A calculus problem by Ankit Nigam

Calculus Level pending

If 0 x f ( t ) d t = x + x 1 t f ( t ) d t \int _{ 0 }^{ x }{ f\left( t \right) } dt=x+\int _{ x }^{ 1 }{ t } f\left( t \right) dt then the value of f ( 1 ) f\left( 1 \right) is

PRACTICE FOR BITSAT HERE

1/2 0 1 -1/2

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1 solution

Ankit Nigam
Apr 22, 2015

0 x f ( t ) d t = x + x 1 t f ( t ) d t \int _{ 0 }^{ x }{ f\left( t \right) dt } =x+\int _{ x }^{ 1 }{ tf\left( t \right) dt }

Applying Newton Leibnitz Theorem

1. { f ( x ) } = 1 { x f ( x ) } . 1 f ( x ) = 1 1 + x f ( 1 ) = 1 2 1.\left\{ f\left( x \right) \right\} =1-\left\{ xf\left( x \right) \right\} .1\\ \therefore f\left( x \right) =\frac { 1 }{ 1+x } \\ \boxed { \therefore f\left( 1 \right) =\frac { 1 }{ 2 } }

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