Question

# If f is continuous function and F(x)=∫x0((2t+3)∫2tf(u)du)dt, then the value of ∣∣∣F′′(2)f(2)∣∣∣

A
3
B
5
C
7
D
9

Solution

## The correct option is C 7∴F(x)=∫x0((2t+3)∫2tf(u)du)dt Differentiating both sides w.r.t. x, then  F′(x)=(2x+3).∫2xf(u)du Again differentiating both sides w.r.t. x, then  F′′(x)=(2x+3).(0−f(x))+∫2xf(u)du×2∴F′′(2)=−7f(2)+0⇒∣∣∣F′′(2)f(2)∣∣∣=7Mathematics

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