Functions Question 373
Question: The domain of the derivative of the function $ f(x)= \begin{cases} & {{\tan }^{-1}}x\ \ \ \ \ ,\ |x|\ \le 1 \\ & \frac{1}{2}(|x|\ -1)\ ,\ |x|\ >1 \\ \end{cases} . $ is
[IIT Screening 2002]
Options:
A) $ R-{0} $
B) $ R-{1} $
C) $ R-{-1} $
D) $ R-{-1,\ 1} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)= \begin{cases} & \frac{1}{2}(-x-1),,x<-1 \\ & {{\tan }^{-1}}x,-1\le x\le 1 \\ & \frac{1}{2}(x+1),,x>1 \\ \end{cases} . $ ; $ {f}’(x)= \begin{cases} & -\frac{1}{2},x<-1 \\ & \frac{1}{1+x^{2}},-1<x<1 \\ & \frac{1}{2},x>1 \\ \end{cases} . $ $ {f}’(-1-0)=-\frac{1}{2};{f}’(-1+0)=\frac{1}{1+{{(-1+0)}^{2}}}=\frac{1}{2} $ $ {f}’(1-0)=\frac{1}{1+{{(1-0)}^{2}}}=\frac{1}{2};{f}’(1+0)=\frac{1}{2} $ \ $ {f}’(-1) $ Does not exist; \ domain of $ {f}’(x)=R-{-1} $ .
BETA
