112年新北國中聯合教甄-資優數學詳解

新北市立國民中學 112 學年度教師聯合甄選

特殊教育資優(數學)科

解答:$$ \sqrt{3+\sqrt 8}+ \sqrt{3-\sqrt 8} =\sqrt{3+2\sqrt 2}+ \sqrt{3-2\sqrt 2} =\sqrt 2+1+\sqrt 2-1=2\sqrt 2，故選\bbox[red, 2pt]{(B)}$$

解答:$$假設長方體的\cases{長a\\ 寬b\\ 高c} \Rightarrow \cases{ab=48\\ bc=72\\ ca=96} \Rightarrow ab\cdot bc\cdot ca =(abc)^2=331776 =576^2 \\ \Rightarrow 體積=abc= 576，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{a,b,c公差為1 \Rightarrow \cases{b=a+1\\ c=a+2} \\ b,c,d成等比 \Rightarrow c^2=bd \\ d=a+8} \Rightarrow (a+2)^2 =(a+1)(a+8) \Rightarrow 4a+4=9a+8 \Rightarrow a=-{4\over 5}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$a=\sqrt{2+ \sqrt{2+\sqrt{\cdots}}} \Rightarrow a^2=2+a \Rightarrow a^2-a-2=0 \Rightarrow (a-2)(a+1)=0 \Rightarrow a=2 \\ \Rightarrow \sqrt{7+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}} = \sqrt{7+a} =\sqrt{7+2} =3，故選\bbox[red, 2pt]{(C)}$$

解答:$$顯然是向左開口的拋物線，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)=x^3-3x^2-5x+9 \Rightarrow f'(x)=3x^2-6x-5 \Rightarrow \cases{f(1)=2\\ f'(1)=-8} \\ \Rightarrow f(1+0.002) \approx f(1)+f'(1)\cdot 0.002=2-8\cdot 0.02= 1.984，故選\bbox[red, 2pt]{(A)}$$

解答:$$二月有五個星期一\Rightarrow 二月有29天且二月一日與二月二十九日都是星期一\\ \Rightarrow 一月三十一日星期日\Rightarrow 一月三日(31-4\times 7=3)星期日 \Rightarrow 一月一日星期五\\，故選\bbox[red, 2pt]{(D)}$$

解答:$${1\over x-1} +{1\over x-2} +{1\over x-3} +{1\over x-4} +{1\over x-5} +{1\over x-6} ={f(x) \over (x-1)(x-2)\cdots (x-6)}=0 \\ \Rightarrow f(x)為5次式，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{x_1=1\\ x_2=2} \Rightarrow \cases{(A)  \sqrt{5\over 2} \\ (B){3\over 2} \\(C) \sqrt 2\\ (D){4\over 3}} \Rightarrow (A)\gt (B)\gt (C) \gt (D)，故選\bbox[red, 2pt]{(A)}$$

解答:

$$\cases{\Gamma_1:圓心O(0,0), 半徑2\\ \Gamma_2:圓心P(5,0),半徑3} \Rightarrow 圓心皆在x軸上\Rightarrow 公切線交點A在x軸上 \Rightarrow 假設A(-a,0) \\ \Rightarrow {\overline{AO} \over \overline{AP}} ={\overline{BO} \over \overline{CP}} \Rightarrow {a\over a+5} ={2\over 3} \Rightarrow a=10 \Rightarrow A(-10,0)，故選\bbox[red, 2pt]{(A)}$$

解答:$$x^2 \equiv n {\mod 23} \Rightarrow x^2=23k+n \Rightarrow  k=2\Rightarrow x^2=46+n \Rightarrow n=3,x=7，故選\bbox[red, 2pt]{(A)}$$

解答:

$$假設\cases{P=\overline{AB}中點\\ Q=\overline{AC}中點 \\ L:2x-y=0} 及\cases{A在\overline{PQ}上的垂足M\\ A在\overline{BC}上的垂足N } \Rightarrow d(A, \overline{PQ}) ={|6-1|\over \sqrt{2^2+1^2}} =\sqrt 5 \\ M\in L \Rightarrow M(a,2a) \Rightarrow \overline{AM} =\sqrt{(a-3)^2+(2a-1)^2} =\sqrt 5 \Rightarrow a^2-2a+1=0 \Rightarrow (a-1)^2=0 \\ \Rightarrow a=1 \Rightarrow M=(1,2) \Rightarrow N=2M-A=(-1,3)，故選\bbox[red, 2pt]{(C)}$$

解答:$${f(1)-f(-1) \over 2} ={1-1\over 2} =0，故選\bbox[red, 2pt]{(B)}$$

解答:$$已知\cases{D\gt B\cdots(1) \\A+B\gt C+D \cdots(2) \\ A+C=B+D \cdots(3)} \Rightarrow \cases{(2)+(3)  \Rightarrow 2A+B+C\gt B+C+2D \Rightarrow A\gt D \\ (2)-(3) \Rightarrow B-C\gt C-B \Rightarrow B\gt C} \\ \Rightarrow A\gt D\gt B\gt C，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設L_1與L_2投影到平面E為兩條相交的直線，其角平分線即為L\\ 又d(L_1,L_2)\gt 0 \Rightarrow L有無限多條，故選\bbox[red, 2pt]{(D)}$$

解答:$$A= \begin{bmatrix}3&0 & -2\\ -7& 0& 4\\ 4& 0& -3 \end{bmatrix} \Rightarrow \det(A-\lambda I ) = -\lambda(\lambda-1)(\lambda+1)=0 \Rightarrow \lambda =0,\pm 1，故選\bbox[red, 2pt]{(D)}$$

解答:$$a+4b\gt a+b \Rightarrow a+b不是斜邊, 剩下兩種可能\\ \textbf{Case I }a+4b為斜邊\Rightarrow (a+b)^2+(3a)^2= (a+4b)^2 \Rightarrow 9a^2-6ab-15b^2=0 \\ \qquad\Rightarrow 3(3a-5b)(a+b)=0  \Rightarrow 3a=5b \Rightarrow \cases{a=5\\ b=3} \\ \textbf{Case II }3a 為斜邊\Rightarrow (a+b)^2+(a+4b)^2=(3a)^2 \Rightarrow 7a^2-10ab-17b^2=0 \\ \qquad\Rightarrow (7a-17b)(a+b)=0  \Rightarrow 7a=17b \Rightarrow \cases{a=17\\ b=7} \\ 因此共有兩組解，故選\bbox[red, 2pt]{(B)}$$

解答:$${1\over b}+{1\over a}={2\over a} \Rightarrow a={2bc\over b+c} \Rightarrow a是b與c的調和平均數\\ 調和平均\le 幾何平均數\Rightarrow a={2bc\over b+c}\le \sqrt{bc} \Rightarrow a^2\le bc \Rightarrow \cos A={b^2+c^2-a^2 \over 2bc} \ge {b^2+c^2-bc\over 2bc} \\ 0\le (b-c)^2 \lt (b-c)^2+bc=b^2+c^2-bc \Rightarrow \cos A\gt 0 \Rightarrow \angle A為銳角，故選\bbox[red, 2pt]{(A)}$$

解答:$$S(n)= \sum_{k=1}^n k= {1\over 2}n(n+1) \Rightarrow  S(44)=1980 \Rightarrow 第1000項是45，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{A(1/2,\sqrt 3/2) \\B(0,0) \\C(1,0)} \Rightarrow \cases{D=(2B+C)/3 =(1/3,0) \\E=(2A+B)/3 =(1/3,\sqrt 3/3)} \Rightarrow \cases{\overrightarrow{AD} =(-1/6,-\sqrt 3/2) \\ \overrightarrow{CE} =(-2/3,\sqrt 3/3)} \\ \Rightarrow \overrightarrow{AD} \cdot \overrightarrow{CE} ={1\over 9} -{1\over 2} = -{7\over 18}，故選\bbox[red, 2pt]{(C)}$$

解答:$$\textbf{Case I }r\gt 0 \Rightarrow 直線y=rx+r 斜率為正且通過(0,r) \Rightarrow 直線通過一、二、三象限\\ \textbf{Case II }r\lt 0 \Rightarrow 直線y=rx+r 斜率為負且通過(0,r)  \Rightarrow 直線通過 二、三、四象限 \\ 無論r值，直線皆通過二、三象限，故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cases{(x+y)^2=x^2+y^2+ 2xy=41+40=81 \Rightarrow x+y= \pm 9\\ (x-y)^2 =x^2+y^2-2xy =41-40=1 \Rightarrow x-y=\pm 1} \Rightarrow 兩圖形交點\cases{A(5,4) \\B(4,5) \\C(-5,-4) \\D(-4,-5)} \\ \Rightarrow ABCD為一矩形\Rightarrow \cases{\overline{AB}=\sqrt 2\\ \overline{AD} =9\sqrt 2} \Rightarrow 面積=\sqrt 2\times 9\sqrt 2=18，故選\bbox[red, 2pt]{(A)}$$

解答:

$$假設正六邊形邊長為2 \Rightarrow \cases{O(0,0) \\ A(-1,-\sqrt 3) \\B(1,-\sqrt 3) \\C(2,0) \\ D(1,\sqrt 3)} \Rightarrow M=\overline{CD}中點\Rightarrow M=({3\over 2},{\sqrt 3\over 2}) \Rightarrow N={1\over 2}(O+M) \\ \Rightarrow N=({3\over 4},{\sqrt 3\over 4}) \Rightarrow \cases{\triangle ABN= \displaystyle {1\over 2}\cdot 2\cdot ({\sqrt 3\over 4}+\sqrt 3) ={5\over 4}\sqrt 3\\ \triangle BCN =\displaystyle {1\over 2} \begin{Vmatrix} 3/4& \sqrt 3/4& 1\\ 2& 0& 1\\ 1& -\sqrt 3& 1\end{Vmatrix} ={3\over 4}\sqrt3} \Rightarrow \triangle ABN:\triangle BCN=5:3\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$ N=\underbrace{8\cdots8}_{50個} a \underbrace{9\cdots9}_{50個}, 從右邊數字每3個一組可得一個三位數的數字\\ \Rightarrow \cases{s_1=s_2=\cdots= s_{16} =999 \\s_{17} =a99 \\s_{18}= s_{19} =\cdots= s_{33}=888\\ s_{34}=88} \Rightarrow  (奇數組和)-(偶數組和)=a99-88 =(100a+99)-88 \\=100a+11 是7的倍數 \Rightarrow a=5，故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cases{\overline{CD}=15\\ 8\overline{CF} =\overline{DF}} \Rightarrow \cases{ \overline{DF} =(8/9) \cdot \overline{CD} =40/3\\ \overline{CF}=(1/9) \cdot \overline{CD} =5/3}, 同理\cases{\overline{AD}=x\\ 5\overline{AE} =4\overline{DE}} \Rightarrow \cases{\overline{AE}=(4/9)x \\ \overline{DE}=(5/9)x}\\ H為切點\Rightarrow 圓冪性質:\overline{CH}^2= \overline{CF}\cdot \overline{CD}={5\over 3}\cdot 15=25 \Rightarrow \overline{CH}=5 \Rightarrow \overline{BH}=x-5\\ 同理,G為切點\Rightarrow \overline{AG}^2= \overline{AE} \cdot \overline{AD} ={4\over 9}x^2 \Rightarrow \overline{AG}={2\over 3}x \Rightarrow \overline{BG}=15-{2\over 3}x \\ \overline{BH}=\overline{BG} \Rightarrow x-5= 15-{2\over 3}x \Rightarrow x=12 \Rightarrow \overline{BH}=12-5=7，故選\bbox[red, 2pt]{(C)}$$

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