Question

If the function
$f\left(x\right)=\{\begin{array}{cc}\frac{A{x}^3+x^2-(A+2)x+A}{x-2}& \text{for}x\ne 2\\ 2& \text{for}x=2\end{array}$
is continuous at $x=2$, then.

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

Answer 1

Answer: (A)

$A=0$

Since $f$ is continuous at $x=2$, the only possibility is that $x-2$ is a factor of $A{x}^3\times x^2-(A+2)x+A.$
$\therefore 8A+4-2(A+2)+A=0\Rightarrow 7A=0$
$\Rightarrow A=0$