114學年度四技二專統測--數學(B)詳解

114 學年度科技校院四年制與專科學校二年制
統 一 入 學 測 驗-數學(B)

解答: $$f(x)=x^n-2x-3 \Rightarrow f(2)=1 \Rightarrow 2^n-4-3=1 \Rightarrow 2^n=8 \Rightarrow n=3，故選\bbox[red, 2pt]{(B)}$$

解答: $${1\over 2}\pi=1.57\lt \theta=3.1 \lt \pi=3.14 \Rightarrow \theta 在第二象限，故選\bbox[red, 2pt]{(C)}$$

解答: $$\overrightarrow{AB} \cdot \overrightarrow{AC} =(2,3) \cdot (3,y) =6+3y \Rightarrow 最大值=6+3\cdot 2=12，故選\bbox[red, 2pt]{(D)}$$

解答:

$$\overrightarrow{AB} +\overrightarrow{BC} =棕色，\overrightarrow{PQ}=紅色，如上圖，只有(A)符合紅色=棕色，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=g(x) \Rightarrow x^2-2x-3=x^2-x-6 \Rightarrow x=3 \Rightarrow f(3)=g(3)=0 \Rightarrow 交點(3,0)，故選\bbox[red, 2pt]{(D)}$$

解答: $${1\over 2}\left(x-{1\over 2} \right)^2+{1\over 8} ={1\over 2}x^2-{1\over 2}x+{1\over 4} =ax^2+bx+c \Rightarrow \cases{a=1/2\\ b=-1/2\\ c=1/4} \\ \Rightarrow {a\over 4}+{b\over 2} +c={1\over 8}-{1\over 4} +{1\over 4} ={1\over 8}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{\sin \theta^\circ=\cos(90^\circ-\theta^\circ) \\ \sin^2 \theta+\cos^2 \theta=1} \Rightarrow \sin^2 10^\circ +2\sin^2 20^\circ+4 \sin^2 30^\circ+ 2\sin^2 70^\circ+ \sin^280^\circ \\=\sin^2 10^\circ +2\sin^2 20^\circ+4 \sin^2 30^\circ+ 2\cos^2 20^\circ+ \cos^2 10^\circ\\ =(\sin^2 10^\circ +\cos^2 10^\circ)+ 2(\sin^2 20^\circ +\cos^2 20^\circ)+4 \sin^2 30^\circ =1+2+4\cdot \left({1\over 2} \right)^2 =4，故選\bbox[red, 2pt]{(D)}$$

解答: $$L_2:{x\over 5}-{y \over 12}=1 且L_1\parallel L_2 \Rightarrow L_2: {x\over 5}-{y\over 12}=k \Rightarrow d(L_1,L_2) ={|k-1|\over \sqrt{1/25+1/144}} =1 \\ \Rightarrow (k-1)^2={1\over 25}+{1\over 144} ={169\over 3600} =\left({13\over 60} \right)^2 \Rightarrow k-1=\pm {13\over 60} \Rightarrow \cases{k=73/60\\ k=47/60} \\ \Rightarrow L_1: {x\over 5}-{y\over 12}={73\over 60} 或{x\over 5}-{y\over 12}={47\over 60} \Rightarrow L_2:y={12\over 5}x-{73\over 5}或y={12\over 5}x-{47\over 5}，故選\bbox[red, 2pt]{(A)}$$

解答: $$一毫米={10^{-3}\over 10^{-9}} =10^6奈米 \Rightarrow {3奈米\over 0.017毫米} ={3\over 0.017\times 10^6} \approx 176.47\times 10^{-6} \approx 1.8\times 10^{-4}\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{L_1:4x-3y+a=0\\ L_2: 8x+by+4=0} \Rightarrow L_1=L_2 \Rightarrow 8x-6y+2a=8x+by+4 \Rightarrow \cases{a=2\\ b=-6} \\ \Rightarrow a+b=-4，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{a=\log_{25} 2 =\log_5 \sqrt 2\\ b=\log_{25} 3 =\log_5\sqrt 3} \Rightarrow 3a+2b= \log_5 2\sqrt 2+ \log_5 3=\log_5 6\sqrt 2 \Rightarrow 3a+2b+1=\log_5 30\sqrt 2 \\ \Rightarrow 5^{3a+2b+1}=30\sqrt 2，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{6\le x\le 8\\ y\ge x/2} \Rightarrow (x,y) =\cases{(6,3-6)共4種\\ (7,4-7) 共4種\\ (8,4-8) 共5種} \Rightarrow 4+4+5=13種，故選\bbox[red, 2pt]{(B)}$$

解答: $$3,4,\alpha,10,\beta \Rightarrow \cases{中位數=\alpha=6\\ 全距=\beta-3=9 \Rightarrow \beta=12} \\\Rightarrow 樣本標準差 \gamma=\sqrt{4^2+3^2+1^2+3^2+5^2\over 5-1} =\sqrt{15} \Rightarrow \alpha+\beta+ \gamma =18+\sqrt{15}，故選\bbox[red, 2pt]{(B)}$$

解答: $${1\over 50}(9\cdot 32+ 8\cdot 10+ 6\cdot 5+ 3\cdot 2+ 1\cdot 1) ={405\over 50} =8.1，故選\bbox[red, 2pt]{(C)}$$

解答: $$(1,4)按A鍵k次後坐標為(1+3k,4-k) \Rightarrow \cases{k=1 \Rightarrow(4,3)下一步(5,3) \\ k=2 \Rightarrow(7,2) \\k=3 \Rightarrow (10,1) \\k=4 \Rightarrow (13,0)\\ k=5 \Rightarrow (16,-1)}，故選\bbox[red, 2pt]{(C)}$$

解答: $$假設\overline{TN} =a \Rightarrow \cos 30^\circ={\sqrt 3\over 2} ={20^2+(20\sqrt 3)^2-a^2 \over 2\cdot 20\cdot 20\sqrt 3} \Rightarrow a=20 \Rightarrow x={20\over 5}=4，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{肉4選1有C^4_1=4種選法\\ 海鮮5選3有C^5_3=10種選法\\ 副餐4選1有C^4_1=4種選法}\Rightarrow 共有4\times 10\times 4=160種搭配，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{35x+46y=2025\\ 65x+54y=-25} \quad 兩式相加\Rightarrow 100x+100y=2000 \Rightarrow x+y=20 \Rightarrow p+q=20\\ \Rightarrow 20x^2+(p+q) x-114=0的兩根和=-{p+q\over 20} =-{20\over 20}=-1，故選\bbox[red, 2pt]{(B)}$$

解答: $$已知\cases{C_1:圓心O_1(0,1), 半徑r_1=1\\ C_2:圓心O_2(0,2), 半徑r_2=2 \\ C_3:圓心O_3(0,3), 半徑r_3=3}\quad 及直線L滿足 \cases{C_1相離\\ C_2相切 \\C_3相割} \Rightarrow \cases{d(O_1, L)\gt r_1\\ d(O_2,L) =r_2\\ d(O_3, L)\lt r_3} \\ (A)\times: L:5x +12y+1=0 \Rightarrow d(O_2,L)= 25/13 \ne 2 \\(B) \times:L:7x-24y-1=0 \Rightarrow d(O_2,L) =49/25 \ne 2 \\(C)\bigcirc: L:3x+4y+2=0 \Rightarrow d(O_2,L) =2 \\ (D)\times: L:8x-15y-5=0 \Rightarrow d(O_2, L)= 35/17 \ne 2\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{L_3\bot L_1 \Rightarrow L_3斜率=1/2 \Rightarrow x截距\times y截距\lt 0\\ L_3\bot L_2 \Rightarrow L_3斜率=-3  \Rightarrow x截距\times y截距\gt 0}，故選\bbox[red, 2pt]{(A)}$$

解答: $$中班與小班共有4個不同節目，順序有4!=24種；\\ 在4個節目中有5個間隔插入大班的3個節目，有C^5_3=10種插入法，每種插入法大班皆有3!=6種順序\\ 因此共有24\times 10\times 6=1440種順序，故選\bbox[red, 2pt]{(C)}$$

解答: $$假設訂購盒裝A有x盒，盒裝B有y盒，則\cases{花費f(x,y)=200x+210y\\ 購得布丁3x+2y\ge 48個\\ 購得奶酪3x+4y \ge 72個}\\ 3x+2y=48與3x+4y=72交於(8,12) \Rightarrow f(8,12)=4120，故選\bbox[red, 2pt]{(A)}$$

解答: $$\overline{CD} =40=\overline{AB} \Rightarrow \overline{AF}= 40/2=20 \Rightarrow \overline{AD} ={\overline{AF} \over \tan 30^\circ} ={20\over 1/\sqrt 3} =20\sqrt 3 \\ \overline{TD}^2= \overline{AD}^2+ \overline{AT}^2 \Rightarrow 4800= 1200+ \overline{AT}^2 \Rightarrow \overline{AT} =60 \Rightarrow \overline{BT} =60-40=20，故選\bbox[red, 2pt]{(C)}$$

解答: $$1000位總平均為a，甲班總分為50\times 80=4000，因此950位同學平均={1000a-4000\over 950}=a-1 \\ \Rightarrow 50a=3050 \Rightarrow a=61，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{a_1,a_3, a_5, \dots為公比為-3的等比數列\\ a_2,a_4,a_6,\dots為公差為3的等差數列} \Rightarrow \cases{a_{99} =a_{2\cdot 50-1}= a_1\cdot (-3)^{49} =5\times 3^{49} \\ a_{100} =a_{2\cdot 50} =a_2 +49\cdot 3=202} \\ \Rightarrow \cases{a_1=-5\\ a_2=55} \Rightarrow a_1+a_2+ \cdots+ a_{10} =-5+55+15+58-45+61+135+64-405+67 \\=0，故選\bbox[red, 2pt]{(D)}$$

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