10 câu Trắc nghiệm Ôn tập Toán 10 Chương 6 có đáp án (Vận dụng)

Đáp án B

$$\begin{gathered} \left\{\begin{array}{c}\left(a+1\right)\left(b+1\right)=2\\ a=\frac{1}{2}\end{array}\right.\iff \left\{\begin{array}{c}b=\frac{1}{3}\\ a=\frac{1}{2}\end{array}\right. \\ \tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x.\tan y} \\ =\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}=1\Rightarrow x+y=\frac{\pi }{4} \end{gathered}$$

Đáp án B

$$\begin{gathered} \left(1+\tan a\right)\left(1+\tan b\right)=1+\tan a+\tan b+\tan a\tan b \\ =1+\tan \left(a+b\right)\left(1-\tan a\tan b\right)+\tan a\tan b \\ =1+\tan \left({20}^0+{25}^0\right)\left(1-\tan {20}^0.\tan {25}^0\right)+\tan {20}^0.\tan {25}^0 \\ =1+\tan {45}^0\left(1-\tan {20}^0.\tan {45}^0\right)+\tan {20}^0.\tan {25}^0 \\ =1+1-\tan {20}^0.\tan {25}^0+\tan {20}^0.\tan {25}^0=2 \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} 0<\alpha <{90}^0\iff 0<\frac{\alpha }{2}<{45}^0\Rightarrow 0<\sin \frac{\alpha }{2}<\frac{\sqrt{2}}{2} \\ \iff 0<\sqrt{\frac{x-1}{2x}}<\frac{\sqrt{2}}{2}\iff x>0 \\ {\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}=1\Rightarrow \cos \frac{\alpha }{2}=\sqrt{1-{\sin }^2\frac{\alpha }{2}} \end{gathered}$$

vì $0<\frac{\alpha }{2}<{45}^0$

$$\begin{gathered} \iff \cos \frac{\alpha }{2}=\sqrt{\frac{x+1}{2x}}\Rightarrow \tan \frac{\alpha }{2}=\sqrt{\frac{x-1}{x+1}} \\ \tan \alpha =\frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}=\frac{2\sqrt{\frac{x-1}{x+1}}}{1-\frac{x-1}{x+1}}=\sqrt{x^2-1} \end{gathered}$$

Đáp án C

Vì $0<x<\frac{\pi }{2}$ nên $\cos \frac{x}{n}>0,\forall n\in N*$

$$\begin{gathered} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos x}}} \\ =\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{2}}} \\ =\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{4}}=\cos \frac{x}{8} \end{gathered}$$

Vậy n = 8

Đáp án A

Gọi R là bán kính đường tròn ngoại tiếp . Ta có:

$$\begin{gathered} \frac{1+\cos B}{1-\cos B}=\frac{2a+c}{2a-c} \\ \iff \frac{1+\cos B}{1-\cos B}=\frac{2.2R\sin A+2R\sin C}{2.2R\sin A-2R\sin C} \\ \iff \frac{1+\cos B}{1-\cos B}=\frac{2\sin A+\sin C}{2\sin A-\sin C} \\ \iff 2\sin A+2\sin A\cos B-\sin C-\sin C\cos B=2\sin A-2\sin A\cos B+\sin C-\sin C\cos B \\ \iff 4\sin A\cos B=2\sin C \\ \iff 4.\frac{a}{2R}.\frac{a^2+c^2-b^2}{2ac}=2.\frac{c}{2R} \\ \iff a^2+c^2-b^2=c^2 \\ \iff a=b \end{gathered}$$

Vậy $\Delta ABC$ cân tại C

Đáp án D

Vì $-1\le \cos x\le 1$, ta có:

${\sin }^{4}x+{\cos }^{7}x\le {\sin }^{4}x+{\cos }^{4}x=1-\frac{1}{2}{\sin }^{2}2x\le 1$

Vậy giá trị lớn nhất của biểu thức ${\sin }^{4}x+{\cos }^{7}x$ là 1