If f(x) = l x^2 - 4x + 3 x^2 - 1 , ; for ;x 1 ; ; ; ; ; ; ; ; ; ;… - Vidyadip

MCQ

If $f(x) = \left\{ \begin{array}{l}\frac{{{x^2} - 4x + 3}}{{{x^2} - 1}},\;{\rm{for}}\;x \ne 1\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;2,\;{\rm{for\, }}x = 1\end{array} \right.$, then

A
$\mathop {\lim }\limits_{x \to 1 + } f(x) = 2$
B
$\mathop {\lim }\limits_{x \to 1 - } f(x) = 3$
✓
$f(x)$ is discontinuous at $x = 1$
D
None of these

✓

Answer

Correct option: C.

$f(x)$ is discontinuous at $x = 1$

c
(c) $f(x) = \left\{ {\frac{{{x^2} - 4x + 3}}{{{x^2} - 1}}} \right\}$, for $x \ne 1$

$ = 2\,$, for $x = 1$

$f(1) = 2,\,\,f(1 + ) = \mathop {\lim }\limits_{x \to 1 + } \,\frac{{{x^2} - 4x + 3}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1 + } \,\frac{{(x - 3)}}{{(x + 1)}} = - 1$

$f(1 - ) = \mathop {\lim }\limits_{x \to 1 - } \,\frac{{{x^2} - 4x + 3}}{{{x^2} - 1}} = - 1\,\, \Rightarrow \,\,f(1) \ne f(1 - )$

Hence the function is discontinuous at $x = 1.$

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