84年大學學測-數學詳解

大學入學考試中心
八十四學年度學科能力測驗

第一部分：選擇題  

壹、單一選擇題 

解答：$$x坐標越大、y坐標越小，則k值較大，因此挑右下角的坐標，故選\bbox[red, 2pt]{(E)}$$

解答：$$由輾轉相除法可得4369與5911的最大公因數為257，即\cases{4369= 257\times 17\\ 5911= 257\times 23} \\ \Rightarrow {1\over 4359} + {1\over 5911} ={ 23+17 \over 257\times 17\times 23} ={40\over 100487}，故選\bbox[red, 2pt]{(A)}$$

解答：$$10\times 5+ 9\times 4+8\times 3+ 7\times 2+6\times 1= 50+36+24+14+6=130，故選\bbox[red, 2pt]{(C)}$$

解答：$$x^2+y^2 +z^2 -2x+4y+2z-19=0 \Rightarrow (x-1)^2 +(y+2)^2 +(z+1)^2 = 25\\ \Rightarrow \cases{圓心P(1,-2,-1)\\ 半徑r=5} \Rightarrow P在平面z=-1上，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{前兩位英文字母有26\times 26種組合\\ 後四位數字有10\times 10\times 10\times 9-1種組合} \Rightarrow 共有26\times 26(10 \times 10 \times 10 \times 9-1)種組合\\，故選\bbox[red, 2pt]{(D)}$$

解答：$$\begin{array}{c|rrrrr|r}X &8 & 9 & 10 & 11 & 12 & 50\\ Y& 11 & 12 & 10 & 8 & 9 & 50\\X^2 &64 & 81 & 100 & 121 & 144 & 510 \\ Y^2 &121 & 144 & 100 & 64 & 81 & 510\\XY &88 & 108 & 100 & 88 & 108 & 492\\\hdashline & & & & & &\sum\end{array} \\ \Rightarrow 相關係數r={n\sum XY -\sum X \sum Y\over \sqrt{n \sum X^2-(\sum X)^2} \cdot \sqrt{n\sum Y^2- (\sum Y)^2}} \\={5\cdot 492-50\cdot 50\over \sqrt{5\cdot 510-50^2} \cdot \sqrt{5\cdot 510-50^2}} ={-40\over 50}=-{4\over 5}，故選\bbox[red, 2pt]{(E)}$$

解答：$$mx^2+10x +m+6 \gt 2 \Rightarrow mx^2+10x+m+4\gt 0 \Rightarrow \cases{m\gt 0\\ 10^2-4m(m+4)\lt 0} \\ \Rightarrow \cases{m\gt 0 \\ (m-(-2+\sqrt{29}))(m-(-2-\sqrt{29})) \gt 0} \Rightarrow m\gt -2+\sqrt{29}，故選\bbox[red, 2pt]{(B)}$$

貳、多重選擇題

解答：$$(A)\bigcirc: f(x)=({1\over 2})^{3x} \Rightarrow f(-x)=({1\over 2})^{-3x} =2^{3x} \\(B)\times: f(x)=2^{3x} \Rightarrow f(-x)=2^{-3x} \ne 3^{2x}\\ (C)\times: f(x)= x^2 \Rightarrow f(-x)=(-x)^2 =x^2 \ne -x^2\\ (D)\bigcirc: f(x)=\log x \Rightarrow f(-x)= \log (-x) \\(E)\times: \cases{f(x)=\cos x \Rightarrow f(-x)= \cos(-x)=\cos x \\\sin(x-\pi/2) = -\cos x} \Rightarrow f(-x) \ne \sin(x-\pi/2)\\，故選\bbox[red, 2pt]{(AD)}$$

解答：$$\cos 74^\circ-\cos 14^\circ = -2\sin{74^\circ+14^\circ\over 2} \sin{74^\circ -14^\circ\over 2} =-2\sin 44^\circ\sin 30^\circ = -\sin 44^\circ \lt 0\\ (A)\times: \cos 60^\circ ={1\over 2}\ne  -\sin 44^\circ\\(B) \times: 2\sin 30^\circ \sin 44^\circ =\sin 44^\circ \ne -\sin 44^\circ \\(C) \times: 2\cos 30^\circ \cos 44^\circ \gt 0\\(D) \bigcirc: \sin 16^\circ-\sin 76^\circ =2\cos {16^\circ +76^\circ \over 2} \sin{16^\circ -76^\circ \over 2} =2\cos 46^\circ \sin (-30^\circ)\\ \qquad \quad =2\cos (90^\circ-44^\circ)\cdot {-1\over 2}=-\sin 44^\circ\\(E)\bigcirc: \sin 164^\circ +\cos 166^\circ = \sin 16^\circ-\cos 14^\circ =\cos 74^\circ-\cos 14^\circ\\，故選\bbox[red, 2pt]{(DE)}$$

解答：$$(A)\bigcirc: 等軸雙曲線的特性之一\\ (B)\times: x+y=0未經過中心點(1,1) \Rightarrow x+y=0不是漸近線 \\(C)\bigcirc: 過中心(1,1)且與坐標軸平行\\(D) \times:{(x-1)^2\over a^2}-{(y-1)^2\over a^2}=1 經過(3,0) \Rightarrow a=\sqrt 3 \Rightarrow 頂點(1\pm \sqrt 3,1) \\(E) \times:a=b=\sqrt 3\Rightarrow c=\sqrt 6 \Rightarrow 焦點(1\pm \sqrt 6,1)\\，故選\bbox[red, 2pt]{(AC)}$$

解答：$$令\cases{A(0,0,0)\\ B(1,0,1)\\ C(0,1,1)\\ D(1,1,0)} \Rightarrow M=(C+D)\div 2=(1/2,1,1/2)\\(A) \bigcirc:\cases{平面ABM的法向量\vec n= \overrightarrow{AB}\times \overrightarrow{AM}= (1,0,1)\times (1/2,1,1/2)= (-1,0,1)\\ \overrightarrow{CD}= (1,0,-1)}\\ \qquad \Rightarrow \vec n\parallel \overrightarrow{CD} \Rightarrow 平面ABM與\overline{CD}垂直 \\(B)\bigcirc: \overrightarrow{AB} \cdot \overrightarrow{CD} =(1,0,1)\cdot (1,0,-1)=0 \Rightarrow \overrightarrow{AB} \bot \overrightarrow{CD} \\(C)\bigcirc: \cases{\cos \angle AMB = {\overrightarrow{MA} \cdot \overrightarrow{MB}\over |\overrightarrow{MA}||\overrightarrow{MB}|} =1/3\\ \cos \angle ADB = {\overrightarrow{DA} \cdot \overrightarrow{DB}\over |\overrightarrow{DA}||\overrightarrow{DB}|} =1/2} \Rightarrow \cos \angle ADB\gt \cos\angle AMB \Rightarrow \angle AMB\gt \angle ADB\\ (D)\bigcirc: \cases{平面ACD的法向量\vec u= \overrightarrow{AC} \times \overrightarrow{AD} =(0,1,1)\times (1,1,0) =(-1,1,-1) \\平面BCD的法向量\vec v= \overrightarrow{BC} \times \overrightarrow{BD} =(-1,1,0)\times (0,1,-1) =(-1,-1,-1)}\\ \qquad \Rightarrow \cos \theta ={\vec u \cdot \vec v\over |\vec u||\vec v|} ={1\over 3} \lt \cos 60^\circ \Rightarrow \theta \gt 60^\circ\\ (E)\times: \cases{\overline{BA}= \sqrt 2\\ \overline{BM}=\sqrt{3\over 2}} \Rightarrow \overline{BA}\ne \overline{BM}\\，故選\bbox[red, 2pt]{(ABCD)}$$

第二部份、填充題

解答：$$令\cases{\Gamma_1: x=y^2+3y-2\\ \Gamma_2: y=x^2+kx +19\\ L:x+y=3},將L代入\Gamma_1 \Rightarrow 3-y=y^2+3y-2 \Rightarrow y^2+4y-5=0\\ \Rightarrow (y+5)(y-1)=0 \Rightarrow \cases{y=1 \Rightarrow x=2\\ y=-5 \Rightarrow x=8} ,將(2,1)及(8,-5)代入\Gamma_2 \Rightarrow \cases{1=4+2k+19\\ -5=64+8k+ 19}\\ \Rightarrow \cases{k=-11\\ k=-11} \Rightarrow k=\bbox[red, 2pt]{-11}$$

解答：$$9= 1+8 =3+6 = 5+4= 7+2=9+0\\ \Rightarrow x^9係數= 2\cdot 9+ 4\cdot 7+ 6\cdot 5+ 8\cdot 3+ 10\cdot 1=18+ 28+30+ 24+ 10 =\bbox[red, 2pt]{110}$$

解答：$$林先生沒射中且陳小姐射中的機率={3\over 5}\times {1\over 2}= \bbox[red, 2pt]{3\over 10}$$

解答：$$10^{n-1}\gt 9^n \Rightarrow n-1\gt n\log 9=2n\log 3 \Rightarrow (1-2\log 3)n\gt 1 \Rightarrow n\gt {1\over 1- 2\log 3} =21.85 \\ \Rightarrow n=\bbox[red, 2pt]{22}$$

解答：

$$四條直線\cases{L_1:x-y=1\\ L_2:x+y=4\\ L_3:8x+y=-10\\ L_4:x=2} 的交點\cases{A(-2,6)\\ B(2,2)\\ C(2,1)\\ D(-1,-2)} \Rightarrow 對角線\cases{\overline{AC}= \sqrt{41}\\ \overline{BD}= \bbox[red, 2pt]{5}}$$

解答：

$$\cos(50^\circ+70^\circ) ={200^2+300^2-x^2 \over 2\cdot 200\cdot 300} \Rightarrow -{1\over 2}= {130000-x^2\over 120000} \Rightarrow x^2=190000 \Rightarrow x= \bbox[red, 2pt]{100\sqrt{19}}$$

解答：$$\cases{a=25\\ ar^{10}=30} \Rightarrow r^{10}={30\over 25}={6\over 5} \Rightarrow r^{20}={36\over 25} \Rightarrow ar^{30}= 30\cdot {36\over 25}={216\over 5}= \bbox[red, 2pt]{43.2}$$

解答：$$f(x)=(\sin x+\cos x+2)^2-4，而\sin x+\cos x=\sqrt 2\sin(x+45^\circ) 最小值為-\sqrt 2\\ 因此f(x)的最小值=(-\sqrt 2+2)^2 -4 = \bbox[red, 2pt]{2-4\sqrt 2}$$

解答：

$$令P'為P(1,2,1)的對稱點(xy-平面為對稱平面)，則P'(1,2,-1)且\overline{OP} =\overline{OP'};\\ 又\overline{OR}=2\overline{PO} \Rightarrow \overrightarrow{OR}=2 \overrightarrow{P'O}= \bbox[red, 2pt]{(-2,-4,2)}$$

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