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 \(tanx=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

1: \(-1<=cosx\le1\)

=>\(-3\le-3\cdot cosx\le3\)

=>\(-3+5\le-3\cdot cosx+5\le3+5\)

=>2<=y<=8

y min=2 khi cosx=1

=>\(x=k2\pi\)

y min=8 khi cosx=-1

=>\(x=\pi+k2\pi\)

3: \(y=cos^2x+2\cdot cos2x\)

\(=\frac{1+cos2x}{2}+2\cdot cos2x=2,5\cdot cos2x+0,5\)

Ta có: \(-1\le cos2x\le1\)

=>\(-2,5\le2,5cos2x\le2,5\)

=>\(-2,5+0,5\le2,5cos2x+0,5\le2,5+0,5\)

=>-2<=y<=3

y min=-2 khi cos2x=-1

=>\(2x=\pi+k2\pi\)

=>\(x=\frac{\pi}{2}+k\pi\)

y max=3 khi cos2x=1

=>\(2x=k2\pi\)

=>\(x=k\pi\)

6: \(y=\sqrt3\cdot\sin x-cosx-2\)

\(=2\left(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx\right)-2=2\cdot\sin\left(x-\frac{\pi}{6}\right)-2\)

Ta có: \(-1\le\sin\left(x-\frac{\pi}{6}\right)\le1\)

=>\(-2\le2\sin\left(x-\frac{\pi}{6}\right)\le2\)

=>\(-2-2\le2\sin\left(x-\frac{\pi}{6}\right)-2\le2-2\)

=>-4<=y<=0

y min=-4 khi \(\sin\left(x-\frac{\pi}{6}\right)=-1\)

=>\(x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{2}+\frac{\pi}{6}+k2\pi=-\frac26\pi+k2\pi=-\frac13\pi+k2\pi\)

y max=0 khi \(\sin\left(x-\frac{\pi}{6}\right)=1\)

=>\(x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac23\pi+k2\pi\)

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left(\frac12\cdot\sin2x\right)^2=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

Ta có: \(0\le\sin^22x\le1\)

=>\(-\frac12\le-\frac12\cdot\sin^22x\le0\)

=>\(-\frac12+5\le-\frac12\cdot\sin^22x+5\le0+5\)

=>\(\frac92\le-\frac12\cdot\sin^22x+5\le5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(4:\frac{3\sqrt2}{2}\ge\frac{4}{\sqrt{-\frac12\cdot sin^22x+5}}\ge\frac{4}{\sqrt5}\)

=>\(\frac{2\sqrt2}{3}\ge y\ge\frac{4\sqrt5}{5}\)

Do đó: \(y_{\max}=\frac{2\sqrt2}{3}\) khi \(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4\sqrt5}{5}\) khi \(\sin^22x=0\)

=>sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\left(cos^2x-\sin^2x\right)-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos^2x+4\cdot\sin^2x-2\)

\(=7\cdot\sin^2x+cos^2x-2=7\cdot\sin^2x+1-\sin^2x-2=6\cdot\sin^2x-1\)

Ta có: \(0\le\sin^2x\le1\)

=>\(0\le6\sin^2x\le6\)

=>\(0-1\le6\sin^2x-1\le6-1\)

=>-1<=f(x)<=5

f(x) min=-1 khi \(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

f(x) max=5 khi \(\sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

Chọn B

Đáp án D

Ta có 

y = sin x = cos 2 x = sin x - 1 - 2 sin 2 x = 2 sin 2 x + sin x - 1

Đặt t = sin(x), - 1 ≤ t ≤ 1

Ta sẽ đi tìm GTLN và GTNN của hàm số y = g t = 2 t 2 + t - 1  trên đoạn [ -1;1 ]

Ta có  g t = - 2 t 3 - t + 1 ,   - 1 ≤ t ≤ 1 2 2 t 3 + t - 1 ,     1 2 ≤ t ≤ 1

* Xét hàm số  h t = - 2 t 3 - t + 1  trên đoạn - 1 ; 1 2

Dễ dàng tìm được 

M a x r ∈ 1 2 ; 1 h t = 9 8 ⇔ t = - 1 4 M i n r ∈ 1 2 ; 1 h t = 0 ⇔ t = 1 2

* Xét hàm số k t = 2 t 3 + t - 1  trên đoạn  1 2 ; 1

Cũng dễ dàng tìm được 

M a x r ∈ 1 2 ; 1 k t = 2 ⇔ t = 1 M i n r ∈ 1 2 ; 1 k t = 0 ⇔ t = 1 2

Qua hai trường hợp trên ta đi đến kết luận

M a x r ∈ - 1 ; 3 g t = 2 ⇔ t = 1 M i n r ∈ - 1 ; 3 g t = 0 ⇔ t = 1 2

Hay 

M = M a x y = 2 ⇔ sin x = - 1 ⇔ x = - π 2 + k 2 π m = Miny = 0 ⇔ sin x = 1 2 ⇔ x = π 6 + k 2 π x = 5 π 6 + k 2 π

Đáp án C