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Câu hỏi: Cho số thực dương $k>0$ thỏa mãn $\int\limits_{0}^{2}{\dfrac{d\text{x}}{\sqrt{{{x}^{2}}+k}}}=\ln \left( 2+\sqrt{5} \right)$. Mệnh đề nào sau đây đúng?
A. $k>\dfrac{3}{2}$
B. $0<k\le \dfrac{1}{2}$
C. $\dfrac{1}{2}<k\le 1$
D. $1<k\le \dfrac{3}{2}$
A. $k>\dfrac{3}{2}$
B. $0<k\le \dfrac{1}{2}$
C. $\dfrac{1}{2}<k\le 1$
D. $1<k\le \dfrac{3}{2}$
Đặt $t=\ln \left( x+\sqrt{{{x}^{2}}+k} \right)\Rightarrow dt=\dfrac{1+\dfrac{x}{\sqrt{{{x}^{2}}+k}}}{x+\sqrt{{{x}^{2}}+k}}d\text{x}\Leftrightarrow dt=\dfrac{1}{\sqrt{{{x}^{2}}+k}}d\text{x}$.
Ta có $\int\limits_{0}^{2}{\dfrac{d\text{x}}{\sqrt{{{x}^{2}}+k}}}=\int\limits_{\ln \left( \sqrt{k} \right)}^{\ln \left( 2+\sqrt{4+k} \right)}{\text{dt}}\Leftrightarrow \ln \left( \dfrac{2+\sqrt{4+k}}{\sqrt{k}} \right)=\ln \left( 2+\sqrt{5} \right)$
${{\left( \dfrac{2+\sqrt{4+k}}{\sqrt{k}} \right)}^{2}}={{\left( 2+\sqrt{5} \right)}^{2}}\Leftrightarrow \left( 1-k \right)\left( 8+4\dfrac{5k+4}{\sqrt{4+k}+\sqrt{5}k} \right)=0\Leftrightarrow k=1$.
Ta có $\int\limits_{0}^{2}{\dfrac{d\text{x}}{\sqrt{{{x}^{2}}+k}}}=\int\limits_{\ln \left( \sqrt{k} \right)}^{\ln \left( 2+\sqrt{4+k} \right)}{\text{dt}}\Leftrightarrow \ln \left( \dfrac{2+\sqrt{4+k}}{\sqrt{k}} \right)=\ln \left( 2+\sqrt{5} \right)$
${{\left( \dfrac{2+\sqrt{4+k}}{\sqrt{k}} \right)}^{2}}={{\left( 2+\sqrt{5} \right)}^{2}}\Leftrightarrow \left( 1-k \right)\left( 8+4\dfrac{5k+4}{\sqrt{4+k}+\sqrt{5}k} \right)=0\Leftrightarrow k=1$.
Đáp án C.