Functions Question 251
Question: Let $ f(x)= \begin{cases} 0, & x<0 \\ x^{2}, & x\ge 0 \\ \end{cases} $ , then for all values of $ x $
[IIT 1984; MP PET 2002]
Options:
A) f is continuous but not differentiable
B) f is differentiable but not continuous
C) $ {f}’ $ is continuous but not differentiable
D) $ {f}’ $ is continuous and differentiable
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)= \begin{cases} \ 0, & x<0 \\ x^{2}, & x\ge 0 \\ \end{cases} . $ ; $ \underset{x\to {0^{-}}}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }} f(0-h)=0 $ and $ \underset{x\to {0^{+}}}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }} f(0+h)=\underset{h\to 0}{\mathop{\lim }} {{(0+h)}^{2}}=0 $
$ \underset{x\to {0^{-}}}{\mathop{\lim }} f(x)=\underset{x\to {0^{+}}}{\mathop{\lim }} f(x)=f(0) $
Hence $ f(x) $ is continuous function at $ x=0 $ .
$ L {f}'(x)=\underset{x\to {0^{-}}}{\mathop{\lim }} \frac{f(x)-f(0)}{x-0}=\underset{h\to 0}{\mathop{\lim }} \frac{f(0-h)-0}{-h}=\underset{h\to 0}{\mathop{\lim }} \frac{0-0}{-h}=0 $
$ R{f}’(x)=\underset{x\to {0^{+}}}{\mathop{\lim }} \frac{f(x)-f(0)}{x-0} $
$ =\underset{h\to 0}{\mathop{\lim }} \frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }} \frac{{{(0+h)}^{2}}-0}{h}=0 $
$ L{f}’(x)=R{f}’(x) $
Hence $ f(x) $ is differentiable at $ x=0 $ .
Now $ {f}’(x)= \begin{cases} 0, & x<0 \\ 2x, & x\ge 0 \\ \end{cases} . $ ;
$ \underset{x\to {0^{-}}}{\mathop{\lim }} {f}’(x)=\underset{h\to 0}{\mathop{\lim }} {f}’(0-h)=0 $ and $ \underset{x\to {0^{+}}}{\mathop{\lim }} {f}’(x)=\underset{h\to 0}{\mathop{\lim }} {f}’(0+h)=\underset{h\to 0}{\mathop{\lim }} 2(0+h)=0 $
$ \underset{x\to {0^{-}}}{\mathop{\lim }} {f}’(x)=\underset{x\to {0^{+}}}{\mathop{\lim }} {f}’(x)=0 $
Hence $ {f}’(x) $ is continuous function at $ x=0 $ .
Now $ L {f}''(x)=\underset{x\to {0^{-}}}{\mathop{\lim }} \frac{f(x)-f(0)}{x-0} $
$ =\underset{h\to 0}{\mathop{\lim }} \frac{f(0-h)-f(0)}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }} \frac{0-0}{-h}=0 $ $ R {f}’’(x)=\underset{x\to {0^{+}}}{\mathop{\lim }} \frac{f(x)-f(0)}{x-0} $
$ =\underset{h\to 0}{\mathop{\lim }} \frac{f(0+h)-f(0)}{h} $
$ =\underset{h\to 0}{\mathop{\lim }} \frac{2(0+h)-0}{h} $
$ =\underset{h\to 0}{\mathop{\lim }} \frac{2h}{h}=2 $ $ L{f}’’(x)\ne R{f}’’(x) $ Hence $ {f}’(x) $ is not differentiable at $ x=0 $ .
BETA
