Functions Question 555
Question: If $ f(x)= \begin{cases} & x,,0\le x\le 1 \\ & 2x-1,,1<x \\ \end{cases} . $ , then
[Orissa JEE 2002]
Options:
A) f is discontinuous at $ x=1 $
B) f is differentiable at $ x=1 $
C) f is continuous but not differentiable at $ x=1 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)= \begin{cases} & ,x,0\le x\le 1 \\ & 2x-1,x>1 \\ \end{cases} . $ $ \underset{x\to {1^{-}}}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},f(1-h)=\underset{h\to 0}{\mathop{\lim }},(1-h)=1 $ $ \underset{x\to {1^{+}}}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},f(1+h)=\underset{h\to 0}{\mathop{\lim 2}},(1+h)-1=1 $ $ \because ,\underset{x\to {1^{-}}}{\mathop{\lim }},f(x)=\underset{x\to {1^{+}}}{\mathop{\lim }},f(x)=1 $ \ Function is continuous at $ x=1 $ . $ Lf’(1)=\underset{h\to 0}{\mathop{\lim }},\frac{f(1-h)-f(1)}{-h}=\underset{h\to 0}{\mathop{\lim }},\frac{(1-h)-1}{-h}=1 $ $ Rf’(1)=\underset{h\to 0}{\mathop{\lim }},\frac{f(1+h)-f(1)}{-h}=\underset{h\to 0}{\mathop{\lim }},\frac{2+2h-1-1}{h}=2 $ \ $ Lf’(1)\ne Rf’(1) $ \ Function is not differentiation at $ x=1 $
BETA
