109年度全國科學班聯合學科資格考-數學科
第壹部分:單選題、多選題及填充題
一、單選題:(共二題,每題5分,共10分)
解答:{2+log2a=k3+log3b=klog6(a+b)=k⇒{a=2k−2b=3k−3a+b=6k⇒1a+1b=a+bab=6k2k−2⋅3k−3=6k2⋅6k−3=632=2162=108,故選(4)解答:假設{p=1/6q=2/6r=3/6p1=p+(1−p)(1−q)(1−r)p+((1−p)(1−q)(1−r))2p+⋯=p1−(1−p)(1−q)(1−r)p2=(1−p)q+(1−p)(1−q)(1−r)(1−p)q+((1−p)(1−q)(1−r))2(1−p)q+⋯=(1−p)q1−(1−p)(1−q)(1−r)p3=(1−p)(1−q)r+(1−p)(1−q)(1−r)(1−p)(1−q)r+(1−p)(1−q)(1−r))2(1−p)(1−q)r+⋯=(1−p)(1−q)r1−(1−p)(1−q)(1−r)由於三者分母相同,僅需考慮分子,即{p=1/6=3/18(1−p)q=5/18(1−p)(1−q)r=5/18,因此p2=p3>p1,故選(4)
二、多選題: (共三題,每題 5 分,共 15 分)
解答:(1)◯:每個禮物有3種分法,共有312種分法(2)◯:12選4給第1人,剩下8個選4個給第2人,剩下4個全給第3人,故C124C84C44種分法(3)×:方法同(2),但需乘上3,3,6的排列數,即C123C93C66×3(4)◯:x+y+z=12的自然數解個數,也就是x+y+z=9的非負整數解個數,即H39=C112(5)◯:x+y+z+w=12的非負整數解再扣除x,y,z其中之一≥8(不可能同時有兩個超過7),因此共有H412−C31H412−8=C153−C31C73故選(1245)解答:(1)◯:sin(A+B)=sinAcosB+sinBcosA=2sinAcosB⇒sinAcosB−sinBcosA=0⇒sin(A−B)=0⇒A=B⇒等腰(2)×:cos(A+B)=cosAcosB−sinAsinB=2cosAcosB⇒cosAcosB+sinAsinB=0⇒cos(A−B)=0⇒A−B=±90∘,不一定是等腰(3)◯:sin(A+B)=sinB⇒180∘−(A+B)=B⇒A+2B=180∘=A+B+C⇒B=C(4)×:{A=30∘B=60∘C=90∘符合sin(2A)=sin(2B),但不是等腰(5)◯:cos(2A)=cos(2B)⇒A=B,故選(135)
解答:f(x)=ax4+bx3+cx2+dx+e⇒f′(x)=4ax3+3bx2+2cx+d(1)◯:f′(x)=0的三根為α,β,γ,又{y=f(x)圖形凹向上⇒a>0|β|<|α|<|γ|⇒α+β+γ=−3b/4a>0⇒b<0(2)◯:αβ+βγ+γα=2c/4a<0⇒c<0(3)×:αβγ=−d/4a<0⇒d>0(4)◯:f″
三、 填充題: (共五題,每題 5 分,共 25 分)
解答:令\cases{\vec u=\overrightarrow{AB} \\ \vec v= \overrightarrow{AC}},則\cases{|\vec u|=|\vec v|=2\\ \vec u\cdot \vec v= |\vec u||\vec v|\cos 60^\circ = 2\cdot 2\cdot {1\over 2} =2} \Rightarrow \cases{\overrightarrow{AP} =20\vec u+\vec v\\ \overrightarrow{AQ}= \vec u+20\vec v} \\ \Rightarrow \cases{ |\overrightarrow{AP}|^2 =400|\vec u|^2 +|\vec v|^2 +40\vec u\cdot \vec v=1600 +4+80 = 1684 \\ |\overrightarrow{AQ}|^2 = |\overrightarrow{AP}|^2 =1684\\ (\overrightarrow{AP} \cdot \overrightarrow{AQ})^2 =20|\vec u|^2 +20|\vec v|^2 +401\vec u\cdot \vec v =80+80+802 = 962} \\ \Rightarrow \triangle APQ 面積= {1\over 2}\sqrt{ |\overrightarrow{AP}|^2 |\overrightarrow{AQ}|^2 -(\overrightarrow{AP} \cdot \overrightarrow{AQ})^2} ={1\over 2}\sqrt {1684^2-962^2} ={1\over 2}\cdot 798 \sqrt 3 =\bbox[red,2pt]{399\sqrt 3}B. 丟擲一個公正骰子(即點數 1,2,3,4,5,6每個點數出現的機率皆為{1\over 6})三次,設依序出現的點數為 a,b,c,定義隨機變數X如下:若a是3的倍數,則X=b+c,若a不是3的倍數,則X=|b-c|,試求P(X=4)的值為_____。
C. 在 xy 平面上有一個圓 C:x^2+y^2=16,O為圓心,且在y軸上有一點 A(0,8),若有一光線自A點射向第一象限中圓C上某一點 P,經反射之後平行 x軸射出經過A'點,且A、A'兩點對稱於\overleftrightarrow{OP},試問 P點的 y座標為___。
解答:P在圓C:x^2+y^2=4^2上\Rightarrow P(4\cos\theta, 4\sin\theta) \Rightarrow \overleftrightarrow{OP}:y=\tan \theta \cdot x\\ 又A'與P有相同的y軸標,取A'(a,4\sin\theta);因此MA=A',其中M為鏡射矩陣\\ 即\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta \end{bmatrix}\begin{bmatrix} 0\\ 8 \end{bmatrix} =\begin{bmatrix} a\\ 4\sin\theta\end{bmatrix} \Rightarrow -8 \cos 2\theta = 4\sin \theta \Rightarrow -8(1-2\sin^2\theta)=4\sin \theta \\ \Rightarrow 4\sin^2\theta- \sin \theta -2=0 \Rightarrow \sin\theta ={1+\sqrt{33}\over 8}(P在第一象限,負值不合)\\ \Rightarrow 4\sin\theta =\bbox[red,2pt]{1+\sqrt{33}\over 2}
z^{12}-z^9+z^6 -z^3+1=0 \Rightarrow (z^3+1)(z^{12}-z^9+z^6 -z^3+1)=0 \Rightarrow z^{15}=-1\\ 又\cases{z^{15}=-1的15根(z_k= \cos(\pi+{2k\pi\over 15}) + i\sin(\pi+{2k\pi\over 15}),k=0-14)代表正15邊形的15個頂點\\ z^3=-1的3根(z_k= \cos(\pi+{2k\pi\over 3}) + i\sin(\pi+{2k\pi\over 3}),k=0-2)代表正三角形的三個頂點} \\\Rightarrow 正15邊形的15個頂點扣除正三角形的三個頂點,即為z^{12}-z^9+z^6 -z^3+1=0的12個根\\也就是上圖15個頂點扣除頂點A,F,K,因此周長為9\overline{IJ}+ 3\overline{JL}\\ 由於\cases{\angle IOJ=24^\circ \Rightarrow \overline{IJ} =2\sin 12^\circ = 2\sqrt{(1-a)/2}\\ \angle JOL=48^\circ \Rightarrow \overline{JL} =2\sin 24^\circ =2 \sqrt{1-a^2}} \Rightarrow 9\overline{IJ}+ 3\overline{JL}= \bbox[red,2pt]{18\sqrt{1- a\over 2} +6\sqrt{1-a^2}}
解答:橢圓\cases{中心(0,0)\\ 2a=4\\ 2b=3} \Rightarrow 橢圓方程式:{x^2\over 4}+{y^2\over 9/4}=1 \Rightarrow y={3\over 4}\sqrt{4-x^2} (只考慮第一象限)\\ \Rightarrow 陰影面積= \int_1^\sqrt 2 {3\over 4}\sqrt{4-x^2}\,dx ={3\over 4}\int_{\pi/3}^{\pi/4} \sqrt {4-4\cos^2\theta}(-2)\sin\theta \,d\theta(\because x=2\cos \theta \Rightarrow dx=-2\sin \theta d\theta)\\ =-3 \int_{\pi/3}^{\pi/4} \sin^2\theta \,d\theta =-{3\over 2}\int_{\pi/3}^{\pi/4} 1-\cos 2\theta\,d\theta =-{3\over 2} \left.\left[ \theta-{1\over 2}\sin 2\theta\right]\right|_{\pi/3}^{\pi/4} \\=-{3\over 2}\left( ({\pi\over 4}-{1\over 2}) -({\pi\over 3} -{\sqrt 3\over 4})\right) = \bbox[red,2pt]{{1\over 8}(6+\pi -3\sqrt 3)}
第貳部分:非選擇題 (數學寫作能力、計算證明題共 50 分)
四、 數學寫作能力: (共二題,共計 12 分)
解答:假設A(x_1,y_1, z_1)在平面E上,且\overleftrightarrow{PA}\bot E,則\overline{PA}即為所求;\\ E的法向量\vec n= (a,b,c) 與\overrightarrow{PA} =(x_1-x_0,y_1-y_0, z_1-z_0)平行 \\,即{ x_1-x_0 \over a} ={ y_1-y_0 \over b} ={z_1-z_0 \over c}=t \Rightarrow \cases{x_1=at+ x_0\\ y_1=bt+y_0 \\ z_1=ct+z_0};\\又A在E上\Rightarrow a(at+x_0 )+ b(bt+y_0) +c(ct+z_0)+ d=0 \Rightarrow t=-{ax_0+ by_0 +cz_0+ d\over a^2+b^2+c^2} \\\Rightarrow \left| \overrightarrow{PA}\right| =\left| t \vec n\right| =\left|-{ax_0+ by_0 +cz_0 +d\over a^2+b^2+c^2}\cdot \sqrt{a^2+b^2 +c^2} \right| =\left|{ax_0+ by_0 +cz_0 +d\over\sqrt{a^2+b^2 +c^2}} \right| \\\bbox[red,2pt]{故得證}解答:(1)(\cos \theta+i\sin\theta)^n = \cos n\theta + i\sin n\theta\\ (2)利用歸納法:\\ n=1時,顯然成立\\ 假設n=k成立,即z^k+{1\over z^k}=2\cos k\theta;\\ 當n=k+1時,z^{k+1}+{1\over z^{k+1}} =(z^k+{1\over z^k}) (z+{1\over z}) - (z^{k-1}+ {1\over z^{k-1}}) =2\cos k\theta \cdot 2\cos \theta-2\cos(k-1)\theta \\ =4\cos k\theta\cos \theta-2\cos (k\theta-\theta) =4\cos k\theta\cos \theta-2(\cos k\theta\cos \theta +\sin k\theta \sin \theta)\\ =2(\cos k\theta \cos\theta -\sin k\theta\sin \theta)=2 \cos(k+1)\theta \Rightarrow z^{k+1}+{1\over z^{k+1}} =2 \cos(k+1)\theta \\ n=k+1時亦成立,\bbox[red,2pt]{故得證}
五、計算證明題: (共 5 題,共計 38 分)
解答:圓:x^2+(y-b)^2=a^2 \Rightarrow \cases{圓心P(0,b) \\ 圓半徑r=a} \Rightarrow \cases{圓面積A=a^2\pi \\ P繞x軸一圈的圓周長=2b\pi} \\ \Rightarrow 圓繞x軸旋轉體積=(a^2\pi)\times (2b\pi)= 2a^2b\pi^2 ,\bbox[red,2pt]{故得證}解答:\cos \angle A= {\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\cdot \overline{AB}\cdot \overline{AC}} ={4^2+ 6^2 - (2\sqrt{19})^2 \over 2\cdot 4\cdot 6} =-{1\over 2} \Rightarrow \angle A=120^\circ \\ 令\angle BAP=\theta,則\angle CAP=120^\circ-\theta \Rightarrow \cases{\triangle ABP ={1\over 2}\cdot \overline{AB}\cdot \overline{AP}\sin \theta=4\sin \theta \\ \triangle ACP ={1\over 2}\cdot \overline{AC}\cdot \overline{AP}\sin (120^\circ-\theta) = 6 \sin (120^\circ-\theta)} \\ \Rightarrow \triangle ABP+\triangle ACP =4\sin\theta +6 \sin (120^\circ-\theta) =4\sin\theta +3\sqrt 3\cos\theta +3\sin\theta\\ =7\sin\theta +3\sqrt 3\cos \theta \Rightarrow 最大值為\sqrt{7^2 +(3\sqrt 3)^2} =\sqrt{76} =\bbox[red,2pt]{2\sqrt{19}}
解答:
(1)由定義可知:\cases{g(2n)=g(n) \\ g(2n-1)=2n-1},n\in \mathbb{N},因此S_n = g(1)+ g(2)+\cdots +g(2^n) \\=\sum_{k=1}^{2^{n-1}} \left(g(2k-1)+g(2k) \right) = \sum_{k=1}^{2^{n-1}} \left(g(2k-1)+g(k) \right) =S_{n-1}+ \sum_{k=1}^{2^{n-1}}(2k-1) =S_{n-1} + {2^n\times 2^{n-1}\over 2}\\ =S_{n-1}+ 4^{n-1}\\ 當n=1時,S_1= g(1)+ g(2^1)= 1+1=2\Rightarrow \bbox[red,2pt]{ S_n= \begin{cases} 2,&n=1\\ S_{n-1}+ 4^{n-1}, &n\ge 2\end{cases} };(2)S_{2019}= S_{2018} +4^{2018} = S_{2017} +4^{2017} +4^{2018} = S_1 +4+ 4^2 +\cdots 4^{2018} \\=1+ (1+ 4+ 4^2 +\cdots 4^{2018})=1+{1-4^{2019}\over 1-4} =1+{1\over 3}(4^{2019}-1)\\ \Rightarrow S_{2019}= \bbox[red,2pt]{{1\over 3}(2+4^{2019})}
解答:36位考生成績依序為:x_1\le x_2\le x_3 \le \cdots \le x_{36},且\cases{\bar x= (x_1+x_2+\cdots +x_{36})/36= 50\\[1ex] \sigma= \sqrt{{1\over 36}\sum_{i=1}^{36}(x_i-\bar x)^2} =10} \\ 令A=\{x_i\mid x_i\ge 65或x_i\le 35,i=1-36\},則我們有(x-\bar x)^2=(x-50)^2 \le 15^2, \forall x\in A\\若A有n個元素,則 \sum_{x_i\in A}(x_i-50)^2 \le 225n \le \sum_{i=1}^{36}(x_i-50)^2 =10^2 \times 36=3600\\ \Rightarrow 225n \le 3600 \Rightarrow n\le 16 \Rightarrow 至少有36-16=\bbox[red,2pt]{20}位成績介於36與64之間
解答:
4.某個班級有36個學生,某次考試成績的算術平均數為50分,標準差為10分,試問最少有多少學生的分數介於35分到65分之間(不包含35分及65分)?
解答:
(1)\cases{P(x_0,y_0)\\ F_1(c,0) \\[1ex]{x^2\over a^2} +{y^2\over b^2}=1} \Rightarrow \overline{PF_1}^2 = (x_0-c)^2 +y_0^2 = (x_0-c)^2 + b^2\left(1-{x_0^2\over a^2} \right)\\ = (x_0-c)^2 + (a^2-c^2)\left(1-{x_0^2\over a^2} \right) = x_0^2-2cx_0+ c^2 +a^2-x_0^2-c^2+{c^2\over a^2}x_0^2 ={c^2\over a^2}x_0^2 -2cx_0+a^2\\ =({c\over a}x_0-a)^2 \Rightarrow \overline{PF_1} =\left|{c\over a}x_0-a \right|\\ 又{x_0^2\over a^2} +{y_0^2\over b^2}=1 \Rightarrow {x_0^2\over a^2}\lt 1 \Rightarrow -1\lt {x_0\over a} \lt 1 \Rightarrow -c\lt c\cdot {x_0\over a} \lt c \lt a \Rightarrow \left|{c\over a}x_0-a \right| =a-{c\over a}x_0\\ \Rightarrow \overline{PF_1} =a-{c\over a}x_0,\bbox[red,2pt]{故得證}(2){x^2\over 25} +{y^2\over 16}=1 \Rightarrow \cases{a=5\\ b=4} \Rightarrow c=3 \Rightarrow \cases{F_1(3,0)\\ F_2(-3,0)} \\ 令P(s,t),由\cases{\overline{PF_1} +\overline{PF_2} =2a=10\\ \overline{PF_1} :\overline{PF_2}= 2:3} \Rightarrow \cases{\overline{PF_1}=4 \\ \overline{PF_2}=6} \Rightarrow \cases{(s-3)^2+ t^2 =16\\ (s+3)^2+t^2=36} \\ \Rightarrow (s+3)^2 -(s-3)^2 =20 \Rightarrow 12s =20 \Rightarrow s=5/3 \Rightarrow {16\over 9}+t^2 =16 \Rightarrow t^2={128\over 9} \\ \Rightarrow t=\pm {8\sqrt 2\over 3} \Rightarrow \bbox[red,2pt]{P\left({5\over 3},\pm {8\sqrt 2\over 3}\right)}
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