Question

giúp mk vs ạ thank nhiều!

Answer 1

1.

Ta thấy: $-1\leq \cos x\leq 1$

$\Leftrightarrow 1\leq 2\cos x+3\leq 5$

$\Leftrightarrow 1\leq \sqrt{2\cos x+3}\leq \sqrt{5}$
$\Leftrightarrow -3\leq \sqrt{2\cos x+3}-4\leq \sqrt{5}-4$
Vậy $y_{\min}=-3$ khi $x=(2k+1)\pi$, $y_{\max}=\sqrt{5}-4$ khi $x=2k\pi$ với $k$ nguyên.

 

 

Answer 2

2.

\(y=\cos ^2x-6\sin x+3=1-\sin ^2x-6\sin x+3\)

\(=-\sin ^2x-6\sin x+4\)

Ta thấy: $\sin ^2x\leq 1\Rightarrow -\sin ^2x\geq -1$

$\sin x\leq 1\Leftrightarrow -6\sin x\geq -6$

$\Rightarrow y=-\sin ^2x-6\sin x+4\geq -1-6+4=-3$

Vậy $y_{\min}=-3$. Giá trị này đạt tại $x=2k\pi +\frac{\pi}{2}$ với $k$ nguyên.

Mặt khác:

\(y=-\sin ^2x-6\sin x+4=9-(\sin x+1)(\sin x+5)\)

$-1\leq \sin x\leq 1\Rightarrow (\sin x+1)(\sin x+5)\geq 0$

$\Rightarrow y=9-(\sin x+1)(\sin x+5)\leq 9$

Vậy $y_{\max}=9$. Giá trị này đạt tại $x=2k\pi -\frac{\pi}{2}$ với $k$ nguyên.

Answer 3

3.

Ta thấy:

\(\cos ^2x+4\cos x+5=(\cos x+1)(\cos x+3)+2\geq 2\) do $\cos x\geq -1$

Do đó: $y=\frac{2}{\cos ^2x+4\cos x+5}\leq \frac{2}{2}=1$

Vậy $y_{\max}=1$. Giá trị này đạt tại $x=(2k+1)\pi$ với $k$ nguyên 

Lại có:

$\cos ^2x+4\cos x+5=(\cos x-1)(\cos x+5)+10\leq 10$ do $-1\leq \cos x\leq 1$

$\Rightarrow y=\frac{2}{\cos ^2x+4\cos x+5}\geq \frac{2}{10}=\frac{1}{5}$

Vậy $y_{\min}=\frac{1}{5}$. Giá trị này đạt tại $y=2k\pi$ với $k$ nguyên.

Answer 4

4.

\(y=\sin ^4x-2\cos ^2x+5=\sin ^4x-2(1-\sin ^2x)+5\)

\(=\sin ^4x+2\sin ^2x+3\)

Ta thấy $\sin ^2x\leq 1$ nên:

$y=\sin ^4x+2\sin ^2x+3\leq 1+2+3=6$

Vậy $y_{\max}=6$

Lại có:

$\sin ^2x\geq 0; \sin ^4x\geq 0$ nên $y=\sin ^4x+2\sin ^2x+3\geq 3$

Vậy $y_{\min}=3$

 

 

Answer 5

5.

\(y=\sin ^2x+2\sin x+5\)

Vì $\sin ^2x\leq 1; \sin x\leq 1$ nên:

$y\leq 1+2+5=8$

Vậy $y_{\max}=8$

Mặt khác:

$y=\sin ^2x+2\sin x+5=(\sin x+1)^2+4\geq 4$ 

$\Rightarrow y_{\min}=4$

 

Answer 6

6.

\(-1\leq \sin x\leq 1\Rightarrow 2\leq \sin x+3\leq 4\)

\(\Rightarrow 2\sqrt{2}\leq 2\sqrt{\sin x+3}\leq 4\)

\(\Leftrightarrow \frac{1}{2\sqrt{2}}\geq \frac{1}{2\sqrt{\sin x+3}}\geq \frac{1}{4}\)

Vậy $y_{\min}=\frac{1}{4}; y_{\max}=\frac{1}{2\sqrt{2}}$

 

Answer 7

7.

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

\(=\cos ^4x+2\cos ^2x-1\)

Vì $\cos ^2x\geq 0$ nên $y\geq 0+2.0-1=-1$

Vậy $y_{\min}=-1$

Mặt khác: $\cos ^2x\leq 1$ nên:

$y=\cos ^4x+2\cos ^2x-1\leq 1+2-1=2$

Vậy $y_{\max}=2$

Answer 8

8.

\(\sin ^2x-2\cos x+5=1-\cos ^2x-2\cos x+5=-\cos ^2x-2\cos x+6\)

Vì $\cos ^2x\leq 1\Rightarrow -\cos ^2x\geq -1$

$\cos x\leq 1\Rightarrow -2\cos x\geq -2$

$\sin ^2x-2\cos x+5\geq -1-2+6=3$

$\Rightarrow y=\frac{1}{\sin ^2x-2\cos x+5}\leq \frac{1}{3}$

Vậy $y_{\max}=\frac{1}{3}$

Mặt khác:
$-\cos ^2x-2\cos x+6=6-(\cos ^2x+2\cos x)$

$=7-(\cos x+1)^2\leq 7$

$\Rightarrow y=\frac{1}{\sin ^2-2\cos x+5}\geq \frac{1}{7}$

Vậy $y_{\min}=\frac{1}{7}$

Answer 9

9.

$-1\leq \cos 2x\leq 1$

$\Rightarrow 1\leq 2+\cos 2x\leq 3$

$\Rightarrow 1\leq \sqrt{2+\cos 2x}\leq \sqrt{3}$

$\Leftrightarrow 1\leq y\leq \sqrt{3}$

Vậy $y_{\min}=1; y_{\max}=\sqrt{3}$

Answer 10

10.

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

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

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

Vậy $y_{\max}=1$

Vì $\sin ^22x\leq 1\Rightarrow y=1-\frac{1}{2}\sin ^22x\geq 1-\frac{1}{2}=\frac{1}{2}$
Vậy $y_{\min}=\frac{1}{2}$