Để phương trình $\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{\tan \left( {x + \frac{\pi }{4}} \right)\tan \left( {x - \frac{\pi }{4}} \right)}} = m$ có nghiệm, tham số m phải thỏa mãn điều kiện:
A. $ - 1 \le m < - \frac{1}{4}.$
B. $ - 2 \le m \le - 1.$
C. $1 \le m \le 2.$
D. $\frac{1}{4} \le m \le 1.$
A. $ - 1 \le m < - \frac{1}{4}.$
B. $ - 2 \le m \le - 1.$
C. $1 \le m \le 2.$
D. $\frac{1}{4} \le m \le 1.$
Chọn A.
ĐK: $\left\{ \begin{array}{l}\sin \left( {x + \frac{\pi }{4}} \right) \ne 0\\\cos \left( {x + \frac{\pi }{4}} \right) \ne 0\\\sin \left( {x - \frac{\pi }{4}} \right) \ne 0\\\cos \left( {x - \frac{\pi }{4}} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + \frac{\pi }{4} \ne \frac{{k\pi }}{2}\\x - \frac{\pi }{4} \ne \frac{{k\pi }}{2}\end{array} \right. \Leftrightarrow x \ne \pm \frac{\pi }{4} + \frac{{k\pi }}{2}$
$\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{\tan \left( {x + \frac{\pi }{4}} \right)\tan \left( {x - \frac{\pi }{4}} \right)}} = m \Leftrightarrow \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)}}{{\frac{{\tan x + 1}}{{1 - \tan x}}.\frac{{\tan x - 1}}{{1 + \tan x}}}} = m$
$ \Leftrightarrow \frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 3{{\sin }^2}x{{\cos }^2}x}}{{ - 1}} = m \Leftrightarrow 1 - \frac{3}{4}{\sin ^2}2x = - m \Leftrightarrow {\sin ^2}2x = \frac{{4m + 4}}{3}$
Phương trình có nghiệm $ \Leftrightarrow \left\{ \begin{array}{l}x \ne \pm \frac{\pi }{4} + \frac{{k\pi }}{2}\\{\sin ^2}2x = \frac{{4m + 4}}{3}\;{\rm{co\`u nghie\"a m}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\sin ^2}\left( {2\left( { \pm \frac{\pi }{4} + \frac{{k\pi }}{2}} \right)} \right) \ne \frac{{4m + 4}}{3}\\0 \le \frac{{4m + 4}}{3} \le 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 \ne \frac{{4m + 4}}{3}\\0 \le 4m + 4 \le 3\end{array} \right.$
$ \Leftrightarrow \left\{ \begin{array}{l}3 \ne 4m + 4\\ - 4 \le 4m \le - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne - \frac{1}{4}\\ - 1 \le m \le - \frac{1}{4}\end{array} \right. \Leftrightarrow - 1 \le m < - \frac{1}{4}$
ĐK: $\left\{ \begin{array}{l}\sin \left( {x + \frac{\pi }{4}} \right) \ne 0\\\cos \left( {x + \frac{\pi }{4}} \right) \ne 0\\\sin \left( {x - \frac{\pi }{4}} \right) \ne 0\\\cos \left( {x - \frac{\pi }{4}} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + \frac{\pi }{4} \ne \frac{{k\pi }}{2}\\x - \frac{\pi }{4} \ne \frac{{k\pi }}{2}\end{array} \right. \Leftrightarrow x \ne \pm \frac{\pi }{4} + \frac{{k\pi }}{2}$
$\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{\tan \left( {x + \frac{\pi }{4}} \right)\tan \left( {x - \frac{\pi }{4}} \right)}} = m \Leftrightarrow \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)}}{{\frac{{\tan x + 1}}{{1 - \tan x}}.\frac{{\tan x - 1}}{{1 + \tan x}}}} = m$
$ \Leftrightarrow \frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 3{{\sin }^2}x{{\cos }^2}x}}{{ - 1}} = m \Leftrightarrow 1 - \frac{3}{4}{\sin ^2}2x = - m \Leftrightarrow {\sin ^2}2x = \frac{{4m + 4}}{3}$
Phương trình có nghiệm $ \Leftrightarrow \left\{ \begin{array}{l}x \ne \pm \frac{\pi }{4} + \frac{{k\pi }}{2}\\{\sin ^2}2x = \frac{{4m + 4}}{3}\;{\rm{co\`u nghie\"a m}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\sin ^2}\left( {2\left( { \pm \frac{\pi }{4} + \frac{{k\pi }}{2}} \right)} \right) \ne \frac{{4m + 4}}{3}\\0 \le \frac{{4m + 4}}{3} \le 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 \ne \frac{{4m + 4}}{3}\\0 \le 4m + 4 \le 3\end{array} \right.$
$ \Leftrightarrow \left\{ \begin{array}{l}3 \ne 4m + 4\\ - 4 \le 4m \le - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne - \frac{1}{4}\\ - 1 \le m \le - \frac{1}{4}\end{array} \right. \Leftrightarrow - 1 \le m < - \frac{1}{4}$