Question

Let f, f', f" be continuous in [0, ln 2] and f(0) = 0, f'(0) = 3, f(In 2) = 6, f'(In 2) = 4 and ${\int }_0^{ln2}{e}^{-2x}f\left(x\right)dx=3$
then ${\int }_0^{ln2}{e}^{-2x}{f}^{\prime \prime }\left(x\right)dx$ is

• A. 10
• B. 13
• C. 12
• D. None of these

Answer 1

The correct option is B 13

${\int }_0^{ln2}{e}^{-2x}{f}^{\prime \prime }\left(x\right)dx$
$={\left[e^{-2x}{f}^{\prime }\right(x\left)\right]}_0^{ln2}+2{\int }_0^{ln2}{e}^{-2x}{f}^{\prime }\left(x\right)dx$
$=\left[\frac{1}{4}{f}^{\prime }\right(In2)-3]+2{\left[e^{-2x}f\right(x)+2\int e^{-2x}f(x\left)dx\right]}_0^{In2}=13$