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Câu hỏi: Có bao nhiêu số thực $a<-1$ để $\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+a}dx}=\dfrac{1}{2\sqrt{-a}}\ln \dfrac{1}{2}$ ?
A. $1$.
B. $2$.
C. $0$.
D. Vô số.
A. $1$.
B. $2$.
C. $0$.
D. Vô số.
Do $a<-1$ nên không mất tổng quát giả sử $a=-b$, ta tìm $b>1$ thỏa mãn yêu cầu.
$\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}-b}dx}=\int\limits_{0}^{1}{\dfrac{1}{\left( x-\sqrt{b} \right)\left( x+\sqrt{b} \right)}dx=\dfrac{1}{2\sqrt{b}}\int\limits_{0}^{1}{\left( \dfrac{1}{x-\sqrt{b}}-\dfrac{1}{x+\sqrt{b}} \right)}dx}$
$=\dfrac{1}{2\sqrt{a}}\operatorname{l}\left. n\left| \dfrac{x-\sqrt{a}}{x+\sqrt{a}} \right| \right|_{0}^{1}=\dfrac{1}{2\sqrt{a}}\ln \left| \dfrac{1-\sqrt{b}}{1+\sqrt{b}} \right|$
Theo giả thiết ta có:
$\left| \dfrac{1-\sqrt{b}}{1+\sqrt{b}} \right|=\dfrac{1}{2}\Leftrightarrow 2\left| 1-\sqrt{b} \right|=\left| 1+\sqrt{b} \right|\Leftrightarrow 3b-10\sqrt{b}+3=0\Leftrightarrow \left[ \begin{aligned}
& \sqrt{b}=\dfrac{1}{3} \\
& \sqrt{b}=3 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& b=\dfrac{1}{9} \\
& b=9 \\
\end{aligned} \right.$
Do $b>1$ nên $b=9\Rightarrow a=-9$.
$\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}-b}dx}=\int\limits_{0}^{1}{\dfrac{1}{\left( x-\sqrt{b} \right)\left( x+\sqrt{b} \right)}dx=\dfrac{1}{2\sqrt{b}}\int\limits_{0}^{1}{\left( \dfrac{1}{x-\sqrt{b}}-\dfrac{1}{x+\sqrt{b}} \right)}dx}$
$=\dfrac{1}{2\sqrt{a}}\operatorname{l}\left. n\left| \dfrac{x-\sqrt{a}}{x+\sqrt{a}} \right| \right|_{0}^{1}=\dfrac{1}{2\sqrt{a}}\ln \left| \dfrac{1-\sqrt{b}}{1+\sqrt{b}} \right|$
Theo giả thiết ta có:
$\left| \dfrac{1-\sqrt{b}}{1+\sqrt{b}} \right|=\dfrac{1}{2}\Leftrightarrow 2\left| 1-\sqrt{b} \right|=\left| 1+\sqrt{b} \right|\Leftrightarrow 3b-10\sqrt{b}+3=0\Leftrightarrow \left[ \begin{aligned}
& \sqrt{b}=\dfrac{1}{3} \\
& \sqrt{b}=3 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& b=\dfrac{1}{9} \\
& b=9 \\
\end{aligned} \right.$
Do $b>1$ nên $b=9\Rightarrow a=-9$.
Đáp án A.