5 Marks Questions

Question 15 Marks

Prove that function $\sin ^2 x(1+\cos x)$, at $\cos x =\frac{1}{3}$ is maximum.

Answer

Suppose $ y=\sin ^2 x(1+\cos x)$
So $,\frac{d y}{d x}=\sin ^2 x(-\sin x)+ (1+\cos x)\ \times 2 \sin x \cos x$
or $ \frac{d y}{d x}=-\sin ^3 x+2 \sin x \cos x+2 \sin x \cos ^2 x$
$\frac{d y}{d x} =\sin x\left(-\sin ^2 x+2 \cos x+2 \cos ^2 x\right)$
$ =\sin x\left(-1+\cos ^2 x+2 \cos x+2 \cos ^2 x\right)$
$ =\sin x\left(3 \cos ^2 x+2 \cos x-1\right)$
For maxima and minima $\frac{d y}{d x}=0$
$ 0=\sin x\left(3 \cos ^2 x+2 \cos x-1\right)=0$
$\Rightarrow \sin x=0,3 \cos ^2 x+2 \cos x-1=0$
$\therefore x=\sin ^{-1}(0)=0 $ and $ \cos x=\frac{-2 \pm \sqrt{4-4 \times 3(-1)}}{2 \times 3}$
$\therefore x=0 $ and $ \cos x=\frac{-2 \pm 4}{6}=\frac{2}{6}, \frac{-6}{6}$
$x=0$ and $\cos x=\frac{1}{3},-1$
now, $\frac{d^2 y}{d x^2}=\sin x[-6 \cos x \sin x-2 \sin x]+$$
\left(3 \cos ^2 x+2 \cos x-1\right) \times \cos x
$or $\frac{d^2 y}{d x^2}=-6 \cos x \sin ^2 x-2 \sin ^2 x+3 \cos ^3 x+$
$2 \cos ^2 x-\cos x$
now at $\cos x=\frac{1}{3}, \frac{d^2 y}{d x^2}$
$=-6 \times \frac{1}{3} \times \frac{2}{9}-2 \times \frac{2}{9}+3 \times \frac{1}{27}+2 \times \frac{1}{9}-\frac{1}{3}$
$=-\frac{12}{27}-\frac{4}{9}+\frac{3}{27}+\frac{2}{9}-\frac{1}{3}$
$=\frac{-12-12+3+6-9}{27}=\frac{-24}{27}<0$
Hence value of function will be maximum at $\cos x=\frac{1}{3}$.

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Question 25 Marks

In interval $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$, Find the difference in maximum value and minimum value of function $f(x)=\sin 2 x-x$.

Answer

Given function
$f(x)=\sin 2 x-x$
So, $\quad f^{\prime}(x)=2 \cos 2 x-1$
For finding critical values :$
\begin{array}{rlrl}
& & f^{\prime}(x) & =0 \\
\Rightarrow & 2 \cos 2 x-1 & =0 \\
\Rightarrow & & \cos 2 x & =\frac{1}{2} \\
\Rightarrow & & 2 x & =\frac{-\pi}{3}, \frac{\pi}{3} \\
& \Rightarrow & x & =\frac{-\pi}{6}, \frac{\pi}{6}
\end{array}
$
now $\begin{aligned} f\left(\frac{-\pi}{2}\right) & =\sin (-\pi)-\left(\frac{-\pi}{2}\right) \\ & =-\sin \pi+\frac{\pi}{2}=\frac{\pi}{2} \\ f\left(\frac{-\pi}{6}\right) & =\sin \frac{-\pi}{3}+\frac{\pi}{6}=\frac{-\sqrt{3}}{2}+\frac{\pi}{6} \\ f\left(\frac{\pi}{6}\right) & =\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6} \\ \text { and } \quad f\left(\frac{\pi}{2}\right) & =\sin \pi-\frac{\pi}{2}=\frac{-\pi}{2}\end{aligned}$
Maximum value $=\frac{\pi}{2}$ and minimum value $=\frac{-\pi}{2}$ Hence such difference $=\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)=\pi \quad$

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Question 35 Marks

Find the maximum and minimum value of this function. $ f(x)=\sec x+\log \cos ^2 x, 0 < x < 2 \pi$

Answer

Given
$f(x) =\sec x+\log \cos ^2 x$
$ =\sec x+2 \log \cos x$
$\Rightarrow f^{\prime}(x) =\sec x \tan x-2 \tan x$
$ =\tan x(\sec x-2)$
for finding critical point $ f^{\prime}(x)=0$
$\Rightarrow \tan x(\sec x-2)=0$
$ \tan x=0 $ or $ \sec x=2$
$\Rightarrow \tan x=0 $ or $ \sec x=\frac{1}{2}$
$\Rightarrow x=\pi, x=\frac{\pi}{3}, \frac{5 \pi}{3} \quad[\because 0 < x <2 \pi]$
now $ f^{\prime}(x)=\tan x(\sec x-2)$
$\Rightarrow f^{\prime \prime}(x)=\sec ^2 x(\sec x-2)+\tan ^2 x \sec x$
$=\sec ^2 x(\sec x-2)+\sec x\left(\sec ^2 x-1\right)$
$\Rightarrow f^{\prime \prime}(x)=2 \sec ^3 x-2 \sec ^2 x-\sec x$
at $ x=\frac{\pi}{3}, f^{\prime \prime}\left(\frac{\pi}{3}\right)=2 \sec ^3 \frac{\pi}{3}-2 \sec ^2 \frac{\pi}{3}-\sec \frac{\pi}{3}$
$=2 \times 8-2 \times 4-2=6>0$
hence minimum point is at $x=\frac{\pi}{3}$ and minimum value
$f\left(\frac{\pi}{3}\right) =\sec \frac{\pi}{3}+\log \cos ^2 \frac{\pi}{3}=2+\log \frac{1}{4}$
$ =2-2 \log 2$
at $x=\pi, f^{\prime \prime}(\pi)=2 \sec ^3 \pi-2 \sec ^2 \pi-\sec \pi$
$=-2-2+1=-3<0$
hence, maximum point is $x=\pi$ and maximum value
$f(\pi) =\sec \pi+\log \cos ^2 \pi$
$ =-1+\log (1)=-1$
$\text { at } x=\frac{5 \pi}{3}, f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=2 \sec ^3\left(\frac{5 \pi}{3}\right)-2 \sec ^2\left(\frac{5 \pi}{3}\right)$
$-\sec \left(\frac{5 \pi}{3}\right)$
$=2(2)^3-2(2)^2+2=10>0$
hence minimum point is at $x=\frac{5 \pi}{3}$ and minimum value$
f\left(\frac{5 \pi}{3}\right) =\sec \frac{5 \pi}{3}+\log \cos ^2 \frac{5 \pi}{3}=2+\log \left(\frac{1}{4}\right)$
$ =2-2 \log 2$

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Mathematics 5 Marks Questions

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