20 câu trắc nghiệm Toán 11 Chân trời sáng tạo Phương trình lượng giác cơ bản có đáp án

Áp dụng công thức:$sinx = sina \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{x = a + k2\pi }\\{x = \pi - a + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)$

$$sinx = sin\frac{\pi }{3} \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{x = }}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + k\pi }}}\\{{\rm{x = }}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{ + k\pi }}}\end{array}} \right.\left( {k \in \mathbb{Z}} \right)$$

Đáp án cần chọn là: D

$${\rm{cosx = cos}}\frac{{\rm{\pi }}}{{{\rm{12}}}} \Rightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{x = }}\frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{ + k2\pi }}}\\{{\rm{x = }} - \frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{ + k2\pi }}}\end{array}} \right.\left( {k \in \mathbb{Z}} \right)$$

Đáp án cần chọn là: B

Ta có: ${\rm{cosx = }} - \frac{{\rm{1}}}{{\rm{2}}} \Leftrightarrow {\rm{cosx = cos}}\left( {\frac{{{\rm{2\pi }}}}{{\rm{3}}}} \right) \Leftrightarrow {\rm{x = \pm }}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{ + k2\pi }}\left( {{\rm{k}} \in {\rm{Z}}} \right)$

Đáp án cần chọn là: A

Lời giải

$$\sqrt {\rm{3}} {\rm{tan2x}} - {\rm{3 = 0}} \Leftrightarrow {\rm{tan2x = }}\sqrt {\rm{3}} \Leftrightarrow {\rm{2x = }}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + k\pi }} \Leftrightarrow {\rm{x = }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{ + k}}\frac{{\rm{\pi }}}{{\rm{2}}}\left( {{\rm{k}} \in \mathbb{Z}} \right)$$

Đáp án cần chọn là: D

Ta có:${\rm{2sinx}} - {\rm{3 = 0}} \Leftrightarrow {\rm{sinx = }}\frac{{\rm{3}}}{{\rm{2}}} > 1$ nên phương trình vô nghiệm.

Đáp án cần chọn là: A

$${\rm{sin}}\left( {{\rm{x + }}\frac{{\rm{\pi }}}{{\rm{4}}}} \right){\rm{ = 1}} \Leftrightarrow {\rm{x + }}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + k2\pi }} \Leftrightarrow {\rm{x = }}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + k2\pi ,}}\,\,{\rm{k}} \in \mathbb{Z}$$

$${\rm{x}} \in \left[ {{\rm{\pi ; 2\pi }}} \right] \Rightarrow {\rm{\pi }} \le \frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + k2\pi }} \le {\rm{2\pi }} \Leftrightarrow \frac{{\rm{3}}}{{\rm{8}}} \le {\rm{k}} \le \frac{{\rm{7}}}{{\rm{8}}}$$

Mà${\rm{k}} \in \mathbb{Z} \Rightarrow {\rm{k}} \in \emptyset$phương trình không có nghiệm trên đoạn$\left[ {{\rm{\pi ; 2\pi }}} \right]$

Đáp án cần chọn là: A

$$\frac{{{\rm{sin3x}}}}{{{\rm{cosx + 1}}}}{\rm{ = 0}} \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{cosx \ne - 1}\\{sin3x = 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{x \ne \pi + k2\pi }\\{x = \frac{{k\pi }}{3}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{x = \pm \frac{\pi }{3} + k\pi }\\{x = k2\pi }\end{array},k \in \mathbb{Z}} \right.$$

Vì ${\rm{x}} \in \left[ {{\rm{2\pi ; 4\pi }}} \right] \Rightarrow 2{\rm{\pi }} \le {\rm{x}} \le 4{\rm{\pi }}$

Xét:${\rm{2\pi }} \le \frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + k\pi }} \le {\rm{4\pi }} \Rightarrow \frac{{{\rm{5\pi }}}}{{\rm{3}}} \le {\rm{k\pi }} \le \frac{{{\rm{11\pi }}}}{{\rm{3}}} \Rightarrow \frac{5}{3} \le {\rm{k}} \le \frac{{11}}{3}$

$$\Rightarrow {\rm{k = 2; k = 3}} \Rightarrow {\rm{x = }}\frac{{{\rm{7\pi }}}}{{\rm{3}}}{\rm{; x = }}\frac{{{\rm{10\pi }}}}{{\rm{3}}}$$

Xét:${\rm{2\pi }} \le - \frac{{\rm{\pi }}}{{\rm{3}}}{\rm{ + k\pi }} \le {\rm{4\pi }} \Rightarrow \frac{{{\rm{7\pi }}}}{{\rm{3}}} \le {\rm{k\pi }} \le \frac{{{\rm{13\pi }}}}{{\rm{3}}} \Rightarrow \frac{7}{3} \le {\rm{k}} \le \frac{{13}}{3}$

$$\Rightarrow {\rm{k = 3; k = 4}} \Rightarrow {\rm{x = }}\frac{{{\rm{8\pi }}}}{{\rm{3}}}{\rm{; x = }}\frac{{{\rm{11\pi }}}}{{\rm{3}}}$$

Xét:${\rm{2\pi }} \le {\rm{k2\pi }} \le {\rm{4\pi }} \Rightarrow 1 \le {\rm{k}} \le 2 \Rightarrow {\rm{k = 1; k = 2}} \Rightarrow {\rm{x = 2\pi ; x = 4\pi }}$

Đáp án cần chọn là: A

$${\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{4x = 1}} \Leftrightarrow {\mathop{\rm c}\nolimits} {\rm{o}}{{\rm{s}}^{\rm{2}}}{\rm{4x = co}}{{\rm{s}}^{\rm{2}}}{\rm{x}} \Leftrightarrow {\mathop{\rm c}\nolimits} {\rm{os8x = cos2x}} \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{x\,{\rm{ = }}\,\frac{{k\pi }}{3}}\\{x\,{\rm{ = }}\,\frac{{l\pi }}{5}}\end{array};\left( {k,l \in \mathbb{Z}} \right)} \right.$$

Mà $x \in \left( {0;\pi } \right) \Rightarrow \left[ {\begin{array}{cccccccccccccccccccc}{0 < k\frac{\pi }{3} < \pi }\\{0 < k\frac{\pi }{5} < \pi }\end{array}} \right. \Rightarrow \left[ {\begin{array}{cccccccccccccccccccc}{0 < k < 3 \Rightarrow k\,{\rm{ = }}1;\,k\,{\rm{ = }}\,2}\\{0 < l < 5 \Rightarrow l = 1;l{\rm{ = }}2;\,l\,{\rm{ = }}\,4}\end{array}} \right.$

$$\Rightarrow {\rm{x}} \in \left\{ {\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{; }}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{; }}\frac{{\rm{\pi }}}{{\rm{5}}}{\rm{; }}\frac{{{\rm{2\pi }}}}{{\rm{5}}}{\rm{; }}\frac{{{\rm{3\pi }}}}{{\rm{5}}}{\rm{; }}\frac{{{\rm{4\pi }}}}{{\rm{5}}}} \right\}$$

Suy ra nghiệm lớn nhất là ${{\rm{x}}_{\rm{0}}}{\rm{ = }}\frac{{{\rm{4\pi }}}}{{\rm{5}}} \Leftrightarrow {\rm{P = 41}}$Đáp án cần chọn là: B

Điều kiện:$\left\{ {\begin{array}{cccccccccccccccccccc}{cos3x \ne 0}\\{cosx \ne 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{cccccccccccccccccccc}{x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}}\\{x \ne \frac{\pi }{2} + k\pi }\end{array}} \right. \Leftrightarrow x \ne \frac{\pi }{6} + \frac{{k\pi }}{3},k \in \mathbb{Z}$

$${\rm{tan3x + tanx = 0}} \Leftrightarrow {\rm{tan3x = tan}}\left( { - {\rm{x}}} \right) \Leftrightarrow {\rm{3x = }} - {\rm{x + k\pi }} \Leftrightarrow {\rm{x = }}\frac{{l{\rm{\pi }}}}{{\rm{4}}}{\rm{,}}\,\,{\rm{k}} \in \mathbb{Z}$$

So sánh điều kiện suy ra nghiệm của phương trình là${\rm{x = k\pi ;}}\,\,{\rm{x = }}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + }}\frac{{{\rm{k\pi }}}}{{\rm{2}}}$

Đáp án cần chọn là: C

Điều kiện : ${\rm{sin2x}} \ne 1 \Leftrightarrow {\rm{x}} \ne \frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + k\pi ,}}\,\,{\rm{k}} \in \mathbb{Z}$

$$\frac{{{\rm{cos2x}}}}{{{\rm{1}} - {\rm{sin2x}}}}{\rm{ = 0}} \Leftrightarrow {\rm{cos2x = 0}} \Leftrightarrow {\rm{x = }}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + }}\frac{{{\rm{k\pi }}}}{{\rm{2}}}{\rm{,}}\,\,{\rm{k}} \in \mathbb{Z}$$

So sánh điều kiện$\Rightarrow {\rm{x = }} - \frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + k\pi ,}}\,\,{\rm{k}} \in \mathbb{Z}$

$${\rm{x}} \in \left[ {{\rm{0; 2\pi }}} \right] \Rightarrow {\rm{x}} \in \left\{ {\frac{{{\rm{3\pi }}}}{{\rm{4}}}{\rm{; }}\frac{{{\rm{7\pi }}}}{{\rm{4}}}} \right\} \Rightarrow {\rm{S = }}\frac{{{\rm{5\pi }}}}{{\rm{2}}}$$

Đáp án cần chọn là: D

Điều kiện :${\rm{4}} - {{\rm{x}}^{\rm{2}}} \ge 0 \Leftrightarrow {\rm{x}} \in \left[ { - 2;2} \right]$

$$\sqrt {{\rm{4}} - {{\rm{x}}^{\rm{2}}}} {\rm{sin2x = 0}} \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{4 - {x^2} = 0}\\{\sin 2x = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{x = \pm 2}\\{x = \frac{{k\pi }}{2},k \in \mathbb{Z}}\end{array}} \right.$$

So sánh điều kiện :${\rm{x}} \in \left[ { - 2;2} \right] \Leftrightarrow - 2 \le \frac{{{\rm{k\pi }}}}{{\rm{2}}} \le 2 \Leftrightarrow - \frac{4}{{\rm{\pi }}} \le {\rm{k}} \le \frac{4}{{\rm{\pi }}}$

Vậy${\rm{x}} \in \left\{ { \pm 2;0; \pm \frac{{\rm{\pi }}}{{\rm{2}}}} \right\}$

Đáp án cần chọn là: B

Điều kiện:${\rm{sin2x}} \ne 0 \Leftrightarrow {\rm{x}} \ne \frac{{{\rm{k\pi }}}}{{\rm{2}}}{\rm{,}}\,\,{\rm{k}} \in \mathbb{Z}$

Cách 1 : ${\rm{cot2x}}{\rm{.cotx = 1}} \Leftrightarrow {\rm{cot2x = tan}}\left( {\frac{{\rm{\pi }}}{{\rm{2}}} - {\rm{x}}} \right) \Leftrightarrow {\rm{x = }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{ + }}\frac{{{\rm{k\pi }}}}{{\rm{3}}}{\rm{,}}\,\,{\rm{k}} \in \mathbb{Z}$

Cách 2 : ${\rm{cot2x}}{\rm{.cotx = 1}} \Leftrightarrow {\rm{cos2xcosx = sin2xsinx}} \Leftrightarrow {\rm{cos3x = 0}}$

$$\Leftrightarrow {\rm{x = }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{ + }}\frac{{{\rm{k\pi }}}}{{\rm{3}}}{\rm{,}}\,\,{\rm{k}} \in \mathbb{Z}$$

So sánh điều kiện$\Rightarrow {\rm{x = \pm }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{ + k\pi ,}}\,{\rm{k}} \in \mathbb{Z}$

Vì${\rm{x}} \in \left[ { - {\rm{\pi ; \pi }}} \right] \Rightarrow {\rm{x}} \in \left\{ {{\rm{ \pm }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{; \pm }}\frac{{{\rm{5\pi }}}}{{\rm{6}}}} \right\}$

Đáp án cần chọn là: C

$${\rm{cos}}\left( {\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{sinx}} - \frac{{{\rm{2\pi }}}}{{\rm{3}}}} \right){\rm{ = 1}} \Leftrightarrow \frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{sinx}} - \frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{ = k2\pi }} \Leftrightarrow {\mathop{\rm s}\nolimits} {\rm{inx}} - {\rm{1 = 3k}}$$

Phương trình có nghiệm khi k = 0.

Khi đó$\sin {\rm{x = 1}} \Leftrightarrow {\rm{x = }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + k2\pi ,}}\,\,{\rm{k}} \in \mathbb{Z}$. Vậy S = 4.

Đáp án cần chọn là: A

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

$${\rm{pt}} \Leftrightarrow \left( {{\rm{1 + cosx}}} \right)\left( {{\rm{2sinx}} - {\rm{cosx}}} \right) - \left( {{\rm{1}} - {\rm{cosx}}} \right)\left( {{\rm{1 + cosx}}} \right){\rm{ = 0}}$$

$${\rm{pt}} \Leftrightarrow \left( {{\rm{1 + cosx}}} \right)\left( {{\rm{2sinx}} - {\rm{cosx}} - {\rm{1 + cosx}}} \right){\rm{ = 0}}$$

$\Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{1 + cosx = 0}}}\\{{\rm{2sinx = 1}}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{c{\rm{osx = }} - 1}\\{sin{\rm{x = }}\frac{1}{2}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{x = \pi + k2\pi }}}\\{{\rm{x = }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{ + k2\pi }}}\\{{\rm{x = }}\frac{{{\rm{5\pi }}}}{{\rm{6}}}{\rm{ + k2\pi }}}\end{array}} \right.,k \in \mathbb{Z}$

$${\rm{x}} \in \left[ {{\rm{0; \pi }}} \right] \Rightarrow {\rm{x}} \in \left\{ {{\rm{\pi ; }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{; }}\frac{{{\rm{5\pi }}}}{{\rm{6}}}} \right\} \Rightarrow {\rm{S = 2\pi }}$$

Đáp án cần chọn là: D

+ Nếu m = 1 thì PTVN

+ Nếu $m \ne 1$thì $\left( {{\rm{m}} - {\rm{1}}} \right){\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x = m}} \Leftrightarrow {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x = }}\frac{{\rm{m}}}{{{\rm{m}} - {\rm{1}}}}$

PT có nghiệm $\Leftrightarrow 0 \le \frac{{\rm{m}}}{{{\rm{m}} - {\rm{1}}}} \le 1 \Leftrightarrow {\rm{m}} \le 0$

Đáp án cần chọn là: A

Vì ${\rm{cosx = 0}} \Rightarrow {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x = 1}} \Rightarrow \left( 1 \right)$ vô lý nên ${\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}} \ne 0$

Ta chia 2 vế của phương trình (1) cho ${\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}$, ta được:

$${\rm{mta}}{{\rm{n}}^{\rm{2}}}{\rm{x + 1 = }}\left( {{\rm{m}} - {\rm{1}}} \right)\left( {{\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right) \Leftrightarrow {\mathop{\rm t}\nolimits} {\rm{a}}{{\rm{n}}^{\rm{2}}}{\rm{x = m}} - {\rm{2}}\,\,\left( 2 \right)$$

(1) có nghiệm ${\rm{x}} \in \left( {{\rm{0; }}\frac{{\rm{\pi }}}{{\rm{4}}}} \right) \Leftrightarrow$ (2) có nghiệm${\rm{x}} \in \left( {{\rm{0; }}\frac{{\rm{\pi }}}{{\rm{4}}}} \right) \Leftrightarrow {\rm{m}} - {\rm{2}} \in \left( {0;1} \right) \Leftrightarrow 2 < {\rm{m}} < 3$

Đáp án cần chọn là: D

$${\rm{1 + sinx + cosx + sin2x + cos2x = 0}}$$

$$\Leftrightarrow \left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x + si}}{{\rm{n}}^{\rm{2}}}{\rm{x + 2sinx}}{\rm{.cosx}}} \right){\rm{ + }}\left( {{\rm{sinx + cosx}}} \right){\rm{ + }}\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right){\rm{ = 0}}$$

$$\Leftrightarrow {\left( {{\rm{sinx + cosx}}} \right)^{\rm{2}}}{\rm{ + }}\left( {{\rm{sinx + cosx}}} \right){\rm{ + }}\left( {{\rm{cosx}} - {\rm{sinx}}} \right){\rm{.}}\left( {{\rm{sinx + cosx}}} \right){\rm{ = 0}}$$

$\Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{sinx + cosx = 0}}}\\{{\rm{2cosx + 1 = 0}}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{sin\left( {x + \frac{\pi }{4}} \right){\rm{ = }}0}\\{cosx\,{\rm{ = }} - \frac{1}{2}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{x{\rm{ = }} - \frac{\pi }{4} + k\pi }\\{x{\rm{ = }} \pm \frac{{2\pi }}{3} + k2\pi }\end{array}} \right.,k \in \mathbb{Z}$

Vì${\rm{x}} \in \left[ {{\rm{0; 2\pi }}} \right] \Rightarrow {\rm{x}} \in \left\{ {\frac{{{\rm{3\pi }}}}{{\rm{4}}}{\rm{; }}\frac{{{\rm{7\pi }}}}{{\rm{4}}}{\rm{; }}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{; }}\frac{{{\rm{4\pi }}}}{{\rm{3}}}} \right\}$

Đáp án cần chọn là: C

$$\left( {{\rm{2sinx}} - {\rm{1}}} \right)\left( {{\rm{2sin2x + 1}}} \right){\rm{ = 3}} - {\rm{4co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}$$

$$\Leftrightarrow \left( {{\rm{2sinx}} - {\rm{1}}} \right)\left( {{\rm{2sin2x + 1}}} \right){\rm{ = 3}} - {\rm{4}}\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right)$$

$$\Leftrightarrow \left( {{\rm{2sinx}} - {\rm{1}}} \right)\left( {{\rm{2sin2x + 1}}} \right){\rm{ = 4si}}{{\rm{n}}^{\rm{2}}}{\rm{x}} - {\rm{1}}$$

$$\Leftrightarrow \left( {{\rm{2sinx}} - {\rm{1}}} \right)\left( {{\rm{2sin2x + 1}}} \right){\rm{ = (2sinx}} - {\rm{1)(2sinx + 1)}}$$

$$\Leftrightarrow \left( {{\rm{2sinx}} - {\rm{1}}} \right)\left( {{\rm{2sin2x + 1}} - {\rm{2sinx}} - {\rm{1}}} \right)$$

$\Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{2sinx - 1\,\,{\rm{ = 0}}}\\{sin2{\rm{x = si}}nx}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{2sinx - 1\,\,{\rm{ = 0}}}\\{2sinx.cos{\rm{x = s}}inx}\end{array}} \right.$

$\Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\mathop{\rm s}\nolimits} {\rm{inx = }}\frac{1}{2}}\\{{\mathop{\rm s}\nolimits} {\rm{inx = }}0}\\{cos{\rm{x = }}\frac{1}{2}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{x = }}\frac{\pi }{6} + k2\pi }\\{{\rm{x = }}\frac{{5\pi }}{6} + k2\pi }\\{{\rm{x = }}k\pi }\\{{\rm{x = }} \pm \frac{\pi }{3} + k2\pi }\end{array}} \right.,k \in \mathbb{Z}$

Vì ${\rm{x}} \in \left[ { - {\rm{\pi ; \pi }}} \right] \Rightarrow {\rm{x}} \in \left\{ { - {\rm{\pi ;}} - \frac{{\rm{\pi }}}{{\rm{3}}}{\rm{; 0; }}\frac{{\rm{\pi }}}{{\rm{6}}}{\rm{; }}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{; }}\frac{{{\rm{5\pi }}}}{{\rm{6}}}{\rm{; \pi }}} \right\} \Leftrightarrow {\rm{S = \pi }}$Đáp án cần chọn là: A

${\rm{h(t) = 29 + 3sin}}\frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{(t}} - {\rm{9)}}$nhỏ nhất khi và chỉ khi

$${\rm{sin}}\frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{(t}} - {\rm{9) = 1}} \Leftrightarrow \frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{(t}} - {\rm{9) = }} - \frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ + k2\pi }} \Leftrightarrow {\rm{t = 3 + 24k, k}} \in \mathbb{Z}$$

Vì$0 \le {\rm{t}} \le 24 \Rightarrow - \frac{1}{8} \le {\rm{k}} \le \frac{7}{8} \Rightarrow {\rm{k}} = 0 \Rightarrow {\rm{t}} = 3$

$${\rm{t = }}3 \Rightarrow {\rm{h(t) = }}29 + 3.\left( { - 1} \right){\rm{ = }}26$$

Đáp án cần chọn là: C

Từ đồ thị ta có$f(x)\,{\rm{ = }}x \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{x = a}} \in \left( { - \infty ;0} \right)}\\{{\rm{x = b}} \in \left( {0;1} \right)}\\{{\rm{x = 2}}}\end{array}} \right.$

Do đó${\rm{f(cosx + 1) = cosx + 1}} \Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{{\rm{cosx + 1 = a}} \in \left( { - \infty ;0} \right)}\\{{\rm{cosx + 1 = b}} \in \left( {0;1} \right)}\\{c{\rm{osx + 1 = 2}}}\end{array}} \right.$

$\Leftrightarrow \left[ {\begin{array}{cccccccccccccccccccc}{co{\rm{sx = a}} - {\rm{1 = }}{{\rm{t}}_{\rm{1}}} \in ( - \infty ; - 1)\,\,(VN)}\\{{\rm{cosx = b}} - {\rm{1 = }}{{\rm{t}}_{\rm{2}}} \in ( - 1;0)\,\,(1)}\\{{\rm{cosx = 1 }}(2)}\end{array}} \right.$

Dựa vào đường tròn lượng giác, phương trình có 3 nghiệm nằm trong$\left( {\frac{{ - {\rm{\pi }}}}{2};3{\rm{\pi }}} \right)$

Phương trình có 2 nghiệm nằm trong$\left( {\frac{{ - {\rm{\pi }}}}{2};3{\rm{\pi }}} \right)$

Vậy phương trình ban đầu có tất cả 5 nghiệm nằm trong $\left( {\frac{{ - {\rm{\pi }}}}{2};3{\rm{\pi }}} \right)$

Đáp án cần chọn là: C