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Integral equation f(x) = x^2 + integral of e^(-t) f(x - t) dt | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
f(x)f(x) is a differentiable function such that
f(x)=x2+∫0xe−tf(x−t) dt,f(x) = x^2+\int_0^x e^{-t}f(x-t)\,dt ,
 then
Af(x)=x33+xf(x) = \dfrac{x^3}{3}+x
Bf(x)=x33+x2f(x) = \dfrac{x^3}{3}+x^2correct
Cf′(x)=x2+2xf'(x) = x^2+2xcorrect
D∫−22(f(x)−x2)dx=0\displaystyle\int_{-2}^{2}\left(f(x)-x^2\right)dx = 0correct
Solution
Step 1: Put u=x−tu = x-t, so t=x−ut = x-u and dt=−dudt = -du; when t=0t=0, u=xu=x and when t=xt=x, u=0u=0.
∫0xe−tf(x−t) dt=∫0xe−(x−u)f(u) du=e−x∫0xeuf(u) du\int_0^x e^{-t}f(x-t)\,dt = \int_0^x e^{-(x-u)}f(u)\,du = e^{-x}\int_0^x e^{u}f(u)\,du
 Step 2:
f(x)=x2+e−x∫0xetf(t) dtf(x) = x^2+e^{-x}\int_0^x e^{t}f(t)\,dt
exf(x)=x2ex+∫0xetf(t) dte^{x}f(x) = x^2e^{x}+\int_0^x e^{t}f(t)\,dt
 Step 3: Differentiating both sides, by the fundamental theorem of calculus on the right,
exf(x)+exf′(x)=2xex+x2ex+exf(x)e^{x}f(x)+e^{x}f'(x) = 2xe^{x}+x^2e^{x}+e^{x}f(x)
 ex≠0e^{x}\ne0, so cancelling exe^{x} and the common term exf(x)e^{x}f(x),
f′(x)=x2+2xf'(x) = x^2+2x
 (3) is correct. Step 4:
f(x)=x33+x2+cf(x) = \frac{x^3}{3}+x^2+c
 Putting x=0x=0 in the given equation, f(0)=0+0=0f(0) = 0+0 = 0, so c=0c = 0 and
f(x)=x33+x2f(x) = \frac{x^3}{3}+x^2
 (2) is correct; (1) is wrong, it has +x+x in place of +x2+x^2. Step 5:
∫−22(f(x)−x2)dx=∫−22x33 dx=0\int_{-2}^{2}\left(f(x)-x^2\right)dx = \int_{-2}^{2}\frac{x^3}{3}\,dx = 0
 since x33\dfrac{x^3}{3} is odd and the interval is symmetric about 00. (4) is correct. Answer: (2), (3) and (4)
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