Question

Let $f\left(x\right)=\left|x\right|$ and $g\left(x\right)=\left[x\right],$ (where, $\left[.\right]$ denotes the greatest integer function). Then, $\left(f o g\right)^{'} \left(- 1\right)$ is

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

Answer 1

Answer: (B)

$ (f \circ g)(x)=f(g(x))=f([x])=|[x]| $
Now, LHL of $(f \circ g)^{\prime}(-1)=\lim _{h \rightarrow 0^{+}} \frac{(f \circ g)(-1-h)-(f \circ g)(-1)}{-h}$
$=\underset{h \rightarrow 0^{+}}{lim} \frac{\left|\left[- 1 - h\right]\right| - \left|\left[- 1\right]\right|}{- h}$
$=\underset{h \rightarrow 0^{+}}{lim} \frac{\left|- 2\right| - \left|- 1\right|}{- h}=\underset{h \rightarrow 0^{+}}{lim} ⁡ \frac{1}{- h} \rightarrow -\in fty$
$\therefore \left(f o g\right)^{'} \left(- 1\right)$ does not exist.