Q105 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Mathematics
If the function defined by $f(x)=\left\{\begin{array}{cl}\frac{2^x-2^{-x}}{x}, & x \neq 0 \text { is continuous at } \\ k, & x=0\end{array}\right.$ $x=0$, then $e^k$ is equal to
- A$\log \left(\frac{2}{e}\right)$
- B$\log 4$
- Correct answerC4
- D1
Solution
Given, $f(x)=\left\{\begin{array}{cc}\frac{2^x-2^{-x}}{x}, & x \neq 0 \\ k, & x=0\end{array}\right.$
Given, $f(x)$ is continuous at $x=0$.
Hence, $\lim _{x \rightarrow 0} f(x)=k$
$$
\Rightarrow k=\lim _{x \rightarrow 0} \frac{2^x-2^{-x}}{x}=\lim _{x \rightarrow 0} \frac{2^x \log 2+2^{-x} \log 2}{1}
$$
(using L-Hospital rule)
$$
\begin{aligned}
& =\log 2\left(2^{\circ}+2^{\circ}\right)=2 \log 2=\log 4 \\
\therefore e^k & =e^{\log 4}=4
\end{aligned}
$$
Hence, the correct answer is option: 'C'.