Question

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x^5-32}{x-2},& x\ne 2\end{array}\\ \begin{array}{cc}k,& x=2\end{array}\end{array}$ is continuous at $x=2$, then the value of $k$ is

• A. $10$
• B. $15$
• C. $35$
• D. $80$

Answer 1

The correct option is D $80$

Function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x^5-32}{x-2},& x\ne 2\end{array}\\ \begin{array}{cc}k,& x=2\end{array}\end{array}$ is continuous at $x=2$

We know that $\underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2}{lim}\left(\frac{x^5-32}{x-2}\right)=\underset{x\to 2}{lim}\left(\frac{x^5-2^5}{x-2}\right)$

We also know that $\underset{x\to a}{lim}\left(\frac{x^n-a^n}{x-a}\right)=n{a}^{n-1}$

Therefore, $\underset{x\to 2}{lim}f\left(x\right)=5\times 2^4=80$

Since $f\left(2\right)=k$ and $f\left(x\right)$ is continuous at $x=2,$ therefore $\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)\Rightarrow k=80.$