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Applications of Derivative — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivative4 Marks
Question
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$\begin{aligned}
f\left(\frac{5 \pi}{3}\right) & =\sec \frac{5 \pi}{3}+\log \cos ^2 \frac{5 \pi}{3}=2+\log \left(\frac{1}{4}\right) \\
& =2-2 \log 2
\end{aligned}$

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