Q52 / 120 · VITEEE 2008 · Mathematics
The value of $\mathrm{f}(0)$ so that $\frac{\left(-\mathrm{e}^{\mathrm{x}}+2^{\mathrm{x}}\right)}{\mathrm{x}}$ may be continuous at $x=0$ is
- A$\log \left(\frac{1}{2}\right)$
- B0
- C4
- Correct answerD$-1+\log 2$
Solution
$f(x)=\frac{-e^{x}+2^{x}}{x}$
$=\frac{1}{x}\left[-\left(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+. .\right)+1+\frac{\log 2}{1 !} x+\right.$
$\left.\quad \frac{(\operatorname{bg} 2)^{2}}{2 !} x^{2}+\frac{(\log 2)^{3}}{3 !} x^{3}+. .\right]$
$\begin{aligned}
\mathrm{f}(\mathrm{x})=& \log 2-1+\frac{\mathrm{x}}{2 !}\left\{(\log 2)^{2}-1\right\} \\
&+\frac{\mathrm{x}^{2}}{3 !}\left\{\left(\log ^{2}\right)^{3}-1\right\}+\ldots .
\end{aligned}$
Putting $x=0$, we get
$\mathrm{f}(0)=\log 2-1+0+0+\ldots .=-1+\log 2$
Hence, the correct answer is option: 'D'.