Q. If \(f(x)\) satisfies the relation
\[
f(x) = e^x + \int_0^1 (y + x e^x) f(y)\,dy,
\]
then \(e + f(0)\) is equal to
Explanation
Given,
\[
f(x) = e^x + \int_0^1 (y + x e^x) f(y)\,dy
\]
Split the integral:
\[
\int_0^1 y f(y)\,dy + x e^x \int_0^1 f(y)\,dy
\]
Let
\[
\int_0^1 y f(y)\,dy = A,\quad \int_0^1 f(y)\,dy = B
\]
Then,
\[
f(x) = e^x + A + x e^x B
\]
Put \(x = 0\):
\[
f(0) = 1 + A
\]
Now integrate \(f(x)\) from \(0\) to \(1\):
\[
\int_0^1 f(x)\,dx = \int_0^1 e^x dx + A + B\int_0^1 x e^x dx
\]
\[
B = (e - 1) + A + B
\]
This gives
\[
A = 1 - e
\]
Hence,
\[
f(0) = 1 + (1 - e) = 2 - e
\]
\[
e + f(0) = e + (2 - e) = 2
\]
Therefore, the correct answer is 2.