Question

For a $aϵ[\pi ,2\pi ],$ the function $f\left(x\right)=\frac{1}{3}\sin a{\tan }^{3}x+(\sin a-1)\tan x+\frac{\sqrt{a-2}}{\sqrt{8-a}}$

• A. $x=n\pi \left(nϵI\right)$ as critical points
• B. no critical points
• C. $x=2n\pi \left(nϵI\right)$ as critical points
• D. $x=(2n+1)\pi \left(nϵI\right)$ as critical points.

Answer 1

The correct option is B no critical points

$f\left(x\right)=\frac{1}{3}\sin a{\tan }^{3}x+(\sin a-1)\tan x+\frac{\sqrt{a-2}}{\sqrt{8-a}}$
$f^{\prime }\left(x\right)=\sin a{\tan }^{2}x{\sec }^{2}x+(\sin a-1){\sec }^{2}x={\sec }^{2}x(\sin a({\sec }^{2}x-1)+(\sin a-1))$
$\Rightarrow f^{\prime }\left(x\right)={\sec }^{2}x(\sin a{\sec }^{2}x-1)$
for critical points $f^{\prime }\left(x\right)=0$
$\Rightarrow {\sec }^{2}x=0$
$\Rightarrow x\in \varnothing$
or ${\sec }^{2}x=\frac{1}{\sin a}$
since, for $aϵ[\pi ,2\pi ],\sin a<0$
Therefore, $x\in \varnothing$
Thus, $f\left(x\right)$ have no critical points.
Ans: B