Question

If $f$ is continuous function and $F\left(x\right)={\int }_0^x\left(\right(2t+3\left){\int }_t^{2}f\left(u\right)du\right)dt$, then the value of $\left|\frac{F^{\prime \prime }\left(2\right)}{f\left(2\right)}\right|$

• A. 3
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (C)

7

$\therefore F\left(x\right)={\int }_0^x\left(\right(2t+3\left){\int }_t^{2}f\left(u\right)du\right)dt$
Differentiating both sides w.r.t. x, then
$F^{\prime }\left(x\right)=(2x+3).{\int }_x^{2}f\left(u\right)du$
Again differentiating both sides w.r.t. x, then
$F^{\prime \prime }\left(x\right)=(2x+3).(0-f(x\left)\right)+{\int }_x^{2}f\left(u\right)du\times 2\therefore F^{\prime \prime }\left(2\right)=-7f\left(2\right)+0\Rightarrow \left|\frac{F^{\prime \prime }\left(2\right)}{f\left(2\right)}\right|=7$