4 Marks

Question 14 Marks

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

Answer

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

View full question & answer→

Question 24 Marks

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

Answer

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

View full question & answer→

Question 34 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 $=\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)=\pi \quad$

View full question & answer→

Maths 4 Marks

Test yourself on this topic