113年台北市聯合轉學考-升高二(普高)-數學詳解

 臺北市高級中等學校 113 學年度聯合轉學考招生考試

升高二數學科試題(普高)

一、 單選題： (共 70 分)

解答: $${2\over 7-\sqrt{41}} ={2(7+\sqrt{41})\over (7-\sqrt{41}) (7+\sqrt{41})} = {14+2\sqrt{41} \over 49-41} = {7+ \sqrt{41} \over 4} ，故選\bbox[red, 2pt]{(A)}$$

解答: $$5\times 2=10，故選\bbox[red, 2pt]{(D)}$$

解答: $$\log 100+\log 0.001 =\log 10^2 +\log ({1\over 10})^3 =2\log 10+3\log{1\over 10} =2\cdot 1+3\cdot(-1) =-1，故選\bbox[red, 2pt]{(C)}$$

解答: $$\log (\log N) =5 \Rightarrow \log N=10^5 \Rightarrow N=10^{10^5} =10^{100000} \Rightarrow N是100001位數，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(-3)=2(-8)^4+15(-8)^3-19(-8)^2-86(-8)+19 \\=16\cdot 8^3-15\cdot 8^3-152\cdot 8+86\cdot 8+19 =8^3-66\cdot 8+19 =(64-66)8+19 \\=-16+19=3，故選\bbox[red, 2pt]{(E)}$$

解答: $$\cases{f(x)=(x^2-2x-35)p(x)+ax-1 =(x-7)(x+5)p(x)+ax-1 \\g(x) =(x-7)q(x)-4} \\ \Rightarrow f(x)+g(x) =(x-7)(x+5)p(x)+ (x-7)q(x)+ax-5 \Rightarrow f(7)+g(7)=7a-5=9 \Rightarrow a=2 \\ \Rightarrow f(x)=(x-7)(x+5)+2x-1 \Rightarrow f(-5)=-10-1=-11，故選\bbox[red, 2pt]{(B)}$$

解答: $$C:x^2+y^2-4x-5=0 \Rightarrow (x-2)^2+y^2=3^2 \Rightarrow \cases{圓心O(2,0)\\ 圓半徑r=3} \\ \Rightarrow \cases{O至L_1距離=d(O,L_1)= 3/\sqrt 2\\ O至L_2距離= d(O,L_2) =2/\sqrt 2} \Rightarrow d(O,L_1)\gt d(O,L_2) \Rightarrow \overline{AB}\lt \overline{CD} \\ \Rightarrow L_1\parallel L_2且\overline{AB}\lt \overline{CD}  \Rightarrow ABCD為梯形，故選\bbox[red, 2pt]{(D)}$$

解答: $$假設第n列有a_n個數字，即a_1=1,a_2=3,a_3=5,a_4=7,則\langle a_n\rangle成等差數列,其中\cases{首項a_1=1\\ 公差d=2}\\ 因此a_1+a_2+ \cdots+a_{11}= \sum_{k=1}^{11}(2k-1)=121 \Rightarrow 第12列第1個數字為122\\ \Rightarrow 第12列有23個數且總和=122+123+ \cdots+144={1\over 2}(122+144)\cdot 23=3059，故選\bbox[red, 2pt]{(E)}$$

解答: $$各組平均數\cases{\mu_甲=6\\ \mu_乙=3\\ \mu_丙=3\\ \mu_丁=6\\ \mu_戊=7 } \Rightarrow \cases{甲-\mu_甲:-5,-5,-5,-5,-5,5,5,5,5,5 \\ 乙-\mu_乙:-2,-2,-1,-1,0,0,1,1,2,2\\ 丙-\mu_丙: -2,-2,-2,-2,-2,2,2,2, 2,2\\ 丁-\mu_丁:-1,-1,-1,0,0,0,0,1,1,1\\ 戊-\mu_戊:0,0,0,0,0,0,0,0,0,0} \\ \Rightarrow \cases{\sigma_甲=5 \\\sigma_乙=\sqrt 2 \\ \sigma_丙=2 \\\sigma_丁= \sqrt{0.6} \\\sigma_戊=0 } \Rightarrow \sigma_甲\gt \sigma_丙 \gt \sigma_乙 \gt \sigma_丁\gt \sigma_戊 \Rightarrow 中位數\sigma_乙，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{\tan \theta=3 \\ \sin \theta\gt 0} \Rightarrow \cases{\sin \theta= 3\sqrt{10}\\ \cos \theta=1/\sqrt{10}} \Rightarrow \sin^2\theta-\cos^2\theta={9\over 10}-{1\over 10}={4\over 5}，故選\bbox[red, 2pt]{(B)}$$


解答: $$\cases{\sin 29^\circ \lt \sin 30^\circ={1\over 2}\\ \tan 45^\circ=1\\ \cos 320^\circ=\cos(360^\circ-40^\circ)=\cos 40^\circ \gt \cos 60^\circ={1\over 2}\\ \cos 200^\circ \lt 0\lt {1\over 2}\\ \cos 1450^\circ =\cos(1450^\circ-360^\circ\times 4)= \cos 10^\circ \gt \cos 60^\circ={1\over 2}} \\ \Rightarrow 大於{1\over 2}:\tan 45^\circ,\cos 320^\circ及\cos 1450^\circ，共3個，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{抽中2紅球的機率=C^3_2/C^9_2=1/12 \\抽中2黃球的機率=C^6_2/C^9_2 =5/12} \Rightarrow 期望值=720\cdot {1\over 12} +120\cdot {5\over 12}=110，故選\bbox[red, 2pt]{(A)}$$

解答: $$1+2+\cdots+ 8=36 \Rightarrow 每列數字和=36\div 2=18\\ \Rightarrow 列數字可為=(1 , 2 , 7 , 8),(1 , 3 , 6 , 8), (1 , 4 , 5 , 8), (1 , 4 , 6 , 7), (2 , 3 , 5 , 8), (2 , 3 , 6 , 7)\\\qquad ,(2 , 4 , 5 , 7), (3 , 4 , 5 , 6),共8組\\ 又需符合每行數字和相等,因此僅有 \begin{array}{|c|c|c|c|}\hline 1&4&6&7\\\hline 8&5&3&2\\\hline\end{array} 符合要求\\ 再加上左右排列數及上下互換，共有4!\times 2=48種填法，故選\bbox[red, 2pt]{(A)}$$

解答: $$擲三次回到A的情形:(不,不,不),(不,右,左),(不,下,上), (右,不,左),(下,不,上), (右,左,不)\\, (下,上,不)， 共有7種情形，每種機率都是{1\over 3^3}={1\over 27}，因此P={7\over 27}，故選\bbox[red, 2pt]{(D)}$$

二、多重選擇題： (共 30 分)

解答: $$(A)\times: 2x+3y=5的斜率為-{2\over 3}\ne -{3\over 2} \\(B) \bigcirc: 3x+y=-6的斜率=-3且x截距=-2\\ (C)\times: \cases{L_1斜率m_1= 2/3\\ L_2斜率m_2=-2/3} \Rightarrow m_1m_2=-{4\over 9} \ne -1 \\(D)\bigcirc: y=mx+m-2=m(x+1)-2 必過(-1,-2)\\ (E) \times: f(x,y)=2x-y+5 \Rightarrow \cases{f(A)=f(-7,-10)=1\\ f(B)=f(18,40) = 1} \Rightarrow f(A)f(B)\gt 0 \Rightarrow A,B同側\\，故選\bbox[red, 2pt]{(BD)}$$


解答: $$(A)\times: 也可能為-1\\(B)\times: 不一定，例如:(x,y)=(-1,0),(0,-1),(3/4,-1/2),(1/4,3/2) \\\Rightarrow \cases{\mu_x= \mu_y=0\\ \sum x_iy_i=0 } \Rightarrow 相關係數=0 \\(C) \times: \cases{Cov(2X,3Y)=6Cov(X,Y)\\ \sigma(2X)=2\sigma(X)\\ \sigma(3Y)=3\sigma(Y)} \Rightarrow \rho(2X,3Y)={Cov(2X,3Y)\over \sigma(2X) \sigma(3Y)}={6Cov(X,Y)\over 6\sigma(X)\sigma(Y)} \\\qquad =\rho(X,Y) =0.1\\ (D)\bigcirc: \cases{Cov(2X+1,-3Y)=-6Cov(X,Y)\\ \sigma(2X+1)=2\sigma(X)\\ \sigma(-3Y)=3\sigma(Y)} \Rightarrow \rho(2X+1,-3Y)={-6Cov(X,Y) \over 6\sigma(X)\sigma(Y)} \\\qquad =-\rho(X,Y)=-0.1\\ (E)\bigcirc: \cases{公頓=1000公斤\\ 公分=0.01公尺} \Rightarrow \rho(公頓,公分)={1000\times 0.01Cov(公斤,公尺) \over 1000\sigma(公斤)\times 0.01\sigma(公尺)} =\rho(公斤,公尺)\\，故選\bbox[red, 2pt]{(DE)}$$

解答: $$(3x+2y)^5 =C^5_02^5y^5+ C^5_1(3x)(2^4y^4) +C^5_2 (3^2x^2) (2^3y^3) +C^5_3 (3^3x^3) (2^2y^2)  +C^5_4 (3^4x^4) (2y)  +C^5_5 (3^6x^5) \\=32y^5+240xy^4 +720x^2y^3 + 1080x^3y^2 + 810x^4y+729 x^5 \\ \Rightarrow \cases{最大係數項:x^3y^2\\ 次大係數項:x^4y}，故選\bbox[red, 2pt]{(CD)}$$

解答: $$(A) \bigcirc: \cos \angle C={3^2+5^2-7^2 \over 2\cdot 3\cdot 5} =-{1\over 2}=-0.5\lt -0.3 \\(B)\times: {3\over \sin \angle A}={5\over \sin \angle B} \Rightarrow \sin \angle A:\sin \angle B=3:5,並非\angle A:\angle B \\(C)\bigcirc: {7\over \sin \angle C} ={7\over \sin 120^\circ} ={7\over \sqrt 3/2}=2R \Rightarrow R={7\over \sqrt 3} \Rightarrow 外接圓面積=R^2\pi ={49\over 3}\pi \lt 17\pi \\(D) \times: 0\lt \angle B\lt 180^\circ \Rightarrow \sin \angle B \gt 0\\ (E) \bigcirc: \triangle ABC面積={1\over 2} \cdot 3\cdot 5\cdot \sin \angle C={15\over 2}\cdot {\sqrt 3\over 2}={15\sqrt 3\over 4} \gt 6\\，故選\bbox[red, 2pt]{(ACE)}$$

解答: $$f(x)的廣域特徵近y=2x^3 \Rightarrow f(x)=2(x+2)^3+p(x+2)^2 +q(x+2)+r\\ 又x=-2附近的一次近似為y=3x+12=3(x+2)+6\\ \Rightarrow f(x)= 2(x+2)^3+p(x+2)^2 +3(x+2)+6 \Rightarrow f'(x)=6(x+2)^2+ 2p(x+2)+3\\ \Rightarrow f''(x)=12(x+2)+2p\\ 對稱中心(-2,k) \Rightarrow \cases{f''(-2)=0\\f(-2)=k} \Rightarrow \cases{p=0\\ k=6} \Rightarrow f(x)=2(x+2)^3+3(x+2)+6\\ (A)\bigcirc: f(-2)=6 \Rightarrow f(x)除以(x+2)的餘式為6 \\(B)\times: 餘式為3(x+2)+6\\ (C) \times: f(0)=2\cdot 2^3+3\cdot 2+6=16+6+6=28 \ne 26 \\(D)\bigcirc: 對稱中心(-2,k)= (-2,6) \\(E)\bigcirc: 圖形向左平移2,再向下平移6就變成f(x)=2x^3+3x\\，故選\bbox[red, 2pt]{(ADE)}$$

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解題僅供參考，轉學考歷年試題及詳解