JEE Main 17 March 2021 Shift 2 Solved Paper

Let I=

10

∫

0

‌

[sin‌2‌π‌x]

ex−[x]

dx
⇒‌‌I=

10

∫

0

‌

[sin‌2‌π‌x]

e{x]

dx [∵x−[x]={x}]
From above integrand, we observe that ‌

[sin‌2‌π‌x]

e[x]

is a periodic function with period ' 1 '.
∴‌‌I=10‌

1

∫

0

‌

[sin‌2‌π‌x]

e[x]

[by the property of definite integral,

nT

∫

0

f(x)dx=n‌

T

∫

0

f(x)dx
where f(x) is a periodic function with period =T]
⇒‌‌I=10(

1/2

∫

0

‌

0

ex

dx+

1

∫

1/2

‌

(−1)

ex

dx)
Comparing Eq. (i) by αe−1+βe‌

−1

2

+γ, we get
α=10,β=−10 and γ=0
Hence, α+β+γ=10−10+0
⇒‌‌α+β+γ=0