Cho f(x)=cos^22x+2(sinx+cosx)^3−3sin2x+m

Hướng dẫn giải:

Đặt  \( t=\sin x+\cos x=\sqrt{2}\cos \left( x-\frac{\pi }{4} \right) \) (điều kiện  \( \left| t \right|\le \sqrt{2} \))

Thì  \( {{t}^{2}}=1+\sin 2x \) và  \( {{\cos }^{2}}2x=1-si{{n}^{2}}2x=1-{{({{t}^{2}}-1)}^{2}}=-{{t}^{4}}+2{{t}^{2}} \).

Vậy f(x) thành  \( g(t)=-{{t}^{4}}+2{{t}^{2}}+2{{t}^{3}}-3({{t}^{2}}-1)+m \).

a) Khi \( m=-3 \) thì \( g(t)=0 \)

 \( \Leftrightarrow -{{t}^{2}}({{t}^{2}}-2t+1)=0\Leftrightarrow \left[ \begin{align}  & t=0 \\  & t=1 \\ \end{align} \right. \) \( \Rightarrow \left[ \begin{align}  & \sqrt{2}\cos \left( x-\frac{\pi }{4} \right)=0 \\  & \sqrt{2}\cos \left( x-\frac{\pi }{4} \right)=1 \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \cos \left( x-\frac{\pi }{4} \right)=0 \\  & \cos \left( x-\frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}=\cos \frac{\pi }{4} \\ \end{align} \right. \)\(\Leftrightarrow \left[ \begin{align}  & x-\frac{\pi }{4}=\frac{\pi }{2}+k\pi  \\ & x-\frac{\pi }{4}=\frac{\pi }{4}+k2\pi \vee x-\frac{\pi }{4}=-\frac{\pi }{4}+k2\pi  \\ \end{align} \right.\)

\(\Leftrightarrow \left[ \begin{align}  & x=\frac{3\pi }{4}+k\pi  \\  & x=\frac{\pi }{2}+k2\pi \vee x=k2\pi  \\ \end{align} \right.,\text{ }k\in \mathbb{Z}\).

b) Ta có: \( {g}'(t)=-4{{t}^{3}}+6{{t}^{2}}-2t=-2t(2{{t}^{2}}-3t+1) \)

Do đó:  \( \left\{ \begin{align}  & {g}'(t)=0 \\  & t\in \left[ -\sqrt{2};\sqrt{2} \right] \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & t=0 \\  & t=1 \\  & t=\frac{1}{2} \\ \end{align} \right. \).

Ta có:  \( g(0)=3+m=g(1),\text{ }g\left( \frac{1}{2} \right)=\frac{47}{16}+m \)

 \( g(-\sqrt{2})=4\sqrt{2}-3+m,\text{ }g(\sqrt{2})=m-3-4\sqrt{2} \).

Vậy:  \( \underset{x\in \mathbb{R}}{\mathop{max}}\,f(x)=\underset{t\in \left[ -\sqrt{2};\sqrt{2} \right]}{\mathop{max}}\,g(t)=m+3 \).

 \( \underset{x\in \mathbb{R}}{\mathop{\min }}\,f(x)=\underset{t\in \left[ -\sqrt{2};\sqrt{2} \right]}{\mathop{\min }}\,g(t)=m-3-4\sqrt{2} \).

Do đó:  \( {{[f(x)]}^{2}}\le 36,\text{ }\forall x\in \mathbb{R}\Leftrightarrow -6\le f(x)\le 6,\forall x\in \mathbb{R} \)

 \( \Leftrightarrow \left\{ \begin{align}  & \underset{\mathbb{R}}{\mathop{max}}\,f(x)\le 6 \\  & \underset{\mathbb{R}}{\mathop{\min }}\,f(x)\ge -6 \\ \end{align} \right. \) \( \Leftrightarrow \left\{ \begin{align}  & m+3\le 6 \\  & m-3-4\sqrt{2}\ge -6 \\ \end{align} \right.\Leftrightarrow 4\sqrt{2}-3\le m\le 3 \).