Cho góc \(\alpha \) thỏa mãn \(\sin \alpha = \frac{{12}}{{13}}\) và \(\frac{\pi }{2} < \alpha < \pi \). Tính \(\cos \alpha \).

Chọn B

Ta có: \(N = 5 + 9 + 12 + 10 + 6 = 42\)

Tứ phân vị thứ nhất nằm trong nhóm thứ \(2\) là \(\left[ {20;40} \right)\)

\({Q_1} = {a_p} + \frac{{\frac{n}{4} - \left( {{m_1} + ... + {m_{p - 1}}} \right)}}{{{m_p}}}\left( {{a_{p + 1}} - {a_p}} \right)\)

\({Q_1} = 20 + \frac{{\frac{{42}}{4} - 5}}{9}\left( {40 - 20} \right) = \frac{{290}}{9}\)

Tứ phân vị thứ ba nằm trong nhóm thứ \(4\) là \(\left[ {60;80} \right)\)

\({Q_3} = {a_p} + \frac{{\frac{{3n}}{4} - \left( {{m_1} + ... + {m_{p - 1}}} \right)}}{{{m_p}}}\left( {{a_{p + 1}} - {a_p}} \right)\)

\({Q_3} = 60 + \frac{{\frac{{3.42}}{4} - 26}}{{10}}\left( {80 - 60} \right) = 71\)

Vậy: \(9{Q_1} - {Q_3} = 9.\frac{{290}}{9} - 71 = 219\)

Chọn A

\(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt {4f\left( x \right) + 9} + 3} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\sqrt x - 1}}.\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {4f\left( x \right) + 9} + 3}}\)

Xét \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\sqrt x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left[ {f\left( x \right) - 10} \right]\left( {\sqrt x + 1} \right)}}{{x - 1}} = 5\left( {1 + 1} \right) = 10\)

Xét \[\mathop {\lim }\limits_{x \to 1} \sqrt {4f\left( x \right) + 9} = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {4\frac{{\left( {f\left( x \right) - 10 + 10} \right)}}{{x - 1}}\left( {x - 1} \right) + 9} } \right)\]

\[ = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {4\left[ {\frac{{f\left( x \right) - 10}}{{x - 1}} + \frac{{10}}{{x - 1}}} \right]\left( {x - 1} \right) + 9} } \right)\]

\[ = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {4\left( {5 + \frac{{10}}{{x - 1}}} \right)\left( {x - 1} \right) + 9} } \right)\]

\[ = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {20\left( {x - 1} \right) + 40 + 9} } \right) = \sqrt {49} = 7\]

Suy ra \[\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {4f\left( x \right) + 9} + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{7 + 3}} = \frac{1}{{10}}\]

\(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt {4f\left( x \right) + 9} + 3} \right)}} = 10.\frac{1}{{10}} = 1\)