Q82 / 125 · BITSAT 2020 · Mathematics

Let $\mathrm{f}$ be the function defined by $f(x)=\left\{\begin{array}{ll}\frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\ 1 / 2, & x=1\end{array}\right.$

  • A
    The function is continuous for all values of x
  • B
    The function is continuous only for $x>1$
  • C
    The function is continuous at $\mathrm{x}=1$
  • Correct answer
    D
    The function is not continuous at $x=1$

Solution

For $x<1, f(x)=\frac{x^{2}-1}{x^{2}+2 x-3}=\frac{x+1}{x+3}$ $\therefore \lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\frac{1}{2}$ For $x>1, f(x)=\frac{x^{2}-1}{x^{2}-2 x+1}=\frac{x+1}{x-1}$ $\therefore \lim _{x \rightarrow 1^{+}} f(x)=\infty$ $\therefore$ The function is not continuous at $\mathrm{x}=1$.

Hence, the correct answer is option: 'D'.

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