Let f be the function defined by f(x)={cases} x²-1/x²-2|x-1|-1, & x ≠ 1 \\ 1/2, & x=1 {cases}

Correct Answer is : The function is not continuous at x=1x=1x=1.

The function is continuous for all values of x

The function is continuous only for x>1

The function is continuous at x=1

The function is not continuous at x=1.

Correct Answer is : The function is not continuous at x=1x=1x=1.

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BITSAT 2024 Shift 1 Paper

Question : 78 of 130

Marks: +1, -0

Let

f

f(x)=\begin{cases} \frac{x^2-1}{x^2-2|x-1|-1}, & x \neq 1 \\ \frac{1}{2}, & x=1 \end{cases}

Solution:

f(x)=\begin{cases} \frac{x^2-1}{x^2-2x+1}, & x>1 \\ \frac{1}{2}, & x=1 \\ \frac{x^2-1}{x^2+2x-3}, & x<1 \end{cases}

\because\; x^2+3x-x-3

= x(x+3)-1(x+3)

Lets Find

\; \text{LHL, } \; \lim\limits_{x \to 1^-} \frac{x^2-1}{x^2-2x+1} \; \text{ and RHL, } \; \lim\limits_{x \to 1^+} \frac{x^2-1}{x^2+2x-3}

\;\Rightarrow \lim\limits_{x \to 1^-} \frac{(x-1)(x+1)}{(x-1)^2} \Rightarrow \lim\limits_{x \to 1^+} \frac{(x-1)(x+1)}{(x+3)(x-1)}

\;\Rightarrow \lim\limits_{x \to 1^-} \frac{x+1}{x-1} = \infty \Rightarrow \lim\limits_{x \to 1^+} \frac{x+1}{x+3} = \frac{2}{4} = \frac{1}{2}

\therefore

LHL

\neq f(1) \Rightarrow f(x)

is not continuous at

x=1

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