CUET Preparation Today
The function $f(x) = \tan x-x$ |
is a decreasing function on $[0,\frac{\pi}{2})$ is an increasing function on $[0,\frac{\pi}{2})$ is a constant function is neither increasing nor decreasing function on $[0,\frac{\pi}{2})$ |
is an increasing function on $[0,\frac{\pi}{2})$ |
The correct answer is Option (2) → is an increasing function on $[0,\frac{\pi}{2})$ $f(x)=\tan x - x$ $f'(x)=\sec^{2}x - 1$ $\sec^{2}x - 1=\tan^{2}x$ $\tan^{2}x \ge 0$ for all $x$ in $\left[0,\frac{\pi}{2}\right)$. Thus $f'(x) > 0$ for all $x>0$, so the function is increasing on the entire interval. Final answer: the function is increasing on $\left[0,\frac{\pi}{2}\right)$ |
