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Câu hỏi: Cho hàm số $f\left(x \right)=\dfrac{x}{{{\cos }^{2}}x}$ trên $\left(-\dfrac{\pi }{2} ; \dfrac{\pi }{2} \right)$ và $F\left(x \right)$ là một nguyên hàm của $x. F'\left(x \right)$ thỏa mãn $F\left(0 \right)=0$. Biết $a\in \left(-\dfrac{\pi }{2} ; \dfrac{\pi }{2} \right)$ thỏa mãn $\tan a=3$. Tính giá trị biểu thức $T=F\left(a \right)-10{{a}^{2}}+3a$.
A. $-\dfrac{1}{2}\ln 10$.
B. $\dfrac{1}{2}\ln 10$.
C. $-\dfrac{1}{4}\ln 10$.
D. $\ln 10$.
A. $-\dfrac{1}{2}\ln 10$.
B. $\dfrac{1}{2}\ln 10$.
C. $-\dfrac{1}{4}\ln 10$.
D. $\ln 10$.
Theo bài ra $F\left( x \right)$ là một nguyên hàm của $x.f'\left( x \right)$ nên ${{\left[ F\left( x \right) \right]}^{\prime }}=x.{f}'\left( x \right)$.
Khi đó lấy tích phân 2 vế ta được
$\begin{aligned}
& \int\limits_{0}^{a}{{{\left[ F\left( x \right) \right]}^{\prime }}\text{d}x}=\int\limits_{0}^{a}{x.{f}'\left( x \right)\text{d}x}\Leftrightarrow \left. F\left( x \right) \right|_{0}^{a}=\int\limits_{0}^{a}{x\text{d}\left[ f\left( x \right) \right]}\Leftrightarrow F\left( a \right)-F\left( 0 \right)=\left. xf\left( x \right) \right|_{0}^{a}-\int\limits_{0}^{a}{f\left( x \right)\text{d}x} \\
& \Leftrightarrow F\left( a \right)=af\left( a \right)-\int\limits_{0}^{a}{\dfrac{x}{{{\cos }^{2}}x}\text{d}x}=af\left( a \right)-\int\limits_{0}^{a}{x\text{d}\left( \tan x \right)}=a.\dfrac{a}{{{\cos }^{2}}a}-\left. x\tan x \right|_{0}^{a}+\int\limits_{0}^{a}{\tan x\text{d}x} \\
& \Leftrightarrow F\left( a \right)={{a}^{2}}\left( {{\tan }^{2}}a+1 \right)-a\tan a+\int\limits_{0}^{a}{\dfrac{\sin x}{\cos x}\text{d}x}=10{{a}^{2}}-3a-\int\limits_{0}^{a}{\dfrac{\text{d}\left( \cos x \right)}{\cos x}}=10{{a}^{2}}-3a-\left. \ln \left| \cos x \right| \right|_{0}^{a}. \\
\end{aligned}$
Suy ra $F\left( a \right)-10{{a}^{2}}+3a=-\ln \left| \cos a \right|$.
Lại có $a\in \left( -\dfrac{\pi }{2} ; \dfrac{\pi }{2} \right)$ nên $\cos a>0$ và $\cos a=\dfrac{1}{\sqrt{1+{{\tan }^{2}}a}}=\dfrac{1}{\sqrt{1+{{3}^{2}}}}=\dfrac{\sqrt{10}}{10}$.
Do đó $T=F\left( a \right)-10{{a}^{2}}+3a=-\ln \left| \dfrac{\sqrt{10}}{10} \right|=\dfrac{1}{2}\ln 10$.
Khi đó lấy tích phân 2 vế ta được
$\begin{aligned}
& \int\limits_{0}^{a}{{{\left[ F\left( x \right) \right]}^{\prime }}\text{d}x}=\int\limits_{0}^{a}{x.{f}'\left( x \right)\text{d}x}\Leftrightarrow \left. F\left( x \right) \right|_{0}^{a}=\int\limits_{0}^{a}{x\text{d}\left[ f\left( x \right) \right]}\Leftrightarrow F\left( a \right)-F\left( 0 \right)=\left. xf\left( x \right) \right|_{0}^{a}-\int\limits_{0}^{a}{f\left( x \right)\text{d}x} \\
& \Leftrightarrow F\left( a \right)=af\left( a \right)-\int\limits_{0}^{a}{\dfrac{x}{{{\cos }^{2}}x}\text{d}x}=af\left( a \right)-\int\limits_{0}^{a}{x\text{d}\left( \tan x \right)}=a.\dfrac{a}{{{\cos }^{2}}a}-\left. x\tan x \right|_{0}^{a}+\int\limits_{0}^{a}{\tan x\text{d}x} \\
& \Leftrightarrow F\left( a \right)={{a}^{2}}\left( {{\tan }^{2}}a+1 \right)-a\tan a+\int\limits_{0}^{a}{\dfrac{\sin x}{\cos x}\text{d}x}=10{{a}^{2}}-3a-\int\limits_{0}^{a}{\dfrac{\text{d}\left( \cos x \right)}{\cos x}}=10{{a}^{2}}-3a-\left. \ln \left| \cos x \right| \right|_{0}^{a}. \\
\end{aligned}$
Suy ra $F\left( a \right)-10{{a}^{2}}+3a=-\ln \left| \cos a \right|$.
Lại có $a\in \left( -\dfrac{\pi }{2} ; \dfrac{\pi }{2} \right)$ nên $\cos a>0$ và $\cos a=\dfrac{1}{\sqrt{1+{{\tan }^{2}}a}}=\dfrac{1}{\sqrt{1+{{3}^{2}}}}=\dfrac{\sqrt{10}}{10}$.
Do đó $T=F\left( a \right)-10{{a}^{2}}+3a=-\ln \left| \dfrac{\sqrt{10}}{10} \right|=\dfrac{1}{2}\ln 10$.
Đáp án C.