22 câu Dạng 3: Tính giá trị lớn nhất – giá trị nhỏ nhất của hàm số lượng giác

Ta có $y={\sin }^{6}x+{\cos }^{6}x=1-\frac{\text{3}}{\text{4}}{\sin }^{2}2x$.

Do $0\le {\sin }^{2}2x\le 1$ nên $-\frac{\text{3}}{\text{4}}.0\ge -\frac{\text{3}}{\text{4}}{\sin }^{2}2x\ge -\frac{\text{3}}{\text{4}}$

$1\ge 1-\frac{\text{3}}{\text{4}}{\sin }^{2}2x\ge 1-\frac{\text{3}}{\text{4}}\iff 1\ge y\ge \frac{1}{4}$.

Vậy  $\min y=\frac{1}{4}\text{ khi }{\sin }^{2}2x=1\iff \cos 2x=0\iff x=\frac{\pi }{4}+\frac{k\pi }{2},k\in \mathbb{Z}$.

$\max y=1\text{ khi }{\sin }^{2}2x=0\iff \sin 2x=0\iff x=\frac{k\pi }{2},k\in \mathbb{Z}$.

Ta có $y={\tan }^{2}x-\tan x+2020={\left(\tan x-\frac{1}{2}\right)}^2+\frac{8079}{4}$.

Hàm số $\tan x$ đồng biến và xác định trên khoảng $\left(-\frac{\pi }{2};\frac{\pi }{2}\right)$

Mà $\left[-\frac{\pi }{4};\frac{\pi }{4}\right]\subset \left(-\frac{\pi }{2};\frac{\pi }{2}\right)$ nên hàm số đồng biến và xác định trên $\left[-\frac{\pi }{4};\frac{\pi }{4}\right]$ .

Do đó $\tan \left(-\frac{\pi }{4}\right)\le \tan x\le \tan \frac{\pi }{4}\iff -1\le \tan x\le 1$

$\Rightarrow -1-\frac{1}{2}\le \tan x-\frac{1}{2}\le 1-\frac{1}{2}\iff -\frac{3}{2}\le \tan x-\frac{1}{2}\le \frac{1}{2}\Rightarrow 0\le {\left(\tan x-\frac{1}{2}\right)}^2\le \frac{9}{4}$.

$\iff \frac{8079}{4}\le {\left(\tan x-\frac{1}{2}\right)}^2+\frac{8079}{4}\le \frac{9}{4}+\frac{8079}{4}\iff \frac{8079}{4}\le y\le 2022$

Vậy $\min y=\frac{8079}{4}\text{khi }\tan x=\frac{1}{2}\iff x=\arctan \frac{1}{2}$;

Đáp án C

Hàm số $y=7-2\cos \left(x+\frac{\pi }{4}\right)$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $-1\le \cos \left(x+\frac{\pi }{4}\right)\le 1\iff -2\le 2\cos \left(x+\frac{\pi }{4}\right)\le 2\iff 5\le 7-2\cos \left(x+\frac{\pi }{4}\right)\le 9$.

Vậy $\min y=5\iff \cos \left(x+\frac{\pi }{4}\right)=1\iff x+\frac{\pi }{4}=k2\pi \iff x=\frac{-\pi }{4}+k2\pi ,k\in \mathbb{Z}$;

$\max y=9\iff \cos \left(x+\frac{\pi }{4}\right)=-1\iff x+\frac{\pi }{4}=-\pi +k2\pi \iff x=\frac{-5\pi }{4}+k2\pi ,k\in \mathbb{Z}$

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Đáp án D

Hàm số $y=4\sqrt{\sin x+3}-1$ có nghĩa $\iff \sin x+3\ge 0\iff \sin x\ge -3\iff \forall x\in ℝ\iff D=ℝ$.

Ta có $-1\le \sin x\le 1\iff 2\le \sin x+3\le 4\iff \sqrt{2}\le \sqrt{\sin x+3}\le 2$

$\iff 4\sqrt{2}\le 4\sqrt{\sin x+3}\le 8\iff 4\sqrt{2}-1\le 4\sqrt{\sin x+3}-1\le 7$.

Vậy $\min y=4\sqrt{2}-1\iff \sin x=-1\iff x=\frac{-\pi }{2}+k2\pi ,k\in \mathbb{Z}$;

$\max y=7\iff \sin x=1\iff x=\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$.

Đáp án B

Hàm số $y={\sin }^{2}x-4\sin x-5$ có nghĩa $\forall x\in ℝ\iff D=ℝ$

Ta có $y={\sin }^{2}x-4\sin x-5={\left(\sin x-2\right)}^2-9$.

 $-1\le \sin x\le 1\iff -3\le \sin x-2\le -1\iff 1\le {\left(\sin x-2\right)}^2\le 9\iff -8\le {\left(\sin x-2\right)}^2-9\le 0$.

Vậy $\min y=-8\iff \sin x-2=-1\iff \sin x=1\iff x=\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$.

Đáp án A

Hàm số $y=\sqrt{2\sin x+3}$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $-1\le \sin x\le 1\iff -2\le 2\sin x\le 2\iff 1\le 2\sin x+3\le 5\iff 1\le \sqrt{2\sin x+3}\le \sqrt{5}$.

Vậy $\min y=1\iff \sin x=-1\iff x=\frac{-\pi }{2}+k2\pi ,k\in \mathbb{Z}$;

$\max y=\sqrt{5}\iff \sin x=1\iff x=\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$.

Đáp án A

Hàm số  $y=\frac{4}{1+2{\sin }^{2}x}$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $-1\le \sin x\le 1\iff 0\le {\sin }^{2}x\le 1\iff 0\le 2{\sin }^{2}x\le 2\iff 1\le 1+2{\sin }^{2}x\le 3$

 $\iff \frac{1}{3}\le \frac{1}{1+2{\sin }^{2}x}\le 1\iff \frac{4}{3}\le \frac{4}{1+2{\sin }^{2}x}\le 4$.

Vậy $\min y=\frac{4}{3}\iff \left[\begin{array}{l}\sin x=-1\iff x=\frac{-\pi }{2}+k2\pi ,k\in \mathbb{Z}\\ \sin x=1\iff x=\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}\end{array}\right.\iff x=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z};$

$\max y=4\iff \sin x=0\iff x=k\pi ,k\in \mathbb{Z}$.

Đáp án D

Hàm số $y=2{\sin }^{2}x+{\cos }^{2}2x$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $y=2{\sin }^{2}x+{\cos }^{2}2x=1-\cos 2x+{\cos }^{2}2x={\left(\cos 2x-\frac{1}{2}\right)}^2+\frac{3}{4}$.

$-1\le \cos 2x\le 1\iff \frac{-3}{2}\le \cos 2x-\frac{1}{2}\le \frac{1}{2}\iff 0\le {\left(\cos 2x-\frac{1}{2}\right)}^2\le \frac{9}{4}\iff \frac{3}{4}\le {\left(\cos 2x-\frac{1}{2}\right)}^2+\frac{3}{4}\le 3$ .

Vậy $\min y=\frac{3}{4}\iff \cos 2x=\frac{1}{2}\iff x=\pm \frac{\pi }{6}+k\pi ,k\in \mathbb{Z}$;

$\max y=3\iff \cos 2x=0\iff x=\frac{\pi }{4}+\frac{k\pi }{2},k\in \mathbb{Z}$.

Đáp án C

Hàm số $y=3\sin x+4\cos x+1$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $y=3\sin x+4\cos x+1=5\left(\frac{3}{5}\sin x+\frac{4}{5}\cos x\right)+1=5\sin \left(x+\alpha \right)+1$ với $\alpha =\arccos \left(\frac{3}{5}\right)+k2\pi$.

$-5\le 5\sin \left(x+\alpha \right)\le 5\iff -4\le 5\sin \left(x+\alpha \right)+1\le 6$ .

Vậy $\min y=-4\iff \sin \left(x+\alpha \right)=-1\iff x+\alpha =\frac{-\pi }{2}+k2\pi \iff x=-\alpha -\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$;

$\max y=6\iff \sin \left(x+\alpha \right)=1\iff x+\alpha =\frac{\pi }{2}+k2\pi \iff x=-\alpha +\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$.

Đáp án A

Hàm số $y=4\sin 6x+3\cos 6x$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $y=4\sin 6x+3\cos 6x=5\left(\frac{4}{5}\sin 6x+\frac{3}{5}\cos 6x\right)=5\sin \left(6x+\alpha \right)$ với $\alpha =\arccos \left(\frac{4}{5}\right)+k2\pi$.

  $-1\le \sin \left(6x+\alpha \right)\le 1\iff -5\le 5\sin \left(6x+\alpha \right)\le 5$.

Vậy $\min y=-5\iff \sin \left(6x+\alpha \right)=-1\iff 6x+\alpha =\frac{-\pi }{2}+k2\pi \iff x=\frac{-2\alpha -\pi }{12}+\frac{k\pi }{3},k\in \mathbb{Z}$ $\max y=5\iff \sin \left(6x+\alpha \right)=1\iff 6x+\alpha =\frac{\pi }{2}+k2\pi \iff x=\frac{-\alpha }{6}+\frac{\pi }{12}+\frac{k\pi }{3},k\in \mathbb{Z}$;

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Đáp án B

Hàm số $y=\sin 2x$ có nghĩa $\forall x\in ℝ\iff D=ℝ$

Khi $x\in \left[-\frac{\pi }{6};\frac{\pi }{3}\right]$ thì  $-\frac{\sqrt{3}}{2}\le \sin 2x\le \frac{\sqrt{3}}{2}$.

Vậy $\min y=-\frac{\sqrt{3}}{2}\iff x=-\frac{\pi }{6};\max y=\frac{\sqrt{3}}{2}\iff x=\frac{\pi }{3}$ .

Đáp án D

Hàm số $y=\sqrt{3}\tan x$ có nghĩa $\iff \cos x\ne 0\iff x\ne \frac{\pi }{2}+k\pi \iff D=ℝ\backslash \left\{\frac{\pi }{2}+k\pi \right\}$.

Khi $x\in \left[-\frac{\pi }{3};\frac{\pi }{4}\right]$ thì hàm số $y=\tan x$ luôn đồng biến.

Suy ra $-\sqrt{3}\le \tan x\le 1\iff -1\le \sqrt{3}\tan x\le \sqrt{3}$.

Vậy $\min y=-1\iff x=-\frac{\pi }{3};\max y=\frac{\sqrt{3}}{2}\iff x=\frac{\pi }{4}$ .

Đáp án D

Hàm số $y=f\left(x\right)=4-3\cos x$ có nghĩa $\forall x\in ℝ\iff D=ℝ$ .

Khi $x\in \left[0;\frac{2\pi }{3}\right]$ thì  $-\frac{1}{2}\le \cos x\le 1\iff \frac{-3}{2}\le 3\cos x\le 3\iff -3\le -3\cos x\le \frac{3}{2}\iff 1\le 4-3\cos x\le \frac{11}{2}$.

Vậy $\min y=1\iff x=0;\max y=\frac{11}{2}\iff x=\frac{2\pi }{3}$.

Đáp án D

Hàm số $y=f\left(x\right)=\sin \left(2x+\frac{\pi }{4}\right)$  có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Khi  $x\in \left[-\frac{\pi }{4};\frac{\pi }{4}\right]$ thì $-\frac{\sqrt{2}}{2}\le \sin \left(2x+\frac{\pi }{4}\right)\le \frac{\sqrt{2}}{2}$.

Vậy $\min y=-\frac{\sqrt{2}}{2}\iff x=\frac{-\pi }{4};\max y=\frac{\sqrt{2}}{2}\iff x=\frac{\pi }{4}$.

Đáp án C

Hàm số $y=\sin x+\sqrt{2-{\sin }^{2}x}$ có nghĩa $\forall x\in ℝ\iff D=ℝ$ .

Ta có $-1\le \sin x\le 1\iff 0\le {\sin }^{2}x\le 1\iff -1\le -{\sin }^{2}x\le 0\iff 1\le 2-{\sin }^{2}x\le 2\iff 1\le \sqrt{2-{\sin }^{2}x}\le \sqrt{2}$ .

Lại có $-1\le \sin x\le 1\Rightarrow 0\le \sin x+\sqrt{2-{\sin }^{2}x}\le 1+\sqrt{2}\iff 0\le y\le 1+\sqrt{2}$.

$y=0\iff \sin x=-1\iff x=-\frac{\pi }{2}+k2\pi$.

Vậy $\min y=0\iff x=-\frac{\pi }{2}+k2\pi$.

 $y=1+\sqrt{2}\iff \left\{\begin{array}{l}\sin x=1\\ \sin x=0\end{array}\right.$ (vô nghiệm).

Áp dụng bất đẳng thức Côsi ta có

$2\sin x\sqrt{2-{\sin }^{2}x}\le {\sin }^{2}x+2-{\sin }^{2}x\iff 2\sin x\sqrt{2-{\sin }^{2}x}\le 2$.

$\iff 2+2\sin x\sqrt{2-{\sin }^{2}x}\le 4\iff y^2\le 4\iff y\le 2$

Dấu “=” khi và chỉ khi $\sin x=\sqrt{2-{\sin }^{2}x}\iff \sin x=1\iff x=\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$.

Vậy $\min y=0\iff x=-\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z};\max y=2\iff x=\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$ .

Đáp án A

Hàm số $y=\frac{\cos x-2\sin x}{2-\sin x}$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $y=\frac{\cos x-2\sin x}{2-\sin x}\Rightarrow 2y-y\sin x=\cos x-2\sin x$

$\iff 2y=y\sin x-2\sin x+\cos x\iff 2y=\left(y-2\right)\sin x+\cos x\Rightarrow 4y^2={\left[\left(y-2\right)\sin x+\cos x\right]}^2$

Sử dụng bất đẳng thức Bunhiacopxki ta có

${\left(y-2\right)}^2+1^2\ge 4y^2\iff 3y^2+4y-5\le 0\iff \frac{-2-\sqrt{19}}{3}\le y\le \frac{-2+\sqrt{19}}{3}$.

Vậy $\min y=\frac{-2-\sqrt{19}}{3};\max y=\frac{-2+\sqrt{19}}{3}$.

Đáp án B

Hàm số ${\left(3\sin x-4\cos x\right)}^2-6\sin x+8\cos x\ge 2m-1$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có ${\left(3\sin x-4\cos x\right)}^2-2\left(3\sin x-4\cos x\right)+1\ge 2m\iff {\left(3\sin x-4\cos x-1\right)}^2\ge 2m$

Để phương trình có nghiệm đúng với mọi $x\in ℝ$  thì $2m\le 0\iff m\le 0$.      

Đáp án A

Hàm số $y={\tan }^{2}x+{\cot }^{2}x+3\left(\tan x+\cot x\right)-1$ có nghĩa $\iff \left\{\begin{array}{l}\cos x\ne 0\iff x\ne \frac{\pi }{2}+k\pi \\ \sin x\ne 0\iff x\ne k\pi \end{array}\right.\iff x\ne k\frac{\pi }{2}$

Ta có $y={\tan }^{2}x+{\cot }^{2}x+3\left(\tan x+\cot x\right)-1$

   $={\tan }^{2}x+2\tan x\cot x+{\cot }^{2}x+3\left(\tan x+\cot x\right)-3={\left(\tan x+\cot x\right)}^2+3\left(\tan x+\cot x\right)-3$        

Đặt $\tan x+\cot x=t=\frac{2}{\sin 2x}\Rightarrow \left|t\right|\ge 2$

Ta có $y=t^2+3t-3.$ Cho $y=0\iff \left[\begin{array}{l}{t}_1=\frac{-3-\sqrt{21}}{2}\\ t_2=\frac{-3+\sqrt{21}}{2}\end{array}\right.$

Vậy $\min y=-5\iff t=\frac{2}{\sin 2x}=-2\iff \sin 2x=-1\iff 2x=\frac{-\pi }{2}+k2\pi \iff x=-\frac{\pi }{4}+k\pi ;\max y=\varnothing$

Đáp án B

Hàm số $y={\cos }^{4}x+{\sin }^{4}x$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

Ta có $y={\cos }^{4}x+{\sin }^{4}x={\left(1-{\sin }^{2}x\right)}^2+{\sin }^{4}x=1-2{\sin }^{2}x+{\sin }^{4}x+{\sin }^{4}x=2{\sin }^{4}x-2{\sin }^{2}x+1$ .

$y=2{\left({\sin }^{2}x\right)}^2-2{\sin }^{2}x+1=2\left({\left({\sin }^{2}x\right)}^2-{\sin }^{2}x+\frac{1}{2}\right)=2{\left({\sin }^{2}x-\frac{1}{2}\right)}^2+\frac{1}{2}$.

$-1\le \sin x\le 1\iff 0\le {\sin }^{2}x\le 1\iff \frac{-1}{2}\le {\sin }^{2}x-\frac{1}{2}\le \frac{1}{2}\iff 0\le {\left({\sin }^{2}x-\frac{1}{2}\right)}^2\le \frac{1}{4}\iff 0\le 2{\left({\sin }^{2}x-\frac{1}{2}\right)}^2\le \frac{1}{2}$.

$\iff \frac{1}{2}\le 2{\left({\sin }^{2}x-\frac{1}{2}\right)}^2+\frac{1}{2}\le 1\iff \frac{1}{2}\le y\le 1$

Vậy $\min y=\frac{1}{2}\iff {\sin }^{2}x=\frac{1}{2}\iff \left[\begin{array}{l}\sin x=\frac{\sqrt{2}}{2}\iff x=\frac{\pi }{4}+k2\pi \\ \sin x=\frac{-\sqrt{2}}{2}\iff x=\frac{-\pi }{4}+k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$;

$\max y=1\iff \sin x=0\iff x=k\pi \left(k\in \mathbb{Z}\right)$

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Đáp án D

Ta có    $\sin 2x+4{\cos }^{2}x+1=\sin 2x+4\frac{1+\cos 2x}{2}+1=\sin 2x+2\cos 2x+3>0$ $\forall x\in ℝ\iff D=ℝ$

$\frac{3\sin 2x+\cos 2x}{\sin 2x+2\cos 2x+3}\le m+1\iff \left(3-y\right)\sin 2x+\left(1-2y\right)\cos 2x=3y$

$\iff 9y^2={\left[\left(3-y\right)\sin 2x+\left(1-2y\right)\cos 2x\right]}^2$

Sử dụng bất đẳng thức Bunhiacopxki ta có

${\left(3-y\right)}^2+{\left(1-2y\right)}^2\ge 9y^2\iff 2y^2+5y-5\le 0\iff \frac{-5-3\sqrt{5}}{4}\le y\le \frac{-5+3\sqrt{5}}{4}$.

Vậy $\max y=\frac{-5+3\sqrt{5}}{4}\Rightarrow \frac{-5+3\sqrt{5}}{4}\le m+1\iff m\ge \frac{3\sqrt{5}-9}{4}$.

Đáp án D

Hàm số $y=\frac{{\cos }^{2}x+\sin x.\cos x}{1+{\sin }^{2}x}$ có nghĩa $\forall x\in ℝ\iff D=ℝ$.

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

Có $y=\frac{1+\cos 2x+\sin 2x}{3-\cos 2x}\iff 3y-y\cos 2x=1+\cos 2x+\sin 2x\iff \left(1+y\right)\cos 2x+\sin 2x=3y-1$

$\iff {\left(3y-1\right)}^2={\left(\left(1+y\right)\cos 2x+\sin 2x\right)}^2$

Sử dụng bất đẳng thức Bunhiacopxki ta có

${\left(1+y\right)}^2+1\ge {\left(3y-1\right)}^2\iff 1+2y+y^2+1\ge 9y^2-6y+1\iff 8y^2-8y-1\le 0\iff \frac{2-\sqrt{6}}{4}\le y\le \frac{2+\sqrt{6}}{4}$.

Vậy $\min y=\frac{2-\sqrt{6}}{4};\max y=\frac{2+\sqrt{6}}{4}$.

Đáp án B

Theo bài ra ${\cos }^{2}x+{\cos }^{2}y+{\cos }^{2}z=1$

Áp dụng bất đẳng thức Bunhiacopxki, ta có

$\sqrt{1+{\cos }^{2}x}+\sqrt{1+{\cos }^{2}y}+\sqrt{1+{\cos }^{2}z}\le \sqrt{1^2+1^2+1^2}.\sqrt{1+{\cos }^{2}x+1+{\cos }^{2}y+1+{\cos }^{2}z}$

.Vậy $\max y=2\sqrt{3}$.