Find the maximum and minimum value of this function. f(x)= x+ ^2 … - Vidyadip

Question

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Applications of Derivative→Applications of Derivative — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If the functions f(x), defined below is continuous at x = 0, find the value of k.
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$

$\text{if} \overrightarrow{\text{r}} = x\hat{\text{i}} + y\hat{\text{j}} + z\hat{\text{k}}, \text{find} \overrightarrow(\text{r} \times \hat{\text{i}}). (\overrightarrow{\text{r}} \times \text{j}) + xy$

Solve the following differential equation
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2\text{y}$

A small firm manufacturers items $A$ and $B.$ The total number of items $A$ and $B$ that it can manufacture in a day is at the most $24.$ Item $A$ takes one hour to make while item $B$ takes only half an hour. The maximum time available per day is $16$ hours. If the profit on one unit of item $A$ be $Rs. 300$ and one unit of item $B$ be $Rs. 160,$ how many of each type of item be produced to maximize the profit? Solve the problem graphically.

If $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}, 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $\hat{\mathrm{i}}-6 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\vec {AB}$ and $\vec {CD}$. Deduce that $\overrightarrow{A B}$ and $\overrightarrow{C D}$ are collinear.

Find all points of discontinuity of $f, $where f is defined by:
$\text{f(x)}= \begin{cases}\text{x}^3 - 3,\ \ \text{if x}\leq 2 \\\text{x}^2 + 1,\ \text{if x}>2\end{cases}$

Evaluate the following intregals:
$\int\frac{1}{3+2\sin\text{x}+\cos\text{x}}\ \text{dx}$

Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is $\text{y}+2(\text{x}+1)=2\text{e}^{2\text{x}}.$

verify that $\text{y}^2=4\text{a}(\text{x}+\text{a})$ is a solution of the differential equation $\Big\{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=2\text{x}\frac{\text{dy}}{\text{dx}}.$