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Câu hỏi: Cho hàm số $f\left( x \right)={{\left( \sin x-m \right)}^{2}}+{{\left( \cos x-n \right)}^{2}}(m,n$ là các tham số nguyên). Có tất cả bao nhiêu bộ $\left( m;n \right)$ sao cho $\underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right)+\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right)=52?$
A. 4
B. 0
C. 8
D. 12
A. 4
B. 0
C. 8
D. 12
Phương pháp:
- Khai triển hằng đẳng thức.
- Sử dụng: $-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin x+b\cos x\le \sqrt{{{a}^{2}}+{{b}^{2}}},$ từ đó tìm $\underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right),\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right).$
- Giải phương trình tìm nghiệm nguyên $m,n.$
Cách giải:
Ta có:
$f\left( x \right)={{\left( \sin x-m \right)}^{2}}+{{\left( \cos x-n \right)}^{2}}$
$f\left( x \right)={{\sin }^{2}}x-2m\sin x+{{m}^{2}}+{{\cos }^{2}}x-2n\cos x+{{n}^{2}}$
$f\left( x \right)=1-2\left( m\sin x+n\cos x \right)+{{m}^{2}}+{{n}^{2}}$
Ta có: $-\sqrt{{{m}^{2}}+{{n}^{2}}}\le m\sin x+n\cos x\le \sqrt{{{m}^{2}}+{{n}^{2}}}$
$\Rightarrow 2\sqrt{{{m}^{2}}+{{n}^{2}}}\ge -2\left( m\sin x+n\cos x \right)\ge -2\sqrt{{{m}^{2}}+{{n}^{2}}}$
$\Rightarrow 1+2\sqrt{{{m}^{2}}+{{n}^{2}}}\ge 1-2\left( m\sin x+n\cos x \right)\ge 1-2\sqrt{{{m}^{2}}+{{n}^{2}}}$
$\Rightarrow 1+2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}\ge f\left( x \right)\ge 1-2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}$
$\Rightarrow \underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right)=1-2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}$
$\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right)=1+2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}$
Theo bài ra ta có:
$\underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right)+\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right)=52$
$\Leftrightarrow 1-2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}+1+2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}=52$
$\Leftrightarrow 2+2{{m}^{2}}+2{{n}^{2}}=52$
$\Leftrightarrow {{m}^{2}}+{{n}^{2}}=25$
Vì $m,n\in \mathbb{Z}\Rightarrow \left( m;n \right)\in \left\{ \begin{aligned}
& \left( 0;5 \right);\left( 0;-5 \right);\left( 5;0 \right);\left( -5;0 \right); \\
& \left( 3;4 \right);\left( 3;-4 \right);\left( -3;4 \right);\left( -3;-4 \right) \\
& \left( 4;3 \right);\left( 4;-3 \right);\left( -4;3 \right);\left( -4;-3 \right) \\
\end{aligned} \right\}$.
Vậy có 12 bộ số $\left( m;n \right)$ thỏa mãn.
- Khai triển hằng đẳng thức.
- Sử dụng: $-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin x+b\cos x\le \sqrt{{{a}^{2}}+{{b}^{2}}},$ từ đó tìm $\underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right),\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right).$
- Giải phương trình tìm nghiệm nguyên $m,n.$
Cách giải:
Ta có:
$f\left( x \right)={{\left( \sin x-m \right)}^{2}}+{{\left( \cos x-n \right)}^{2}}$
$f\left( x \right)={{\sin }^{2}}x-2m\sin x+{{m}^{2}}+{{\cos }^{2}}x-2n\cos x+{{n}^{2}}$
$f\left( x \right)=1-2\left( m\sin x+n\cos x \right)+{{m}^{2}}+{{n}^{2}}$
Ta có: $-\sqrt{{{m}^{2}}+{{n}^{2}}}\le m\sin x+n\cos x\le \sqrt{{{m}^{2}}+{{n}^{2}}}$
$\Rightarrow 2\sqrt{{{m}^{2}}+{{n}^{2}}}\ge -2\left( m\sin x+n\cos x \right)\ge -2\sqrt{{{m}^{2}}+{{n}^{2}}}$
$\Rightarrow 1+2\sqrt{{{m}^{2}}+{{n}^{2}}}\ge 1-2\left( m\sin x+n\cos x \right)\ge 1-2\sqrt{{{m}^{2}}+{{n}^{2}}}$
$\Rightarrow 1+2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}\ge f\left( x \right)\ge 1-2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}$
$\Rightarrow \underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right)=1-2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}$
$\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right)=1+2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}$
Theo bài ra ta có:
$\underset{x\in \mathbb{R}}{\mathop{\min }} f\left( x \right)+\underset{x\in \mathbb{R}}{\mathop{\max }} f\left( x \right)=52$
$\Leftrightarrow 1-2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}+1+2\sqrt{{{m}^{2}}+{{n}^{2}}}+{{m}^{2}}+{{n}^{2}}=52$
$\Leftrightarrow 2+2{{m}^{2}}+2{{n}^{2}}=52$
$\Leftrightarrow {{m}^{2}}+{{n}^{2}}=25$
Vì $m,n\in \mathbb{Z}\Rightarrow \left( m;n \right)\in \left\{ \begin{aligned}
& \left( 0;5 \right);\left( 0;-5 \right);\left( 5;0 \right);\left( -5;0 \right); \\
& \left( 3;4 \right);\left( 3;-4 \right);\left( -3;4 \right);\left( -3;-4 \right) \\
& \left( 4;3 \right);\left( 4;-3 \right);\left( -4;3 \right);\left( -4;-3 \right) \\
\end{aligned} \right\}$.
Vậy có 12 bộ số $\left( m;n \right)$ thỏa mãn.
Đáp án D.