Functions Question 68
Question: If $ f(x)= \begin{cases} & \frac{x^{2}-4x+3}{x^{2}-1},\ for\ x\ne 1 \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2,\ \text{for }x=1 \\ \end{cases} . $ , then
[IIT 1972]
Options:
A) $ \underset{x\to 1^+}{\mathop{\lim }}f(x)=2 $
B) $ \underset{x\to 1^-}{\mathop{\lim }}f(x)=3 $
C) $ f(x) $ is discontinuous at $ x=1 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)={ \frac{x^{2}-4x+3}{x^{2}-1} } $ , for $ x\ne 1 $ $ =2, $ , for $ x=1 $ $ f(1)=2,f(1+)=\underset{x\to 1+}{\mathop{\lim }}\frac{x^{2}-4x+3}{x^{2}-1}=\underset{x\to 1+}{\mathop{\lim }}\frac{(x-3)}{(x+1)}=-1 $ $ f(1-)=\underset{x\to 1-}{\mathop{\lim }}\frac{x^{2}-4x+3}{x^{2}-1}=-1\Rightarrow f(1)\ne f(1-) $ Hence the function is discontinuous at $ x=1. $
BETA
