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2024年7月30日 星期二

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:{A(1,2)B(3,4)C(5,3){AB=(2,2)AC=(6,5){|AB|2=8|AC|2=61ABAC=22|AB+AC|2=(AB+AC)(AB+AC)=|AB|2+2ABAC+|AC|2=844+61=25|AB+AC|=5(D)
解答:{(A)sin20cos20=12sin40(B)sin35cos35=12sin70(C)sin50cos50=12sin100=12sin80(D)sin65cos65=12sin130=12sin50(C)(C)

解答:西:((x+5)2+(y1)2+(z3)2)(22+(6)2+32)(2(x+5)6(y1)+3(z3))2((x+5)2+(y1)2+(z3)2)49(2x6y+3z+7)2=142(x+5)2+(y1)2+(z3)4(x+5)2+(y1)2+(z3)2(B)



解答:{p=2/3q=3/4:(1,2,1,2)=(1,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,0),(0,1,0,1),(0,0,1,1)p2(1q)2+4(1p)pq(1q)+(1p)2q2=1144(4+24+9)=37144(D)

解答:{y=f(x)=tanxy=g(x)=πx/100tanx(kπ2,(k+2)π2):x(π2,π2),(π2,3π2),(3π2,5π2){f(π)=0g(π)<0x[π,π2),(32π,π);{f(3π)=0g(3π)>0x(52π,3π],(3π,72π)33(B)
解答:log3(x+1)+log3(x7)=log3(x+1)(x7)=log320(x+1)(x7)=20x26x27=0(x9)(x+3)=0x=9(x=3,x+1>0)(C)

解答:

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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解題僅供參考, 其他轉學考歷年試題及詳解


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