Cho phương trình: $\displaystyle \frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x}=2m.\tan 2x$, trong đó m là tham số. Để phương trình có nghiệm, các giá trị thích hợp của m làA. $\displaystyle

ngoisaoxanhnhoemnhieu

2 năm trước

Cho phương trình: $\displaystyle \frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x}=2m.\tan 2x$, trong đó m là tham số. Để phương trình có nghiệm, các giá trị thích hợp của m là
A. $\displaystyle m<-\frac{1}{8}\ hay\ m>\frac{1}{8}$ .
B. $\displaystyle m\le -\frac{1}{4}\ hay\ m\ge \frac{1}{4}$.
C. $\displaystyle m\le -\frac{1}{2}\ hay\ m\ge \frac{1}{2}$.
D. $\displaystyle m\le -1\ hay\ m\ge 1$.

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hoanganh2002vt

Đáp án đúng: A
ĐK: $\displaystyle \cos 2x
e 0$
$\displaystyle \frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x}=2m.\tan 2x\Leftrightarrow \frac{{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}-3{{\sin }^{2}}x{{\cos }^{2}}x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}{\cos 2x}=2m\tan 2x$
$\displaystyle \Leftrightarrow \frac{1-\frac{3}{4}{{\sin }^{2}}2x}{\cos 2x}=2m\tan 2x\Leftrightarrow 1-\frac{3}{4}{{\sin }^{2}}2x=2m\sin 2x\Leftrightarrow 3{{\sin }^{2}}2x+8m\sin 2x-4=0.$
Đặt$\displaystyle \sin 2x=t\,\left( t\in \left( -1;1 \right) \right)$ .Khi đó phương trình trở thành:$\displaystyle 3{{t}^{2}}+8mt-4=0\,\left( * \right)$
Phương trình đã cho có nghiệm khi phương trình$\displaystyle \left( * \right)$có nghiệm$\displaystyle t\in \left( -1;1 \right)$
TH1:$\displaystyle \left( * \right)$có 1 nghiệm:
$\displaystyle \begin{array}{l}t\in \left( -1;1 \right)\Leftrightarrow f\left( 1 \right)f\left( -1 \right)<0\Leftrightarrow \left( 8m-1 \right)\left( -8m-1 \right)<0\Leftrightarrow \left[ \begin{array}{l}m>\frac{1}{8}\\m<-\frac{1}{8}\end{array} \right.\\.\end{array}$
TH2:$\displaystyle \left( * \right)$có 2 nghiệm
$\displaystyle t\in \left( -1;1 \right)\Leftrightarrow \left\{ \begin{array}{l}{\Delta }'=16{{m}^{2}}+12>0\\f\left( 1 \right)=8m-1>0\\f\left( -1 \right)=-8m-1>0\\-1<\frac{s}{2}=-\frac{4m}{3}<1\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}m>\frac{1}{8}\\m<-\frac{1}{8}\\-\frac{3}{4}<m<\frac{3}{4}\end{array} \right.\left( VN \right).$
Chọn A

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