1.Tìm max,min của hàm số: a) y=2sin(x+π/4)+1 b) y=2√cosx+1 -3 c) y=4sin^2x-4sinx+3 d) y=cos^2x+2sinx+2

Đáp án:

Giải thích các bước giải:

$a)y=2sin(x+\dfrac{\pi}{4})+1$

Ta có:$-1≤sin (x+ \dfrac{\pi}{4})≤1\\\Leftrightarrow -2 ≤2sin(x+\dfrac{\pi}{4})≤2\\\Leftrightarrow-1≤2sin(x+\dfrac{\pi}{4})+1≤3\\\rightarrow Miny= -1\\\rightarrow Maxy =3$

$b) y =2\sqrt{cosx+1}+3$

Ta có:$-1≤cosx≤1\\\Leftrightarrow 0 ≤cosx+1≤2\\\Leftrightarrow0 ≤\sqrt{cosx+1}≤\sqrt{2}\\\Leftrightarrow 0≤2\sqrt{cosx+1}≤2\sqrt{2}\\\Leftrightarrow-3≤2\sqrt{cosx+1}-3≤2\sqrt{2}-3\\\rightarrow Miny=-3\\\rightarrow Mayx=2\sqrt{2}-3$

$c)y=4sin^2x-4sinx+3=(2sin^2x)-2.2sinx .1 +1^2 +2=(2sinx-1)^2+2$

Ta có:$-1\leq sinx \leq 1 \\\Leftrightarrow -2 \leq 2sinx \leq 2\\\Leftrightarrow -3 \leq 2sinx -1 \leq 1\\\Leftrightarrow 9\geq (2sinx-1)^2 \geq 1\\\Leftrightarrow 11 \geq (2sinx -1)^2 +2 \geq 3 \\\rightarrow Miny =3\\\rightarrow Mayx =11$

$d)y=cos^2x+2sinx+2=(1-sin^2x)+2sinx+2\\=-sin^2x+2sinx+1+2=-(sin^2x-2sinx+1)+4\\=-(sinx-1)^2+4$

Ta có:$-1 \leq sinx\leq 1\\\Leftrightarrow -2 \leq sinx-1 \leq0\\\Leftrightarrow4 \geq (sinx-1)^2 \geq 0\\\Leftrightarrow-4 \leq -(sinx-1)^2\leq 0\\\Leftrightarrow 0 \leq -(sinx-1)^2+4 \leq 4\\\rightarrow Miny =0\\\rightarrow Maxy =4$