Interval in which the function \(f\) given by \(f(x) = \tan x - 4x, x \in \left(0,\displaystyle\frac{\pi}{2}\right)\) is strictly decreasing is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0 \lt x \lt \displaystyle\frac{\pi}{3}\)
\(f'(x) = \sec^2 x - 4\)
f is strictly decreasing where \(f'(x) \lt 0\):
\(\sec^2 x \lt 4\)
\(\displaystyle \cos^2 x \gt \frac{1}{4}\)
On \(\left(0, \frac{\pi}{2}\right)\), \(\cos x \gt 0\), so this means
\(\displaystyle \cos x \gt \frac{1}{2}\)
cos x is decreasing on this interval and \(\cos\frac{\pi}{3} = \frac{1}{2}\), so
\(\displaystyle 0 \lt x \lt \frac{\pi}{3}\)
On \(\frac{\pi}{3} \lt x \lt \frac{\pi}{2}\), \(f'(x) \gt 0\) and f is strictly increasing.