Let f :R R be a twice differentiable function such that f (2 )=1. If F (x )= xf (x ) for all x R, 0.05emx F^ ' (x )dx=6 and 0.05emx^2 F^ ' ' (x )dx=40, then F^ ' (2 )+ 0.05em F (x )dx is equal to:

Question

Accepted Answer

Correct answer: A. (a) _0^2 0.05em xF^ ' (x ) dx=6 = ( xF (x ) )_0^2- 0.05emf (x ) dx=6 =2 F (b )- 0.05em xF (x ) dx=6 [ ∵f (b )=2 F (b )=2 ] 0.05em xF (x ) dx=-2 … (a ) ⇨ 0.05emF (x ) dx=-2 … (b ) Also _0^2 0.05emx^2F^ ' ' (x ) dx= (x^2F^ ' (x ) )_0^2-2 _0^2 0.05emx^ 'F^ ' (x ) dx=40 =4F^ ' (b )-2 6=40 F^ ' (b )=13 ∴F^ ' (b )+ _0^2 0.05emF (x )=13-2=11

$Let f : R \to R be a twice differentiable function such that f (2) = 1 . If F (x) = xf (x) for all x R, \sum x F^{’} (x) d x = 6 and \sum x^{2} F^{’ ’} (x) d x = 40, then F^{’} (2) + \sum F (x) d x is equal to:$

Solution

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