The function f(x) = tan x - x

The correct answer is Option (2) → is an increasing function on \(\left[0, \displaystyle\frac{\pi}{2}\right)\)

The sign of the derivative decides:

    \(f'(x) = \sec^2 x - 1 = \tan^2 x\)

For \(0 \lt x \lt \frac{\pi}{2}\), \(\tan x \gt 0\), so \(f'(x) \gt 0\). The derivative is 0 only at the single point x = 0.

Since f is continuous on \(\left[0, \frac{\pi}{2}\right)\) and \(f'(x) \gt 0\) on \(\left(0, \frac{\pi}{2}\right)\), f is strictly increasing on \(\left[0, \frac{\pi}{2}\right)\).