Question

If $f(x)=\begin{cases}\cos 2 x, & 0 \leq x \leq \frac{\pi}{4} \\ \lambda(\sin x-\cos x), & \frac{\pi}{4}< x \leq \frac{\pi}{2}\end{cases}$ is continuous in $\left[0, \frac{\pi}{2}\right]$ and differentiable in $\left(0, \frac{\pi}{2}\right)$ where $\lambda \in R$, then

• A. $f ( x )$ is non-monotonic function in $\left(0, \frac{\pi}{2}\right)$.
• B. there exist more than one $c \in\left(0, \frac{\pi}{2}\right)$ such that $f ^{\prime}( c )=-2$.
• C. there exist atleast one $c \in\left(0, \frac{\pi}{2}\right)$ such that $f ^{\prime}( c )=\frac{-2(\sqrt{2}+1)}{\pi}$.
• D. there cannot exist any $c \in\left(0, \frac{\pi}{2}\right)$ such that $f ^{\prime}( c )=\frac{-2(\sqrt{2}+1)}{\pi}$

Answer 1

Answer: (C)

Since, $f(x)$ is differentiable at $x=\frac{\pi}{4}$, so $\lambda=-\sqrt{2}$.
$\therefore f ( x )=\begin{cases}\cos 2 x ; & 0 \leq x \leq \frac{\pi}{4} \\-\sqrt{2}(\sin x -\cos x ) ; & \frac{\pi}{4}< x \leq \frac{\pi}{2}\end{cases}$
Clearly $f ^{\prime}( x )<0 \quad \forall x \in\left(0, \frac{\pi}{2}\right)$.
Hence, $f(x)$ is decreasing in $\left(0, \frac{\pi}{2}\right)$.
Applying L.M.V.T. to $f ( x )$ in $\left[0, \frac{\pi}{2}\right]$, there exist some $c \in\left(0, \frac{\pi}{2}\right)$ such that
$f^{\prime}(c)=\frac{f\left(\frac{\pi}{2}\right)-f(0)}{\frac{\pi}{2}-0}=\frac{-2}{\pi}(\sqrt{2}+1) $
Also, $f ^{\prime}\left(\frac{\pi}{4}\right)=-2 \quad \Rightarrow \quad f ^{\prime}( x )=-2$
for unique value of $c \in\left(0, \frac{\pi}{2}\right)$.