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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& {{x}^{2}}+x+1\ \ \text{khi}\ x\ge 0 \\
& 2x+1\ \ \ \ \ \ \ \text{khi}\ x\le 0 \\
\end{aligned} \right. $ $ $
Biết $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x+\int\limits_{e}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}}=\dfrac{a}{b}$ với $\dfrac{a}{b}$ là phân số tối giản. Giá trị của a.b bằng
A. $60$.
B. $92$.
C. $174$.
D. $132$.
& {{x}^{2}}+x+1\ \ \text{khi}\ x\ge 0 \\
& 2x+1\ \ \ \ \ \ \ \text{khi}\ x\le 0 \\
\end{aligned} \right. $ $ $
Biết $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x+\int\limits_{e}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}}=\dfrac{a}{b}$ với $\dfrac{a}{b}$ là phân số tối giản. Giá trị của a.b bằng
A. $60$.
B. $92$.
C. $174$.
D. $132$.
+ Đặt $t=2\sin x-1\Rightarrow \text{d}t=2\cos x\text{d}x\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( t \right)\text{d}t}=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)\text{d}x}$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{0}{f\left( x \right)\text{d}x}+\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{1}{2}\int\limits_{-1}^{0}{\left( 2x+1 \right)\text{d}x}+\dfrac{1}{2}\int\limits_{0}^{1}{\left( {{x}^{2}}+x+1 \right)\text{d}x}=\dfrac{11}{12}$
+ Đặt $u=\ln x\Rightarrow \text{d}u=\dfrac{\text{d}x}{x}\Rightarrow \int\limits_{e}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}}dx=\int\limits_{1}^{2}{f\left( u \right)\text{d}x}=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\int\limits_{1}^{2}{\left( {{x}^{2}}+x+1 \right)\text{d}x}=\dfrac{29}{6}$ $$
$\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x+\int\limits_{e}^{{{e}^{2}}}{\frac{f\left( \ln x \right)}{x}}=\frac{23}{4}$
$\Rightarrow a=23,b=4\Rightarrow a.b=92$.
$$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{0}{f\left( x \right)\text{d}x}+\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{1}{2}\int\limits_{-1}^{0}{\left( 2x+1 \right)\text{d}x}+\dfrac{1}{2}\int\limits_{0}^{1}{\left( {{x}^{2}}+x+1 \right)\text{d}x}=\dfrac{11}{12}$
+ Đặt $u=\ln x\Rightarrow \text{d}u=\dfrac{\text{d}x}{x}\Rightarrow \int\limits_{e}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}}dx=\int\limits_{1}^{2}{f\left( u \right)\text{d}x}=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\int\limits_{1}^{2}{\left( {{x}^{2}}+x+1 \right)\text{d}x}=\dfrac{29}{6}$ $$
$\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{f\left( 2\sin x-1 \right)}.\cos x\text{d}x+\int\limits_{e}^{{{e}^{2}}}{\frac{f\left( \ln x \right)}{x}}=\frac{23}{4}$
$\Rightarrow a=23,b=4\Rightarrow a.b=92$.
$$
Đáp án B.