Giải va biện luận theo $m$ phương trình   $y' = 0$ với

$y= \frac{-1}{2}\sin2x - (2m-5)cosx + 2(2 - m)x + 1$
$y' =-\cos 2x +(2m-5)\sin x +4 -2m=0$

$\Leftrightarrow 2\sin^2 x +(2m-5)\sin x +4-2m=0$ Đặt $\sin x = t ;\ t \in [-1;\ 1]$

Pt $\Leftrightarrow 2t^2 +(2m-5)t +3-2m=0 \ (*);\ t \in [-1;\ 1]$

Ta có $\Delta = (2m-5)^2 -4.2.(3-2m) =(2m-1)^2$

+ Nếu $2m-1=0 \Leftrightarrow  m=\dfrac{1}{2}$ pt $(*)$ có nghiệm kép $t_1=t_2=\dfrac{5-2m}{4}$

+ Nếu $\dfrac{5-2m}{4} \in [-1;\ 1]$ thì pt $y'=0$ có nghiệm và ngược lại (dài quá tự lắp các trường hợp giải tìm $m$ nhé)


+ Nếu $2m-1 \ne 0$ thì pt $(*)$ có 2 nghiệm phân biệt $t_1 = -1;\ t_2 = m-\dfrac{3}{2}$

+ Nếu $m-\dfrac{3}{2} \in [-1;\ 1]$ thì pt ban đầu có nghiệm và ngược lại... tự giải
cảm ơn nhìu nha ^^ –  kutequata_vu 03-04-14 10:54 AM

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