An extreme value of f(x)=frac{4}{sin x}+frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.To find the extreme value,we differentiate $f(x)$ with respect to $x$:$f'(x) = -\frac{4 \cos

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Given $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.To find the extreme value,we differentiate $f(x)$ with respect to $x$:$f'(x) = -\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2}$.Setting $f'(x) = 0$ for critical points:$-\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2} = 0$.Since $x \in (0, \pi/2)$,$\cos x 
eq 0$,so we can divide by $\cos x$:$-\frac{4}{\sin^2 x} + \frac{1}{(1-\sin x)^2} = 0$.$\frac{1}{(1-\sin x)^2} = \frac{4}{\sin^2 x}$.Taking the square root on both sides:$\frac{1}{1-\sin x} = \frac{2}{\sin x}$ (since $\sin x > 0$ and $1-\sin x > 0$ in the given interval).$\sin x = 2 - 2 \sin x$.$3 \sin x = 2 \Rightarrow \sin x = \frac{2}{3}$.Now,substitute $\sin x = \frac{2}{3}$ into $f(x)$:$f(x) = \frac{4}{2/3} + \frac{1}{1-2/3} = 4 	imes \frac{3}{2} + \frac{1}{1/3} = 6 + 3 = 9$.

An extreme value of $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in $(0, \pi / 2)$ is

Similar Questions

Maximum value of $x(1 - x)^2$ when $0 \le x \le 2$,is

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$,then

The difference between the maximum and minimum values of $f(x) = x^4e^{-x^2}$ for all $x \in R$ is:

Find the maximum and minimum values of $f,$ if any,of the function given by $f(x)=|x|, x \in R.$

Consider a cuboid of sides $2x$,$4x$,and $5x$ and a closed hemisphere of radius $r$. If the sum of their surface areas is a constant $k$,then the ratio $x:r$,for which the sum of their volumes is maximum,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module