Question

If $n$ is the number of real solutions of the equation $min(e^{-|x|},1-e^{-|x|})=\frac{1}{4}$ and $L=\underset{x\to 0^{-}}{lim}(\frac{e^{2x}-1}{x}+\frac{e^{3x}-1}{x}+\frac{e^{4x}-1}{x}+\cdots \text{upto}n\text{terms}),$ then the value of $L$ is

• A. $5$
• B. $7$
• C. $10$
• D. $14$

Answer 1

The correct option is D $14$

For intersection point,
$1-e^{-|x|}=e^{-|x|}$
$\Rightarrow e^{-|x|}=\frac{1}{2}$
$\Rightarrow -|x|=-\ln 2$
$\Rightarrow x=\ln 2,-\ln 2$


Using the graph,
Number of solutions $=4$

$L=\underset{x\to 0^{-}}{lim}(\frac{e^{2x}-1}{x}+\frac{e^{3x}-1}{x}+\frac{e^{4x}-1}{x}+\frac{e^{5x}-1}{x})$
$=\underset{x\to 0^{-}}{lim}(2\cdot \frac{e^{2x}-1}{2x}+3\cdot \frac{e^{3x}-1}{3x}+4\cdot \frac{e^{4x}-1}{4x}+5\cdot \frac{e^{5x}-1}{5x})$
Since $\underset{x\to 0}{lim}\frac{e^{ax}-1}{ax}=1$
$\therefore L=2+3+4+5=14$