Application of DerivativesmediumFree

Solve the Integral Equation f(x) = 3x^2 + integral of e^-t f(x-t) dt | JEE

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

Question
If
f(x)=3x2+∫0xe−tf(x−t) dt,f(x) = 3x^{2} + \int_{0}^{x} e^{-t} f(x-t)\,dt ,
 then
Af(x)=0f(x) = 0 has 33 solutions
Bf(x)f(x) is monotonic increasing
Cf(x)=4f(x) = 4 has 22 solutionscorrect
D∫−11f(x) dx=2\displaystyle\int_{-1}^{1} f(x)\,dx = 2correct
Solution
Step 1: Tidy the integral by substitution. Put u=x−tu = x-t, so t=x−ut = x-u and dt=−dudt = -du; the limits t=0,xt = 0, x become u=x,0u = x, 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: Rewrite the given equation.
f(x)=3x2+e−x∫0xetf(t) dt....(i)f(x) = 3x^{2} + e^{-x}\int_{0}^{x}e^{t}f(t)\,dt . \qquad \text{...(i)}
 Step 3: Differentiate, using the product rule on the second term and the fundamental theorem of calculus on the integral.
f′(x)=6x−e−x∫0xetf(t) dt+e−x⋅exf(x).f'(x) = 6x - e^{-x}\int_{0}^{x}e^{t}f(t)\,dt + e^{-x}\cdot e^{x}f(x) .
 Step 4: Replace the integral term using (i), which says e−x∫0xetf(t) dt=f(x)−3x2e^{-x}\int_{0}^{x}e^{t}f(t)\,dt = f(x) - 3x^{2}.
f′(x)=6x−[f(x)−3x2]+f(x)=3x2+6x.f'(x) = 6x - \left[f(x) - 3x^{2}\right] + f(x) = 3x^{2} + 6x .
 Step 5: Integrate.
f(x)=x3+3x2+C.f(x) = x^{3} + 3x^{2} + C .
 Step 6: Fix the constant. Putting x=0x = 0 in (i) makes the integral vanish, so f(0)=0f(0) = 0, giving C=0C = 0:
f(x)=x3+3x2.f(x) = x^{3} + 3x^{2} .
 Question attachment Step 7: Test (1).
x3+3x2=0  ⟹  x2(x+3)=0  ⟹  x=0 or x=−3,x^{3}+3x^{2} = 0 \implies x^{2}(x+3) = 0 \implies x = 0 \ \text{or}\ x = -3 ,
 which is 22 distinct solutions, not 33. False. Step 8: Test (2).
f′(x)=3x2+6x=3x(x+2)<0  for x∈(−2,0),f'(x) = 3x^{2}+6x = 3x(x+2) < 0 \ \text{ for } x \in (-2, 0) ,
 so ff decreases there and is not monotonic increasing. False. Step 9: Test (3).
x3+3x2=4  ⟹  x3+3x2−4=0.x^{3}+3x^{2} = 4 \implies x^{3}+3x^{2}-4 = 0 .
 Since x=1x = 1 works, (x−1)(x-1) is a factor:
(x−1)(x2+4x+4)=(x−1)(x+2)2=0  ⟹  x=1 or x=−2,(x-1)\left(x^{2}+4x+4\right) = (x-1)(x+2)^{2} = 0 \implies x = 1 \ \text{or}\ x = -2 ,
 which is 22 distinct solutions. True. Step 10: Test (4). The odd term integrates to zero over a symmetric interval:
∫−11(x3+3x2)dx=0+3[x33]−11=[x3]−11=1−(−1)=2.\int_{-1}^{1}\left(x^{3}+3x^{2}\right)dx = 0 + 3\left[\frac{x^{3}}{3}\right]_{-1}^{1} = \left[x^{3}\right]_{-1}^{1} = 1-(-1) = 2 .
 True. Answer: (3) and (4).
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