Let f be the function defined by f(x)=\ array cc (x^2-1)/(x^2-2|x-1|-1), & x ≠ 1 (1)/(2), & x=1 array .

Correct answer: D. The function is not continuous at x=1 f(x)=\ array cc (x^2-1)/(x^2-2 x+1) ; & x>1 (1)/(2) ; & x=1 (x^2-1)/(x^2+2 x-3) ; & x < 1 array . x^2+3 x-x-3=x(x+3)-1(x+3) Lets Find LHL, _ x arrow 1^ - (x^2-1)/(x^2-2 x+1) and RHL, _ x arrow 1^ + (x^2-1)/(x^2+2 x-3) aligned & _ x arrow 1^ - ((x-1)(x+1))/((x-1)^2) _ x arrow 1^ + ((x-1)(x+1))/((x+3)(x-1)) & _ x arrow 1^ - (x+1)/(x-1)=∞ _ x arrow 1^ + (x+1)/(x+3)=(2)/(4)=(1)/(2) aligned LHL ≠ f(1) f(x) is not continuous at x=1

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Mathematics
Chapter: Continuity and Differentiability

BITSAT 2024 · Mathematics · Continuity and Differentiability

Let f be the function defined by f(x)=\ array cc...

Let

f

be the function defined by

f(x)=\left\{

\begin{array}{cc} \frac{x^2-1}{x^2-2|x-1|-1}, & x \neq 1 \\ \frac{1}{2}, & x=1 \end{array}

\right.

Options

AThe function is continuous for all values of $x$
BThe function is continuous only for $x>1$
CThe function is continuous at $x=1$
DThe function is not continuous at $x=1$

Correct answer

D. The function is not continuous at $x=1$

Step-by-step solution

f(x)=\left\{

\begin{array}{cc}\frac{x^2-1}{x^2-2 x+1} ; & x>1 \\ \frac{1}{2} \quad ; & x=1 \\ \frac{x^2-1}{x^2+2 x-3} ; & x < 1\end{array}

\right.\because \quad x^2+3 x-x-3=x(x+3)-1(x+3)

Lets Find LHL,

\lim _{x \rightarrow 1^{-}} \frac{x^2-1}{x^2-2 x+1}

and RHL,

\lim _{x \rightarrow 1^{+}} \frac{x^2-1}{x^2+2 x-3}

\begin{aligned} & \Rightarrow \lim _{x \rightarrow 1^{-}} \frac{(x-1)(x+1)}{(x-1)^2} \Rightarrow \lim _{x \rightarrow 1^{+}} \frac{(x-1)(x+1)}{(x+3)(x-1)} \\ & \Rightarrow \lim _{x \rightarrow 1^{-}} \frac{x+1}{x-1}=\infty \Rightarrow \lim _{x \rightarrow 1^{+}} \frac{x+1}{x+3}=\frac{2}{4}=\frac{1}{2} \end{aligned}

\therefore

LHL

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

is not continuous at

x=1

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