Question

lf $f\left(x\right)=a{e}^{2x}+b{e}^x+cx$ and $f\left(0\right)=-1,f^{\prime }\left(\log 2\right)=31$,

${\int }_0^{\log 4}\left(f\right(x)-cx)dx=\frac{39}{2}$, then

• A. $a=5$
• B. $b=6$
• C. $c=2$
• D. $a=3$

Answer 1

The correct option is A $a=5$

$f\left(x\right)=a{e}^{2x}+b{e}^x+cx\Rightarrow f\left(0\right)=a+b=-1$ ...(1)
$f^{\prime }\left(x\right)=2a{e}^{2x}+b{e}^x+c\Rightarrow f^{\prime }\left(\log 2\right)=8a+2b+c=31$ ...(2)
${\int }_0^{\log 4}(f\left(x\right)-cx)dx=\frac{39}{2}\Rightarrow {\int }_0^{\log 4}(a{e}^{2x}+b{e}^x)dx=\frac{39}{2}$
$\Rightarrow {[\frac{1}{2}a{e}^{2x}+b{e}^x]}_0^{\log 4}=\frac{39}{2}\Rightarrow 15a+6b=39$ ...(3)
Solving (1), (2) and (3), we get
$a=5,b=-6,c=3$