20 câu trắc nghiệm Toán 11 Chân trời sáng tạo Giá trị lượng giác của một góc lượng giác có đáp án

Ở góc phần tư thứ I thì ${\rm{sin\alpha > 0, cos\alpha > 0, tan\alpha > 0, cot\alpha > 0}}$

Đáp án cần chọn là: A

Góc α thuộc góc phần tư thứ II thì${\rm{sin\alpha > 0, cos\alpha < 0,t an\alpha < 0, cot\alpha < 0}}$

Đáp án cần chọn là: C

Chọn đáp án A.

Đáp án cần chọn là: A

$\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < \alpha < \pi }} \Leftrightarrow - \frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < }}\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{\alpha < 0}}$ nên $\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{\alpha }}$ thuộc góc phần tư thứ IV.

Chỉ I đúng; II và III sai.

Chọn đáp án C.

Đáp án cần chọn là: C

${\rm{\pi < \alpha < }}\frac{{{\rm{3\pi }}}}{{\rm{2}}} \Leftrightarrow {\rm{0 < }}\frac{{{\rm{3\pi }}}}{{\rm{2}}} - {\rm{\alpha < }}\frac{{\rm{\pi }}}{{\rm{2}}}$ nên$\frac{{{\rm{3\pi }}}}{{\rm{2}}} - {\rm{\alpha }}$thuộc góc phần tư thứ I.

$\Rightarrow {\rm{tan}}\left( {\frac{{{\rm{3\pi }}}}{{\rm{2}}} - {\rm{\alpha }}} \right){\rm{ > 0}}$

Chọn đáp án B.

Đáp án cần chọn là: B

$\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < \alpha < \pi }} \Leftrightarrow 0 < - \frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + \alpha < }}\frac{{\rm{\pi }}}{{\rm{2}}}$ nên$- \frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + \alpha }}$thuộc góc phần tư thứ I.

$\Rightarrow \cos \left( { - \frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + \alpha }}} \right) > 0$

$\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < \alpha < \pi }} \Leftrightarrow 0 < \pi - {\rm{\alpha < }}\frac{{\rm{\pi }}}{{\rm{2}}}$$\pi - {\rm{\alpha }}$ </>

</>$\pi - {\rm{\alpha }}$thuộc góc phần tư thứ I.

$\Rightarrow \tan \left( {{\rm{\pi }} - {\rm{\alpha }}} \right) > 0$

$${\rm{M = cos}}\left( { - \frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + \alpha }}} \right){\rm{.tan}}\left( {{\rm{\pi }} - {\rm{\alpha }}} \right){\rm{ > 0}}$$

Chọn đáp án B.

Đáp án cần chọn là: B

Ta có:${\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\alpha = 1}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\alpha = 1}} - \frac{{\rm{5}}}{{\rm{9}}}{\rm{ = }}\frac{{\rm{4}}}{{\rm{9}}} \Leftrightarrow {\rm{sin\alpha = \pm }}\frac{{\rm{2}}}{{\rm{3}}}$

Do${\rm{\pi < \alpha < }}\frac{{{\rm{3\pi }}}}{{\rm{2}}}$nên ${\rm{sin\alpha < 0}}$. Vậy${\rm{sin\alpha = }} - \frac{2}{3}$

Chọn đáp án D.

Đáp án cần chọn là: D

Ta có:${\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\alpha = 1}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\alpha = 1}} - \frac{{\rm{9}}}{{{\rm{25}}}}{\rm{ = }}\frac{{{\rm{16}}}}{{{\rm{25}}}} \Leftrightarrow {\rm{sin\alpha = \pm }}\frac{{\rm{4}}}{{\rm{5}}}$

Do ${\rm{0 < \alpha < }}\frac{\pi }{2}$ nên ${\rm{sin\alpha > 0}}$. Do đó${\rm{sin\alpha = }}\frac{{\rm{4}}}{{\rm{5}}} \Rightarrow {\rm{tan\alpha = }}\frac{{{\rm{sin\alpha }}}}{{{\rm{cos\alpha }}}}{\rm{ = }}\frac{{\rm{3}}}{{\rm{4}}}$</>

Chọn đáp án A.

Đáp án cần chọn là: A

Ta có:${\rm{tan\alpha }}{\rm{.cot\alpha = 1}} \Rightarrow {\rm{cot\alpha = }}\frac{{\rm{1}}}{{\frac{{\rm{3}}}{{\rm{4}}}}}{\rm{ = }}\frac{{\rm{4}}}{{\rm{3}}}$

Ta lại có:$\frac{{\rm{1}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\alpha }}}}{\rm{ = 1 + co}}{{\rm{t}}^{\rm{2}}}{\rm{\alpha = 1 + }}{\left( {\frac{{\rm{4}}}{{\rm{3}}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{25}}}}{{\rm{9}}} \Rightarrow {\rm{sin\alpha = \pm }}\frac{{\rm{3}}}{{\rm{5}}}$

Do$- \pi {\rm{ < \alpha < }} - \frac{\pi }{2}$nên ${\rm{sin\alpha }} < 0$. Vậy${\rm{sin\alpha = }} - \frac{3}{5}$

Chọn đáp án B.

Đáp án cần chọn là: B

$${\rm{D = sin}}\left( {\frac{{{\rm{5\pi }}}}{{\rm{2}}} - {\rm{\alpha }}} \right){\rm{ + cos}}\left( {{\rm{13\pi + \alpha }}} \right) - {\rm{3sin}}\left( {{\rm{\alpha }} - {\rm{5\pi }}} \right)$$

$${\rm{ = sin}}\left( {\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{\alpha }}} \right){\rm{ + cos}}\left( {{\rm{\pi + \alpha }}} \right){\rm{ + 3sin}}\left( {{\rm{\pi }} - {\rm{\alpha }}} \right)$$

$${\rm{ = cos\alpha }} - {\rm{cos\alpha + 3sin\alpha = 3sin\alpha = 3}}{\rm{.}}\frac{{\rm{3}}}{{\rm{5}}}{\rm{ = }}\frac{{\rm{9}}}{{\rm{5}}}$$

Chọn đáp án A.

Đáp án cần chọn là: A

Ta có: ${\rm{cos\alpha = }} - {\rm{cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{5\pi }}}}{{\rm{5}}}} \right){\rm{; cos}}\left( {{\rm{\alpha + }}\frac{{\rm{\pi }}}{{\rm{5}}}} \right){\rm{ = }} - {\rm{cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{6\pi }}}}{{\rm{5}}}} \right){\rm{; cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{2\pi }}}}{{\rm{5}}}} \right){\rm{ = }} - {\rm{cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{7\pi }}}}{{\rm{5}}}} \right){\rm{;}}$ ${\rm{cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{3\pi }}}}{{\rm{5}}}} \right){\rm{ = }} - {\rm{cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{8\pi }}}}{{\rm{5}}}} \right){\rm{; cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{4\pi }}}}{{\rm{5}}}} \right){\rm{ = }} - {\rm{cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{9\pi }}}}{{\rm{5}}}} \right)$

$$\Rightarrow {\mathop{\rm c}\nolimits} {\rm{os\alpha + cos}}\left( {{\rm{\alpha + }}\frac{{\rm{\pi }}}{{\rm{5}}}} \right){\rm{ + cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{2\pi }}}}{{\rm{5}}}} \right){\rm{ + }}...{\rm{ + cos}}\left( {{\rm{\alpha + }}\frac{{{\rm{9\pi }}}}{{\rm{5}}}} \right){\rm{ = 0}}$$

Chọn đáp án C.

Đáp án cần chọn là: C

$${\rm{P = }}\frac{{{\rm{sin}}\left( { - {\rm{23}}{{\rm{4}}^{\rm{0}}}} \right) - {\rm{cos21}}{{\rm{6}}^{\rm{0}}}}}{{{\rm{sin14}}{{\rm{4}}^{\rm{0}}} - {\rm{cos12}}{{\rm{6}}^{\rm{0}}}}}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}} = \frac{{ - sin\left( {{{180}^0} + {{54}^0}} \right) - cos\left( {{{180}^0} + {{36}^0}} \right)}}{{sin\left( {{{180}^0} - {{36}^0}} \right) - cos\left( {{{90}^0} + {{36}^0}} \right)}}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}}$$

$$= \frac{{sin{{54}^0} + cos{{36}^0}}}{{sin{{36}^0} + sin{{36}^0}}}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}} = \frac{{sin\left( {{{90}^0} - {{36}^0}} \right) + cos{{36}^0}}}{{sin{{36}^0} + sin{{36}^0}}}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}}$$

$$= \frac{{cos{{36}^0} + cos{{36}^0}}}{{sin{{36}^0} + sin{{36}^0}}}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}} = 2\frac{{cos{{36}^0}}}{{sin{{36}^0}}}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}} = 2\cot {36^0}{\rm{.tan3}}{{\rm{6}}^{\rm{0}}} = 2$$

Chọn đáp án B.

Đáp án cần chọn là: B

$${\rm{A = co}}{{\rm{s}}^{\rm{2}}}{\rm{xco}}{{\rm{t}}^{\rm{2}}}{\rm{x + 3co}}{{\rm{s}}^{\rm{2}}}{\rm{x}} - {\rm{co}}{{\rm{t}}^{\rm{2}}}{\rm{x + 2si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}$$

$${\rm{ = co}}{{\rm{s}}^{\rm{2}}}{\rm{xco}}{{\rm{t}}^{\rm{2}}}{\rm{x}} - {\rm{co}}{{\rm{t}}^{\rm{2}}}{\rm{x + 2si}}{{\rm{n}}^{\rm{2}}}{\rm{x + 2co}}{{\rm{s}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}$$

$${\rm{ = }} - {\rm{co}}{{\rm{t}}^{\rm{2}}}{\rm{x}}\left( {{\rm{1}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right){\rm{ + 2}}\left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}$$

$${\rm{ = }} - {\rm{co}}{{\rm{t}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{x + 2 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x = }} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x + 2 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x = 2}}$$

Chọn đáp án A.

Đáp án cần chọn là: A

$${\rm{B = }}\frac{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x - si}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}} - {\rm{co}}{{\rm{t}}^{\rm{2}}}{\rm{xco}}{{\rm{t}}^{\rm{2}}}{\rm{y = }}\frac{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{y}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{y}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}$$

$${\rm{B = }}\frac{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{y}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{\rm{ = }}\frac{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x(1}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{y)}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}$$

$${\rm{B = }}\frac{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{\rm{ = }}\frac{{{\rm{ - si}}{{\rm{n}}^{\rm{2}}}{\rm{y(1}} - {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x)}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{\rm{ = }}\frac{{ - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xsi}}{{\rm{n}}^{\rm{2}}}{\rm{y}}}}{\rm{ = }} - {\rm{1}}$$

Chọn đáp án D.

Đáp án cần chọn là: D

Ta có:

$${\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x = 1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}} \Rightarrow {\rm{3si}}{{\rm{n}}^{\rm{4}}}{\rm{x}} - {\rm{co}}{{\rm{s}}^{\rm{4}}}{\rm{x = 3si}}{{\rm{n}}^{\rm{4}}}{\rm{x}} - {\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}}$$

$$\Rightarrow {\rm{3si}}{{\rm{n}}^{\rm{4}}}{\rm{x}} - {\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}} \Leftrightarrow {\rm{2si}}{{\rm{n}}^{\rm{4}}}{\rm{x + 2si}}{{\rm{n}}^{\rm{2}}}{\rm{x}} - \frac{{\rm{3}}}{{\rm{2}}}{\rm{ = 0}}$$

$$\Leftrightarrow \left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}} - \frac{{\rm{1}}}{{\rm{2}}}} \right)\left( {{\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{x + 3}}} \right){\rm{ = 0}} \Leftrightarrow {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x = }}\frac{{\rm{1}}}{{\rm{2}}}$$

Vậy${\rm{si}}{{\rm{n}}^{\rm{4}}}{\rm{x + 3co}}{{\rm{s}}^{\rm{4}}}{\rm{x = si}}{{\rm{n}}^{\rm{4}}}{\rm{x + 3}}{\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + 3}}{\left( {{\rm{1}} - \frac{{\rm{1}}}{{\rm{2}}}} \right)^{\rm{2}}}{\rm{ = 1}}$

Chọn đáp án C.

Đáp án cần chọn là: C

$${\rm{A = sin}}\left( {{\rm{x + }}\frac{{{\rm{85\pi }}}}{{\rm{2}}}} \right){\rm{ + cos}}\left( {{\rm{2023\pi + x}}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{x + 33\pi }}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{x}} - \frac{{{\rm{5\pi }}}}{{\rm{2}}}} \right)$$

${\rm{A = sin}}\left( {{\rm{x + 42\pi + }}\frac{\pi }{2}} \right){\rm{ + cos}}\left( {{\rm{2022\pi + \pi + x}}} \right){\rm{ + sin2}}\left( {{\rm{32\pi + \pi + x}}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{x}} - {\rm{2\pi }} - \frac{\pi }{2}} \right)$${\rm{A = sin}}\left( {{\rm{x + }}\frac{\pi }{2}} \right){\rm{ + cos}}\left( {{\rm{\pi + x}}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{\pi + x}}} \right){\rm{ + si}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{x}} - \frac{\pi }{2}} \right)$

$${\rm{A = cosx}} - {\rm{cosx + }}{\left( { - {\rm{sinx}}} \right)^{\rm{2}}}{\rm{ + }}{\left( { - {\rm{cosx}}} \right)^{\rm{2}}}{\rm{ = si}}{{\rm{n}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{x = 1}}$$

Chọn đáp án C.

Đáp án cần chọn là: C

$${\rm{si}}{{\rm{n}}^{\rm{4}}}{\rm{x + co}}{{\rm{s}}^{\rm{4}}}{\rm{x = }}{\left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x = 1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x}}$$

$$\Rightarrow {\rm{co}}{{\rm{s}}^{\rm{8}}}{\rm{x + si}}{{\rm{n}}^{\rm{8}}}{\rm{x = }}{\left( {{\rm{si}}{{\rm{n}}^{\rm{4}}}{\rm{x + co}}{{\rm{s}}^{\rm{4}}}{\rm{x}}} \right)^{\rm{2}}} - {\rm{2si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x}}$$

$${\rm{ = }}{\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}} - {\rm{2si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x = 2si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x}} - {\rm{4si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x + 1}}$$

$${\rm{M = 2}}{\left( {{\rm{si}}{{\rm{n}}^{\rm{4}}}{\rm{x + co}}{{\rm{s}}^{\rm{4}}}{\rm{x + si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}} - \left( {{\rm{co}}{{\rm{s}}^{\rm{8}}}{\rm{x + si}}{{\rm{n}}^{\rm{8}}}{\rm{x}}} \right)$$

$${\rm{M = 2}}{\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x + si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}} - \left( {{\rm{2si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x}} - {\rm{4si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x + 1}}} \right)$$

$${\rm{M = 2}}{\left( {1 - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right)^{\rm{2}}} - \left( {{\rm{2si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x}} - {\rm{4si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x + 1}}} \right)$$

$${\rm{M = 2}}\left( {{\rm{si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x + 1}}} \right) - \left( {{\rm{2si}}{{\rm{n}}^{\rm{4}}}{\rm{xco}}{{\rm{s}}^{\rm{4}}}{\rm{x}} - {\rm{4si}}{{\rm{n}}^{\rm{2}}}{\rm{xco}}{{\rm{s}}^{\rm{2}}}{\rm{x + 1}}} \right){\rm{ = 1}}$$

Chọn đáp án A.

Đáp án cần chọn là: A

$${\rm{C = 6co}}{{\rm{s}}^{\rm{2}}}{\rm{x + 5si}}{{\rm{n}}^{\rm{2}}}{\rm{x = 6}}\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right){\rm{ + 5si}}{{\rm{n}}^{\rm{2}}}{\rm{x = 6}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}$$

Ta có: $0 \le sin2x \le 1 \Leftrightarrow 0 \ge - sin2x \ge - 1 \Leftrightarrow 6 \ge 6 - sin2x \ge 5 \Leftrightarrow 6 \ge C \ge 5$

Vậy giá trị lớn nhất của biểu thức C là 6.

Chọn đáp án C.

$${\rm{F = co}}{{\rm{s}}^{\rm{2}}}{\rm{x + 2sinx + 2 = 1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x + 2sinx + 2 = }} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x + 2sinx + 3}}$$

$${\rm{ = }} - \left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}} - {\rm{2sinx + 1}}} \right){\rm{ + 4 = }} - {\left( {{\rm{sinx}} - {\rm{1}}} \right)^{\rm{2}}}{\rm{ + 4}}$$

$$- 1 \le sinx \le 1 \Rightarrow - 2 \le sinx - 1 \le 0 \Rightarrow 0 \le (sinx - 1)2 \le 4$$

$$\Rightarrow - 4 \le - (sinx - 1)2 \le 0 \Rightarrow 0 \le F \le 4$$

Vậy giá trị nhỏ nhất của biểu thức F là 0.

Chọn đáp án B.

Đáp án cần chọn là: B

$${\rm{K = }}\frac{{{\rm{1 + ta}}{{\rm{n}}^{\rm{3}}}{\rm{x}}}}{{{{\left( {{\rm{1 + tanx}}} \right)}^{\rm{3}}}}}{\rm{ = }}\frac{{\left( {{\rm{1 + tanx}}} \right)\left( {{\rm{1}} - {\rm{tanx + ta}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right)}}{{{{\left( {{\rm{1 + tanx}}} \right)}^{\rm{3}}}}}{\rm{ = }}\frac{{{\rm{1}} - {\rm{tanx + ta}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}{{{\rm{1 + 2tanx + ta}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}$$

Đặt${\rm{tanx = t}}\left( {{\rm{t}} \ne - 1} \right)$

$$\Rightarrow {\rm{K = }}\frac{{{\rm{1}} - {\rm{t + }}{{\rm{t}}^{\rm{2}}}}}{{{\rm{1 + 2t + }}{{\rm{t}}^{\rm{2}}}}} \Rightarrow \left( {{\rm{K}} - {\rm{1}}} \right){{\rm{t}}^{\rm{2}}}{\rm{ + }}\left( {{\rm{2K + 1}}} \right){\rm{t + K}} - {\rm{1 = 0}}$$

Với K = 1 thì phương trình có nghiệm t = 1

Với ${\rm{K}} \ne 1$ thì phương trình phải có nghiệm${\rm{t}} \ne - 1$

$\Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{\Delta \ge 0}\\{\left( {K - 1} \right){{\left( { - 1} \right)}^2} + \left( {2K + 1} \right)\left( { - 1} \right) + K - 1 \ne 0}\end{array}} \right.$

$\Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{{{\left( {2K + 1} \right)}^2} - 4{{\left( {K - 1} \right)}^2}{\rm{ = }}12K - 3 \ge 0}\\{K \ne - 4}\end{array}} \right.$

$\Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{K \ge \frac{1}{4}}\\{K \ne - 4}\end{array}} \right.$

Vậy giá trị nhỏ nhất của biểu thức K là$\frac{1}{4}$

Chọn đáp án D.

Đáp án cần chọn là: D