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\(y = {\sin ^6}x + c{\rm{o}}{{\rm{s}}^6}x + {\rm{a}}\sin x\cos x =- \frac{3}{4}{\sin ^2}2x + \frac{a}{2}\sin 2x + 1\)
Đặt \(t = \sin 2x\left( { - 1 \le t \le 1} \right)\), khi đó:
\(y = f\left( t \right) = - \frac{3}{4}{t^2} + \frac{a}{2}t + 1\,\,\,\left( {t \in \left[ { - 1,\,1} \right]} \right)\)
\(\Rightarrow f'\left( t \right) = - \frac{3}{2}t + \frac{a}{2}\,\,\\\,f'\left( t \right) = 0 \Leftrightarrow t = \frac{a}{3}\)
Nếu \(\frac{a}{3} \le -1 \left( {a \le- 3} \right)\) thì $f(t)$ nghịch biến trong \(\left[ { - 1,\,1} \right]\)
\( \Rightarrow \max f = f\left( { - 1} \right) = \frac{{1 - 2a}}{4};\,\,\min \,f = f\left( 1 \right) = \frac{{1 + 2a}}{4}\)
Nếu \(\frac{a}{3} \ge 1 \left( {a \ge 3} \right)\) thì $f(t)$ đồng biến trong \(\left[ { - 1,\,1} \right]\)
\( \Rightarrow \max f = f\left( 1 \right) = \frac{{1 + 2a}}{4};\,\,\min \,f = f\left( { - 1} \right) = \frac{{1 - 2a}}{4}\)
Nếu \( - 1 < \frac{a}{3} < 1 \left( { - 3 < a < 3} \right)\)
Ta có bảng biến thiên:
Nên \(m{\rm{ax}}\,f = 1 + \frac{{{a^2}}}{{12}};\,\,\min \,f = \frac{{1 - 2|a|}}{4}\)
Kết luận \(\min y = \dfrac{{1 - 2|a|}}{4};\,\,\max y = \left\{ \begin{array}{l}
\dfrac{{1 + 2|a|}}{4}\,\,\,khi\,|a| \ge 3\\
1 + \dfrac{{{a^2}}}{{12}}\,\,\,khi\,|a| < 3
\end{array} \right.\)
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Đăng bài 02-05-12 08:40 AM
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