The function \( f(x)=\log_e(\cos x) \) increases on which of the following intervals for \( x\in\left(-\displaystyle\frac{\pi}{2},\displaystyle\frac{\pi}{2}\right) \)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \left(-\displaystyle\frac{\pi}{2},0\right) \)
By the chain rule,
\( \displaystyle f'(x)=\frac{1}{\cos x}\cdot(-\sin x) \)
\( \displaystyle =-\tan x \)
f is increasing where \( f'(x)\gt 0 \), i.e. where \( \tan x\lt 0 \). In \( \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \) this happens for \( -\frac{\pi}{2}\lt x\lt 0 \).
So f increases on \( \left(-\frac{\pi}{2},0\right) \).
Why the other options are incorrect
Option (1) \( \left(-\displaystyle\frac{\pi}{2},\displaystyle\frac{\pi}{2}\right) \) — Incorrect. On \( \left(0,\frac{\pi}{2}\right) \), \( f'(x)=-\tan x\lt 0 \), so f is not increasing on the whole interval.
Option (2) \( \left(0,\displaystyle\frac{\pi}{4}\right) \) — Incorrect. On \( \left(0,\frac{\pi}{4}\right) \), \( \tan x\gt 0 \), so f is decreasing there.
Option (3) \( \left(0,\displaystyle\frac{\pi}{2}\right) \) — Incorrect. This is the interval where f is decreasing (the minus sign from the derivative of cos x is lost), since \( -\tan x\lt 0 \) there.