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2024年1月21日 星期日

113年大學學測-數學A詳解

113學年度學科能力測驗試題-數學A考科

第 壹 部 分 、 選 擇 ( 填 ) 題 ( 占 85 分 )
一 、 單 選 題 ( 占 30 分 )

解答:$$假設原始藥物殘留量為A,二小時後變為{1\over 2}A \Rightarrow f(h)=({1\over 2})^{h/2}A \\(1)\times: f(3)=({1\over 2})^{3/2}A \ne {1\over 3}A\\ (2)\bigcirc: f(4)=({1\over 2})^{2}A ={1\over 4}A\\ (3)\times:f(6)= ({1\over 2})^{3}A ={1\over 8}A \ne {1\over 6}A \\(4) \times:f(8) =({1\over 2})^{4}A={1\over 16}A \ne {1\over 8}A \\(5) \times: f(10) =({1\over 2})^{5}A ={1\over 32}A \ne {1\over 10}A\\ ,故選\bbox[red, 2pt]{(2)}$$


解答:$$假設\cases{O(0,0,0)\\ C(1,0,0)\\ A(0,1,0)\\ B(1,1,0)\\ D(0,0,1)\\ E(0,1,1) \\ F(1,1,1) \\G(1,0,1)} \Rightarrow \cases{\overrightarrow{AD}=(0,-1,1) \\ \overrightarrow{AG} =(1,-1,1)} \Rightarrow \vec u= \overrightarrow{AD} \times\overrightarrow{AG} = (0,1,1) \\ (1)\times: \overrightarrow{AE}= (0,0,1) \not \parallel \vec u \\(2) \times: \overrightarrow{BE}=(-1,0,1) \not \parallel \vec u \\(3) \times: \overrightarrow{CE} =(-1,1,1) \not \parallel \vec u \\(4) \times: \overrightarrow{DE}=(0,1,0)  \not \parallel \vec u \\ (5)  \bigcirc: \overrightarrow{OE} =(0,1,1) \parallel \vec u \\,故選\bbox[red, 2pt]{(5)}$$
解答:$$f(x)=a(x+7)(x+7-a)(x+7-2a) \Rightarrow f(0)=7a(7-a)(7-2a)\\ \Rightarrow \cases{a=-6 \Rightarrow f(0)=-42\times 13\times 19 \lt 0\\ a=-4 \Rightarrow f(0)=-28\times 11\times 15 \lt 0 \\ a=-2 \Rightarrow f(0)=-14\times 9\times 11 \lt 0\\ a=2 \Rightarrow f(0)=14\times 5\times 3 \gt 0\\ a=4 \Rightarrow f(0)=28\times 3\times (-1) \lt 0\\ a=6 \Rightarrow f(0)=42\times 1\times (-5) \lt 0} \Rightarrow 只有一個a(a=2)使得f(0)\gt 0\\,故選\bbox[red, 2pt]{(1)}$$
解答:$$ \sin(x+{\pi\over 6}) =  \sin x+\sin {\pi\over 6}  \Rightarrow \sin(x+{\pi\over 6})-\sin x=\sin{\pi \over 6} ={1\over 2}\\ \Rightarrow 2\sin({\pi \over 12})\cos(x+{\pi \over 12})={1\over 2} \Rightarrow 2\cdot {\sqrt 6-\sqrt 2\over 4} \cos(x+{\pi \over 12})={1\over 2} \\ \Rightarrow \cos(x+{\pi \over 12})={1\over \sqrt 6-\sqrt 2} ={\sqrt 6+\sqrt 2\over 4} \Rightarrow x+{\pi \over 12}={\pi \over 12},(2\pi-{\pi\over 12}) \\ \Rightarrow x=0,(2\pi-{\pi\over 6}) \Rightarrow 有兩個實數x,故選\bbox[red, 2pt]{(2)}\\ 註:\sin{\pi \over 12}={\sqrt 6-\sqrt 2\over 4}, \cos {\pi\over 12} ={\sqrt 6+\sqrt 2\over 4}, \sin x-\sin y=2\sin((x-y)/2)\cos((x+y)/2)$$
解答:$$甲的中位數是k \Rightarrow 乙的中位數是k+1 \Rightarrow \cases{在k的左邊有24個數\\在k+1的右邊也有24個數 } \Rightarrow k=25 \\ \Rightarrow 24個數均分成2群有C^{24}_{12}種分法,因此共有(C^{24}_{12})^2分法,故選\bbox[red, 2pt]{(4)}$$

解答

$$\cos \angle A=\cos {\theta}={7^2+9^2-9^2\over 2\cdot 7\cdot 9} ={7\over 18} \Rightarrow \cos {\theta\over 2} = \sqrt{\cos \theta+1\over 2}={5\over 6}\\ \Rightarrow \cos \angle QAB= \cos {\theta\over 2}={9^2+7^2-\overline{BQ}^2 \over 2\cdot 7\cdot 9} ={130-\overline{BQ}^2\over 126} ={5\over 6} \Rightarrow \overline{BQ}=5,故選\bbox[red, 2pt]{(3)}$$

二 、 多 選 題 ( 占 30 分 )

解答:$$y=\log x,其中x\gt 0 \\ (1)\times: y+{1\over 2}=\log(5x)= \log 5+\log x \Rightarrow y=\log x+\log 5-{1\over 2} \ne \log x\\ (2) \times: 2y=\log(x^2) \Rightarrow x可以為負值,其圖形與\Gamma不同 \\ (3)\bigcirc: 3y=\log(x^3) =3\log x \Rightarrow y=\log x\\ (4)\bigcirc: x=10^y \Rightarrow\log x=y \\(5)\times: x^3=10^{(y^3)} \Rightarrow 3\log x=y^3 \Rightarrow y=\sqrt[3]{3\log x} \ne \log x\\,故選\bbox[red, 2pt]{(3,4)}$$
解答:$$(1)\times: 最大角為\theta \Rightarrow \cos\theta ={n^2+(n+1)^2-(n+2)^2 \over 2n(n+1)} ={(n-1)^2-4\over 2n(n+1)}\\\qquad 若n=2,則\cos \theta \lt 0 \Rightarrow \theta 為鈍角\\(2)\bigcirc: T_n的周長=n+(n+1)+(n+2)=3n+3=3(n+1) \Rightarrow \langle T_n\rangle 為公差為3的等差數列 \\(3)\bigcirc: n越大,三邊長皆越大,面積變大\\(4) \times: T_5三邊長為5,6,7 \Rightarrow s=9 \Rightarrow 三角形面積=\sqrt{9\cdot 4\cdot 3\cdot 2}=6\sqrt 6\\\qquad \Rightarrow 三高\cases{h_5=12\sqrt 6/5\\ h_6=2\sqrt 6\\ h_7=12\sqrt 6/7} \Rightarrow h_5,h_6,h_7非等差數列\\ (5)\times:  \cases{T_3最大角\theta_3\\ T_2最大角\theta_2} \Rightarrow \cases{\cos\theta_3=0 \Rightarrow \theta_3=90^\circ \\ \cos \theta_2\lt 0 \Rightarrow \theta_2 \gt 90^\circ} \Rightarrow \theta_3 \not \gt \theta_2\\,故選\bbox[red, 2pt]{(2,3)}$$
解答:$$(1)\times: \cases{L_A經過(\bar x_A,\bar y_A) \\ L_B經過(\bar x_B,\bar y_B)} \Rightarrow \cases{\bar y_A= 2\times 5.2-0.6=9.8\\ \bar y_B= 1.5\times 6+0.4=9.4} \Rightarrow \bar y_A \not \lt \bar y_B \\(2)\times: 斜率=相關係數\times {\sigma_y\over \sigma_x} \Rightarrow \cases{2=0.6\cdot {\sigma_{y_A}\over 0.3} \\ 1.5=0.3\cdot {\sigma_{y_B}\over 0.1}} \Rightarrow \cases{\sigma_{y_A}=1 \\ \sigma_{y_B}=0.5} \Rightarrow 1\not \lt 0.5 \\(3)\bigcirc: |8.6-9.8|=1.2 \gt 1\\ (4)\times: \cases{d((5.6,8.6),L_A)=|11.2-8.6-0.6|/\sqrt 5 =2/\sqrt 5\\ d((5.6,8.6),L_B) =|7.1-8.6+0.4| /\sqrt 5=1.1/\sqrt 5} \Rightarrow {2\over \sqrt 5}\not \lt {1.1\over \sqrt 5} \\(5)\times: \cases{d((5.6,8.6),(5.2,9.8))^2 = 0.4^2+1.2^2\\ d((5.6,8.6),(6,9.4))^2 =0.4^2+0.8^2} \Rightarrow d((5.6,8.6),(\bar x_A,\bar y_A )) \gt d((5.6,8.6),(\bar x_B, \bar y_B))\\,故選\bbox[red, 2pt]{(3)}$$
解答

$$(1)\times: \cases{d(B,x軸)= d(P,x軸)=2\sqrt 3 \gt 3\\ d(A,x軸) =6/2=3} \Rightarrow d(A,x軸)\lt d(B,x軸) \\(2) \times:正六邊形邊長=2\sqrt 3\times {2\over \sqrt 3}=4  \\(3)\bigcirc: \cases{A(3,3)\\ B(-4,2\sqrt 3)} \Rightarrow \overrightarrow{BA}=(7,3-2\sqrt 3) \\(4)\times: \overline{AP} =\sqrt{3^2+(3-2\sqrt 3)^2} =\sqrt{30-12\sqrt 3} =\sqrt{30-12\cdot 1.732} =\sqrt{9.216}\lt \sqrt{10} \\(5)\bigcirc: \overleftrightarrow{AP} 斜率=-{2\sqrt 3-3\over 3} \gt -{1\over \sqrt 3}\\,故選\bbox[red, 2pt]{(35)}$$
解答:$$(1)\times: \cases{P(a=b=1)={1\over 6}\times {1\over 2}={1\over 12}\\ P(a=b=2)={1\over 6}\times {1\over 2}={1\over 12}} \Rightarrow P(a=b)={1\over 12}+ {1\over 12}={1\over 6}\ne {1\over 3}\\ (2)\bigcirc: \begin{vmatrix}a & 6 \\1 & b\end{vmatrix}=ab-6=0 \Rightarrow \cases{a=6,b=1 \Rightarrow 無限多解\\ a=3,b=2 \Rightarrow 無解} \Rightarrow P(a=3,b=2)=1/12 \\(3) \bigcirc: \cases{P(無限多解)=1/12\\ P(無解)=1/12} \Rightarrow P(唯一解)=1-{1\over 12}-{1\over 12}={5\over 6} \\(4)\times: 有解\Rightarrow b=2且a\ne 3 \Rightarrow P(b=2,a\ne 3)={1\over 2}\times {5\over 6} ={5\over 12} \ne {1 \over 2} \\(5) \times: b=2 \Rightarrow \cases{ax+6y=6\\ x+2y=1} \Rightarrow x={3\over a-3}\gt 0 \Rightarrow a=4,5,6\\\qquad \Rightarrow P(a=4,5,6且b=2)={1\over 2}\cdot {1\over 2}={1\over 4}\\,故選\bbox[red, 2pt]{(2,3)}$$

解答:$$T=\begin{bmatrix}3 & 0 \\a & 1 \end{bmatrix} \Rightarrow \cases{T(A)=A'=(3,a)\\ T(B)=B'=(0,1)\\ T(C)=C'=(-3,-a)} \\ (1)\times: a=0 \Rightarrow \cases{A'(3,0)\\ B'(0,1)\\ C'(-3,0)} \Rightarrow \overline{B'A'}=\overline{B'C'}但\angle A'B'C \ne 90^\circ \\(2)\bigcirc: \cases{B=B'\\ T(O)=O=(0,0)在\overline{AB}上}\Rightarrow 至少原點O與B兩點不變\\ (3)\times: a=0 \Rightarrow A',B',C'皆不在第四象限\\ (4)\bigcirc: \cases{P(1/3,-a/3)\\ Q(-1/3,a/3) }\Rightarrow \cases{T(P)=(1,0)=A\\ T(B)=B\\ T(Q)=(-1,0)=C }, 可取\Omega=\triangle PBQ \\ (5)\bigcirc: \triangle A'B'C' ={1\over 2}\begin{Vmatrix}3& a& 1\\ 0& 1& 1\\ -3& -a& 1\end{Vmatrix} =3為常數\\,故選\bbox[red, 2pt]{(245)}$$

三 、 選 填 題 ( 占 25 分 )

解答:$$\cases{{100A+400B+ 240C \over A+B+C}=260 \cdots(1) \\{100A+ 400B\over A+B} =280 \cdots(2)},由(2)可得100A+400B=280A+280B \Rightarrow A={2\over 3}B\\ 將A={2\over 3}B代入(1) \Rightarrow C={5\over 3}B \Rightarrow A:B:C ={2\over 3}:1:{5\over 3} = \bbox[red, 2pt]{2:3:5}$$
解答:$$\cases{f(x)=p_1(x) (x^2-2x+3)+x+1\\ g(x)=p_2(x)(x^2-2x+3)+x-3\\ h(x)= p_3(x)(x^2-2x+3)-2} \\ \Rightarrow xf+ag+bh= (x^2-2x+3)(xp_1+ap_2+ bp_3)+x^2+x+ax-3a-2b \\ \Rightarrow 餘式:  x^2+x+ax-3a-2b=x^2+(a+1)x-3a-2b=x^2-2x+3\\ \Rightarrow \cases{a+1=-2\\ -3a-2b=3} \Rightarrow \bbox[red, 2pt]{a=-3,b=3}$$
解答:$$假設報名人數N,中獎機率0.4\% \Rightarrow N\times 0.4\%=10 \Rightarrow N=2500\\ 前100人抽完後,剩下5000元禮券4張,8000元禮券5張;\\ 因此第101人\cases{抽中5000元禮券的機率=4/(2500-100) =4/2400\\ 抽中8000元禮券的機率=5/(2500-100)=5/2400} \\ \Rightarrow 期望值=5000\times {4\over 2400}+ 8000\times {5\over 2400}={150\over 6}=\bbox[red, 2pt]{25}$$
解答

$$假設\cases{過原點方向向量為(2,-3)的直線L_1:3x+2y=0\\ 過原點方向向量為(3,2)的直線L_2:2x=3y \\過原點方向向量為\vec v的直線L_3 \Rightarrow P在L_3上},由於(2,-3)\cdot (3,2)=0 \Rightarrow L_1\bot L_2\\ 令R,S分別是P在L_1及L_2上的垂足,依題意\cases{\overline{PR} =\overline{OP}-1 \\\overline{PS}=\overline{OP}-2}, \\也就是直角三角形OSP的三邊長為\overline{OP}-2,\overline{OP}-1,\overline{OP},顯然\overline{OP}=5\\令\cases{\theta_1為L_2與\overline{OP}的夾角\\ \theta_2為L_2與\overline{OQ}的夾角}\Rightarrow \cases{\cos \theta_1 =3/5 \\ \cos\theta_2= {(4,7)\cdot (3,2)\over |(4,7)||(3,2)|}={26\over 13\sqrt{5}} ={2\over \sqrt 5}} \\ \Rightarrow \cos(\theta_1+ \theta_2)= \cos \theta_1\cos\theta_2-\sin\theta_1\sin \theta_2 ={3\over 5}\cdot {2\over \sqrt 5}-{4\over 5}\cdot {1\over \sqrt 5} ={2\over 5\sqrt 5}=\bbox[red, 2pt]{2\sqrt 5\over 5}$$

解答

$$\text{Case I: }x\ge y \Rightarrow \cases{P(x,y)至右邊界(x=1)的距離\ge |x-y| \\ P(x,y)至下邊界(y=0)的距離\ge |x-y|} \Rightarrow \cases{1-x\ge x-y\\ y\ge x-y} \\ \qquad \Rightarrow \cases{2x-y\le 1\\ 2y\ge x}\\\text{Case II: }x\le y \Rightarrow \cases{P(x,y)至左邊界(x=0)的距離\ge |x-y| \\ P(x,y)至上邊界(y=1)的距離\ge |x-y|} \Rightarrow \cases{x\ge y-x\\ 1-y\ge y-x} \\\qquad \Rightarrow \cases{2x\ge y\\ 2y-x\le 1}\\四條直線所圍菱形如上圖,其中\cases{P= (y=2x)\cap (2y-x=1) =({1\over 3},{2\over 3})\\ Q=(x=2y)\cap (2x-y=1) =({2\over 3},{1\over 3})}\\ 因此\cases{兩平行直線距離={1\over \sqrt 5} \\ \overline{PB}={\sqrt 5\over 3}} \Rightarrow 菱形面積={1\over \sqrt 5}\times {\sqrt 5\over 3} =\bbox[red, 2pt]{1\over 3}$$

第 貳 部 分 、 混 合 題 或 非 選 擇 題 ( 占 15 分 )

18-20 題 為 題 組
解答:$$平面E的法向量\vec n=(1,0,-1) \Rightarrow 過O且方向向量為\vec n的直線L:(t,0,-t),t\in \mathbb R \\ \Rightarrow Q=E\cap L \Rightarrow t-(-t)=4 \Rightarrow t=2 \Rightarrow Q=(2,0,-2)\\ \Rightarrow \cos \alpha ={(2,0,-2)\cdot (1,0,0) \over \Vert(2,0,-2)\Vert \Vert (1,0,0)\Vert} ={2\over 2\sqrt 2 } ={\sqrt 2\over 2},故選\bbox[red, 2pt]{(4)}$$
解答:$$\cos \theta ={(a,b,c)\cdot (1,0,0)\over \Vert (a,b,c)\Vert \Vert (1,0,0)\Vert} ={a\over \sqrt{a^2+b^2+c^2}} \ge \cos {\pi\over 6} ={\sqrt 3\over 2} \\ \Rightarrow {a^2\over a^2+b^2+c^2} \ge {3\over 4} \Rightarrow 4a^2 \ge 3a^2+3b^2+3c^2 \Rightarrow a^2\ge 3(b^2+c^2) ,\bbox[red, 2pt]{故得證}$$
解答:$$P(a,0,c)在E上 \Rightarrow a-c=4 \Rightarrow a=c+4 代入a^2\ge 3(b+c^2) \Rightarrow (c+4)^2 \ge 3c^2\\ \Rightarrow c^2-4c-8\le 0 \Rightarrow (c-(2+ 2\sqrt 3))(c-(2-2\sqrt 3))\le 0 \Rightarrow \bbox[red, 2pt]{2-2\sqrt 3\le c\le 2+2\sqrt 3} \\ 又\overline{OP} =\sqrt{(c+4)^2 +c^2} =\sqrt{2(c+2)^2+8} \Rightarrow \overline{OP}的最小值發生在c=2-2\sqrt 3\\ 此時\overline{OP} =\sqrt{2(4-2\sqrt 3)^2+8} =\sqrt{64-2\sqrt{768}} =\sqrt{48}-\sqrt{16} = \bbox[red, 2pt]{4\sqrt 3-4}$$
 

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解題僅供參考, 其他歷年試題及詳解


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