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Let  f:(0,)R and let

F(x)=0xf(t)dt

If Fx2=x2(1+x) then f (4) equals

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detailed solution

Correct option is C

F(x)=∫0x f(t)dt⇒F′(x)=f(x)∴f(4)=F′(2) Also, 2xF′x2=2x(1+x)+x2⇒F′x2=1+3x/2∴f(4)=F′22=1+3=4


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