Question

Let $f\left(x\right)=\cos x$ and $g\left(x\right)=[x+2]$, where $[.]$ denotes the greatest integer function. Then, ${\left(gof\right)}^{\prime }\left(\frac{\pi }{2}\right)$ is?

• A. $1$
• B. $0$
• C. $-1$
• D. Does not exist

Answer 1

The correct option is D Does not exist

$L{\left(gof\right)}^{\prime }\left(\frac{\pi }{2}\right)$

$=\underset{h\to 0}{lim}\frac{\left(gof\right)(\frac{\pi }{2}-h)-\left(gof\right)\left(\frac{\pi }{2}\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{\left[\cos (\frac{\pi }{2}-h)+2\right]-[\cos \frac{\pi }{2}+2]}{-h}$

$=\underset{h\to 0}{lim}\frac{[\sin h+2]-\left[2\right]}{-h}=\underset{h\to 0}{lim}\frac{2-2}{-h}=0$

$R{\left(fog\right)}^{\prime }\left(\frac{\pi }{2}\right)=\underset{h\to 0}{lim}\frac{{\left(gof\right)}^{\prime }(\frac{\pi }{2}+h)-\left(gof\right)\left(\frac{\pi }{2}\right)}{h}$

$=\underset{h\to 0}{lim}\frac{[\cos (\frac{\pi }{2}+h)+2]-[\cos \frac{\pi }{2}+2]}{h}$

$=\underset{h\to 0}{lim}\frac{[-\sin h+2]-\left[2\right]}{h}$

$=\underset{h\to 0}{lim}\frac{1-2}{h}\to -\infty$

$\therefore \left(gof\right)$ is not differentiable at $x=\frac{\pi }{2}.$