JEE Main 2026 — Mathematics Calculus

$$f(x) = e^x + \displaystyle\int_0^1 (y + xe^x)f(y)\, dy$$f(x)=ex+∫01​(y+xex)f(y)dy

Splitting the integral:

$$f(x) = e^x + \displaystyle\int_0^1 yf(y)\, dy + xe^x \int_0^1 f(y)\, dy$$f(x)=ex+∫01​yf(y)dy+xex∫01​f(y)dy

Let $\displaystyle A = \displaystyle\int_0^1 yf(y)\, dy$A=∫01​yf(y)dy and $\displaystyle B = \displaystyle\int_0^1 f(y)\, dy$B=∫01​f(y)dy

$$\Rightarrow f(x) = (1 + Bx)e^x + A$$⇒f(x)=(1+Bx)ex+A

At $x = 0$x=0: $f(0) = 1 + A$f(0)=1+A

Now compute $B$B using the expression for $f(y)$f(y):

$$B = \displaystyle\int_0^1 \left[(1 + By)e^y + A\right] dy$$B=∫01​[(1+By)ey+A]dy

$$B = \displaystyle\int_0^1 e^y\, dy + B\int_0^1 ye^y\, dy + \int_0^1 A\, dy$$B=∫01​eydy+B∫01​yeydy+∫01​Ady

Evaluating each integral:

$$\displaystyle\int_0^1 e^y\, dy = e - 1$$∫01​eydy=e−1

$$\displaystyle\int_0^1 ye^y\, dy = [ye^y]_0^1 - \int_0^1 e^y\, dy = e - (e - 1) = 1$$∫01​yeydy=[yey]01​−∫01​eydy=e−(e−1)=1

$$\displaystyle\int_0^1 A\, dy = A$$∫01​Ady=A

$$\Rightarrow B = (e - 1) + B(1) + A$$⇒B=(e−1)+B(1)+A

$$\Rightarrow 0 = e - 1 + A$$⇒0=e−1+A

$$\Rightarrow A = 1 - e$$⇒A=1−e

$$f(0) = 1 + A = 1 + (1 - e) = 2 - e$$f(0)=1+A=1+(1−e)=2−e

$$e + f(0) = e + (2 - e) = 2$$e+f(0)=e+(2−e)=2

Answer: $2$2