Question

If the function $f\left(x\right)=\frac{x^2-(A+2)x+A}{x-2}$, for $x\ne 2$ and $f\left(2\right)=2$, is continuous at $x=2$, then find the value of A.

Answer 1

Here, we given that

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

For Remainder,

Let put $x=2$ in $x^2-(A+2)x+A$.

as we know for continuous function remainder will be zero.

$\therefore f\left(x\right)=x^2-(A+2)x+A$

$f\left(2\right)=(2{)}^2-(A+2\left)\right(2)+A$

$=4-2A-4+A$

$\therefore A=0$

So, function $f\left(x\right)$ is written as

$f\left(x\right)=\frac{x^2-(0+2)x+0}{x-2}$

$=\frac{x^2-2x}{x-2}$

$=\frac{x(x-2)}{(x-2)}$

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