Question

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

• A. $A=0$
• B. $A=\frac{1}{8}$
• C. $A=0,\frac{1}{8}$
• D. $A\ne 0,\frac{1}{8}$

Answer 1

The correct option is A $A=0$

Given f(x) is continuous at x = 2, the only possible is that x - 2 is a factor of $A^2{x}^3+x^2-(A+2)x+A$
$\therefore 8A^2+4-(A+2).2+A=0\Rightarrow 8A^2-A=0\Rightarrow A=0,\frac{1}{8}But\underset{x\to 2}{lim}f\left(x\right)=2\Rightarrow A=0\therefore A=0$