Question

If the function $f:R\to R$ is defined by $f\left(x\right)=\tan x-x$, then $f\left(x\right)$

• A. increases
• B. decreses
• C. remains constant
• D. becomes zero

Answer 1

The correct option is A increases

$f\left(x\right)=\tan x-x$

On differentiating, we get

$f^{\prime }\left(x\right)={\sec }^{2}x-1=\frac{1}{{\cos }^{2}x}-1=\frac{1-{\cos }^{2}x}{{\cos }^{2}x}$

Since $0\le {\cos }^{2}x\le 1$ for all values of $x,$

$\therefore f^{\prime }\left(x\right)>0$ for all values of $x.$

Thus, $f\left(x\right)$ always increases.