JEE Main 2022 — Mathematics Calculus

Given, $\displaystyle {\int }_{-3}^{101}([\mathrm{sin}\pi x]+{e}^{[\mathrm{cos}2\pi x]})dx$∫−3101​([sinπx]+e[cos2πx])dx

Now let $\displaystyle {I}_{1}={\int }_{-3}^{101}[\mathrm{sin}\pi x]dx$I1​=∫−3101​[sinπx]dx and $\displaystyle {I}_{2}={\int }_{-3}^{101}{e}^{[\mathrm{cos}2\pi x]}dx$I2​=∫−3101​e[cos2πx]dx

Now checking periodicity of $[\mathrm{sin}\pi x]$[sinπx] we get,

$$0<x<1\Rightarrow [\mathrm{sin}\pi x]=0$$0<x<1⇒[sinπx]=0

$$1<x<2\Rightarrow [\mathrm{sin}\pi x]=-1$$1<x<2⇒[sinπx]=−1

$[\mathrm{sin}\pi x]\rightarrow$[sinπx]→ periodic with period 2

So, $\displaystyle {I}_{1}=52{\int }_{0}^{2}[\mathrm{sin}\pi x]dx$I1​=52∫02​[sinπx]dx

$$=52[{\int }_{0}^{1}0+{\int }_{1}^{2}-1dx]=-52$$=52[∫01​0+∫12​−1dx]=−52

Now checking periodicity of $[\mathrm{cos}2\pi x]$[cos2πx] we get,

$$0<x<\frac{1}{4}\Rightarrow [\mathrm{cos}2\pi x]=0$$0<x<41​⇒[cos2πx]=0

$$\frac{1}{4}<x<\frac{1}{2}\Rightarrow [\mathrm{cos}2\pi x]=-1$$41​<x<21​⇒[cos2πx]=−1

$$\frac{1}{2}<x<\frac{3}{4}\Rightarrow [\mathrm{cos}2\pi x]=-1$$21​<x<43​⇒[cos2πx]=−1

$$\frac{3}{4}<x<1\Rightarrow [\mathrm{cos}2\pi x]=0$$43​<x<1⇒[cos2πx]=0

So, $[\mathrm{cos}2\pi x]$[cos2πx] has periodicity of $1$1

So, $\displaystyle {I}_{2}=104({\int }_{0}^{\frac{1}{4}}{e}^{0}dx+{\int }_{\frac{1}{4}}^{\frac{3}{4}}{e}^{-1}dx+{\int }_{\frac{3}{4}}^{1}{e}^{0}dx)$I2​=104(∫041​​e0dx+∫41​43​​e−1dx+∫43​1​e0dx)

$$\Rightarrow {I}_{2}=104(\frac{1}{4}+\frac{1}{e}\times \frac{1}{2}+\frac{1}{4})$$⇒I2​=104(41​+e1​×21​+41​)

$$\Rightarrow {I}_{2}=52+\frac{52}{e}$$⇒I2​=52+e52​

So, ${I}_{1}+{I}_{2}=\frac{52}{e}$I1​+I2​=e52​ or $\displaystyle {\int }_{-3}^{101}([\mathrm{sin}\pi x]+{e}^{[\mathrm{cos}2\pi x]})dx=\frac{52}{e}$∫−3101​([sinπx]+e[cos2πx])dx=e52​