Question

Let $y' = e^{-2x}$ and $y = 0$ when $x =e$. Then the value of $x$ when $y =\frac{1}{2}$ is

• A. $e - 1$
• B. $\frac{1}{2} (e - 1)$
• C. $\frac{1}{2} (3e - 1)$
• D. none of these

Answer 1

Answer: (D)

none of these

Answer 2

$y' = e^{-2x}$ ......(i)
Integrating both sides, we get
$y =\frac{e^{-2x}}{- 2} +c$ ......(ii)
We have, $y (e) = 0$
$\therefore \:\: 0 =\frac{e^{-2e}}{-2} + c \Rightarrow c= \frac{e^{-2e}}{2}$
Putting in (ii), we get
$y = \frac{e^{-2x}}{-2} + \frac{e^{-2x}}{2}$
$y\left(x\right)= \frac{1}{2} \Rightarrow \frac{1}{2} = \frac{e^{-2x}}{-2} + \frac{e^{-2e}}{2}$
$\Rightarrow 1=-e^{-2x} + e^{-2e}$
$\Rightarrow e^{-2x} =e^{-2e} - 1$
Taking log on both sides, we get
$-2x = \log\left(e^{-2e} -1\right)$
$x =-\frac{1}{2}\log\left(e^{-2e} -1\right)$