Question

lf $f\left(x\right)=\{\begin{array}{cc}-x-\frac{\pi }{2},& x\le -\frac{\pi }{2}\\ -cosx,& -\frac{\pi }{2}<x\le 0\\ x-1,& 0<x\le 1\\ lnx,& x>1\end{array}$ then

• A. $f\left(x\right)$ ls contlnuous at $x=-\frac{\pi }{2}$
• B. $f\left(x\right)$ is not differentiable at $x=0$
• C. $f\left(x\right)$ is differentiable at $x=1$
• D. $f\left(x\right)$ is differentiable at $x=-\frac{3}{2}$

Answer 1

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

The correct options are
A $f\left(x\right)$ ls contlnuous at $x=-\frac{\pi }{2}$
B $f\left(x\right)$ is differentiable at $x=1$
C $f\left(x\right)$ is differentiable at $x=-\frac{3}{2}$
D $f\left(x\right)$ is not differentiable at $x=0$

Given: $f\left(x\right)=\{\begin{array}{cc}-x-\frac{\pi }{2},& x\le -\frac{\pi }{2}\\ -cosx,& -\frac{\pi }{2}<x\le 0\\ x-1,& 0<x\le 1\\ lnx,& x>1\end{array}$

At $x=-\frac{\pi }{2}$,
$LHL={lim}_{x\to -\frac{\pi }{2}}f\left(x\right)={lim}_{h\to 0}-(-\frac{\pi }{2}-h)-\frac{\pi }{2}=0$
$RHL={lim}_{x\to {-\frac{\pi }{2}}^{+}}f\left(x\right)={lim}_{h\to 0}\cos (-\frac{\pi }{2}+h)={lim}_{h\to 0}\cos (\frac{\pi }{2}-h)={lim}_{h\to 0}\sin h=0$
Also, $f(-\frac{\pi }{2})=-(-\frac{\pi }{2})-\frac{\pi }{2}=0$
$\Rightarrow LHL=RHL=f(-\frac{\pi }{2})=0$

$\Rightarrow f\left(x\right)$ is continuous at $x=-\frac{\pi }{2}$
Now, $f^{\prime }\left(x\right)=\{\begin{array}{cc}-1& x\le -\frac{\pi }{2}\\ \sin x& -\frac{\pi }{2}<x\le 0\\ 1& 0<x\le 1\\ \frac{1}{x},& x>1\end{array}$
At $x=0$

$LHL=0,RHL=1$

$LHL\ne RHL$

$\Rightarrow f\left(x\right)$ is not differentiable at $x=0$

At $x=1$

$LHD=RHD=1$

$\Rightarrow f\left(x\right)$ is differentiable at $x=1$
Now, at $x=-\frac{3}{2}$

$LHD={lim}_{x\to -{\frac{3}{2}}^{-}}={lim}_{h\to 0}\sin (-\frac{3}{2}-h)=-\sin \frac{3}{2}$

$RHD={lim}_{x\to -{\frac{3}{2}}^{+}}={lim}_{h\to 0}\sin (-\frac{3}{2}+h)=-\sin \frac{3}{2}$

$LHD=RHD$

$\Rightarrow f\left(x\right)$ is differentiable at $x=-\frac{3}{2}$