Q47 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
If $f(x)$ satisfies the relation $f(x)=e^{x}+\int_{0}^{1}\left(y+x e^{x}\right) f(y) d y$, then $e+f(0)$ is equal to $\_\_\_\_$.
Correct answer: 2
Solution
$f(x) = e^x + \displaystyle\int_0^1 (y + xe^x)f(y)\, dy$
Splitting the integral:
$f(x) = e^x + \displaystyle\int_0^1 yf(y)\, dy + xe^x \int_0^1 f(y)\, dy$
Let $A = \displaystyle\int_0^1 yf(y)\, dy$ and $B = \displaystyle\int_0^1 f(y)\, dy$
$\Rightarrow f(x) = (1 + Bx)e^x + A$
At $x = 0$: $f(0) = 1 + A$
Now compute $B$ using the expression for $f(y)$:
$B = \displaystyle\int_0^1 \left[(1 + By)e^y + A\right] dy$
$B = \displaystyle\int_0^1 e^y\, dy + B\int_0^1 ye^y\, dy + \int_0^1 A\, dy$
Evaluating each integral:
$\displaystyle\int_0^1 e^y\, dy = e - 1$
$\displaystyle\int_0^1 ye^y\, dy = [ye^y]_0^1 - \int_0^1 e^y\, dy = e - (e - 1) = 1$
$\displaystyle\int_0^1 A\, dy = A$
$\Rightarrow B = (e - 1) + B(1) + A$
$\Rightarrow 0 = e - 1 + A$
$\Rightarrow A = 1 - e$
$f(0) = 1 + A = 1 + (1 - e) = 2 - e$
$e + f(0) = e + (2 - e) = 2$
Answer: $2$
Splitting the integral:
$f(x) = e^x + \displaystyle\int_0^1 yf(y)\, dy + xe^x \int_0^1 f(y)\, dy$
Let $A = \displaystyle\int_0^1 yf(y)\, dy$ and $B = \displaystyle\int_0^1 f(y)\, dy$
$\Rightarrow f(x) = (1 + Bx)e^x + A$
At $x = 0$: $f(0) = 1 + A$
Now compute $B$ using the expression for $f(y)$:
$B = \displaystyle\int_0^1 \left[(1 + By)e^y + A\right] dy$
$B = \displaystyle\int_0^1 e^y\, dy + B\int_0^1 ye^y\, dy + \int_0^1 A\, dy$
Evaluating each integral:
$\displaystyle\int_0^1 e^y\, dy = e - 1$
$\displaystyle\int_0^1 ye^y\, dy = [ye^y]_0^1 - \int_0^1 e^y\, dy = e - (e - 1) = 1$
$\displaystyle\int_0^1 A\, dy = A$
$\Rightarrow B = (e - 1) + B(1) + A$
$\Rightarrow 0 = e - 1 + A$
$\Rightarrow A = 1 - e$
$f(0) = 1 + A = 1 + (1 - e) = 2 - e$
$e + f(0) = e + (2 - e) = 2$
Answer: $2$
Hence, the correct answer is 2.