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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{8}$ và $f'(x)=\dfrac{2x+\sin 2x}{2(1+\cos 2x)}$. Khi đó $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{f\left( x \right)}{x}}dx=a\ln 2+b\ln 3+c$ ( $a,b,c$ là các số hữu tỉ). Tính $S=8a+b+c$.
A. $S=16$.
B. $S=8$.
C. $S=-\dfrac{7}{4}$.
D. $S=-\dfrac{7}{8}$.
A. $S=16$.
B. $S=8$.
C. $S=-\dfrac{7}{4}$.
D. $S=-\dfrac{7}{8}$.
Ta có: ${f}'\left( x \right)=\dfrac{2x+\sin 2x}{2\left( 1+\cos 2x \right)}=\dfrac{2x+2\sin x.\cos x}{4{{\cos }^{2}}x}=\dfrac{1}{2}\left( \dfrac{x}{{{\cos }^{2}}x}+\tan x \right)={{\left( \dfrac{1}{2}x.\tan x \right)}^{\prime }}$.
$\Rightarrow f\left( x \right)=\dfrac{1}{2}x.\tan x+C$. Mà $f\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{8}\Rightarrow \dfrac{1}{2}.\dfrac{\pi }{4}.\tan \dfrac{\pi }{4}+C=\dfrac{\pi }{8}\Rightarrow C=0$ $\Rightarrow f\left( x \right)=\dfrac{1}{2}x.\tan x$.
Khi đó: $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{f\left( x \right)}{x}}dx=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\tan xdx=-\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{1}{\cos x}}d\left( \cos x \right)=-\dfrac{1}{2}\ln \left| \cos x \right|\left| _{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} \right.}$.
$=-\dfrac{1}{2}\left( \ln \dfrac{\sqrt{2}}{2}-\ln \dfrac{\sqrt{3}}{2} \right)=-\dfrac{1}{4}\ln \dfrac{2}{3}=-\dfrac{1}{4}\ln 2+\dfrac{1}{4}\ln 3$.
Vậy $a=-\dfrac{1}{4}; b=\dfrac{1}{4}; c=0\Rightarrow S=-2+\dfrac{1}{4}+0=-\dfrac{7}{4}$.
$\Rightarrow f\left( x \right)=\dfrac{1}{2}x.\tan x+C$. Mà $f\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{8}\Rightarrow \dfrac{1}{2}.\dfrac{\pi }{4}.\tan \dfrac{\pi }{4}+C=\dfrac{\pi }{8}\Rightarrow C=0$ $\Rightarrow f\left( x \right)=\dfrac{1}{2}x.\tan x$.
Khi đó: $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{f\left( x \right)}{x}}dx=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\tan xdx=-\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\dfrac{1}{\cos x}}d\left( \cos x \right)=-\dfrac{1}{2}\ln \left| \cos x \right|\left| _{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} \right.}$.
$=-\dfrac{1}{2}\left( \ln \dfrac{\sqrt{2}}{2}-\ln \dfrac{\sqrt{3}}{2} \right)=-\dfrac{1}{4}\ln \dfrac{2}{3}=-\dfrac{1}{4}\ln 2+\dfrac{1}{4}\ln 3$.
Vậy $a=-\dfrac{1}{4}; b=\dfrac{1}{4}; c=0\Rightarrow S=-2+\dfrac{1}{4}+0=-\dfrac{7}{4}$.
Đáp án C.