Question

Tìm GTLN và GTNN:

1.\(y=\sqrt{5-2cos^2x.sin^2x}\)

2.\(y=1+\dfrac{1}{2}sin2x.cos2x\)

3.\(y=\sqrt{1+sinx}-3\)

4.\(y=\sqrt{2+sin^22x}\)

Answer 1

1.

\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{1}{2}(2\cos x\sin x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)

Dễ thấy:

$\sin ^22x\geq 0\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$

Vậy $y_{\max}=\sqrt{5}$

$\sin ^22x\leq 1\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\geq \sqrt{5-\frac{1}{2}}=\frac{3\sqrt{2}}{2}$

Vậy $y_{\min}=\frac{3\sqrt{2}}{2}$

Answer 2

2.

$y=1+\frac{1}{2}\sin 2x\cos 2x=1+\frac{1}{4}.2\sin 2x\cos 2x$

$=1+\frac{1}{4}\sin 4x$

Vì $-1\leq \sin 4x\leq 1$

$\Rightarrow \frac{5}{4}\leq 1+\frac{1}{4}\sin 4x\leq \frac{3}{4}$

$\Leftrightarrow \frac{5}{4}\leq y\leq \frac{3}{4}$
Vậy $y_{\max}=\frac{5}{4}; y_{\min}=\frac{3}{4}$

Answer 3

3.

$\sin x\geq -1\Rightarrow \sqrt{1+\sin x}\geq 0$

$\Rightarrow y\geq -3$

Vậy $y_{\min}=-3$

$\sin x\leq 1\Rightarrow \sqrt{1+\sin x}\leq \sqrt{2}$

$\Rightarrow y\leq \sqrt{2}-3$
Vậy $y_{\max}=\sqrt{2}-3$

 

Answer 4

4.

$\sin ^22x\geq 0\Rightarrow y=\sqrt{2+\sin ^22x}\geq \sqrt{2}$Vậy $y_{\min}=\sqrt{2}$$\sin ^22x\leq 1\Rightarrow y=\sqrt{2+\sin ^22x}\leq \sqrt{3}$Vậy $y_{\max}=\sqrt{3}$