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Let $f_{1} (x) = \int_{0}^{x} f (t) d t, f_{2} (x) = \int_{0}^{x} f_{1} (t) d t$ and $f_{3} (x)$
$= \int_{0}^{x} f_{2} (t) d t$ if $f_{3} (x) = A \int_{0}^{x} f (t) (x - t)^{2} d t$ then the value of $A$ is

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Correct option is B

Note that ddxf3(x)=f2(x),ddxf2(x)=f(x) and ddxf1(x)=f(x). Integrating by partsf3(x)=tf2(t)0x−∫0x tf1(t)dt=xf2(x)−∫0x tf1(t)dt=∫0x (x−t)f1(t)dt=−(x−t)22f1(t)0x+12∫0x (x−t)2f(t)dt=12∫0x (x−t)2f(t)dt. Thus A=12.

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Let f1(x)=∫0x f(t)dt,f2(x)=∫0x f1(t)dt and f3(x)=∫0x f2(t)dt if f3(x)=A∫0x f(t)(x−t)2dt then the value of A is

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Let $f_{1} (x) = \int_{0}^{x} f (t) d t, f_{2} (x) = \int_{0}^{x} f_{1} (t) d t$ and $f_{3} (x)$
$= \int_{0}^{x} f_{2} (t) d t$ if $f_{3} (x) = A \int_{0}^{x} f (t) (x - t)^{2} d t$ then the value of $A$ is