JEE Main 2021 — Mathematics Calculus

Let $\displaystyle I={\int }_{0}^{10}\frac{[\mathrm{sin}2\pi x]}{{e}^{x-[x]}}dx={\int }_{0}^{10}\frac{[\mathrm{sin}2\pi x]}{{e}^{{x}}}dx,$I=∫010​ex−[x][sin2πx]​dx=∫010​ex[sin2πx]​dx, where $x-[x]={x}$x−[x]=x is the fractional part function.

We know that the ${x}$x is a periodic function with period $1$1 and also that $\mathrm{sin}2\pi x$sin2πx is a periodic function with period $\frac{2\pi }{2\pi }=1.$2π2π​=1.

Thus, the function $f(x)=\frac{[\mathrm{sin}2\pi x]}{{e}^{{x}}}$f(x)=ex[sin2πx]​ is periodic with period $1.$1.

If $f(x)$f(x) is a periodic function with period $T$T then $\displaystyle {\int }_{0}^{nT}f(x)dx=n{\int }_{0}^{T}f(x)dx,n\in N$∫0nT​f(x)dx=n∫0T​f(x)dx,n∈N

Therefore $\displaystyle I=10{\int }_{0}^{1}\frac{[\mathrm{sin}2\pi x]}{{e}^{{x}}}dx$I=10∫01​ex[sin2πx]​dx

$$\Rightarrow I=10{\int }_{0}^{1}\frac{[\mathrm{sin}2\pi x]}{{e}^{x}}dx$$⇒I=10∫01​ex[sin2πx]​dx

Now, for $0\leq x\leq 1\Rightarrow -1\leq \mathrm{sin}2\pi x\leq 1,$0≤x≤1⇒−1≤sin2πx≤1, hence $[\mathrm{sin}2x]$[sin2x] can take values $0$0 and $-1$−1

$$\Rightarrow I=10({\int }_{0}^{1/2}\frac{[\mathrm{sin}2\pi x]}{{e}^{x}}dx+{\int }_{1/2}^{1}\frac{[\mathrm{sin}2\pi x]}{{e}^{x}}dx)$$⇒I=10(∫01/2​ex[sin2πx]​dx+∫1/21​ex[sin2πx]​dx)

$$\Rightarrow I=10(0+{\int }_{1/2}^{1}\frac{(-1)}{{e}^{x}}dx)$$⇒I=10(0+∫1/21​ex(−1)​dx)

$$\Rightarrow I=-10{\int }_{1/2}^{1}{e}^{-x}dx$$⇒I=−10∫1/21​e−xdx

$$\Rightarrow I=10{[{e}^{-x}]}_{1/2}^{1}$$⇒I=10[e−x]1/21​

$$\Rightarrow I=10({e}^{-1}-{e}^{-1/2})$$⇒I=10(e−1−e−1/2)

Given, $I=\alpha {e}^{-1}+\beta {e}^{-\frac{1}{2}}+\gamma ,$I=αe−1+βe−21​+γ, thus $\alpha =10,\beta =-10,\gamma =0.$α=10,β=−10,γ=0.

And, $\alpha +\beta +\gamma =10-10=0.$α+β+γ=10−10=0.