Question

If $f\left(x\right)=\{\begin{array}{ccc}\frac{e^{3x}-1}{4x}& for& x\ne 0\\ \frac{k+x}{4}& for& x=0\end{array}$ is continuous at $x=0$, then $k=$

• A. $5$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is B $3$

$f\left(x\right)$ is continuous at $x=0$

$\therefore \underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\frac{e^{3x}-1}{4x}=\frac{k+0}{4}$

$\Rightarrow \underset{x\to 0}{lim}\frac{e^{3x}-1}{x}=k$

$\Rightarrow \underset{x\to 0}{lim}3\frac{(e^{3x}-1)}{3x}=k$

$\Rightarrow 3.1=k$

$\Rightarrow k=3$