114年大學學測-數學A詳解

財團法人大學入學考試中心基金會

114學年度學科能力測驗試題

數學A考科

解答:$$\cases{A:抽到藍色球的事件\\ B:抽到1號球的事件} \Rightarrow \cases{P(A)=5/(k+9) \\ P(B) =6/(k+9) \\ P(A\cap B) =2/(k+9)} \Rightarrow P(A\cap B)=P(A)P(B) \\ \Rightarrow {2\over k+9} ={5\over k+9}\cdot {6\over k+9} \Rightarrow 2(k+9)=6\cdot 5 \Rightarrow k=6,故選\bbox[red, 2pt]{(5)}$$

解答:$$\cases{L_1: y=-4(x-a)/3\\ L_2: y=-3(x-a)/2\\ P(a,0)\\ O(0,0)} \Rightarrow \cases{L_1與y軸交於A(0,4a/3) \\ L_2與y軸交於B(0,3a/2)} \Rightarrow \overline{AB} ={3\over 2}a-{4\over 3}a= {1\over 6}a\\ \Rightarrow \triangle PAB面積=3 \Rightarrow {1\over 2}\overline{AB}\times \overline{OP} ={1\over 2}\cdot {a\over 6}\cdot a={a^2\over 12}=3 \Rightarrow a=6,故選\bbox[red, 2pt]{(2)}$$

解答:$$\cases{P:鋼琴表演\\ V:小提琴表演 \\ S:歌唱表演} \Rightarrow 三種表演的排列順序:PVS, VPS, PSV, VPS,共四種\\ 每一種都有5!4!3!表演順序，因此共有4\times 5!4!3!種，故選\bbox[red, 2pt]{(4)}$$

解答:$$f(x)=\log_2 x \Rightarrow \begin{cases}f(x)=0 & x=1 & 格子點0\\ f(x)=1 & x=2& 格子點0,(2,1)邊界點\\ f(x)\lt 2& x=3& 格子點1\\ f(x)=2& x=4& 格子點1,(4,2)邊界點 \\ f(x)\lt 3 & x=5-7& 格子點2\times3=6\\ f(x)=3& x=8& 格子點2,(8,3)邊界點 \\ f(x)\lt 4 & x=9-15& 格子點3\times 7=21\\ f(x)=4 & x=16& 格子點3,(16,4)邊界點\\ f(x)\lt 5 & x=17-30& 格子點4\times 14=56\\ \end{cases} \\ \Rightarrow 格子點共有1+1+ 6+ 2+21+3+56=90，故選\bbox[red, 2pt]{(3)}$$

解答:$$\cases{\sin 2\theta \gt \sin \theta  \Rightarrow 0\lt x\lt {\pi\over 3}, \pi \lt x\lt {5\pi\over 3} \\ \cos 2\theta \gt \cos \theta  \Rightarrow {2\pi\over 3} \lt x\lt {4\pi\over 3}} \Rightarrow \pi\lt x\lt {4\pi\over 3} \Rightarrow \cases{b=4/3\\ a=1} \Rightarrow b-a={1\over 3},故選\bbox[red, 2pt]{(1)}$$

解答:$$\vec u,\vec v,\vec w兩兩互垂\Rightarrow \cases{\vec u\cdot \vec w=0\\ \vec w\cdot \vec v=0\\ \vec u\cdot \vec v=0} \\\Rightarrow (\vec u-\vec v) \cdot (\vec v-\vec w) =(2,-1,0) \cdot (-1,2,3) \Rightarrow \vec u\cdot \vec v-\vec u\cdot \vec w-|\vec v|^2+\vec v\cdot \vec w=-|\vec v|^2=-4 \Rightarrow |\vec v|=2 \\ \vec u-\vec v=(2,-1,0) \Rightarrow |\vec u-\vec v|^2 =(\vec u-\vec v)\cdot (\vec u-\vec v)=  5 \Rightarrow |\vec u|^2-\vec u\cdot \vec v-\vec v\cdot \vec u+|\vec v|^2=5\\ \Rightarrow |\vec u|^2 +|\vec v|^2=5 \Rightarrow |\vec u|^2=1 \Rightarrow |\vec u|=1\\    同理, \vec v-\vec w=(-1,2,3 ) \Rightarrow (\vec v-\vec w) \cdot (\vec v-\vec w) =|\vec v|^2+ |\vec w|^2 =14 \Rightarrow |\vec w|=\sqrt{10}\\ \Rightarrow 體積= |\vec v| \cdot |\vec u| \cdot |\vec w| =2\cdot 1\cdot \sqrt{10}=2\sqrt{10}, 故選\bbox[red, 2pt]{(3)}$$

二、多選題(占30分)

解答:$$3a_{n+1}=a_n+n \Rightarrow a_n={1\over 3}a_{n-1}+{1\over 3}(n-1) ={1\over 3}\left( {1\over 3}a_{n-2} + {1\over 3}(n-2)\right)+{1\over 3}(n-1) \\ ={1\over 3^2}a_{n-2}+{1\over 3^2}(n-2)+{1\over 3}(n-1) =\cdots ={1\over 3^{n-1}}a_1+ {1\over 3^{n-1}}\cdot 1+ {1\over 3^{n-2}}\cdot 2+ \cdots + {1\over 3^{1}}\cdot (n-1) \\ \Rightarrow a_n={1\over 4} \left( {1\over 3^{n-3}}-1\right) +{n-1\over 2} \\(1)\times: 3a_2=a_1+1=2+1=3 \Rightarrow a_2=1 \ne 2\\ (2) \bigcirc:b_2=a_2-{2\over 2}+{3\over 4} =1-1+{3\over 4}={3\over 4}\\ (3) \times: b_1=a_1-{1\over 2}+{3\over 4}=2-{1\over 2}+{3\over 4} ={9\over 4} \Rightarrow {b_2\over b_1}={3/4 \over 9/4} ={1\over 3} \ne {2\over 3} \\(4)\bigcirc:a_n={1\over 3}a_{n-1}+{1\over 3}(n-1) ={1\over 3}\left({1\over 3}a_{n-2}+ {1\over 3}(n-2) \right)+{1\over 3}(n-1) \\\qquad={1\over 3^2}a_{n-2}+{1\over 3^2}(n-2)+ {1\over 3}(n-1) \\\qquad ={1\over 3^{n-1}}a_1 +{1\over 3^{n-1}}1 +{1\over 3^{n-2}}2 + \cdots+{1\over 3}(n-1)\\\qquad  \Rightarrow 3^na_n=3a_1+3+3^2\cdot 2+\cdots + 3^{n-1}(n-1) 為正整數\\(5)\times: a_n={1\over 4} \left( {1\over 3^{n-3}}-1\right) +{n-1\over 2} \Rightarrow b_{10}=a_{10}-{10\over 5}+{3\over 4}= {1\over 4}\cdot {1\over 3^7} ={1\over 8748} \gt {1\over 10000}\\ 故選\bbox[red, 2pt]{(24)}$$

解答:$$(1)\times: {2^{x^2} \over 8}={4^x \over 2^{y^2}} \Rightarrow 2^{x^2+y^2} =2^{2x+3} \Rightarrow x^2+y^2=2x+3 \Rightarrow (x-1)^2+y^2 =2^2 為一圓\\ \qquad x=3 \Rightarrow y=0 \Rightarrow 交點只有一點(3,0) \\ (2) \times: (3,0)在圓上, 但(-3,0)不在圓上\\ (3)\bigcirc: (x-1)^2+y^2 =2^2 為一圓 \\(4) \times:圓心(1,0)至直線x+y=4的距離={3\over \sqrt 2} \gt 2(圓半徑) \Rightarrow 直線與圓無交點 \\(5) \bigcirc:\cases{x=2\cos \theta+1\\ y=2\sin \theta} \Rightarrow x-y=2(\cos \theta-\sin \theta)+1 =2\sqrt 2\sin(45^\circ-\theta)+1 最大值為2\sqrt 2+1 \\故選\bbox[red, 2pt]{(35)}$$

解答:$$(1)\times:\cases{x^2+bx+c=0有實根\\ x^2+(b+2)x+c=0沒有實根} \Rightarrow \cases{b^2-4c\ge0\\ (b+2)^2-4c\lt 0} \Rightarrow 4c\gt(b+2)^2\ge 0\Rightarrow c\gt 0 \\(2)\bigcirc: b^2\ge 4c\gt (b+2)^2 \Rightarrow b^2 \gt (b+2)^2 \Rightarrow 0\gt 4b+4\Rightarrow b \lt -1 \\(3)\bigcirc:\cases{b=-3\\ c=1}符合原要求 \Rightarrow x^2+(b+1)x+c=x^2-2x+1=0有實根x=1 \\(4)\bigcirc: 判別式: (b+2)^2+4c \gt 0 有實根(c\gt 0且(b+2)^2\ge 0) \\(5) \bigcirc: \cases{b=-2\\c=1} 符合原要求 \Rightarrow x^2+(b-2)x+c= x^2-4x+1 =0 \Rightarrow 判別式= 12\gt 0有實根$$

解答:

$$(1)\bigcirc: 當0\lt x\lt 1時,y=f(x)=\sin(\pi x) \gt 0,因此該圖形與y=k有交點, 顯然k\gt 0\\ (2)\times: 可能有四交點,如上圖\\ (3)\times: 當0\lt x\lt 1時, 圖形對稱x={1\over 2}, 因此x_1+x_2=1\\ (4)\bigcirc: k={1\over 2} \Rightarrow \sin(\pi x)={1\over 2} \Rightarrow \cases{P(1/6,1/2) \\Q(5/6,1/2) \\R(13/6,1/2)} \Rightarrow \cases{\overline{PQ}=4/6\\ \overline{QR}=8/6} \Rightarrow 2\overline{PQ} =\overline{QR} \\(5) \bigcirc: 假設L與\Gamma交於P,Q,R,S四點,則\cases{ P,Q對稱x=1/2 \\R,S對稱x=5/2} \Rightarrow \cases{x_1+x_2=1\\ x_3+x_4=5} \Rightarrow x_1+x_2+x_3+x_4 =6\\ 故選\bbox[red, 2pt]{(145)}$$

解答:$$(1)\times: \overline{BP}為\angle B的角平分線\Rightarrow {\overline{CP} \over \overline{DP}} ={\overline{BC} \over \overline{BD}}={4\over 3} \Rightarrow \overline{CP}= {4\over 7}\overline{CD} \\(2) \times:\overrightarrow{AP}= \overrightarrow{AC}+ \overrightarrow{CP} =\overrightarrow{AC}+ {4\over 7 } \overrightarrow{CD} =\overrightarrow{AC}+ {4\over 7 } (\overrightarrow{CB}+ \overrightarrow{BD}) =\overrightarrow{AC}+ {4\over 7 } (\overrightarrow{CA} +\overrightarrow{AB}+ {1\over 2}\overrightarrow{BA}) \\\qquad =\overrightarrow{AC}+ {4\over 7 } (-\overrightarrow{AC} + \overrightarrow{AB}- {1\over 2}\overrightarrow{AB})  ={3\over 7} \overrightarrow{AC}+ {2\over 7}\overrightarrow{AB} \\(3) \bigcirc:\cos \angle BAC={5^2+6^2-4^2\over 2\cdot 5\cdot 6} ={45\over 60} ={3\over 4} \\(4)\bigcirc: {\triangle ACP \over \triangle ACD} ={\overline{CP} \over \overline{CD}} ={4\over 7} \Rightarrow \triangle ACP ={4\over 7} \triangle ACD ={4\over 7}\cdot {1\over 2}\triangle ABC ={2\over 7}\triangle ABC \\ \triangle ABC面積={1\over 2}\overline{AC}\cdot \overline{AB}\sin \angle BAC ={1\over 2}\cdot 5 \cdot 6 \cdot {\sqrt 7\over 4} ={15\over 4}\sqrt 7 \Rightarrow \triangle ACP={15\over 4}\sqrt 7 \cdot {2\over 7} ={15\over 14}\sqrt 7 \\(5)\bigcirc: 假設\cases{A(0,0)\\ D(3,0)\\ C(5\cos A,5\sin A)} \Rightarrow P={4D+3C\over 7} =({12+15\cos A\over 7}, {15\sin A\over 7}) \\\quad \Rightarrow \cases{\overrightarrow{AC} =(5\cos A,5\sin A) \\ \overrightarrow {AP} =({12+15\cos A\over 7}, {15\sin A\over 7})} \Rightarrow \overrightarrow {AP} \cdot \overrightarrow {AC}={1\over 7}(60\cos A+75\cos^2 A+75 \sin^2A) \\\quad ={1\over 7}(60\cos A+75)={1\over 7}(60\cdot {3\over 4}+75) ={120\over 7}\\ 故選\bbox[red, 2pt]{(345)}$$



解答:$$(1) \bigcirc: x_k+u_k=100  \Rightarrow u_k=100-x_k\\ (2)\times: 1微米=10^3奈米\Rightarrow 1奈米= 10^{-3}微米 \Rightarrow v_k=10^{-3}y_k \\(3)\bigcirc: u_k=100-x_k \Rightarrow Var(u)=Var(100-x)= (-1)^2Var(x)= Var(x) \Rightarrow \sigma(u)=\sigma(x) \\(4) \bigcirc: y=21.3x-40 \Rightarrow 10^{-3}y=0.0213x-0.04=-0.0213(100-x)+2.09 \\\qquad \Rightarrow v=-0.0213u+2.09  \Rightarrow b=2.09 \\(5) \bigcirc:增減迴歸直線上的樣本不會改變欲求之迴歸直線\\ 故選\bbox[red, 2pt]{(1345)}$$

三、選填題（占 25 分）

解答:$$q(x)在x=-6有最大值8 \Rightarrow q(x)=-a(x+6)^2+8 \Rightarrow f(x)=(-a(x+6)^2+8)(x+6)+3 \\ =-a(x+6)^3+8(x+6)+3 \Rightarrow f'(x)=-3a(x+6)^2+8 \Rightarrow f''(x)=-6a(x+6)\\ 因此f''(x)=0 \Rightarrow x=-6 \Rightarrow 對稱中心坐標(-6,f(-6)) =\bbox[red, 2pt]{(-6,3)}$$

解答:$$A=(a,b,c)\Rightarrow \cases{d(A,E_1) =|4b+3c-2|=30\\ d(A,E_2) =|3b+4c+5|=30\\ d(A,E_3)= |a+2b+2c+2| =18} \Rightarrow \cases{4b+3c=-28\\ 3b+4c=-35 \\ a+2+2+2=-20} \\ \Rightarrow \cases{a=-2\\b=-1\\ c=-8} \Rightarrow a+b+c= \bbox[red, 2pt]{-11}$$


解答:$$\cases{前3次皆正面,只有1種情形, 機率為{1\over 2^3}={1\over 8} \\前3次為二正一反,第4次為正的情形共有3種,機率為{3\over 2^4} ={3\over 16} \\前4次為二正二反,第5次為正的情形共有C^4_2=6種,機率為{6\over 2^5} ={3\over 16} \\ 不在前3種情形的機率為1-{1\over 8}-{3\over 16}-{3\over 16} ={1\over 2}} \\ \Rightarrow 期望值=240\cdot {1\over 8}+ 320\cdot {3\over 16}+400\cdot {3\over 16}+ 480 \cdot {1\over 2} =\bbox[red, 2pt]{405}$$

解答:$$\cases{L_1:y=m(x-3)+1 \\ L_2:y=-m(x-3)+1} \Rightarrow \cases{d(O,L_1)=1 \\ d(O,L_2)=R(圓半徑)} \\\Rightarrow \cases{{|1-3m|\over \sqrt{m^2+1}}=1 \cdots(1)\\ {|1+3m|\over \sqrt{m^2+1}}=R \cdots(2)}, 由(1)可得(1-3m)^2=m^2+1 \Rightarrow  8m^2-6m=0 \\ \Rightarrow m(4m-3)=0\Rightarrow m=3/4,(0不合,否則L_1=L_2) \Rightarrow R={13/4 \over 5/4} ={13\over 5} \\ \Rightarrow \overline{AB}=2\sqrt{R^2-1} =2\cdot \sqrt{{169 \over 25}-1} =\bbox[red, 2pt]{24\over 5}$$

解答:$$\cos \angle ABC ={\overline{AB}^2+ \overline{BC}^2-\overline{AC}^2 \over 2\cdot \overline{AB}\cdot \overline{BC}} \Rightarrow -{1\over 8} ={3^2+3^2-\overline{AC}^2\over 2\cdot 3\cdot 3} \Rightarrow \overline{AC} ={9\over 2} \\\triangle ABC: 2R={\overline{AC} \over \sin \angle ABC} \Rightarrow {9/2\over 3\sqrt 7/8} = {12\over \sqrt 7} \\ \triangle BCD: 2R={\overline{BD} \over \sin \angle BCD} \Rightarrow {12\over \sqrt 7}={4 \over \sin \angle BCD} \Rightarrow \sin \angle BCD={\sqrt 7\over 3} \Rightarrow \cos \angle BCD={\sqrt 2\over 3} \\ \Rightarrow \cos \angle BCD={\overline{BC}^2+ \overline{CD}^2- \overline{BD}^2 \over 2\cdot \overline{BC} \cdot \overline{CD}} \Rightarrow {\sqrt 2\over 3} ={x^2-7\over 6x} \quad (x=\overline{BD}) \\ \Rightarrow x^2-2\sqrt 2 x-7=0 \Rightarrow x=\bbox[red, 2pt]{3+\sqrt 2}$$

第貳部分、混合題或非選擇題（占 15 分）

解答:$$\begin{bmatrix} 0& c\\ 1& d\end{bmatrix} =\begin{bmatrix} \cos \theta& -\sin \theta\\ \sin \theta& \cos \theta\end{bmatrix} \Rightarrow c=-\sin \theta=-1,故選\bbox[red, 2pt]{(2)}$$

解答:$$A=\begin{bmatrix} \cos \alpha& -\sin \alpha\\ \sin \alpha& \cos \alpha\end{bmatrix} \Rightarrow A^2=\begin{bmatrix} \cos 2\alpha& -\sin 2\alpha\\ \sin 2\alpha& \cos 2\alpha\end{bmatrix} =\begin{bmatrix} 0& c\\ 1& d\end{bmatrix}  \Rightarrow \alpha={\pi\over 4} \\ \Rightarrow A^3=\begin{bmatrix} \cos 3\alpha& -\sin 3\alpha\\ \sin 3\alpha& \cos 3\alpha\end{bmatrix} =\begin{bmatrix} -\sqrt 2/2& -\sqrt 2/2\\ \sqrt 2/2& -\sqrt 2/2\end{bmatrix} \Rightarrow Q=\begin{bmatrix} -\sqrt 2/2& -\sqrt 2/2\\ \sqrt 2/2& -\sqrt 2/2\end{bmatrix} \begin{bmatrix} 1\\ 1\end{bmatrix} =\begin{bmatrix} -\sqrt 2\\ 0\end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{Q=(-\sqrt 2,0)} \\B=\begin{bmatrix} \cos \beta& -\sin \beta\\ \sin \beta & \cos \beta\end{bmatrix} \Rightarrow B^3= \begin{bmatrix} \cos 3\beta& -\sin 3\beta\\ \sin 3\beta & \cos 3\beta\end{bmatrix} =\begin{bmatrix} 0& c\\ 1& d\end{bmatrix}  \Rightarrow \beta={\pi\over 6} \\ \Rightarrow B^4= \begin{bmatrix} \cos 4\beta& -\sin 4\beta\\ \sin 4\beta & \cos 4\beta\end{bmatrix} =\begin{bmatrix} -1/2& -\sqrt 3/2\\ \sqrt 3/2& -1/2\end{bmatrix} \Rightarrow R=\begin{bmatrix} -1/2& -\sqrt 3/2\\ \sqrt 3/2& -1/2\end{bmatrix} \begin{bmatrix} -\sqrt 2\\ 0\end{bmatrix} =\begin{bmatrix} \sqrt 2/2\\ -\sqrt 6/2\end{bmatrix} \\假設\vec a=(1,0) 及\overrightarrow{OR}與\vec a 的夾角為\gamma ,則\cos \gamma ={\vec a\cdot \overrightarrow{OR} \over |\vec a||\overrightarrow{OR}|} ={\sqrt  2/2 \over \sqrt 2\cdot 1} ={1\over 2} \Rightarrow \gamma=\bbox[red, 2pt]{\pi\over 3}$$

解答:$$\cases{P(1,1)\\ Q(-\sqrt 2,0) \\ R(\sqrt 2/2,-\sqrt 6/2)} \Rightarrow \cases{L:y=1 \\ \overleftrightarrow{OR}:y=-\sqrt 3x} \Rightarrow S=L\cap \overleftrightarrow{OR} \Rightarrow \bbox[red, 2pt]{S=(-\sqrt 3/3,1)} \\ \Rightarrow \bbox[red, 2pt]{ \angle OSP ={\pi\over 3}}$$

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