Đáp án:
$17)\\ a)min_y=\dfrac{3}{4}; max_y=\dfrac{5}{4}\\ b) min_y=-2; max_y=0\\ 18)\\ a)min_y=\sqrt{2}; max_y=\sqrt{3}\\ b)min_y=\dfrac{1}{2}; max_y=1$
Giải thích các bước giải:
$17)\\ a)y=1+\dfrac{1}{2}\sin x.\cos x\\ =1+\dfrac{1}{4}.2\sin x.\cos x\\ =1+\dfrac{1}{4}\sin 2x\\ \sin 2x \in [-1;-1] \Rightarrow 1+\dfrac{1}{4}\sin 2x \in \left[\dfrac{3}{4};\dfrac{5}{4}\right]\\ \Rightarrow min_y=\dfrac{3}{4}; max_y=\dfrac{5}{4}\\ b)y=2\cos^2x-2\\ =2(\cos^2x-1)\\ =-2\sin^2x\\ \sin^2x \in [0;1] \Rightarrow -2\sin^2x \in \left[-2;0\right]\\ \Rightarrow min_y=-2; max_y=0\\ 18)\\ a)y=\sqrt{2+\cos^2x}\\ \cos^2x \in [0;1] \Rightarrow \sqrt{2+\cos^2x} \in \left[\sqrt{2};\sqrt{3}\right]\\ \Rightarrow min_y=\sqrt{2}; max_y=\sqrt{3}\\ b)y=\sin^4x+\cos^4x\\ =\sin^4x+2\sin^2x\cos^2x+\cos^4x-2\sin^2x\cos^2x\\ =\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cos^2x\\ =1-2\sin^2x\cos^2x\\ =1-2.\dfrac{1-\cos 2x}{2}.\dfrac{1+\cos 2x}{2}\\ =1-\dfrac{(1-\cos 2x)(1+\cos 2x)}{2}\\ =\dfrac{2-(1-cos^22x)}{2}\\ =\dfrac{cos^22x+1}{2}\\ =\dfrac{\dfrac{1+\cos 4x}{2}+1}{2}\\ =\dfrac{3+\cos 4x}{4}\\ \cos 4x \in [-1;1] \Rightarrow \dfrac{3+\cos 4x}{4} \in \left[\dfrac{1}{2};1\right]\\ \Rightarrow min_y=\dfrac{1}{2}; max_y=1$
