Question

Let $f\left(x\right)=4x+8\cos x-4\ln \left\{\cos x\right(1+\sin x\left)\right\}+\tan x-2\sec x-6$. If $f\left(x\right)$ is strictly increasing $\forall x\in (0,a)$ then

• A. $a=\frac{\pi }{6}$
• B. $a=\frac{\pi }{3}$
• C. $a=\frac{\pi }{2}$
• D. None of these

Answer 1

The correct option is A $a=\frac{\pi }{6}$

$f^{\prime }\left(x\right)=4-8\sin x-4\frac{-\sin x+{\cos }^{2}x-{\sin }^{2}x}{\cos x(1+\sin x)}+{\sec }^{2}x-\sec x\tan x=4(1-2\sin x)-4\sec x(1-2\sin x)+{\sec }^{2}x(1-2\sin x)\Rightarrow f^{\prime }\left(x\right)=(\sec x-2{)}^2(1-2\sin x)$
If $f^{\prime }\left(x\right)>0\forall x\in (0,a)$ then $f\left(x\right)$ is increasing in $(0,a)\Rightarrow a=\frac{\pi }{6}$