Question

If ${\int }_0^a{e}^{x-[x]}dx=10e-9$, then the value of $a$ is (where $[.]$ is greatest integer function)

• A. $9+\ln 2$
• B. $10+\ln 2$
• C. $10$
• D. $9$

Answer 1

The correct option is B

$10+\ln 2$

Explanation for the correct answer:

Let $a$ be any real number which lies between two successive whole numbers

$\Rightarrow n\le a\le n+1$

$a=\left[a\right]+\left\{a\right\}$, where $\left[a\right]$ is the greatest integer less than or equal to $a$ and $\left\{a\right\}$ is the fractional part.

Here $\left[a\right]=n$

The given integral can be split as

$\Rightarrow$${\int }_0^a{e}^{x-[x]}dx={\int }_0^n{e}^{x-[x]}dx+{\int }_n^a{e}^{x-[x]}dx$

$\Rightarrow$${\int }_0^a{e}^{x-[x]}dx={\int }_0^n{e}^{\left\{x\right\}}dx+{\int }_n^a{e}^{x-[x]}dx$

We know that$x-\left[x\right]=\left\{x\right\}$, and the function $\left\{x\right\}$ repeats after every integer

$\Rightarrow$${\int }_0^n{e}^{\left\{x\right\}}dx=n{\int }_0^1{e}^{\left\{x\right\}}dx$

$\Rightarrow$${\int }_0^a{e}^{x-[x]}dx=n{\int }_0^1{e}^{\left\{x\right\}}dx+{\int }_n^a{e}^{x-[x]}dx$

$\left[x\right]=n$ in the interval $[n,n+1)$

$\Rightarrow$${\int }_0^a{e}^{x-[x]}dx=n{\int }_0^1{e}^{x}dx+{\int }_n^a{e}^{x-n}dx$

$\Rightarrow$ $=n{\left[e^x\right]}_0^1+{\left[e^{x-n}\right]}_n^a$

$\Rightarrow$${\int }_0^a{e}^{x-[x]}dx=n\left(e-1\right)+\left(e^{a-n}-1\right)$

${\int }_0^a{e}^{x-[x]}dx=10e-9$

$\Rightarrow$$n\left(e-1\right)+\left(e^{a-n}-1\right)=10e-9$

$\Rightarrow$$\left(ne+e^{a-n}\right)-\left(n+1\right)=10e-9$

$\Rightarrow$ $\left(ne\right)+e^{\left\{a\right\}}-\left(n+1\right)=10e-9$

Comparing coefficients we get,

$\Rightarrow$ $n=10$

$\Rightarrow$$e^{\left\{a\right\}}=-9+11$

$\Rightarrow$$e^{\left\{a\right\}}=2$

$\Rightarrow$ $\left\{a\right\}=\ln 2$

$\Rightarrow$ $a=n+\left\{a\right\}$

$\Rightarrow$ $a=10+\ln 2$

Hence, the value of $a$ is $10+\ln 2$,

Hence, option(B) is the correct answer.