74年大學聯考-自然組數學詳解

 74年升大學聯考數學科(自然組)試題

解答：$$令\cases{O(0,0,0)\\ P(-1,0,2)\\ Q(1,-1,0)} \Rightarrow \cases{\overrightarrow{OP} =(-1,0,2)\\ \overrightarrow{OQ}= (1,-1,0)} \Rightarrow \vec n= \overrightarrow{OP} \times \overrightarrow{OQ} =(2,2,1) \Rightarrow 平面\pi:2x+2y+z=0\\ 過球心A(4,3,4)且方向向量為\vec n的直線L:{x-4\over 2} ={y-3\over 2} ={z-4\over 1} \Rightarrow 切點B(2t+4, 2t+3,t+4)\\ \Rightarrow \overline{AB}=  d(A,\pi) \Rightarrow \sqrt{4t^2+ 4t^2+ t^2} = \left|{ 8+ 6+4\over \sqrt{2^2+2^2 +1^2}} \right| =6 \Rightarrow t=-2 (2不合) \\\Rightarrow B(0,-1,2) =(x,y,z) \Rightarrow \cases{x=0 \Rightarrow 第1題選 \bbox[red,2pt]{ (C)}\\ y=-1 \Rightarrow 第2題選 \bbox[red,2pt]{ (B)}\\z=2 \Rightarrow 第3題選 \bbox[red,2pt]{ (E)}\\}$$

解答：$$\cases{z\omega =-2+11i\\ z^2+\omega^2 =4+28i} \Rightarrow \cases{(z+\omega)^2 =4+28i+ 2(-2+11i) = 50i =50(\cos {\pi\over 2} + i\sin {\pi\over 2})\\ (z-\omega)^2 = 4+28i -2(-2+11i) = 8+6i =10(\cos \theta+ i\sin \theta)} \\ \Rightarrow \cases{z+ \omega= \pm 5\sqrt 2(\cos {\pi\over 4} +i\sin {\pi \over 4}) = \pm (5+5i) \\z-\omega= \pm \sqrt{10} (\cos {\theta\over 2}+ i\sin {\theta\over 2} ) =\pm (3+i)}， 其中\cases{\cos \theta = 4/5 \\ \sin \theta = 3/5} \Rightarrow \cases{\cos \theta/2 = 3/\sqrt{10}\\ \sin \theta/2= 1/\sqrt{10}} \\ \Rightarrow (z,\omega)=(4+3i,1+2i),(-1-2i,-4,-3i), (1+2i, 4+3i),(-4-3i,-1-2i)\\ \Rightarrow 共有4組解，故選\bbox[red,2pt]{(D)}$$

解答：$$由4.可知:z=4+3i、1+2i、-1-2i、-4-3i，故選\bbox[red,2pt]{(AE)}$$

第二部分：非選擇題

    在本部分中第一題為填充題，第二題至第五題為計算與證明題，請都在「非選擇題試卷」上作答。
一、填充題：
    本題共有八個空格，每個空格5分，共40分，請答在「非選擇題試卷」上的第一欄，務必寫上格號(子，丑，‧‧‧，未)後，再寫答案。(為節省「非選擇題試卷」空間，本題作答，請不要寫出演算過程。)

解答：$${x^2+2\over x^2+4x+1} +{x^2+4x +1\over x^2+2} ={5\over 2} \Rightarrow 2((x^2+2)^2 +(x^2+4x+1)^2)= 5((x^2+2) (x^2+4x+1))\\ \Rightarrow x^4+4x^3 -29x^2+24 x =0 \Rightarrow (x+8)x(x-1)(x-3) =0 \Rightarrow x=-8,0,1,3\\ \Rightarrow 最大者為\bbox[red,2pt]{3}，最小者為\bbox[red, 2pt]{-8}$$

解答：$$\cos 4\theta = 2\cos^2 2\theta-1 = 2(2\cos^2 \theta -1)^2-1 =8\cos^4\theta -8\cos^2\theta +1 \\= 8(1-\sin^2 \theta)^2- 8(1-\sin^2 \theta) +1 =8\sin^4\theta-8\sin^2\theta +1 =\sin \theta \Rightarrow \bbox[red, 2pt]{8x^4-8x^2-x+1=0}\\ \Rightarrow 8x^2(x+1)(x -1)-(x-1) =0 \Rightarrow (x-1)(8x^3+8x^2-1)=0 \\\Rightarrow (x-1)(2x+1)( 4x^2+2x-1)=0 \Rightarrow x=1,-{1\over 2},{-1\pm \sqrt 5\over 4} \Rightarrow 最小實根=\bbox[red, 2pt]{-1- \sqrt 5\over 4}$$

解答：$$\cases{x=\cos^3\theta \\ y=\sin^3\theta +\sin \theta} \Rightarrow x^2 +y^2 = \color{blue}{\cos^6\theta +\sin^6\theta} +2\sin^4\theta +\sin^2\theta \\ =\color{blue}{(\cos^2\theta +\sin^2 \theta)(\cos^4\theta -\cos^2\theta \sin^2\theta+ \sin^4\theta)} +2\sin^4\theta +\sin^2\theta \\=\cos^4\theta-\cos^2\theta \sin^2\theta+ 3\sin^4\theta+\sin^2\theta =(1-\sin^2\theta)^2 -(1-\sin^2\theta)\sin^2\theta+ 3\sin^4\theta+\sin^2\theta\\ = 5\sin^4\theta -2\sin^2\theta+1\\ 令f(\theta)= 5\sin^4\theta -2\sin^2\theta+1 \Rightarrow f'(\theta) = 20\sin^3\theta\cos\theta -4\sin\theta \cos \theta = 4\sin\theta \cos\theta(5\sin^2\theta-1)\\ =2\sin 2\theta(5\sin^2\theta-1)，因此f'(\theta)=0 \Rightarrow \cases{\sin 2\theta =0 \Rightarrow \sin\theta =1(\theta=\pi/2) \Rightarrow f(\theta)=4\\ \sin^2\theta =1/5 \Rightarrow f(\theta)=4/5 } \\ \Rightarrow \cases{\sqrt{x^2+y^2}最小值=\sqrt{4\over 5} =\bbox[red,2pt]{2\sqrt 5\over 5} \\\sqrt{x^2+y^2}最大值= \sqrt 4 =\bbox[red,2pt]{2}}$$

解答：$$先算1-9999，再扣除9877-9999的結果\\ \begin{array}{} 數字& 樣式 &個數\\\hline 二位數& X0 & 9\\\hline 三位數 & XX0 & 9\cdot 9=81\\ & X0X & 9\cdot 9=81\\ & X00 & 9\\\hline 四位數 & XXX0 & 9^3= 729\\ & XX0X & 729\\ & X0XX & 729\\ & XX00 & 9\cdot 9=81\\ & X0X0 & 81 \\ & X00X & 81\\ &X000 & 9\\ \hline\end{array} \Rightarrow 共有 9\cdot 3+ 81\cdot 5+729\cdot 3=2619\\ 需扣除9880,9890,9900,99[1-9]0,990[1-9]，共1+1+1+9+9= 21個\\ 因此共有2619-21=\bbox[red,2pt]{2598}個數字有0;\\0的數量需再加上數字上有2個0以上的數字:X00、XX00、X0X0、X00X、X000，\\各有9+81\cdot 3+9\cdot 2=270，因此1-9999共有2619+270=2889個0，\\需扣除9880,9890,9900,99[1-9]0,990[1-9]，共1+1+2+9+9= 22個，即2889-22=\bbox[red,2pt]{2867}$$

解答：

(1)$$u^2+{1\over u^2} =(u+{1\over u})^2-2 =x^2-2\Rightarrow u^3+{1\over u^3} =(u+{1\over u})(u^2-u\cdot {1\over u}+{1\over u^2}) =x(x^2-2-1) \\= \bbox[red,2pt]{x^3-3x}$$(2)$$x^3-3x+1 = u^3+{1\over u^3}+1 = 0 \Rightarrow u^6+u^3+ 1=0 \Rightarrow (u^3-1)(u^6+u^3+ 1)=0 \\ \Rightarrow u^9=1 \Rightarrow u= \cos{2k\pi\over 9}+ i\sin{2k\pi \over 9},k=0-8\\ 需扣除u^3=1的增根，因此u=\cos{2k\pi\over 9}+ i\sin{2k\pi \over 9}=c^k,k=1,2,4,5,7,8\\ \Rightarrow x=c+c^{-1},c^2+c^{-2},c^4+c^{-4},c^5+c^{-5},c^7+c^{-7},c^8+c^{-8}\\ 而c+c^{-1}=c^8+c^{-8}、c^2+c^{-2}=c^7+c^{-7}、c^4+c^{-4}=c^5+c^{-5}\\ 因此x=\bbox[red,2pt]{c+c^{-1},c^2+c^{-2},c^4+c^{-4}}$$

解答：$$假設\Psi: y^2=4cx 且\cases{P_1(cp_1^2,2cp_1)\\ P_2(cp_2^2,2cp_2)\\ P_3(cp_3^2,2cp_3)}，其中p_1,p_2,p_3\in \mathbb{R}; \\又y^2=4cx \Rightarrow 2yy'=4c \Rightarrow y'={2c\over y} \Rightarrow \cases{L_1斜率=1/p_1\\ L_2斜率=1/p_2 \\ L_3斜率=1/p_3} \Rightarrow \cases{L_1: y=(x-cp_1^2)/p_1+2cp_1 \\ L_2: y=(x-cp_2^2)/p_2+2cp_2 \\L_3: y=(x-cp_3^2)/p_3+2cp_3 } \\ \Rightarrow \cases{L_1:p_1y = x+cp_1^2 \\L_2:p_2y = x+cp_2^2 \\L_3:p_3y = x+cp_3^2 } \Rightarrow \cases{Q_1= L_2\cap L_3 = (cp_2p_3, c(p_2+p_3)) \\Q_2= L_1\cap L_3 = (cp_1p_3, c(p_1+p_3)) \\Q_3 = L_1\cap L_2 = (cp_1p_2, c(p_1+p_2)) } \\ \Rightarrow \triangle P_1P_3P_3 = {1\over 2}\begin{Vmatrix} cp_1^2 &2cp_1& 1\\ cp_2^2 &2cp_2 & 1 \\cp_3^2 & 2cp_3 & 1\end{Vmatrix} =c^2 \begin{Vmatrix} p_1^2 &p_1& 1\\ p_2^2 &p_2 & 1 \\p_3^2 & p_3 & 1 \end{Vmatrix} = c^2 \begin{Vmatrix} p_1^2 &p_1& 1\\ p_2^2-p_1^2 &p_2-p_1 & 0 \\p_3^2-p_1^2 & p_3-p_1 & 0 \end{Vmatrix} \\ =c^2 |(p_2^2-p_1^2)(p_3-p_1)-(p_3^2-p_1^2)(p_2-p_1)| =c^2|(p_1-p_2)(p_2-p_3)(p_3-p_1)|\\ 又\triangle Q_1Q_2Q_3 = {1\over 2} \begin{Vmatrix} cp_2p_3 & c(p_2+p_3) & 1\\ cp_1p_3 & c(p_1+p_3) & 1 \\cp_1p_2 & c(p_1+p_2) & 1\end{Vmatrix} ={c^2\over 2} \begin{Vmatrix} p_2p_3 & (p_2+p_3) & 1\\ p_1p_3 & (p_1+p_3) & 1 \\p_1p_2 & (p_1+p_2) & 1\end{Vmatrix} \\ ={c^2\over 2} \begin{Vmatrix} p_2p_3 & (p_2+p_3) & 1\\ p_3(p_1-p_2) & (p_1-p_2) & 0 \\p_2(p_1-p_3) & (p_1-p_3) & 0\end{Vmatrix} ={c^2\over 2} |p_3(p_1-p_2)(p_1-p_3)-p_2(p_1-p_2)(p_1-p_3)| \\ ={c^2\over 2}|(p_1-p_2)(p_2-p_3)(p_3-p_1)|\\ 因此{\triangle P_1P_2P_3 \over \triangle Q_1Q_2Q_3} ={c^2 \over c^2/2}= 2 \Rightarrow \triangle P_1P_2P_3: \triangle Q_1Q_2 Q_3 = 2:1，\bbox[red,2pt]{故得證}$$

解答：$$由於x^2-x+1\gt 0,\forall x\in \mathbb{R}，因此 {(a+1)x^2 +(a-2)x +(a+1)\over x^2-x+1} \gt b \\\Rightarrow (a-b+1)x^2 +(a+b-2)x +(a-b+1)\gt 0 \Rightarrow \cases{a-b+1 \gt 0\\ (a+b-2)^2-4(a-b+1)^2 \lt 0} \\ \Rightarrow \cases{a-b+1\gt 0 \\ (3a-b)(a-3b+4)\gt 0}，如下圖斜線區域，其中水平方向為a值，垂直方向為b值$$

解答：$$3x^2-4xy+6y^2-2x-8y-9=0 \Rightarrow \cases{6x-4y-2=0\\ -4x+12y-8=0} \Rightarrow 中心點(x,y)=\bbox[red,2pt]{(1,1)}\\ 將中心點(1,1)平移至(0,0)，即\cases{x'=x-1\\ y'=y-1} \Rightarrow \cases{x=x'+1\\ y= y'+1} 代回原式 \\\Rightarrow 3(x'+1)^2-4(x'+1)(y'+1)+ 6(y'+1)^2-2(x'+1)-8(y'+1)-9=0 \\ \Rightarrow 3x'^2-4x'y'+6y'^2=14，接下來旋轉\theta ，使長軸與x軸平行；\tan 2\theta = {-2\over 3-6} ={2\over 3} \Rightarrow \cases{\sin \theta = 1/\sqrt 5\\ \cos \theta =2/\sqrt 5}\\ 假設旋轉後方程式: ax''^2+by''^2=14 \Rightarrow \cases{a+c= 3+6=9\\ a-c= -\sqrt{(3-6)^2 +(-4)^2}=-5} \Rightarrow \cases{a=2\\ c=7} \\ \Rightarrow 2x''^2+7y''^2=14 \Rightarrow {x''^2\over 7}+{y''^2\over 2}=1 \Rightarrow \cases{半長軸長=\bbox[red, 2pt]{\sqrt 7}\\ 半短軸長= \bbox[red, 2pt]{\sqrt 2}}\\ \Rightarrow 長軸直線通過中心(1,1)且斜率\tan\theta =1/2，即y={1\over 2}(x-1)+1 \Rightarrow \bbox[red, 2pt]{x-2y+1=0};\\ 短軸直線通過中心(1,1)且斜率=-2，即y=-2(x-1)+1 \Rightarrow \bbox[red, 2pt]{2x+y=3}$$