Question

If the function $f\left(x\right)=\{\begin{array}{c}\frac{x^2-(A+2)x+2A}{x-2}forx\ne 2\\ 2forx=2\end{array}$ is continuous at $x=2$, then-

• A. $A=0$
• B. $A=1$
• C. $A=-1$
• D. None of these

Answer 1

The correct option is A $A=0$

Given $f\left(x\right)=\{\begin{array}{c}\frac{x^2-(A+2)x+2A}{x-2},x\ne 2\\ 2,x=2\end{array}$
Since, f(x) is continuous at $x=2$
$\underset{h\to 0}{lim}f(2+h)=\underset{h\to 0}{lim}f(2-h)=f\left(2\right)$ ...(1)
$RHL=\underset{x\to 2^{+}}{lim}$
$=\underset{h\to 0}{lim}f(2+h)$
$=\underset{h\to 0}{lim}\frac{{(2+h)}^2-(A+2)(2+h)+2A}{h}$
$=\underset{h\to 0}{lim}\frac{4+4h+h^2-2A-4-Ah-2h+2A}{h}$
$=\underset{h\to 0}{lim}\frac{2h+h^2-Ah}{h}$
$=\underset{h\to 0}{lim}h+2-A$
$\Rightarrow 2-A$
Given $f\left(2\right)=2$
So, by eqn (1)
$2-A=2$
$\Rightarrow A=0$