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Question

Giá trị lớn nhất của hàm số $(\sin x-\cos x)^2+2\cos 2x +3\sin x\cos x =?$

score 1 · Answer 1

1 cách khác

$y=(\sin x -\cos x)^2 +2\cos 2x +3\sin x \cos x = 1+2\cos 2x +\dfrac{1}{2}\sin 2x$

$\Leftrightarrow\dfrac{1}{2}\sin 2x +2\cos 2x +1 -y = 0$

Pt có nghiệm $\Leftrightarrow (\dfrac{1}{2})^2 +2^2 \ge (1-y)^2$

$\Leftrightarrow \dfrac{17}{4} \ge y^2 -2y+1$

$\Leftrightarrow -4y^2 +8y +13 \ge 0$

$\Leftrightarrow \dfrac{2-\sqrt{17}}{2} \le y \le \dfrac{2+\sqrt{17}}{2}$

Xong

score 0 · Answer 2

$A=(sinx-cosx)^{2}+2cos2x+3sinx.cosx$
$=1-2sinx.cosx+2cos2x+3sinx.cosx$

$=1+2cos2x+\frac{1}{2}sin2x$

$=1+\frac{\sqrt{17}}{2}(\frac{4}{\sqrt{17}}cos2x+\frac{1}{\sqrt{17}}sin2x)$

$=1+\frac{\sqrt{17}}{2}sin(\alpha +2x)$ ( voi $sin \alpha=\frac{4}{\sqrt{17}},cos \alpha=\frac{1}{\sqrt{17}}$)

Ta co : $-1\leq sin(\alpha+2x)\leq 1$

$\Rightarrow 1-\frac{\sqrt{17}}{2}\leq A\leq 1+\frac{\sqrt{17}}{2}$

Vay $Max A=1+\frac{\sqrt{17}}{2}$

2 Đáp án