Question

Let $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{(x^3+x^2-16x+20)}{{(x-2)}^2},& x\ne 2\end{array}\\ \begin{array}{cc}k& x=2\end{array}\end{array}$.

If $f\left(x\right)$ is continuous for all $x$, then $k$ is equal to

• A. $7$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

The correct option is A $7$

$f\left(2\right)=k$

$\underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2}{lim}\frac{x^3+x^2-16x+20}{{(x-2)}^2}$

$=\underset{x\to 2}{lim}\frac{{(x-2)}^2(x+5)}{{(x-2)}^2}=\underset{x\to 2}{lim}(x+5)=7$

As $f\left(x\right)$ is continuous

$\Rightarrow k=7$