Let f:R R be defined as f(x)=ae^2x+be^x+cx.

Question

Let $f:R	o R$ be defined as $f(x)=a{e}^{2x}+b{e}^{x}+cx$. If $f(0)=-1,{f}^{'}({\log }_{e}2)=21$ and ${\int }_{{\log }_{e}4}^{0}(f(x)-cx)dx=\frac{39}{2}$, then the value of $|a+b+c|$ equals

Accepted Answer

$\begin{gathered} f(x)=a{e}^{2x}+b{e}^{x}+cx f(0)=-1 \ \ a+b=-1 \ \ {f}^{'}(x)=2a{e}^{2x}+b{e}^{x}+c {f}^{'}(\ln 2)=21 \ \ 8a+2b+c=21 \ \ \int _{0}^{\ln 4}(a{e}^{2x}+b{e}^{x})dx=\frac{39}{2} \end{gathered}$

$\begin{gathered} {[\frac{{ae}^{2x}}{2}+{be}^{x}]}_{\ln 4}^{0}=\frac{39}{2}\Rightarrow 8a+4b-\frac{a}{2}-b=\frac{39}{2} \ 15a+6b=39 \ 15a-6a-6=39 \ 9a=45\Rightarrow a=5 \ b=-6 \ c=21-40+12=-7 \ a+b+c-8 \ |a+b+c|=8 \end{gathered}$

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