Q113 / 180 · COMEDK 2012 · Mathematics

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

  • A
    5
  • Correct answer
    B
    3
  • C
    2
  • D
    0

Solution

$\because f(x)$ is continuous at $x=0$ $$ \begin{aligned} &\text { So, } \lim _{x \rightarrow 0} f(x)=f(0) \\ &\Rightarrow \quad \lim _{x \rightarrow 0} \frac{e^{3 x-1}}{4 x}=\frac{k+0}{4} \\ &\Rightarrow \quad \frac{1}{4} \lim _{x \rightarrow 0} 3\left(\frac{e^{3 x}-1}{3 x}\right)=\frac{\bar{k}}{4} \\ &\Rightarrow \quad 3 \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{3 x}=k \Rightarrow 3 \times 1=k \Rightarrow k=3 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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