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Câu hỏi: Cho ${f}'(x)=\sin 2x-5\sin x{{\cos }^{4}}x,\forall x\in \mathbb{R}$, $f\left( \dfrac{\pi }{2} \right)=0$ và $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)\text{d}x=a+b\pi }$ với $a,b\in \mathbb{Q}.$ Đặt $T=\dfrac{1}{a}+b.$ Mệnh đề nào sau đây đúng?
A. $T\in \left( 1;2 \right).$
B. $T\in \left( 0;1 \right).$
C. $T\in \left( 2;3 \right).$
D. $T\in \left( -2;0 \right).$
A. $T\in \left( 1;2 \right).$
B. $T\in \left( 0;1 \right).$
C. $T\in \left( 2;3 \right).$
D. $T\in \left( -2;0 \right).$
$f(x)=\int{\left( \sin 2x-5\sin x{{\cos }^{4}}x \right)}dx=-\dfrac{1}{2}\cos 2x+{{\cos }^{5}}x+C$
$f\left( \dfrac{\pi }{2} \right)=0\Rightarrow C=-\dfrac{1}{2}$. Do đó $f(x)=-\dfrac{1}{2}\cos 2x+{{\cos }^{5}}x-\dfrac{1}{2}={{\cos }^{5}}x-{{\cos }^{2}}x.$
Suy ra $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{5}}xdx-}\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}={{I}_{1}}-{{I}_{2}}.$
${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{5}}xdx=}\int\limits_{0}^{\dfrac{\pi }{2}}{{{(1-{{\sin }^{2}}x)}^{2}}d(\sin x)=\left. (\sin x-\dfrac{2}{3}{{\sin }^{3}}x+\dfrac{1}{5}{{\sin }^{5}}x) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{8}{15}.}$
${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx=}\dfrac{\pi }{4}.$ Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx=\dfrac{8}{15}-\dfrac{\pi }{4}}\Rightarrow a=\dfrac{8}{15},b=-\dfrac{1}{4}\Rightarrow T=\dfrac{15}{8}-\dfrac{1}{4}=\dfrac{13}{8}\in (1;2).$
$f\left( \dfrac{\pi }{2} \right)=0\Rightarrow C=-\dfrac{1}{2}$. Do đó $f(x)=-\dfrac{1}{2}\cos 2x+{{\cos }^{5}}x-\dfrac{1}{2}={{\cos }^{5}}x-{{\cos }^{2}}x.$
Suy ra $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{5}}xdx-}\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}={{I}_{1}}-{{I}_{2}}.$
${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{5}}xdx=}\int\limits_{0}^{\dfrac{\pi }{2}}{{{(1-{{\sin }^{2}}x)}^{2}}d(\sin x)=\left. (\sin x-\dfrac{2}{3}{{\sin }^{3}}x+\dfrac{1}{5}{{\sin }^{5}}x) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{8}{15}.}$
${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx=}\dfrac{\pi }{4}.$ Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx=\dfrac{8}{15}-\dfrac{\pi }{4}}\Rightarrow a=\dfrac{8}{15},b=-\dfrac{1}{4}\Rightarrow T=\dfrac{15}{8}-\dfrac{1}{4}=\dfrac{13}{8}\in (1;2).$
Đáp án A.