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2023年6月20日 星期二

112年高中運動績優生甄試--數學科詳解

112學年度高級中等以上學校運動成績優良學生
升學輔導甄試學科考試

說明:單選題共40題,請在「答案卡」上劃記。每題2.5 分,共100分。

解答:$$\frac{2}{\sqrt 5+\sqrt 3} =\frac{2(\sqrt 5-\sqrt 3)}{(\sqrt 5+\sqrt 3)(\sqrt 5-\sqrt 3)} =\frac{2(\sqrt 5-\sqrt 3)}{2}= \sqrt 5-\sqrt 3,故選\bbox[red, 2pt]{(B)}$$
解答:$$\left| x-(-2))\right|=6 \Rightarrow |x+2|=6,故選\bbox[red, 2pt]{(E)}$$
解答:$$100\times 10^{-7} =10^{-5},故選\bbox[red, 2pt]{(A)}$$
解答:$$\log 10000 \lt \log 2345 \lt \log 100000 \Rightarrow 4\lt \log 2345 \lt 5,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{a_1=3\\ a_{n+1}=2(a_n-1), n\in \mathbb N} \Rightarrow a_2=2(a_1-1)=2(3-1)=4 \Rightarrow a_3=2(a_2-1) =2(4-1)=6\\,故選\bbox[red, 2pt]{(C)}$$
解答:$$S=1+3+9+\cdots +243=1+3+3^2+ \cdots+3^5\\ \Rightarrow 3S=3+3^2+3^3+ \cdots +3^6 \Rightarrow S-3S=1-3^6 \Rightarrow S={3^6-1\over 2} ={729-1\over 2}= 364,故選\bbox[red, 2pt]{(B)}$$
解答:$$f(x)=(x^2+3x-4)P(x)+ (x+2) =(x+4)(x-1)P(x)+(x+2) \\ \Rightarrow f(1)= 1+2=3,故選\bbox[red, 2pt]{(D)}$$
解答:$$y=f(x)=x^2 \Rightarrow 向右平移112 \Rightarrow y=f(x-112)= (x-112)^2\\ \Rightarrow 向上平移2023 \Rightarrow y=f(x-112)+2023 =(x-112)^2+2023,故選\bbox[red, 2pt]{(E)}$$
解答:$$f(x)=x^3-3x^2+2x+1=(x-h)^3 +p(x-h)+k \Rightarrow f'(x)=3x^2-6x+2=3(x-h)^2+ p\\ \Rightarrow f''(x) = 6x-6=6(x-h) \Rightarrow h=1,故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)= (x-2)^3+3(x-2)+1 \Rightarrow f'(x)= 3(x-2)^2+3 \Rightarrow f'(2)=3\\ \Rightarrow 切線方程式:y=f'(2)(x-2)+f(2) =3(x-2)+1,故選\bbox[red, 2pt]{(A)}$$
解答:$$(A)斜率=\frac{1-(-2)}{1-0} =2\\ (B)斜率=\frac{1-0}{1-0} =1\\ (C)斜率=\frac{1-1}{1-0} =0\\ (D)斜率=\frac{1-2}{1-0} =-1\\ (E)斜率=\frac{1-3}{1-0} =-2\\  ,故選\bbox[red, 2pt]{(E)}$$
解答:$$(x,y)關於y軸的對稱點為(-x,y),因此P(-3,2)的對稱點為(3,2),故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{x-y=-1\\ x-y=-3} \Rightarrow 兩直線距離={|-1-(-3)|)\over \sqrt{1^2+1^2}} ={2\over \sqrt 2}=\sqrt 2,故選\bbox[red, 2pt]{(D)}$$
解答:$$令f(x,y)=2x+3y-6 \\(A) \times:f(2,2)=4+6-6\gt 0 \not \in S \\(B)\bigcirc: f(1,1)=-1\lt 0 \\(C)\times: f(0,3)=3\gt 0 \\(D)\times: f(4,4)=14\gt 0\\ (E)\times: f(5,-1)=1\gt 0 \\,故選\bbox[red, 2pt]{(B)}$$
解答:$$x^2+y^2-2x+6y+6=0 \Rightarrow (x^2-2x+1)+(y^2+6y+9)=4 \Rightarrow (x-1)^2+(y+3)^2= 2^2\\ \Rightarrow \cases{h=1\\ k=-3\\ r=2} \Rightarrow h+k+r =0,故選\bbox[red, 2pt]{(A)}$$
解答:$$P在圓外,故選\bbox[red, 2pt]{(C)}$$
解答:$$將兩極端的值刪除後,變異數變小,標準差同時變小,故選\bbox[red, 2pt]{(B)}$$
解答:$$標準化後\sigma_Y=\sigma_X,迴歸直線斜率=相關係數\times {\sigma_Y\over \sigma_X}=0.7,故選\bbox[red, 2pt]{(C)}$$
解答:$$C^9_2=36,故選\bbox[red, 2pt]{(A)}$$
解答:$$已經甲甲,剩下可能為\cases{甲\\ 乙甲\\ 乙乙甲\\ 乙乙乙},共四種情形,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{偶數點機率=1/2\\ 奇數點機率=1/2} \Rightarrow 期望值=12\times{1\over 2}+ 10\times {1\over 2}=11,故選\bbox[red, 2pt]{(D)}$$
解答:$$(x+y)^4 = \sum_{n=0}^4 C^4_nx^ny^{n-4} \Rightarrow x^2y^2係數=C^4_2=6,故選\bbox[red, 2pt]{(E)}$$
解答:$$兩球都沒中的機率=0.2^2=0.04 \Rightarrow 至少罰進一球的機率=1-0.04=0.96=96\%,故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{\theta為銳角\\ \cos \theta=1/3} \Rightarrow \sin \theta =\sqrt{1-\cos^2 \theta} ={2\sqrt 2\over 3} \Rightarrow \tan \theta ={\sin \theta\over \cos \theta} ={2\sqrt 2\over 1}=2\sqrt 2,故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{x=r\cos \theta = 2\cdot \cos 240^\circ=2\cdot (-{1\over 2})=-1 \\ y=r\sin \theta= 2\cdot \sin 240^\circ = 2\cdot (-{\sqrt 3\over 2})=-\sqrt 3},故選\bbox[red, 2pt]{(E)}$$
解答:$$正弦定理:{\overline{BC}\over \sin \angle BAC}=2R \Rightarrow {4\over 1/2}=2R \Rightarrow R=4,故選\bbox[red, 2pt]{(C)}$$
解答:$$分針從12轉動到8相當於旋轉{8\over 12}\times 2\pi={4\over 3}\pi,故選\bbox[red, 2pt]{(D)}$$
解答:$$\sin x的週期為2\pi \Rightarrow 3\sin x的週期仍為2\pi,故選\bbox[red, 2pt]{(A)}$$
解答:$$2\log x=1 \Rightarrow \log x={1\over 2} \Rightarrow x=\sqrt{10} \Rightarrow x^2=10,故選\bbox[red, 2pt]{(B)}$$
解答:$$(A)\times: 2^1=2\ne 0 \Rightarrow 不通過(1,0) \\(B) \times: 2^{-1}={1\over 2} \Rightarrow 通過(-1,{1\over 2})為第二象限,在y軸的左方 \\ (C)\times: 2^x \gt 0 \Rightarrow 圖形不通過第三、四象限,例:與y=-1無交點\\ (D)\times: f'(x)=\ln 2\cdot 2^x \gt 0 \Rightarrow f(x)為嚴格遞增 \Rightarrow f(2023)\gt f(2022)\\,故選\bbox[red, 2pt]{(E)}$$
解答:$$\vec c=x\vec a+y\vec b  \Rightarrow (3,8)=(2x,3x) +(-y,2y) =(2x-y,3x+2y) \\ \Rightarrow \cases{2x-y=3 \\ 3x+2y=8} \Rightarrow \cases{x=2\\ y=1},故選\bbox[red, 2pt]{(A)}$$
解答:$$\vec a\cdot \vec b=0 \Rightarrow 3x-6=0 \Rightarrow x=2,故選\bbox[red, 2pt]{(B)}$$
解答:$$\overline{AC} =\sqrt{\overline{AB}^2+ \overline{BC}^2}=\sqrt{3^2+2^2} =\sqrt{13} \Rightarrow \overline{AG}=\sqrt{\overline{AC}^2+\overline{CG}^2} =\sqrt{13+1} = \sqrt{14},故選\bbox[red, 2pt]{(D)}$$
解答:$$\sqrt{3^2+(-4)^2} =5,故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\times: \overleftrightarrow{AB} 與 \overleftrightarrow{DH} 不在同平面上\\(B)\times: \overleftrightarrow{AB} 與 \overleftrightarrow{DH} 不在同平面上,若是\overrightarrow{AB}與\overrightarrow{DH}則垂直 \\(C)\times: \overleftrightarrow{AB} 與 \overleftrightarrow{DH} 不在同平面上\\ (D) \times: A,C是交點,顯然不平行\\,故選\bbox[red, 2pt]{(E)}$$
解答:$$\cases{2x+3y=6\\ 4x+5y=7} \Rightarrow \begin{bmatrix}2 & 3 \\4 & 5\end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} = \begin{bmatrix} 6 \\7\end{bmatrix},故選\bbox[red, 2pt]{(D)}$$
解答:$$\begin{bmatrix}9 & 8 \\7 & -9\end{bmatrix}=x \begin{bmatrix}1 & 1 \\1 & -1 \end{bmatrix} -y \begin{bmatrix} -1 & 0 \\1 & 1\end{bmatrix} = \begin{bmatrix}x & x \\x & -x \end{bmatrix}-\begin{bmatrix}-y & 0 \\y & y\end{bmatrix} =\begin{bmatrix}x+y & x \\x-y & -x-y\end{bmatrix}\\ \Rightarrow \cases{x=8\\ y=1},故選\bbox[red, 2pt]{(A)}$$
解答:$$\begin{bmatrix}1 & 3 \\5 & 7\end{bmatrix} \begin{bmatrix}x & y \\z & w\end{bmatrix}=\begin{bmatrix}2 & 4 \\6 & 8\end{bmatrix} \Rightarrow \begin{bmatrix}1 & 3 \\5 & 7\end{bmatrix}^{-1}\begin{bmatrix}1 & 3 \\5 & 7\end{bmatrix} \begin{bmatrix}x & y \\z & w\end{bmatrix}= \begin{bmatrix}1 & 3 \\5 & 7\end{bmatrix}^{-1} \begin{bmatrix}2 & 4 \\6 & 8\end{bmatrix} \\ \Rightarrow  \begin{bmatrix}x & y \\z & w\end{bmatrix}= \begin{bmatrix}1 & 3 \\5 & 7\end{bmatrix}^{-1} \begin{bmatrix}2 & 4 \\6 & 8\end{bmatrix},故選\bbox[red, 2pt]{(B)}$$
解答:$$P(A\cup B)=P(A)+ P(B)-P(A\cap B) \Rightarrow {3\over 4}= {1\over 4}+ P(B)-{1\over 4}P(B) \Rightarrow {3\over 4}P(B)={1\over 2} \\ \Rightarrow P(B)={2\over 3},故選\bbox[red, 2pt]{(C)}$$
解答:$$\frac{甲廠瑕疵品}{甲廠瑕疵品+乙廠瑕疵品} =\frac{75\%\times 4\% }{75\%\times 4\%+ 25\%\times 3\%} =\frac{300}{300+75} =80\%,故選\bbox[red, 2pt]{(E)}$$



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