Question

If $f(x)=\{\begin{array}{ll}\cos 2x,& 0\le x\le \frac{\pi }{4}\\ \lambda (\sin x-\cos x),& \frac{\pi }{4}<x\le \frac{\pi }{2}\end{array}$ 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 $\lambda =-\sqrt{2}$.
$\therefore f(x)=\{\begin{array}{ll}\cos 2x;& 0\le x\le \frac{\pi }{4}\\ -\sqrt{2}(\sin x-\cos x);& \frac{\pi }{4}<x\le \frac{\pi }{2}\end{array}$
Clearly $f^{\prime }(x)<0\forall x\in \left(0,\frac{\pi }{2}\right)$.
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, $f^{\prime }\left(\frac{\pi }{4}\right)=-2\Rightarrow f^{\prime }(x)=-2$
for unique value of $c\in \left(0,\frac{\pi }{2}\right)$.