let f:R→R be non-constant differentiable function and satisfies. f(x)=x2−∫01 (x+f(t))2dt. Then f(4) is equal to

Given that f(x)=x2−∫01 (x+f(t))2dt⇒f(x)=x2−∫01 x2+f2(t)+2xf(t)dt           =x2−x2(t)01−∫01 f2(t)dt+2x∫01 f(t)dt⇒f(x)=−∫01 f2(t)dt−2x∫01 f(t)dt∴f'(x)=−2∫01 f(t)dt……..(i)               = constant ⇒ f must be linear. So, let f(x)=ax+b∴a=−2∫01 (at+b)dt&b=−∫01 (at+b)2dt      since f0=a0+b=b⇒a=3,b=−3⇒f(x)=3(x−1) So, f(4)=3(4−1)=9