Question

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

• A. $10$
• B. $13$
• C. $12$
• D. $8$

Answer 1

The correct option is B $13$

$\underset{0}{\overset{\ln 2}{\int }}{e}^{-2x}{f}^{\prime \prime }\left(x\right)dx={\left[e^{-2x}f{}^{\prime }\right(x\left)\right]}_0^{\ln 2}+2{\int }_0^{\ln 2}{e}^{-2x}{f}^{{}^{\prime }}\left(x\right)dx$

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

$=13$