JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, f(x)=

x

∫

0

etf(t)dt+ex . . . (i)
Since, f(x) is differentiable function, differentiate Eq. (i)
f′(x)=exf(x)+ex[Using Newton Leibnitz theorem ]
f′(x)=ex(f(x)+1)⇒‌

f′(x)

f(x)+1

=ex
Integrating it, ∫‌

f′(x)

f(x)+1

dx=∫exdx+C
Let f(x)+1=u, then f′(x)dx=du
∫‌

du

u

=ex+C⇒log‌u=ex+C
log(f(x)+1)=ex+C‌‌ ‌‌[∵u=f(x)+1]... (ii)
Now
f(x)=

x

∫

0

etf(t)dt+ex⇒f(0)=e0=1
Put x=0, in Eq. (ii), we get
log(2)=e∘+C⇒C=log(2)−1
From Eq. (ii), we get
log(f(x)+1)‌‌=ex+log‌2−1
f(x)+1‌‌=eex+log‌2−1=eex⋅elog‌2⋅e−1
f(x)+1‌‌=2eex⋅e−1=2eex−1
∴‌‌f(x)‌‌=2eex−1−1