Cho các số thực a, b khác 0. Xét hàm số f(x) = a/(x+1)^3 + bxe^x với

* Hướng dẫn giải

Đáp án A

Ta có:

$$\begin{gathered} \underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}\frac{a}{{\left(x+1\right)}^3}dx+\underset{0}{\overset{1}{\int }}bx{e}^{x}dx \\ ={\left.-\frac{a}{2}\frac{1}{{\left(x+1\right)}^2}\right|}_0^1+\underset{0}{\overset{1}{\int }}bx{e}^{x}dx=\frac{3\text{a}}{8}+\underset{0}{\overset{1}{\int }}bx{e}^{x}dx \end{gathered}$$

Đặt:

$$\begin{gathered} \left\{\begin{array}{l}u=x\\ dv=e^{x}d\text{x}\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=dx\\ v=e^x\end{array}\right. \\ \Rightarrow \underset{0}{\overset{1}{\int }}bx{e}^{x}dx={\left.bx{e}^x\right|}_0^1-\underset{0}{\overset{1}{\int }}b{e}^{x}dx={\left.bx{e}^x\right|}_0^1-{\left.b{e}^x\right|}_0^1=b \end{gathered}$$

Suy ra $\underset{0}{\overset{1}{\int }}f\left(x\right)d\text{x}=\frac{3\text{a}}{8}+b=5\left(1\right)$

Mặt khác $f\text{'}\left(x\right)=-\frac{3\text{a}}{{\left(x+1\right)}^4}+b{e}^x+b\text{x}{e}^x\Rightarrow f\text{'}\left(0\right)=-3\text{a}+b=-22\left(2\right)$

Từ $\left(1\right),\left(2\right)$ suy ra $a=8;b=2\Rightarrow a+b=10$