113年高雄聯合轉學考-升高三-數學詳解

 高雄區公立高中 113 學年度聯合招考轉學生
《升高三數學》科試卷

一、 單選題(60 分)：

解答: $$\cases{A(1,-2)\\ B(3,-4) \\ C(-5,3)}  \Rightarrow \cases{\overrightarrow{AB}= (2,-2) \\ \overrightarrow{AC} = (-6,5)} \Rightarrow \cases{ |\overrightarrow{AB}|^2 =8\\ |\overrightarrow{AC}|^2 = 61 \\ \overrightarrow{AB} \cdot \overrightarrow{AC} =-22}\\ \Rightarrow |\overrightarrow{AB}+ \overrightarrow{AC}|^2= (\overrightarrow{AB}+ \overrightarrow{AC}) \cdot (\overrightarrow{AB} +\overrightarrow{AC}) = |\overrightarrow{AB}|^2+ 2\overrightarrow{AB} \cdot \overrightarrow{AC}+ |\overrightarrow{AC}|^2 \\=8-44+61=25 \Rightarrow |\overrightarrow{AB} +\overrightarrow{AC}|=5，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{(A) \sin 20^\circ \cos 20^\circ={1\over 2}\sin 40^\circ\\ (B) \sin 35^\circ \cos 35^\circ={1\over 2}\sin 70^\circ\\ (C) \sin 50^\circ \cos 50^\circ ={1\over 2}\sin 100^\circ={1\over 2}\sin 80^\circ\\ (D) \sin65^\circ \cos 65^\circ={1\over 2}\sin 130^\circ={1\over 2}\sin 50^\circ} \Rightarrow (C)最大，故選\bbox[red, 2pt]{(C)}$$

解答: $$柯西不等式:\left((x+5)^2+(y-1)^2+(z-3)^2 \right)(2^2+(-6)^2+3^2) \ge (2(x+5)-6(y-1)+3(z-3))^2 \\ \Rightarrow \left((x+5)^2+(y-1)^2+(z-3)^2 \right) \cdot 49\ge (2x-6y+3z+7)^2=14^2 \\ \Rightarrow (x+5)^2+(y-1)^2+(z-3)\ge 4 \Rightarrow \sqrt{(x+5)^2+(y-1)^2+(z-3)} \ge 2，故選\bbox[red, 2pt]{(B)}$$

解答: $$令\cases{p=2/3\\ q=3/4} \Rightarrow 靶面中兩發情形:(甲1,甲2,乙1,乙2)=\\(1,1,0,0),(1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1),(0,0,1,1) \\ \Rightarrow 機率為p^2(1-q)^2+ 4(1-p)pq(1-q)+ (1-p)^2 q^2={1\over 144}(4+24+9)={37\over 144}，故選\bbox[red, 2pt]{(D)}$$

解答: $$假設兩圖形\cases{y=f(x)=\tan x\\ y=g(x)=\pi x/100}，並考慮\tan x在\left({k\pi\over 2},{(k+2)\pi\over 2} \right)週期的遞增特性:\\ x在區間 (-{\pi\over 2},{\pi\over 2}), ({\pi\over 2},{3\pi\over 2}),({3\pi\over 2},{5\pi\over 2}) 各有一交點\\ \cases{f(-\pi)=0\\ g(-\pi)\lt 0} \Rightarrow x在區間[-\pi, -{\pi\over 2})沒有交點,交點在(-{3\over 2}\pi, -\pi )內;\\  又\cases{f(3\pi)=0\\ g(3\pi)\gt 0} \Rightarrow x在區間({5\over 2}\pi, 3\pi]沒有交點,交點在(3\pi, {7\over 2}\pi )內\\ 因此共有3個交點，即3個解，故選\bbox[red, 2pt]{(B)}$$

解答: $$\log_3(x+1)+ \log_3(x-7)=\log_3 (x+1)(x-7)=\log_3 20 \Rightarrow (x+1)(x-7)=20 \\ \Rightarrow x^2-6x-27=0 \Rightarrow (x-9)(x+3)=0 \Rightarrow x=9 \;(x=-3不合,違反x+1\gt 0)，故選\bbox[red, 2pt]{(C)}$$

解答: $$\begin{Vmatrix}1 & 1& -1 \\0&-3 & -1 \\2& 4& k\end{Vmatrix} =2 \Rightarrow |-3k-4|=2 \Rightarrow k=-2,(-{2\over 3}非整數,不合)，故選\bbox[red, 2pt]{(A)}$$

解答: $$(B) B=\begin{bmatrix}-2 & 3 & 1 & -7\\1 & 2 & 3 & 7\\3 & 2 & -1 & 5 \end{bmatrix} \xrightarrow{2R_2+R_1\to R_1,-3R_2+R_3\to R_3} \begin{bmatrix}0 & 7 & 7 & 7\\1 & 2 & 3 & 7\\0 & -4 & -10 & -16 \end{bmatrix} \\ \xrightarrow{R_1/7\to R_1} \begin{bmatrix}0 & 1 & 1 & 1\\1 & 2 & 3 & 7\\0 & -4 & -10 & -16 \end{bmatrix} \xrightarrow{R_2-2R_1\to R_2,R_4+4R_1\to R_4} \begin{bmatrix}0 & 1 & 1 & 1\\1 & 0 & 1 & 5\\0 & 0 & -6 & -12 \end{bmatrix} \\\xrightarrow{R_1 \leftrightarrow R_2, R_3/(-6)\to R_3} \begin{bmatrix}1 & 0 & 1 & 5\\0 & 1 & 1 & 1\\0 & 0 & 1 & 2 \end{bmatrix} \xrightarrow{3R_1+4R_2-10R_3 \to R_1} \begin{bmatrix}3 & 4 & -3 & -1\\0 & 1 & 1 & 1\\0 & 0 & 1 & 2 \end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答: $$\log 0.57=\log 57-\log 100=\log 57-2=-0.2441 \Rightarrow \log 57=1.7559\\\log 57^{100}=100\log 57 =100\times  1.7559=175.59 \Rightarrow 57^{100}=10^{175.59}，故選\bbox[red, 2pt]{(B)}$$

解答: $$(E) \overrightarrow{AB} \cdot (1,-2,2)=(2,-1,-2) \cdot (1,-2,2)=0  \Rightarrow \overline{AB} \parallel (x-2y+2z=3) \\ \quad \Rightarrow 投影長最長，故選\bbox[red, 2pt]{(E)}$$

解答: $$x+y+z=3 \Rightarrow (x^2+y^2+ z^2)(1^2+1^2 +1^2) \ge (x+y+z)^2=9\\ \Rightarrow x^2+y^2+z^2 \ge 3 \Rightarrow \triangle PDE+ \triangle PFG+ \angle PHI={1\over 2}(x^2+y^2+z^2) \ge {3\over 2}，故選\bbox[red, 2pt]{(B)}$$

解答: $${(a,4)\cdot (3,4) \over 5} \cdot {(3,4)\over 5} ={(6,b)\cdot (3,4)\over 5} \cdot {(3,4)\over 5} \Rightarrow (a,4)\cdot (3,4)= (6,b)\cdot (3,4) \\ \Rightarrow 3a+16=18+4b \Rightarrow 3a=4b+2，故選\bbox[red, 2pt]{(B)}$$

解答: $$\begin{vmatrix}a & b \\c & d \end{vmatrix} =20 \Rightarrow ad-bc=20\\ \begin{vmatrix}2a+5b & a+2b \\2c+5d & c+2d \end{vmatrix} =(2a+5b)(c+2d)-(a+2b)(2c+5d) =-ad+bc=-20，故選\bbox[red, 2pt]{(C)}$$

解答: $$(\vec a+\vec c)(\vec b\times \vec c)= \vec a\cdot (\vec b\times \vec c)+ \vec c\cdot (\vec b\times \vec c)=\vec a\cdot (\vec b\times \vec c)+ 0=18，故選\bbox[red, 2pt]{(D)}$$

解答: $$令\cases{f(x)=2^x\\ g(x)=3^x\\ p(x)=\log_2 x\\ q(x)=\log_3 x} \Rightarrow \cases{g(x) \gt p(x)\gt q(x), x\in [1,\infty)\\ y=5-x圖形為左上右下}  \Rightarrow d\gt c\gt b \\ 只需考慮a,b的大小，2^x+x=6 \Rightarrow x=2 \Rightarrow a=2,而\cases{g(2)=9\\ 5-2=3} \Rightarrow b\lt a，故選\bbox[red, 2pt]{(D)}$$

二、 多選題(40 分)：

解答: $$(A) \times: (1,-1,1)\cdot (1,1,0)=0 但(1,2,1)在L上也在E上，即L與E重疊 \\(B) \times:(1,-1,1)\cdot (1,-1,1)=3 \ne 0 \\(C) \bigcirc: (1,-1,1)\cdot (1,2,1)=0 \\ (D) \times: (1,-1,1)  \cdot (2,-3,-5)=0 但(1,2,1)在L上也在E上，即L與E重疊 \\(E) \bigcirc: (1,-1,1)\cdot (2,-3,-5)=0\\，故選\bbox[red, 2pt]{(CE)}$$

解答: $$(A) \times:A^2=\begin{bmatrix}0 &-1 \\1 & 0 \end{bmatrix} \begin{bmatrix}0 &-1 \\1 & 0 \end{bmatrix} =\begin{bmatrix}-1 &0 \\0 & -1 \end{bmatrix} \ne I \\(B)\times: B^2=\begin{bmatrix}1/2 & \sqrt 3/2 \\\sqrt 3/2 & -1/2 \end{bmatrix} \begin{bmatrix}1/2 & \sqrt 3/2 \\\sqrt 3/2 & -1/2 \end{bmatrix} =\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} =I\ne -I \\(C)\times: \cases{AB= \begin{bmatrix}0 &-1 \\1 & 0 \end{bmatrix} \begin{bmatrix}1/2 & \sqrt 3/2 \\\sqrt 3/2 & -1/2 \end{bmatrix} =\begin{bmatrix}-\sqrt 3/2 & 1/2 \\1/2 & \sqrt 3/2 \end{bmatrix} \\[1ex]BA= \begin{bmatrix}1/2 & \sqrt 3/2 \\\sqrt 3/2 & -1/2 \end{bmatrix}  \begin{bmatrix}0 &-1 \\1 & 0 \end{bmatrix} =  \begin{bmatrix}\sqrt 3/2 &-1/2 \\-1/2 & -\sqrt 3/2 \end{bmatrix}} \Rightarrow AB\ne BA \\(D)\bigcirc: A^4=\begin{bmatrix}-1 &0 \\0 & -1 \end{bmatrix} \begin{bmatrix}-1 &0 \\0 & -1 \end{bmatrix} =I \Rightarrow \cases{A^4B=B\\BA^4=B} \Rightarrow A^4B=BA^4=B \\(E)\bigcirc: \cases{A^3B^4A= A^3IA=A^4=I\\ B^3A^4B= B^3IB=B^4=I} \Rightarrow A^3B^4A= B^3A^4B\\，故選\bbox[red, 2pt]{(DE)}$$

解答: $$x=-30^\circ時有最大值2 \Rightarrow \cases{f(-30^\circ)=2\\ f'(-30^\circ)=0} \Rightarrow \cases{-{a\over 2}+{\sqrt 3 b\over 2}=2 \\ a\cos(-30^\circ)-b\sin(-30^\circ)=0} \\ \Rightarrow \cases {-a+\sqrt 3b=4\\ \sqrt 3a+b=0} \Rightarrow \cases{a=-1\\ b=\sqrt 3} \Rightarrow f(x)=-\sin x+\sqrt 3\cos x \\(A)\bigcirc:  f(370^\circ)= f(360^\circ+10^\circ )=f(10^\circ) =k\\ (B)\bigcirc: f(190^\circ)= -\sin 190^\circ+ \sqrt 3\cos 190^\circ= \sin 10^\circ -\sqrt 3\cos 10^\circ=-f(10^\circ)=-k \\(C)\bigcirc: f(x)=-\sin x+\sqrt 3\cos x=-2\sin(x-\alpha)週期為2\pi \\(D) \bigcirc: f(150^\circ)=-\sin 150^\circ +\sqrt 3\cos 150^\circ = -{1\over 2}+\sqrt 3\cdot (-{\sqrt 3\over 2})=-2為最小值\\(E) \bigcirc: f(x)=-2\sin(x-\alpha) 最小值=-2，故選\bbox[red, 2pt]{(ABCDE)}$$

解答: $$(A)\times: L_1與L_2可能是歪斜\\ (C)\times: 任取B,C,則\angle BAC非唯一\\ (D)\times:可能三平面兩兩相交直線相互平行，任兩平面皆不平行\\，故選\bbox[red, 2pt]{(BE)}$$

解答:

$$\cases{P(0,2)\\ Q(3,0-2)\\ R(-2,1)} \Rightarrow \cases{L_1= \overleftrightarrow{PQ} : y=-{4\over 3}x+2\\ L_2=\overleftrightarrow{PR}: 3x+5y+1=0} \Rightarrow \cases{B=L_1\cap x軸=({3\over 2},0)\\ C=L_2\cap y軸= (0,-{1\over 5})} \\(A) \bigcirc: \sin \alpha={\overline{OB}\over \overline{PB}}={3/2\over 5/2}={3\over 5} \\(B) \bigcirc: \cos \beta={\overline{PA} \over \overline{PR} } ={1\over \sqrt 5} \\(C) \times:\sin(\alpha+\beta)= \sin \alpha \cos \beta+ \sin \beta\cos \alpha={3\over 5}\cdot {1\over \sqrt 5}+{2\over \sqrt 5}\cdot {4\over 5}={11\over 5\sqrt 5} \\(D) \times: \cases{\sin \alpha=3/5\\ \sin \beta =2/\sqrt 5=2\sqrt 5/5} \Rightarrow \beta \gt \alpha \\(E) \bigcirc: \triangle PQR={1\over 2}\cdot \overline{PQ} \cdot \overline{PR}\sin (\alpha+\beta) ={1\over 2}\cdot 5\cdot \sqrt 5\cdot {11\over 5\sqrt 5}=5.5\gt 5\\，故選\bbox[red, 2pt]{(ABE)}$$

解答: $$\cases{A=\{(1-2,1-6,1-6)\} \\B=\{(1-6,3-4,1-6)\} \\C=\{(1-6,1-6,5-6)\} \\} \Rightarrow \cases{\#(A)=2\cdot 6\cdot 6=72 \\\#(B)=6\cdot 2\cdot 6=72 \\\#(C)=6\cdot 6\cdot 2=72 } \\(A) \bigcirc:  P(A)=P(B) =P(C) ={72\over 6^3} ={1\over 3} \\(B)\times: P(A\mid B)={P(A\cap B)\over P(B)} ={2\cdot 2\cdot 6\over 72} ={1\over 3}\ne {1\over 6} \\(C)\bigcirc: \cases{P(A \cap B)=24/6^3=1/9 \\ P(A)P(B)=(1/3)\cdot (1/3)=1/9} \Rightarrow P(A\cap B)=P(A)\cdot P(B) \Rightarrow 獨立\\(D)\times: B\cap C \ne \varnothing\\(E)\bigcirc: 1-{4^3\over 6^3}=1-{8\over 27}={19\over 27}\\，故選\bbox[red, 2pt]{(ACE)}$$

解答: $$\\令f(x)=144(1-2^{-0.3x}), g(x)=144(1-3^{-0.2x})\\ (A)\bigcirc: f(10)=144(1-2^{-3})=144\cdot {7\over 8}=126 \\ (B)\times: \cases{f(20)=144(1-{1\over 64})\\ g(20)=144(1-{1\over 729})} \Rightarrow g(20)\gt f(20),乙比較多 \\ (C) \bigcirc: 1-2^{-0.3n} \lt 1, n\in \mathbb N\\ (D) \times: f(1)=144(1-2^{-0.3}) <144(1-2^{-1})=72 \\(E)\bigcirc: \cases{-0.3n \log 2=-0.0903n\\ -0.2n \log 3 =-0.09542n}  \Rightarrow -0.3n\log 2\gt -0.2n\log 3 \Rightarrow 2^{-0.3n} \gt 3^{-0.2n}\\\qquad \Rightarrow 1-2^{-0.3n} \lt 1-3^{-0.2n} \Rightarrow g(x)\gt f(x)，故選\bbox[red, 2pt]{(ACE)}$$

解答: $$(A)\bigcirc: ab\begin{bmatrix}{3\over b} & {2\over b} \\{2\over a} & {3\over a} \end{bmatrix} =\begin{bmatrix}{3\over b} \cdot ab & {2\over b} \cdot ab \\{2\over a} \cdot ab & {3\over a} \cdot ab \end{bmatrix} =\begin{bmatrix}3a & 2a \\2b & 3b \end{bmatrix} =A\\ (B)\times: \begin{bmatrix}3 & 2 \\2 & 3 \end{bmatrix} \begin{bmatrix}a & 0 \\0 & b \end{bmatrix} =\begin{bmatrix}3a & 2b \\2a & 3b \end{bmatrix} \ne A \\(C) \times: \begin{bmatrix}a & a \\b & b \end{bmatrix} \begin{bmatrix}3 & 0 \\0& 2 \end{bmatrix} =\begin{bmatrix}3a & 2a \\3b & 2b \end{bmatrix} \ne A \\(D) \bigcirc:  \begin{bmatrix}\frac{3}{5a} & \frac{-2}{5b} \\\frac{-2}{5a} & \frac{3}{5b} \end{bmatrix} A=A\begin{bmatrix}\frac{3}{5a} & \frac{-2}{5b} \\\frac{-2}{5a} & \frac{3}{5b} \end{bmatrix}=I \Rightarrow A^{-1}= \begin{bmatrix}\frac{3}{5a} & \frac{-2}{5b} \\\frac{-2}{5a} & \frac{3}{5b} \end{bmatrix} \\ (E)\bigcirc: A為轉移矩陣\Rightarrow a=b={1\over 5} \Rightarrow \begin{vmatrix}3a & 2a \\2b & 3b \end{vmatrix} =5ab={1\over 5}=a\\，故選\bbox[red, 2pt]{(ADE)}$$

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