Question

If $f\left(x\right)={\cos }^{2}x\cdot e^{\tan x},x\in (-\frac{\pi }{2},\frac{\pi }{2})$, then

• A. $f^{\prime }\left(x\right)$ has a point of local minimum at $x=\frac{\pi }{4}$
• B. $f^{\prime }\left(x\right)$ has a point of local maximum in $(-\frac{\pi }{4},0)$
• C. $f^{\prime }\left(x\right)$ has exactly two points of extrema.
• D. $f^{\prime \prime }\left(x\right)=0$ has no real roots.

Answer 1

Answer: (A), (B), (C)

$f^{\prime }\left(x\right)$ has exactly two points of extrema.

$f\left(x\right)={\cos }^{2}x\cdot e^{\tan x}\Rightarrow f^{\prime }\left(x\right)=-2\cos x\sin x{e}^{\tan x}+{\cos }^{2}x{e}^{\tan x}{\sec }^{2}x\Rightarrow f^{\prime }\left(x\right)=e^{\tan x}\cdot (1-\sin 2x)\Rightarrow f^{\prime \prime }\left(x\right)=e^{\tan x}{\sec }^{2}x(1-\sin 2x)-2e^{\tan x}\cos 2x\Rightarrow f^{\prime \prime }\left(x\right)=e^{\tan x}[1+{\tan }^{2}x-2\tan x+\frac{2({\tan }^{2}x-1)}{1+{\tan }^{2}x}]\Rightarrow f^{\prime \prime }\left(x\right)=e^{\tan x}\left[\right(\tan x-1{)}^2+\frac{2({\tan }^{2}x-1)}{1+{\tan }^{2}x}]\Rightarrow f^{\prime \prime }\left(x\right)=e^{\tan x}(\tan x-1)\left[\right(\tan x-1)+\frac{2(\tan x+1)}{1+{\tan }^{2}x}]\Rightarrow f^{\prime \prime }\left(x\right)=\frac{e^{\tan x}(\tan x-1)}{1+{\tan }^{2}x}[\tan x+{\tan }^{3}x-1-{\tan }^{2}x+2\tan x+2]\Rightarrow f^{\prime \prime }\left(x\right)=\frac{e^{\tan x}(\tan x-1)}{1+{\tan }^{2}x}[{\tan }^{3}x-{\tan }^{2}x+3\tan x+1]$

Now, let $g\left(x\right)={\tan }^{3}x-{\tan }^{2}x+3\tan x+1$
$g^{\prime }\left(x\right)=[3{\tan }^{2}x-2\tan x+3]{\sec }^{2}x$
For $x\in (-\frac{\pi }{2},\frac{\pi }{2}),$
$g^{\prime }\left(x\right)>0$,
so $g\left(x\right)$ is an increasing function.

Now, $f^{\prime \prime }\left(\frac{\pi }{4}\right)=0$
$f^{\prime \prime }(-\frac{\pi }{4})f^{\prime \prime }\left(0\right)<0$
So, $f^{\prime \prime }\left(x\right)=0$ has one root in $(-\frac{\pi }{4},0)$
$f^{\prime }\left(x\right)$ has exactly two points of local maxima/minima.

$f^{\prime \prime }\left(x\right)=0$ has exactly two real roots in $(-\frac{\pi }{2},\frac{\pi }{2})$

At $x=\frac{\pi }{4},$
$f^{\prime \prime }\left({\frac{\pi }{4}}^{+}\right)>0$ and $f^{\prime \prime }\left({\frac{\pi }{4}}^{-}\right)<0$
So, $f^{\prime }\left(x\right)$ has local minimum at $x=\frac{\pi }{4}$.