JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given esin‌y‌cos‌y‌

dy

dx

+esin‌y‌cos‌x=cos‌x . . . (i)
Let esin‌y=t, then esin‌y⋅cos‌y⋅‌

dy

dx

=‌

dt

dx′

Putting in Eq. (i),
cos‌x ... (ii) (Linear form)
Then, IF =e∫cos‌x‌d‌x=esin‌x
Solution of differential Eq. (ii) is,
t⋅IF=∫cos‌x⋅IFdx+C
t⋅esin‌x=∫cos‌x⋅esin‌xdx+C
=eu‌ i.e. let ‌sin‌x=u‌ then ‌cos‌x‌d‌x=du
⇒‌‌t⋅esin‌x=∫eudu+C=eu+C
Put u=sin‌x and t=esin‌y
⇒‌‌esin‌y⋅esin‌x=esin‌x+C
Given, y(0)=0, this gives C=0
⇒‌‌esin‌y⋅esin‌x=esin‌x
⇒‌‌esin‌y+sin‌x=esin‌x
⇒‌‌sin‌y+sin‌x=sin‌x
⇒‌‌sin‌y=0
⇒‌‌y=0
∴y(π∕6)=y(π∕3)=y(π∕4)=0
Hence, 1+y(‌

π

6

)+‌

√3

2

y(‌

π

3

)+‌

1

√2

y(‌

π

4

)
=1+0+0+0=1