Question

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

• A. 10
• B. 13
• C. 12
• D. 8

Answer 1

The correct option is B 13

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