Question

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

• A. $f$ is monotonic increasing in $\left(0,\frac{\pi }{4}\right)$.
• B. $f$ is monotonic decreasing in $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$.
• C. $\exists$ some $c\in \left(0,\frac{\pi }{2}\right)$ for which $f^{\prime }(c)=0$
• D. The equation $f(x)=0$ has two roots in $\left[0,\frac{\pi }{2}\right]$.

Answer 1

Answer: (C), (D)

We have $f^{\prime }(x)=5{\sin }^{4}x\cos x-5{\cos }^{4}x\sin x=5\sin x\cos x(\sin x-\cos x)(1+\sin x\cos x)$
$\therefore f^{\prime }(x)=0$ at $x=\frac{\pi }{4}.$ Also $f^{\prime }(0)=f^{\prime }\left(\frac{\pi }{2}\right)=0$
Hence $\exists$ some $c\in \left(0,\frac{\pi }{2}\right)$ for which $f^{\prime }(c)=0$ (By Rolle's Theorem) $\Rightarrow$ (C) is correct.
Also in $\left(0,\frac{\pi }{4}\right)f$ is decreasing and in $\left(\frac{\pi }{4},\frac{\pi }{2}\right)f$ is increasing $\Rightarrow$ minimum at $x=\frac{\pi }{4}$
As $f(0)=f\left(\frac{\pi }{2}\right)=0\Rightarrow 2$ roots $\Rightarrow$ (D) is correct.