Question

Consider the function $f\left(x\right)={\sin }^{5}x+{\cos }^{5}x-1,$ $x\in [0,\frac{\pi }{2}].$ Which of the following is (are) CORRECT?

• A. $f$ is strictly decreasing in $[0,\frac{\pi }{4}]$
• B. $f$ is strictly increasing in $[\frac{\pi }{4},\frac{\pi }{2}]$
• C. There exists a number $c$ in $(0,\frac{\pi }{2})$ such that $f^{\prime }\left(c\right)=0$
• D. The equation $f\left(x\right)=0$ has only two roots in $[0,\frac{\pi }{2}]$

Answer 1

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

The correct options are
A $f$ is strictly decreasing in $[0,\frac{\pi }{4}]$
B $f$ is strictly increasing in $[\frac{\pi }{4},\frac{\pi }{2}]$
C There exists a number $c$ in $(0,\frac{\pi }{2})$ such that $f^{\prime }\left(c\right)=0$
D The equation $f\left(x\right)=0$ has only two roots in $[0,\frac{\pi }{2}]$

$f\left(x\right)={\sin }^{5}x+{\cos }^{5}x-1$
$f^{\prime }\left(x\right)=\left(5\sin x\cos x\right)({\sin }^{3}x-{\cos }^{3}x)$
$\therefore f^{\prime }\left(x\right)<0$ for all $x\in (0,\frac{\pi }{4})$
and $f^{\prime }\left(x\right)>0$ for $x\in (\frac{\pi }{4},\frac{\pi }{2})$

Since, $f\left(0\right)=0=f\left(\frac{\pi }{2}\right)$
Applying Rolles theorem to $f$ on $(0,\frac{\pi }{2})$
$f^{\prime }\left(c\right)=0$ for at least one $c$ in $(0,\frac{\pi }{2})$

Also, ${\sin }^{5}x+{\cos }^{5}x\le {\sin }^{2}x+{\cos }^{2}x=1$ for $x\in [0,\frac{\pi }{2}]$
Equality holds only if ${\sin }^{5}x={\sin }^{2}x$ and ${\cos }^{5}x={\cos }^{2}x$
$\Rightarrow x=0,\frac{\pi }{2}$