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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& 4x-\sqrt{4x+9} \text{khi} x>0 \\
& 4a+{{\tan }^{2}} x \text{khi} x\le 0 \\
\end{aligned} \right. $, đồng thời $ I=\int\limits_{-\dfrac{\pi }{4}}^{4}{f\left( x \right)dx}=\dfrac{50}{3} $. Tính $ a$.
A. $a=1.$
B. $a=\dfrac{1}{2}.$
C. $a=\dfrac{3}{4}.$
D. $a=\dfrac{1}{4}.$
& 4x-\sqrt{4x+9} \text{khi} x>0 \\
& 4a+{{\tan }^{2}} x \text{khi} x\le 0 \\
\end{aligned} \right. $, đồng thời $ I=\int\limits_{-\dfrac{\pi }{4}}^{4}{f\left( x \right)dx}=\dfrac{50}{3} $. Tính $ a$.
A. $a=1.$
B. $a=\dfrac{1}{2}.$
C. $a=\dfrac{3}{4}.$
D. $a=\dfrac{1}{4}.$
Ta có
$I=\int\limits_{-\dfrac{\pi }{4}}^{4}{f\left( x \right)dx}=\int\limits_{-\dfrac{\pi }{4}}^{0}{f\left( x \right)dx}+\int\limits_{0}^{4}{f\left( x \right)dx}=\int\limits_{-\dfrac{\pi }{4}}^{0}{\left( 4a+{{\tan }^{2}} x \right)dx}+\int\limits_{0}^{4}{\left( 4x-\sqrt{4x+9} \right)dx}$
$=\int\limits_{-\dfrac{\pi }{4}}^{0}{\left( 4a-1 \right)dx}+\int\limits_{-\dfrac{\pi }{4}}^{0}{\left( 1+{{\tan }^{2}} x \right)dx}+\int\limits_{0}^{4}{\left( 4x-\sqrt{4x+9} \right)dx}$
$=\left. \left( 4a-1 \right) \right|_{-\dfrac{\pi }{4}}^{0}+\left. \tan x \right|_{-\dfrac{\pi }{4}}^{0}+\left. \left( 2{{x}^{2}}-\dfrac{\sqrt{{{\left( 4x+9 \right)}^{3}}}}{6} \right) \right|_{0}^{4}$ $=\left( 4a-1 \right)\pi +1+\dfrac{47}{3}=\dfrac{50}{3}\Leftrightarrow a=\dfrac{1}{4}$.
$I=\int\limits_{-\dfrac{\pi }{4}}^{4}{f\left( x \right)dx}=\int\limits_{-\dfrac{\pi }{4}}^{0}{f\left( x \right)dx}+\int\limits_{0}^{4}{f\left( x \right)dx}=\int\limits_{-\dfrac{\pi }{4}}^{0}{\left( 4a+{{\tan }^{2}} x \right)dx}+\int\limits_{0}^{4}{\left( 4x-\sqrt{4x+9} \right)dx}$
$=\int\limits_{-\dfrac{\pi }{4}}^{0}{\left( 4a-1 \right)dx}+\int\limits_{-\dfrac{\pi }{4}}^{0}{\left( 1+{{\tan }^{2}} x \right)dx}+\int\limits_{0}^{4}{\left( 4x-\sqrt{4x+9} \right)dx}$
$=\left. \left( 4a-1 \right) \right|_{-\dfrac{\pi }{4}}^{0}+\left. \tan x \right|_{-\dfrac{\pi }{4}}^{0}+\left. \left( 2{{x}^{2}}-\dfrac{\sqrt{{{\left( 4x+9 \right)}^{3}}}}{6} \right) \right|_{0}^{4}$ $=\left( 4a-1 \right)\pi +1+\dfrac{47}{3}=\dfrac{50}{3}\Leftrightarrow a=\dfrac{1}{4}$.
Đáp án D.