Bài 4: Ôn tập chương Hàm số lượng giác và phương trình lượng giác

\(f\left(-x\right)=\dfrac{-tanx+x\cdot cos\left(-x\right)\left(-3x+pi\right)}{\left|x\right|}\)

\(=\dfrac{-tanx+x\cdot cos\left(x\right)\cdot\left(-3x+pi\right)}{\left|x\right|}\)

=>f(-x)<>f(x) và f(-x)<>-f(x)

=>f(x) ko chẵn, ko lẻ

ĐKXĐ: x<>pi/2+kpi và \(\dfrac{1}{2}sin2x+1< >0\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}x< >\dfrac{pi}{2}+kpi\\sin2x\cdot\dfrac{1}{2}< >-1\end{matrix}\right.\Leftrightarrow x< >\dfrac{pi}{2}+kpi\)

ĐKXĐ: \(\sqrt{2}sinx-sin2x< >0\)

\(\Leftrightarrow sinx\left(\sqrt{2}-2cosx\right)< >0\)

=>sinx<>0và cosx<>căn 2/2

=>x<>kpi và \(x\ne\pm\dfrac{pi}{4}+k2pi\)

ĐKXĐ: x<>pi/2+kpi và 2x<>k2pi

=>x<>pi/2+kpi và x<>kpi

=>x<>kpi/2

\(y=\left(2sinx+1\right)^2+1\ge1\)

\(y_{min}=1\) khi \(sinx=-\dfrac{1}{2}\)

\(y=4\left(sin^2x+sinx-2\right)+8=4\left(sinx-1\right)\left(sinx+2\right)+8\)

Do \(\left\{{}\begin{matrix}sinx-1\le0\\sinx+2>0\end{matrix}\right.\) ;\(\forall x\Rightarrow\left(sinx-1\right)\left(sinx+2\right)\le0\)

\(\Rightarrow y\le8\)

\(y_{max}=8\) khi \(sinx=1\)

\(y=2\sqrt{3}\left(\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\right)+2=2\sqrt{3}sin\left(x+\dfrac{\pi}{6}\right)+2\)

Do \(-1\le sin\left(x+\dfrac{\pi}{6}\right)\le1\)

\(\Rightarrow-2\sqrt{3}+2\le y\le2\sqrt{3}+2\)

\(y=3sinx+\sqrt{3}cosx+2\)

=>\(2-\sqrt{3^2+3}< =y< =2+\sqrt{3^2+3}\)

=>\(2-2\sqrt{3}< =y< =2+2\sqrt{3}\)

\(y_{min}\) xảy ra khi \(3sinx+\sqrt{3}cosx=-2\sqrt{3}\)

=>\(\dfrac{3}{2\sqrt{3}}sinx+\dfrac{\sqrt{3}}{2\sqrt{3}}cosx=-1\)

=>sin(x+a)=-1(với \(sina=\dfrac{1}{2}\))

=>sin(x+pi/6)=-1

=>x+pi/6=-pi/2+k2pi

=>x=-2/3pi+k2pi

\(y_{max}\) xảy ra khi sin(x+pi/6)=1

=>x+pi/6=pi/2+k2pi

=>x=pi/3+k2pi

5.

Gọi G là trọng tâm tam giác PQR, theo công thức trọng tâm: \(\left\{{}\begin{matrix}x_G=\dfrac{-3+1+2}{3}=0\\y_G=\dfrac{2+1-4}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow G\left(0;-\dfrac{1}{3}\right)\)

\(V_{\left(O;-\dfrac{1}{3}\right)}\left(G\right)=G'\left(x';y'\right)\Rightarrow\left\{{}\begin{matrix}x'=-\dfrac{1}{3}.0=0\\y'=-\dfrac{1}{3}.\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\end{matrix}\right.\)

\(\Rightarrow G'\left(0;\dfrac{1}{9}\right)\)

6.

Theo công thức phép vị tự, \(V_{\left(A;k\right)}\left(B\right)=C\Rightarrow\left\{{}\begin{matrix}-1-0=k\left(2-0\right)\\5-3=k\left(-1-3\right)\end{matrix}\right.\)

\(\Rightarrow k=-\dfrac{1}{2}\)

8.

\(V_{\left(I;-2\right)}\left(d\right)=d'\Rightarrow d'\) cùng phương d

Hay pt d' có dạng: \(2x+y+c=0\) (1)

Lấy \(A\left(0;4\right)\in d\), gọi \(V_{\left(I;-2\right)}\left(A\right)=A'\left(x';y'\right)\Rightarrow A'\in d'\)

\(\left\{{}\begin{matrix}x'=-2\left(0+1\right)-1=-3\\y'=-2\left(4-2\right)+2=-2\end{matrix}\right.\)

Thế vào (1) 

\(\Rightarrow2.\left(-3\right)-2+c=0\Rightarrow c=8\)

\(\Rightarrow2x+y+8=0\)

9.

\(V_{\left(O;-\dfrac{2}{3}\right)}\left(d\right)=d'\Rightarrow d'\) cùng phương d hay pt d' có dạng:

\(3x-y+c=0\) (1)

Chọn \(A\left(0;-5\right)\)  là 1 điểm thuộc d

\(V_{\left(O;-\dfrac{2}{3}\right)}\left(A\right)=A'\left(x';y'\right)\Rightarrow A'\in d'\)

\(\left\{{}\begin{matrix}x'=-\dfrac{2}{3}.0=0\\y'=-\dfrac{2}{3}.\left(-5\right)=\dfrac{10}{3}\end{matrix}\right.\)

Thế vào (1):

\(3.0-\dfrac{10}{3}+c=0\Rightarrow c=\dfrac{10}{3}\)

\(\Rightarrow3x-y+\dfrac{10}{3}=0\)

\(\Leftrightarrow9x-3y+10=0\)

Câu 11: B

Câu 12: 

ĐK: \(\left(-m\right)^2+5^2>=m^2+2m+1\)

=>2m<=24

=>m<=12

Câu 13: 
ĐK:m^2+25>=m^2+2m+1

=>2m+1<=25

=>m<=12

Câu 14: 
ĐK: m^2+1>=m^2-4m+4

=>-4m+4<=1

=>-4m<=-3

=>m>=3/4

\(\Leftrightarrow\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}+2sin\dfrac{x}{2}.cos\dfrac{x}{2}\right)^2=cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\)

\(\Leftrightarrow\left(sin\dfrac{x}{2}+cos\dfrac{x}{2}\right)^4=\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)\left(cos\dfrac{x}{2}+sin\dfrac{x}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{x}{2}+sin\dfrac{x}{2}=0\Leftrightarrow tan\dfrac{x}{2}=-1\Leftrightarrow...\\\left(sin\dfrac{x}{2}+cos\dfrac{x}{2}\right)^3=cos\dfrac{x}{2}-sin\dfrac{x}{2}\left(1\right)\end{matrix}\right.\)

Xét (1):

Với \(cos\dfrac{x}{2}=0\) không là nghiệm

Với \(cos\dfrac{x}{2}\ne0\) chia 2 vế cho \(cos^3\dfrac{x}{2}\)

\(\Rightarrow\left(tan\dfrac{x}{2}+1\right)^3=\dfrac{1}{cos^2\dfrac{x}{2}}-tan\dfrac{x}{2}.\dfrac{1}{cos^2\dfrac{x}{2}}\)

\(\Leftrightarrow tan^3\dfrac{x}{2}+3tan^2\dfrac{x}{2}+3tan\dfrac{x}{2}+1=1+tan^2\dfrac{x}{2}-tan\dfrac{x}{2}\left(1+tan^2\dfrac{x}{2}\right)\)

\(\Leftrightarrow2tan^3\dfrac{x}{2}+2tan^2\dfrac{x}{2}+4tan\dfrac{x}{2}=0\)

\(\Leftrightarrow2tan\dfrac{x}{2}\left(tan^2\dfrac{x}{2}+tan\dfrac{x}{2}+2\right)=0\)

\(\Leftrightarrow tan\dfrac{x}{2}=0\)

\(\Leftrightarrow...\)