Question

Let $f\left(x\right)=\sin x,g\left(x\right)=[x+1]$ and $g\left(f\right(x\left)\right)=h\left(x\right)$, where $[.]$ is the greatest integer function. The $h^{\prime }\left(\frac{\pi }{2}\right)$ is

• A. Non existent
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is A Non existent

$h\left(x\right)=g\left(f\left(x\right)\right)=[f\left(x\right)+1]=\left[\sin \left(x\right)+1\right]{\bigcup }_{k⟶0}h(\frac{\pi }{2}+k)={\bigcup }_{k⟶0}1+{\bigcup }_{k⟶0}\left[\sin (\frac{\pi }{2}+k)\right]=1+0=1{\bigcup }_{k⟶0}h(\frac{\pi }{2}-k)={\bigcup }_{k⟶0}1+{\bigcup }_{k⟶0}\left[\sin (\frac{\pi }{2}-k)\right]=1+0=1h\left(\frac{\pi }{2}\right)=1+\left[\sin \left(\frac{\pi }{2}\right)\right]=1+1=2$

where $[.]$ represents greatest integer function

$\because$ left hand limit $=$ Right hand lmit

$\ne$ value of fnction at that point

$\therefore h\left(x\right)$ is not continous at $x=\frac{\pi }{2}$

Here$h\left(x\right)$ s not differentiable at $x=\frac{\pi }{2}$

$\Rightarrow h^{{}^{\prime }}\left(x\right)$ doe not exst

Answer A