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# If the function $f(x)=\left\{\begin{array}{1 1}\large\frac{x^2-(A+2)x+A}{x-2}&x\neq 2\\2&x=2\end{array}\right.$ is continuous at $x=2$ then

$(a)\;A=0\qquad(b)\;A=1\qquad(c)\;A=-1\qquad(d)\;None\;of\;these$

Since $f(x)$ is continuous at $x=2$
$f(2)=\lim\limits_{x\to 2}f(x)$
$\Rightarrow 2=\lim\limits_{x\to 2}\large\frac{x^2-(A+2)x+A}{x-2}$
$\Rightarrow$ This is satisfied if $A=0$
Hence (a) is the correct answer.