Let [f: R R] be defined as f(x)=a e^2 x+b e^x+c x.

Question

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

Accepted Answer

$$\[\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_\limits0^{\ln 4}\left(a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}=\frac{39}{2} & \end{array}$$\]

$$\begin{aligned}
& {\left[\frac{\mathrm{ae}^{2 \mathrm{x}}}{2}+\mathrm{be}^{\mathrm{x}}\right]_0^{\ln 4}=\frac{39}{2} \Rightarrow 8 \mathrm{a}+4 \mathrm{~b}-\frac{\mathrm{a}}{2}-\mathrm{b}=\frac{39}{2}} \\
& 15 a+6 b=39 \\
& 15 a-6 a-6=39 \\
& 9 \mathrm{a}=45 \Rightarrow \mathrm{a}=5 \\
& b=-6 \\
& c=21-40+12=-7 \\
& \mathrm{a}+\mathrm{b}+\mathrm{c}-8 \\
& |\mathrm{a}+\mathrm{b}+\mathrm{c}|=8 \\
\end{aligned}$$

Solutions · 1