For the function \( f(x)=\tan x-4x,\ x\in\left(0,\displaystyle\frac{\pi}{2}\right) \), which of the following statements are correct?
A. \( f(x) \) is increasing in \( \left(0,\displaystyle\frac{\pi}{3}\right) \)
B. \( f(x) \) is increasing in \( \left(\displaystyle\frac{\pi}{3},\frac{\pi}{2}\right) \)
C. \( f(x) \) is decreasing in \( \left(0,\displaystyle\frac{\pi}{3}\right) \)
D. \( f(x) \) is decreasing in \( \left(\displaystyle\frac{\pi}{3},\frac{\pi}{2}\right) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B and C only
\( \displaystyle f'(x)=\sec^2 x-4 \)
On \( \left(0,\frac{\pi}{2}\right) \), cos x decreases from 1 to 0, so \( \sec^2 x \) increases from 1 without bound, and \( \sec^2 x=4 \) at \( \cos x=\frac{1}{2} \), that is, \( x=\frac{\pi}{3} \).
\( \displaystyle 0\lt x\lt\frac{\pi}{3}:\ \sec^2 x\lt 4,\ f'(x)\lt 0 \)
\( \displaystyle \frac{\pi}{3}\lt x\lt\frac{\pi}{2}:\ \sec^2 x\gt 4,\ f'(x)\gt 0 \)
So f is decreasing on \( \left(0,\frac{\pi}{3}\right) \) and increasing on \( \left(\frac{\pi}{3},\frac{\pi}{2}\right) \).
Checking the statements
(A) \( f(x) \) is increasing in \( \left(0,\frac{\pi}{3}\right) \) — Incorrect. f′(x) < 0 there.
(B) \( f(x) \) is increasing in \( \left(\frac{\pi}{3},\frac{\pi}{2}\right) \) — Correct. f′(x) > 0 there.
(C) \( f(x) \) is decreasing in \( \left(0,\frac{\pi}{3}\right) \) — Correct. f′(x) < 0 there.
(D) \( f(x) \) is decreasing in \( \left(\frac{\pi}{3},\frac{\pi}{2}\right) \) — Incorrect. f′(x) > 0 there.