MCQ
If the function$f(x)=\left\{\begin{array}{cc}\frac{x^2-(A+2) x+A}{x-2}, & \text { for } x \neq 2 \\2, & \text { for } x=2\end{array}\right. $is continuous at $x=2$, then
  • $A=0$
  • B
    $A=1$
  • C
    $A=-1$
  • D
    $A=2$

Answer

Correct option: A.
$A=0$
(A)
Since $f (x)$ is continuous at $x=2$.
$\therefore f (2)=\lim _{x \rightarrow 2} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 2} \frac{x^2-( A +2) x+ A }{x-2}$
$\Rightarrow 2=\lim _{x \rightarrow 2} \frac{x(x-2)- A (x-1)}{x-2}$,
which is true if $A =0$

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