin ^2 x=\sin x \\
& \Rightarrow 2 \sin ^2 x+\sin x-1=0 \\
& (2 \sin x-1)(\sin x+1)=0 \\
& \Rightarrow \quad \sin x=\frac{1}{2} \text { or } \sin x=-1 \\
& \Rightarrow \quad x=\frac{\pi}{6} \text { or } \pi \\
& \text { but } x \in\left[0, \frac{\pi}{2}\right] \\
& \text { So } x=\frac{\pi}{6} \\
& \text { Now } f(0)=2 \\
& f\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{2} \\
& \text { and } f\left(\frac{\pi}{2}\right)=0 \\
&
\end{aligned}
$
so absolute maximum $=\frac{3 \sqrt{3}}{2}$ and absolute minimum $=0$

Question 2.
Find the intervals of monotonicities and hence find the local extremum for the following functions
(i) $f(x)=2 x^3+3 x^2-12 x$
(ii) $f(x)=\frac{x}{x-5}$
(iii) $f(x)=\frac{e^x}{1-e^x}$
(iv) $f(x)=\frac{x^3}{3}-\log x$
(v) $f(x)=\sin x \cos x+5, x \in(0,2 \pi)$
Solution:
$
\begin{aligned}
& \text { (i) } f(x)=2 x^3+3 x^2-12 x \\
& f^{\prime}(x)=6 x^2+6 x-12 \\
& f^{\prime}(x)=0 \Rightarrow 6\left(x^2+x-2\right)=0 \\
& \text { (i.e.) } 6(x+2)(x-1)=0 \\
& \Rightarrow x=-2 \text { or } 1
\end{aligned}
$
Taking the points in the number line

The intervals are $(-\infty,-2),(-2,1),(1, \infty)$
when $x \in(-\infty,-2), f^{\prime}(x)=6(-1)(-4)=+v e$
say $x=-3$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing in the interval $(-\infty,-2)$ when $\mathrm{x} \in(-2,1), \mathrm{f}^{\prime}(\mathrm{x})=6(2)(-1)=-v e$
say $x=0$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly decreasing in the interval $(-2,1)$
when $x \in(1, \infty), f^{\prime}(x)=6(4)(+1)=+$ ve say $x=2$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing in $(1, \infty)$
Since $f(x)$ changes from +ve to - ve when passing through $-2$, the first derivative, test tells us there is a local maximum at $\mathrm{x}=-2$ and the local maximum value is $\mathrm{f}(-2)=20$.
Again $f^{\prime}(x)$ changes from - ve to +ve when passing through $1 \Rightarrow$ there is a local minimum at $x=1$ and the local minimum value is $f(1)=-7$. So $(1) f(x)$ is strictly increasing on $(-\infty,-2)$ and $(1, \infty)$. And $(2) f(x)$ is strictly decreasing on $(-2,1)$
The local maximum $=20$ and the local minimum $=-7$

(ii)
$
\begin{aligned}
& f(x)=\frac{x}{x-5} \\
& f^{\prime}(x)=\frac{(x-5)(1)-x(1)}{(x-5)^2}=\frac{x-5-x}{(x-5)^2} \\
&=\frac{-5}{(x-5)^2}<0 \\
&(\text { when } x \neq 5) .
\end{aligned}
$
$\mathrm{f}(\mathrm{x})$ is strictly decreasing on $(-\infty, 5)$ and $(5, \infty)$
And there is no local extremum

(iii)
$
\begin{aligned}
f(x) & =\frac{e^x}{1-e^x} \\
f^{\prime}(x) & =\frac{\left(1-e^x\right) e^x-e^x\left(-e^x\right)}{\left(1-e^x\right)^2}=\frac{e^x-e^{2 x}+e^{2 x}}{\left(1-e^x\right)^2}>0 \\
& =\frac{e^x}{\left(1-e^x\right)^2}>0
\end{aligned}
$
For all $\mathrm{x}$ values, so $\mathrm{f}(\mathrm{x})$ is strictly increasing in $(-\infty, \infty)$ and there is no local extremum.

(iv)
$
\begin{aligned}
f(x) & =\frac{x^3}{3}-\log x \\
f^{\prime}(x) & =\frac{3 x^2}{3}-\frac{1}{x}=x^2-\frac{1}{x} \\
f^{\prime}(x) & =0 \Rightarrow x^2=\frac{1}{x} \Rightarrow x^3=1 \Rightarrow x=1
\end{aligned}
$
So the intervals are $(0,1)$ and $(1, \infty)$
when $x \in(0,1), f^{\prime}(x)=\frac{1}{4}-\frac{1}{2}=-\mathrm{ve}$
say $x=\frac{1}{2}$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly decreasing in $(0,1)$
when $x \in(1, \infty), f^{\prime}(x)=4-\frac{1}{2}=+\mathrm{ve}$
say $x=2$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing in $(1, \infty)$
since $\mathrm{f}^{\prime}(\mathrm{x})$ changes from -ve to +ve at $\mathrm{x}=1$, there is a local minimum at $\mathrm{x}=1$ and the local minimum values is $f(1)=\frac{1}{3}-0=\frac{1}{3}$
So the function is strictly decreasing on $(0,1)$ and strictly increasing on $(1, \infty)$ and the local minimum value is $\frac{1}{3}$

(v)
$
\begin{aligned}
& \text { Now } f(x)= \sin x \cos x+5 \\
&= \frac{1}{2}(\sin 2 x)+5 \\
& \Rightarrow \quad f^{\prime}(x)= \frac{1}{2}(\cos 2 x) 2=\cos 2 x \\
& f^{\prime}(x)= 0 \Rightarrow \cos 2 x=0 \\
& 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \ldots \\
& \Rightarrow \quad x= \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \\
&(\therefore x \in(0,2 \pi))
\end{aligned}
$
So the intervals are $\left(0, \frac{\pi}{4}\right),\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right),\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right),\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$ and $\left(\frac{7 \pi}{4}, 2 \pi\right)$ when $x \in\left(0, \frac{\pi}{4}\right), f^{\prime}(x)=\cos \frac{\pi}{3}=\frac{1}{2}=+\mathrm{ve}$ say $x=\frac{\pi}{6}$ $\Rightarrow f(x)$ is increasing in $\left(0, \frac{\pi}{4}\right)$ when $x \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
$
\text { say } x=\frac{\pi}{3}
$
$
f^{\prime}(x)=\cos \frac{2 \pi}{3}=-\mathrm{ve}
$
$\Rightarrow \quad f^{\prime}(x)$ is strictly decreasing in $\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
similarly, $f(x)$ is strictly increasing in $\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$f(x)$ is strictly decreasing in $\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$

and $f(x)$ is strictly increasing in $\left(\frac{7 \pi}{4}, 2 \pi\right)$
Thus $f(x)$ is strictly increasing in $\left(0, \frac{\pi}{4}\right),\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right),\left(\frac{7 \pi}{4}, 2 \pi\right)$ and strictly decreasing in $\left(\frac{\pi}{4}-\frac{3 \pi}{4}\right),\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$
So there is a local maximum at $x=\frac{\pi}{4}, \frac{5 \pi}{4}$ and the local maximum value is $\frac{11}{2}$. There is a local minimum at $x=\frac{3 \pi}{4}$ and $\frac{7 \pi}{4}$ and the local minimum value is $\frac{9}{2}$.