Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-(a+2)x+a}{x-2}& x\ne 2\\ 2& x=2\end{array}$ is continuous at $x=2$, then the value of $a$ is

• A. $-6$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

$0$

Given,

$f\left(x\right)=\frac{x^2-(a+2)x+a}{x-2}$ when $x\ne 2$

$f\left(2\right)=2$

For $f\left(x\right)$ to be continuous at $x=2,$

$\underset{x\to 2}{lim}\text{f}\left(x\right)=\text{f}\left(2\right)$

$\underset{x\to 2}{lim}\frac{x^2-(a+2)x+a}{x-2}=2$

Using L'Hospital rule,

$\underset{x\to 2}{lim}\frac{2x-(a+2)}{1}=2$

$\Rightarrow 2\times 2-a-2=2$

$\Rightarrow a=0$