F(x) = log |sin x|,where x in (0, pi),is stri | vedclass

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

Question

(C) Given the function $f(x) = \log |\sin x|$ for $x \in (0, \pi)$.Since $x \in (0, \pi)$,$\sin x$ is always positive,so we can write $f(x) = \log(\si

Vedclass · Accepted Answer

$\left(0, \frac{\pi}{2}\right)$ only

Vedclass · Answer

(C) Given the function $f(x) = \log |\sin x|$ for $x \in (0, \pi)$.Since $x \in (0, \pi)$,$\sin x$ is always positive,so we can write $f(x) = \log(\sin x)$.To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.For the function to be strictly increasing,we must have $f'(x) > 0$.Therefore,$\cot x > 0$.In the interval $(0, \pi)$,$\cot x$ is positive when $x \in \left(0, \frac{\pi}{2}\right)$.Thus,the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.

$F(x) = \log |\sin x|$,where $x \in (0, \pi)$,is strictly increasing on

Similar Questions

If $\log (1+x)-\frac{2x}{2+x}$ is increasing,then

The function $f(x) = x + \frac{1}{x}, (x \neq 0)$ is a non-increasing function in the interval

For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.

Show that $y=\log (1+x)-\frac{2 x}{2+x}, x>-1,$ is an increasing function of $x$ throughout its domain.

If $f(x) = \frac{\log x}{x}$ $(x > 0)$,then it is increasing in

Vedclass Test Series

Exam Paper Generator

Online Exam Module