113年高中運動績優生甄試--數學科詳解

113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: $$0.\bar 3+0.\bar 7={3\over 9}+{7\over 9}={10\over 9}，故選\bbox[red, 2pt]{(D)}$$

解答: $$|x-3|=113 \Rightarrow \cases{x-3=113\\ x-3=-113} \Rightarrow \cases{x_1=116\\ x_2=-110} \Rightarrow x_1+x_2=6，故選\bbox[red, 2pt]{(C)}$$

解答: $$1公斤=1000公克=10^3公克 \Rightarrow 1000公斤=1000\times 10^3=10^6公克，故選\bbox[red, 2pt]{(B)}$$

解答: $$\log 1+\log 10+\log 100=0+1+2=3，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{P(1,2)\\ Q(3,k)\\ R(4,8)  }\Rightarrow \cases{\overrightarrow{PQ} =(2,k-2)\\ \overrightarrow{PR}=(3,6)} \Rightarrow \overrightarrow{PQ} \parallel \overrightarrow{PR} \Rightarrow {2\over 3}={k-2\over 6} \Rightarrow k-2=4 \Rightarrow k=6\\，故選\bbox[red, 2pt]{(E)}$$

解答: $$\cases{(A)斜率=0\\ (B)斜率=-1\\ (C)斜率=-2\\ (D)斜率=-3\\  (E)斜率=-4} \Rightarrow (A)斜率最大，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{L_1:2x+3y=5\\ L_2:3x-2y=5} \Rightarrow \cases{L_1斜率-2/3\\ L_2斜率3/2} \Rightarrow 兩直線不平行也不重合，故選\bbox[red, 2pt]{(C)}$$

解答: $$(x-1)^2+(y+3)^2=16 \Rightarrow 圓心P(1,-3)代入直線2x+y=k \Rightarrow k=2-3=-1，故選\bbox[red, 2pt]{(B)}$$

解答: $$圓x^2+y^2=1 \Rightarrow \cases{圓心P(0,0)\\ 圓半徑r=1}, 圓與直線相切代表圓心至直線距離=r \\ (A) d(P,L)={2\over \sqrt 2} =\sqrt 2\ne 1\\ (B)d(P,L)={1\over \sqrt 2} \ne 1\\ (C) d(P,L)=0 \ne 1\\ (D) d(P,L)={1\over \sqrt 2}\ne 1\\ (E)d(P,L)= {\sqrt 2\over \sqrt 2}=1\\，故選\bbox[red, 2pt]{(E)}$$

解答: $$f(x)=2(x-1)^3+3(x-1)^2+4(x-1)+5=(x-1)^2p(x)+ax+b\\ \Rightarrow f'(x)=6(x-1)^2+4(x-1)+4=2(x-1)p(x)+ (x-1)^2p'(x)+a \\ \Rightarrow \cases{f(1)=5=a+b\\ f'(1)=4=a} \Rightarrow b=1 \Rightarrow 餘式為4x+1=4(x-1)+5，故選\bbox[red, 2pt]{(C)}$$

解答: $$y=f(x)=-x^2+8x+k =-(x^2-8x+16)+k+16 =-(x-4)^2+k+16\\ \Rightarrow 對稱軸:x=4，故選\bbox[red, 2pt]{(D)}$$

解答: $$y=f(x)=3x^3+2x^2+x-4 \Rightarrow 廣域特徵: y=3x^3，故選\bbox[red, 2pt]{(A)}$$

解答: $$a_3=a_2+a_1= 2+1=3 \Rightarrow a_4=a_3+a_2=3+2=5，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{a_1=1\\ a_7=a_1r^6=8} \Rightarrow {a_7\over a_1}={a_1r^6\over a_1} ={8\over 1} \Rightarrow r^6=8 \Rightarrow r=\sqrt 2，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x,y)=2x-y \Rightarrow f(3,-1)\cdot f(1,t)\lt 0 \Rightarrow 7\cdot (2-t)\lt 0 \Rightarrow t\gt 2，故選\bbox[red, 2pt]{(E)}$$

解答: $$\sigma(X)=\sqrt{10} \Rightarrow \sigma(2X+3)=2\sigma(X)=2\sqrt{10}，故選\bbox[red, 2pt]{(B)}$$

解答: $$剛好一直線\Rightarrow r=1，故選\bbox[red, 2pt]{(E)}$$

解答: $$y={5\over 4}\cdot 56+100=170，故選\bbox[red, 2pt]{(C)}$$

解答: $$50-(22+20-10)=50-32=18，故選\bbox[red, 2pt]{(A)}$$

解答: $$4!=24，故選\bbox[red, 2pt]{(D)}$$

解答: $$C^5_3\times C^5_3=10\times 10=100，故選\bbox[red, 2pt]{(E)}$$

解答: $$60\% \times 70\%=42\%，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{出現兩個正面的機率=1/4 \\出現兩個反面的機率=1/4 } \Rightarrow 期望值=10\times {1\over 4}+2\cdot {1\over 4}=3，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sin(90^\circ-\alpha)= \cos \alpha =4/5，故選\bbox[red, 2pt]{(C)}$$

解答: $$\sin^2 \theta+\cos^2\theta =1 且\sin \theta \lt 0 \Rightarrow \sin \theta=-{1\over 2}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cos 60^\circ={5^2+8^2-\overline{BC}^2 \over 2\cdot 5\cdot 8} \Rightarrow {1\over 2}={89-\overline{BC}^2\over 80} \Rightarrow \overline{BC}=7，故選\bbox[red, 2pt]{(B)}$$

解答: $$12^2 \pi \cdot {\theta\over 2\pi}=24\pi \Rightarrow \theta={\pi\over 3}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\sin x={\sqrt 3\over 2} \Rightarrow x={\pi\over 3}，故選\bbox[red, 2pt]{(C)}$$

解答: $$\log_2 16+\log_3 27=\log_2 2^4+\log_3 3^3=4+3=7，故選\bbox[red, 2pt]{(E)}$$

解答: $$y=f(x)=2^x \gt 0 且為遞增函數，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{P(1,0)\\ Q(2,1)\\ R(3,3)} \Rightarrow \cases{\overrightarrow{PQ}=(1,1)\\ \overrightarrow {PR} =(2,3)} \Rightarrow |\overrightarrow {PQ}+ \overrightarrow {PR}|= |(3,4)| =\sqrt{3^2+4^2} =5，故選\bbox[red, 2pt]{(E)}$$

解答: $$\overrightarrow{AB} \cdot \overrightarrow{AC} =|\overrightarrow{AC}| |\overrightarrow{AC}| \cos \angle BAC =1\cdot \sqrt 2\cdot {\sqrt 2\over 2}=1，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sqrt{8^2+5^2+2^2} =\sqrt{93}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{\overline{PS}= \overline{OA}=6\\ \overline{BS}= \overline{OC} =5} \Rightarrow \overline{PB}=\sqrt{6^2+ 5^2} =\sqrt{61}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{(A)\overleftrightarrow{EH} 與\overleftrightarrow{AE}交於E \\(B)\overleftrightarrow{EH} \parallel \overleftrightarrow{BC} \\(D)\overleftrightarrow{EH} 與\overleftrightarrow{DH}交於H\\ (E)\overleftrightarrow{EH}與\overleftrightarrow{EG}交於E} \Rightarrow 故選\bbox[red, 2pt]{(C)}$$

解答: $${P(反,正)\over P(正,正)+P(正,反)+P(反,正)} ={1\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答: $$40\%\times 5\%+60\%\times 10\%=8\%，故選\bbox[red, 2pt]{(C)}$$

解答: $$\begin{bmatrix}113 & k \\5 & w \end{bmatrix} =2\begin{bmatrix}x & 1 \\y & -1 \end{bmatrix} +\begin{bmatrix}-1 & 0 \\1 & 1 \end{bmatrix} =\begin{bmatrix}2x & 2 \\2y & -2 \end{bmatrix} +\begin{bmatrix}-1 & 0 \\1 & 1 \end{bmatrix} =\begin{bmatrix}2x-1 & 2 \\2y+1 & -1 \end{bmatrix} \\ \Rightarrow 2x-1=113 \Rightarrow 57，故選\bbox[red, 2pt]{(D)}$$

解答: $$\begin{bmatrix}9 & -8 \\6 & 5 \end{bmatrix} \begin{bmatrix}x \\y  \end{bmatrix} = \begin{bmatrix}9x-8y \\6x+5y  \end{bmatrix} =\begin{bmatrix}7 \\4 \end{bmatrix} \Rightarrow \cases{9x-8y=7\\ 6x+5y=4}，故選\bbox[red, 2pt]{(E)}$$

解答: $$\left[ \begin{array}{rr|rr}4& 3& 1& 0\\ 3& 2& 0 & 1 \end{array}\right ] \xrightarrow{R_1/4\to R_1} \left[ \begin{array}{rr|rr}1& 3/4& 1/4& 0\\ 3& 2& 0 & 1 \end{array}\right ] \xrightarrow{R_2-3R_1} \left[ \begin{array}{rr|rr}1& 3/4& 1/4& 0\\ 0& -1/4& -3/4 & 1 \end{array}\right ] \\ \xrightarrow{-4R_2\to R_2} \left[ \begin{array}{rr|rr}1& 3/4& 1/4& 0\\ 0& 1& 3 & -4 \end{array}\right ] \xrightarrow{R_2-(3/4)R_2 \to R_1}\left[ \begin{array}{rr|rr}1& 0& -2& 3\\ 0& 1& 3 & -4 \end{array}\right ] \\ \Rightarrow 反方陣=\begin{bmatrix}-2 & 3 \\3 & -4 \end{bmatrix} \Rightarrow a=-2，故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考,運優甄試歷年試題及詳解