Question

If function be $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-1}{x-1}& \text{when}x\ne 1\\ k& \text{when}x=1\end{array}$ is continuous at $x=1$, then the value of $k$ is?

• A. -1
• B. 2
• C. -3
• D. -2

Answer 1

Answer: (B)

If a function $f\left(x\right)$ is continuous at $x=a$ then

${lim}_{x\to a^{+}}f\left(x\right)={lim}_{x\to a^{-}}f\left(x\right)=f\left(a\right)$

Given: $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-1}{x-1}& \text{when}x\ne 1\\ k& \text{when}x=1\end{array}$

${lim}_{x\to 1^{+}}f\left(x\right)=\frac{x^2-1}{x-1}$

$\Rightarrow {lim}_{x\to 1^{+}}f\left(x\right)=\frac{(x+1)(x-1)}{x-1}$

$\Rightarrow {lim}_{x\to 1^{+}}f\left(x\right)=x+1$

$\Rightarrow {lim}_{x\to 1^{+}}f\left(x\right)=1+1$

$\Rightarrow {lim}_{x\to 1^{+}}f\left(x\right)=2$---(1)

${lim}_{x\to 1^{-}}f\left(x\right)=\frac{x^2-1}{x-1}$

$\Rightarrow {lim}_{x\to 1^{-}}f\left(x\right)=\frac{(x+1)(x-1)}{x-1}$

$\Rightarrow {lim}_{x\to 1^{-}}f\left(x\right)=x+1$

$\Rightarrow {lim}_{x\to 1^{-}}f\left(x\right)=1+1$

$\Rightarrow {lim}_{x\to 1^{-}}f\left(x\right)=2$---(2)

$f\left(1\right)=k$---(3)

It is given that, $f\left(x\right)$ is continuous at $x=a$.

${lim}_{x\to 1^{+}}f\left(x\right)={lim}_{x\to 1^{-}}f\left(x\right)=f\left(1\right)$

$\Rightarrow f\left(1\right)={lim}_{x\to 1^{+}}f\left(x\right)$

$\Rightarrow k=2$

Hence, Option C is correct.