a.

\(\Leftrightarrow m-cosx\ge0\) ; \(\forall x\)

\(\Leftrightarrow m\ge max\left(cosx\right)\)

\(\Leftrightarrow m\ge1\)

b.

\(\Leftrightarrow2sinx-m\ge0\) ; \(\forall x\)

\(\Leftrightarrow m\le2sinx\) ; \(\forall x\)

\(\Leftrightarrow m\le\min\limits_{x\in R}\left(2sinx\right)\)

\(\Leftrightarrow m\le-2\)

c.

\(\Leftrightarrow cosx+m\ne0\) ; \(\forall x\)

\(\Leftrightarrow\left[{}\begin{matrix}m>\max\limits_R\left(cosx\right)\\m< \min\limits_R\left(cosx\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

1. \(D=R\)

2. \(sinx\ne0\Leftrightarrow x\ne k\pi\Rightarrow D=R\backslash\left\{k\pi|k\in R\right\}\)

3. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\Rightarrow D=R\backslash\left\{\dfrac{\pi}{4}+\dfrac{k\pi}{2}|k\in R\right\}\)

4. \(cos\left(x+\dfrac{\pi}{4}\right)\ne0\Leftrightarrow x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\Rightarrow D=R\backslash\left\{\dfrac{\pi}{4}+k\pi|k\in R\right\}\)

cảm ơn ạ

1. Không dịch được đề

2.

\(-1\le cos2x\le1\Rightarrow1\le y\le3\)

3.

a. \(-2\le2sinx\le2\Rightarrow-1\le y\le3\)

\(y_{min}=-1\) khi \(sinx=-1\Rightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=3\) khi \(sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

b.

\(0\le cos^2x\le1\Rightarrow-1\le y\le2\)

\(y_{min}=-1\) khi \(cos^2x=1\Rightarrow x=k\pi\)

\(y_{max}=2\) khi \(cosx=0\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

4.

\(y=\left(tanx-1\right)^2+2\ge2\)

\(y_{min}=2\) khi \(tanx=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

Đáp án D

Ta có y = s inx + 2 cos x + 1 s inx + cos x + 2 ⇔ y − 1 s inx + y − 2 cos x = 1 − 2 y    1 .  

PT (1) có nghiệm ⇔ y − 1 2 + y − 2 2 ≥ 1 − 2 y 2 ⇔ 2 y 2 + 2 y − 4 ≤ 0 ⇔ − 2 ≤ y ≤ 1 ⇒ M = 1.  

1. Do \(\cos x+2>0\forall x\in R\) \(\Rightarrow\) Hàm số xác định \(\forall x\in R\)

\(y=\dfrac{\sin x+1}{\cos x+2}\)

\(\Leftrightarrow\)\(y\cos x-\sin x=1-2y\)

pt có nghiệm \(\Leftrightarrow y^2+\left(-1\right)^2\ge\left(1-2y\right)^2\)

\(\Leftrightarrow3y^2-4y\le0\)

\(\Leftrightarrow0\le y\le\dfrac{4}{3}\)

2. \(y=\dfrac{\cos x+2\sin x+3}{2\cos x-\sin x+4}\)

\(\Leftrightarrow\left(2y-1\right)\cos x-\left(y+2\right)\sin x=3-4y\)

pt có nghiệm \(\Leftrightarrow\left(2y-1\right)^2+\left(y+2\right)^2\ge\left(3-4y\right)^2\)

\(\Leftrightarrow11y^2-24y+4\le0\)

\(\Leftrightarrow\dfrac{2}{11}\le y\le2\)

kiểm tra giúp mình xem có sai sót gì không...

bạn ơi tsao chỗ pt có nghiệm chỗ câu 1 lại ra bất pt vậy

1.

Hàm số xác định khi: \(1-2sinx\ne0\Leftrightarrow sinx\ne\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

2.

Đặt \(t=cosx\left(t\in\left[-1;1\right]\right)\)

Hàm số xác định trên R khi:

\(m-1+2cosx\ge0\forall x\in R\)

\(\Leftrightarrow m\ge f\left(t\right)=1-2t\forall x\in R\)

\(\Leftrightarrow m\ge maxf\left(t\right)=f\left(-1\right)=3\)

Vậy \(m\ge3\)

Xét  − sin x + 2 cos x + 4 = 0

Ta thấy − 1 2 + 2 2 < 4 2  nên phương trình vô nghiệm.

Do đó − sin x + 2 cos x + 4 ≠ 0 .

Như vậy,  y = 2 sin x + cos x + 3 − sin x + 2 cos x + 4

⇔ y − sin x + 2 cos x + 4 = 2 sin x + cos x + 3

⇔ sin x 2 + y + cos x 1 − 2 y + 3 − 4 y = 0

Để phương trình có nghiệm thì  2 + y 2 + 1 − 2 y 2 ≥ 3 − 4 y 2

⇔ 5 y 2 + 5 ≥ 16 y 2 − 24 y + 9

⇔ 11 y 2 − 24 y + 4 ≤ 0

⇔ 2 11 ≤ y ≤ 2

Chọn đáp án D.

Đáp án A

Ta có:  y = cos x + 2 sin x + 3 2 cos x − sin x + 4

⇒ y 2 cos x − sin x + 4 = cos x + 2 sin x + 3

⇔ 2 + y sin x + 1 − 2 y cos x = 4 y − 3    1

PT (1) có nghiệm  ⇔ 2 + y 2 + 1 − 2 y 2 ≥ 4 y − 3 2

⇔ 11 y 2 − 24 y + 4 ≤ 0 ⇔ 2 11 ≤ y ≤ 2

Suy ra M = 2 m = 2 11 ⇒ M . m = 4 11