JEE Main 2021 — Mathematics Calculus

Given:$a>0$a>0

Let $n\leq a<n+1,n\in W$n≤a<n+1,n∈W

Then,

$$a=[a]+{a}$$a=[a]+a

Here, $[a]=n$[a]=n

Now,

$${\int }_{0}^{a}{e}^{x-[x]}dx=10e-9$$∫0a​ex−[x]dx=10e−9

$$\Rightarrow {\int }_{0}^{n}{e}^{{x}}dx+{\int }_{n}^{a}{e}^{x-[x]}dx=10e-9$$⇒∫0n​exdx+∫na​ex−[x]dx=10e−9

$$\Rightarrow n{\int }_{0}^{1}{e}^{x}dx+{\int }_{n}^{a}{e}^{x-n}dx=10e-9$$⇒n∫01​exdx+∫na​ex−ndx=10e−9

$$\Rightarrow n(e-1)+({e}^{a-n}-1)=10e-9$$⇒n(e−1)+(ea−n−1)=10e−9

$$\Rightarrow ne-n+{e}^{a-n}-1=10e-9$$⇒ne−n+ea−n−1=10e−9

$$\Rightarrow ne-n-1+{e}^{a}{e}^{-n}=10e-9$$⇒ne−n−1+eae−n=10e−9

On comparing, we get

$$n=10$$n=10

and,

$$-n-1+{e}^{a}{e}^{-n}=-9$$−n−1+eae−n=−9

$$\Rightarrow -10-1+{e}^{a-10}=-9$$⇒−10−1+ea−10=−9

$$\Rightarrow {e}^{a-10}=2$$⇒ea−10=2

$$\Rightarrow (a-10){\mathrm{log}}_{e}(e)={\mathrm{log}}_{e}(2)$$⇒(a−10)loge​(e)=loge​(2)

$$\Rightarrow a=(10+{\mathrm{log}}_{e}2)$$⇒a=(10+loge​2)