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Find f(4) given F(x²)=x²(1+x), F=∫f | JEE

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Question
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} and F(x)=∫0xf(t) dtF(x)=\displaystyle\int_{0}^{x}f(t)\,dt. If F ⁣(x2)=x2(1+x)F\!\big(x^{2}\big)=x^{2}(1+x), then f(4)f(4) equals
A54\dfrac54
B77
C44correct
D22
Solution
Step 1: By the Fundamental Theorem of Calculus, F′(x)=f(x)F'(x)=f(x). The given relation is
F ⁣(x2)=x2+x3.F\!\big(x^{2}\big)=x^{2}+x^{3}.
 Step 2: Differentiate both sides with respect to xx, using the chain rule on the left:
F′ ⁣(x2)⋅2x=2x+3x2.F'\!\big(x^{2}\big)\cdot 2x=2x+3x^{2}.
 Step 3: Solve for F′ ⁣(x2)=f ⁣(x2)F'\!\big(x^{2}\big)=f\!\big(x^{2}\big):
f ⁣(x2)=2x+3x22x=1+32x.f\!\big(x^{2}\big)=\dfrac{2x+3x^{2}}{2x}=1+\dfrac{3}{2}x.
 Step 4: To obtain f(4)f(4), set x2=4x^{2}=4, i.e. x=2x=2:
f(4)=1+32(2)=1+3=4.f(4)=1+\dfrac{3}{2}(2)=1+3=4.
 Correct answer: (3)
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