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

Question

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

Infinity Learn · Accepted Answer

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, $$f(4)=3(4$$−$$1)=9$$

$let f : R \to R be non-constant differentiable function and satisfies.$ $f (x) = x^{2} - \int_{0}^{1} (x + f (t))^{2} d t . Then f (4) is equal to$

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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

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$let f : R \to R be non-constant differentiable function and satisfies.$ $f (x) = x^{2} - \int_{0}^{1} (x + f (t))^{2} d t . Then f (4) is equal to$