Let f be a differentiable function such that f(x) = x2+∫0xe−t f(x−t) dt then f(1)=

f(x)=x2+∫0xe−tf(x−t) dtf(x) = x2+∫0xe−(x−t) f(t)dtf(x)=x2+e−x∫0xetf(t)dt −−−−−−(1) f1(x) = 2x−e−x∫0xet(t)dt+e−x.exf(x) f1(x) = 2x−e−x∫0xetf(t)dt + f(x)∴ f(x) = x33+x2+c (1)+(2)⇒f1(x)=x2+2x ∴ f(x) =x33+x2 ⇒f(1) = 43

Let f be a differentiable function such that f(x) = x2+∫0xe−t f(x−t) dt then f(1)=

f(x)=x2+∫0xe−tf(x−t) dtf(x) = x2+∫0xe−(x−t) f(t)dtf(x)=x2+e−x∫0xetf(t)dt −−−−−−(1) f1(x) = 2x−e−x∫0xet(t)dt+e−x.exf(x) f1(x) = 2x−e−x∫0xetf(t)dt + f(x)∴ f(x) = x33+x2+c (1)+(2)⇒f1(x)=x2+2x ∴ f(x) =x33+x2 ⇒f(1) = 43

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Let $f$ be a differentiable function such that $f (x) = x^{2} + \int_{0}^{x} e^{- t} f (x - t) d t$ then $f (1) =$

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$\frac{4}{3}$

$\frac{1}{3}$

answer is D.

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Detailed Solution

$f (x) = x^{2} + \int_{0}^{x} e^{- t} f (x - t) d t$

$f (x) = x^{2} + \int_{0}^{x} e^{- (x - t)} f (t) d t$

$f (x) = x^{2} + e^{- x} \int_{0}^{x} e^{t} f (t) d t - - - - - - (1) f^{1} (x) = 2 x - e^{- x} \int_{0}^{x} e^{t} (t) d t + e^{- x} . e^{x} f (x) f^{1} (x) = 2 x - e^{- x} \int_{0}^{x} e^{t} f (t) d t + f (x) ∴ f (x) = \frac{x^{3}}{3} + x^{2} + c (1) + (2) \Rightarrow f^{1 (x)} = x^{2} + 2 x$