Let \mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} be defined as f(x)=a e^{2 x}+b e^x+c x. If f(0)=-1, f^{\prime}\left(\log _e 2\right)=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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Answer: C. 8 \begin{array}{ll} \mathrm{f}(\mathrm{x})=a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{cx} & \mathrm{f}(0)=-1 \\ & \mathrm{a}+\mathrm{b}=-1 \\ \mathrm{f}^{\prime}(\mathrm{x})=2 a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{c} & \mathrm{f}^{\prime}(\ln 2)=21 \\ & 8 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=21 \\ \int\limits_0^{\ln 4}\left(a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}ight) \mathrm{dx}=\frac{39}{2} & \end{array} \begin{aligned} & {\left[\frac{\mathrm{ae}^{2 \mathrm{x}}}{2}+\mathrm{be}^{\mathrm{x}}ight]_0^{\ln 4}=\frac{39}{2} \Rightarrow 8 \mathrm{a}+4 \mathrm{~b}-\frac{\mathrm{a}}{2}-\mathrm{b}=\frac{39}{2}} \\ & 15 a+6 b=39 \\ & 15 a-6 a-6=39 \\ & 9 \mathrm{a}=45 \Rightarrow \mathrm{a}=5 \\ & b=-6 \\ & c=21-40+12=-7 \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}-8 \\ & |\mathrm{a}+\mathrm{b}+\mathrm{c}|=8 \\ \end{aligned}

Let $$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$$ be defined as $$f(x)=a e^{2 x}+b e^x+c x$$. If $$f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$$ 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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