Question

Let $a$ be a positive real number such that ${\int }_0^a{e}^{x-\left[x\right]}dx=10e-9$ where $\left[x\right]$ is the greatest integer less than or equal to $x$. Then $a$ is equal to

• A. $10+{\log }_{e}3$
• B. $10-{\log }_e(1+e)$
• C. $10+{\log }_e(1+e)$
• D. $10+{\log }_{e}2$

Answer 1

The correct option is D $10+{\log }_{e}2$

${\int }_0^a{e}^{x-\left[x\right]}dx=10e-9$
Here $e^{x–\left[x\right]}=e^{\left\{x\right\}}$ is periodic function of period $1$
$\therefore {\int }_0^{\left[a\right]+\left\{a\right\}}{e}^{\left\{x\right\}}dx=10e-9$
$\Rightarrow {\int }_0^{\left[a\right]}{e}^{\left\{x\right\}}dx+{\int }_{\left[a\right]}^{\left\{a\right\}+\left[a\right]}{e}^{\left\{x\right\}}dx=10e-9$
$\because {\int }_k^{k+nT}f\left(x\right)dx={\int }_0^{nT}f\left(x\right)dx=n{\int }_0^{T}f\left(x\right)dx,$ where $T$ is the period of the $f\left(x\right),n\in N,k\in R$
$\Rightarrow \left[a\right]{\int }_0^1{e}^{x}dx+{\int }_0^{\left\{a\right\}}{e}^{x}dx=10e-9$
$\Rightarrow \left[a\right](e-1)+(e^{\left\{a\right\}}-1)=10e-9$
$\Rightarrow \left[a\right]e+e^{\left\{a\right\}}-\left[a\right]-1=10e-9$
$\therefore$ Possible value of $a=10+{\log }_{e}2$