Let f: R rightarrow R be defined as f(x)=a e^ | vedclass

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(D) Given $f(x)=a e^{2 x}+b e^x+c x$.$f(0)=-1 \Rightarrow a+b=-1 \quad (1)$$f^{\prime}(x)=2 a e^{2 x}+b e^x+c$.$f^{\prime}(\ln 2)=2 a(4)+b(2)+c=8 a+2

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$8$

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(D) Given $f(x)=a e^{2 x}+b e^x+c x$.$f(0)=-1 \Rightarrow a+b=-1 \quad (1)$$f^{\prime}(x)=2 a e^{2 x}+b e^x+c$.$f^{\prime}(\ln 2)=2 a(4)+b(2)+c=8 a+2 b+c=21 \quad (2)$Given $\int_0^{\ln 4}(f(x)-c x) d x=\frac{39}{2}$.$\int_0^{\ln 4}(a e^{2 x}+b e^x) d x=\left[\frac{a e^{2 x}}{2}+b e^x\right]_0^{\ln 4}=\frac{39}{2}$.$\left(\frac{a(16)}{2}+b(4)\right)-\left(\frac{a}{2}+b\right)=\frac{39}{2}$.$8 a+4 b-\frac{a}{2}-b=\frac{39}{2} \Rightarrow \frac{15 a}{2}+3 b=\frac{39}{2} \Rightarrow 15 a+6 b=39 \Rightarrow 5 a+2 b=13 \quad (3)$.From $(1)$,$b=-1-a$. Substitute into $(3)$:$5 a+2(-1-a)=13 \Rightarrow 3 a-2=13 \Rightarrow 3 a=15 \Rightarrow a=5$.Then $b=-1-5=-6$.From $(2)$,$8(5)+2(-6)+c=21 \Rightarrow 40-12+c=21 \Rightarrow 28+c=21 \Rightarrow c=-7$.Thus,$a+b+c=5-6-7=-8$.Therefore,$|a+b+c|=|-8|=8$.

Let $f: R \rightarrow R$ be defined as $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1$,$f^{\prime}(\log _e 2)=21$,and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$,then the value of $|a+b+c|$ equals:

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