114年台南市國中教甄聯招-數學詳解

 臺南市 114 學年度市立國民中學正式教師聯合甄選

以下題目共 40 題，為四選一單選選擇題(每題 2.5 分，共 100 分)

解答:$$\sqrt{13+4\sqrt 3}+\sqrt{12-6\sqrt 3}-\sqrt{4+2\sqrt 3} =\sqrt{13+2\sqrt {12}}+\sqrt{12-2\sqrt {27}}-\sqrt{4+2\sqrt 3} \\= \sqrt{12}+1+\sqrt 9-\sqrt 3-(\sqrt 3+1) =2\sqrt 3+1+3-\sqrt 3-\sqrt 3-1=3，故選\bbox[red, 2pt]{(A)}$$

解答:$$0.997^{50} =(1-0.003)^{50} =\sum_{n=0}^{50}C^{50}_n(-0.003)^n =1-50\cdot 0.003+ 1225\cdot 0.003^2 -\cdots \\\approx 0.85+0.011-\cdots=0.0861，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x) =\begin{cases} \sqrt{x^2+1},& x\lt 0\\ x+1, & x\ge 0\end{cases} \Rightarrow \lim_{x\to 0^-} f(x) = \lim_{x\to 0^+} f(x) =0 \Rightarrow f(x)在x=0是連續的 \\ f'(x)=\begin{cases}{x\over\sqrt{x^2+1}},& x \lt 0\\ 1,& x\ge 0\end{cases} \Rightarrow \cases{\lim_{x\to 0^-}f'(x)=0\\ \lim_{x\to 0^+}f'(x)=1} \Rightarrow \lim_{x\to 0^-} f'(x) \ne \lim_{x\to 0^+} f'(x) \\\Rightarrow f(x)在x=0不可微分，故選\bbox[red, 2pt]{(B)}$$

解答:

$$\overline{AP}為\angle BAC的角平分線 \Rightarrow {\overline{PC} \over \overline{PB}} ={\overline{AC} \over \overline{AB}} ={2\over 1}, 又\overline{PQ} \parallel \overline{AB} \Rightarrow {\overline{PQ} \over \overline{AB}} ={\overline{PC} \over \overline{BC}}  \Rightarrow {\overline{PQ} \over 2} ={2 \over 3} \\ \Rightarrow \overline{PQ}={4\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答:

$$假設\cases{圓心O\\ 在\overline{AB}的切點P \\在\overline{AC}的切點Q \\s=(13+14+15)/2=21}  \Rightarrow \triangle ABC面積=\sqrt{s(s-13) (s-14)(s-15)} =\sqrt{21\cdot 8\cdot 7\cdot 6}=84\\ \Rightarrow {1\over 2}\overline{AB}\cdot R+{1\over 2} \overline{AC} \cdot R ={1\over 2}(13R+14R)=84 \Rightarrow R={56\over 9}=6.22，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{k=1 \Rightarrow 餘數10\\ k=2  \Rightarrow 餘數9\\  k=3\Rightarrow 餘數12\\  k=4\Rightarrow 餘數3\\  k=5\Rightarrow 餘數4\\  k=6\Rightarrow 餘數1\\  k=7\Rightarrow 餘數10\\  k=8\Rightarrow 餘數9\\ \cdots} \Rightarrow 餘數有6種，故選\bbox[red, 2pt]{(C)}$$

解答:$$\lim_{x\to \infty}x\left( e^{1/x}-1\right) =\lim_{x\to \infty} {e^{1/x}-1 \over 1/x}  =\lim_{x\to \infty} ({e^{1/x}-1)' \over (1/x)'}  =\lim_{x\to \infty} {(-1/x^2)e^{1/x} \over -1/x^2}  =\lim_{x\to \infty}e^{1/x} =e^0=1\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{r_1=\sqrt{2-2\sqrt 3i} \\r_2=-\sqrt{2-2\sqrt 3}i} \Rightarrow r_1-r_2=2\sqrt{2-2\sqrt 3i} =\sqrt{8-8\sqrt 3 i} \\ (A)\bigcirc: (-2\sqrt 3+2i)^2= 8-8\sqrt 3 i \\(B)\times: (2\sqrt 3+2i)^2 \ne 8-8\sqrt 3 i \\(C) \times 0^2 \ne 8-8\sqrt 3 i\\ (D)2^2 \ne 8-8\sqrt 3 i\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$\log 14.4\gt \log 4.4 \gt \log 0.4 \Rightarrow -\log 14.4\lt -\log 4.4 \lt -\log 0.4 \Rightarrow b\gt c\gt a，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{a^3=9\\ b=25^{1/4}=\sqrt 5 \Rightarrow b^3=5\sqrt 5\\ c^3=\sqrt{79} \lt \sqrt{81}=9} \Rightarrow b\gt a\gt c   ，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)為3次\Rightarrow [f(x)]^2為6 \Rightarrow [f(x)]^2\cdot f(x)為9次，故選\bbox[red, 2pt]{(B)}$$

解答:$$(A) \cos{x\over 2} 週期4\pi\\ (B) \sin 3x週期{2\pi\over 3} \\(C) \cos x週期2\pi\\ (D) \cos x+|\cos x|週期2\pi，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{a=2.1^5(2.1^2-2.1) =2.1^5\times 2.31\\ b=2.1^5(2.1-1)=2.1^5\times 1.1} \Rightarrow a\gt b，故選\bbox[red, 2pt]{(A)}$$

解答:

$$假設P為\triangle ABC的外心，Q為\overline{BC}中點 \Rightarrow \overline{CQ}=6\\\cases{\angle B=85^\circ\\ \angle C=35^\circ} \Rightarrow \angle A=180^\circ-85^\circ-35^\circ=60^\circ \Rightarrow 外接圓半徑\overline{PC} =6\times {2\over \sqrt 3}=4\sqrt 3，故選\bbox[red, 2pt]{(C)}\\直接用正弦定理:{12\over \sin 60^\circ}=2R \Rightarrow R=4\sqrt 3$$

解答:$$\cos(2\theta-30^\circ), \cos 2\theta, \cos(2\theta+30^\circ) 成等差\Rightarrow \cos(2\theta-30^\circ)+\cos(2\theta+30^\circ)=2\cos 2\theta\\ \Rightarrow {\sqrt 3\over 2}\cos 2\theta+{1\over 2}\sin 2\theta +{\sqrt 3\over 2}\cos 2\theta-{1\over 2}\sin 2\theta  =2\cos 2\theta \\ \Rightarrow \sqrt 3\cos 2\theta=2\cos 2\theta \Rightarrow \cos 2\theta=0, \theta={\pi\over 4},{3\pi\over 4}有兩個解，故選\bbox[red, 2pt]{(B)}$$

(A)7  (B)8  (C)9   (D)10

解答:$$令\cases{O(0,0) \\A(2,3) \\B(4,-2) \\C(1,-3)} \Rightarrow \cases{L_1=\overleftrightarrow{AC} :6x-y=9 \\L_2=\overleftrightarrow{OB}: x+4y=0} \Rightarrow \cases{d(O,L_1)\lt d(B,L_1) \Rightarrow \triangle ABC\gt \triangle AOC\\d(A,L_2) \gt d(C,L_2) \Rightarrow \triangle OAB \gt \triangle OBC} \\ \Rightarrow \cases{\triangle ABC= \overline{AC}\cdot d(B,L_1)/2 =8\\ \triangle OAB=\overline{OB}\cdot d(A,L_2)/2=7} \Rightarrow 最大三角形\triangle ABC面積為8，故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cases{\triangle PMU為等腰直角三角形\Rightarrow \overline{PM} =2/\sqrt 2=\sqrt 2\\ \triangle QMN為等腰直角三角形\Rightarrow \overline{PM} =4/\sqrt 2=2\sqrt 2} \Rightarrow 正方形PQRS邊長為2\sqrt 2+\sqrt 2=3\sqrt 2\\ \Rightarrow 面積為3\sqrt 2\cdot 3\sqrt 2=18，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{f(x)=ax^2+bx+c圖形通過(0,-1) \Rightarrow f(0)=-1 \Rightarrow c=-1\\圖形與x軸相切 \Rightarrow 判別式:b^2-4ac=0} \Rightarrow 圖形為凹向下 \Rightarrow a\lt 0 \\ \Rightarrow f(1)\le 0 \Rightarrow a+b+c\le 0，故選\bbox[red, 2pt]{(C)}$$

解答:$$共有4\times 3\times 2\times 1=24種不同的四位數\Rightarrow 四位數字和皆是(4+3+2+1)\times 6=60 \\ \Rightarrow 總和=60(1000+100+10+1)=60\times 1111 =5\times 12\times 1111 \\ \Rightarrow 商數=12\times 1111=13332，故選\bbox[red, 2pt]{(D)}$$

解答:

$$假設\cases{\triangle AOE面積為a\\ \triangle AOD面積為b} \Rightarrow {\triangle OBE\over \triangle OBC } ={\overline{OE} \over \overline{OC}} ={\triangle AOE\over \triangle AOC} \Rightarrow {5\over 10}={a\over b+8} \Rightarrow 2a-b=8 \cdots(1)\\ 同理,{\triangle OBC\over \triangle OCD } ={\overline{OE} \over \overline{OD}} ={\triangle AOB\over \triangle AOD} \Rightarrow {10\over 8} ={a+5\over b} \Rightarrow 4a-5b+20=0\cdots(2) \\ 由(1)及(2)可得 \cases{a=10\\ b=12} \Rightarrow a+b=22，故選\bbox[red, 2pt]{(C)}$$

解答:$$內角總和:180(n-2), n要最大,內角就要越大，但內角最大的整數為179度 \\ \Rightarrow 179+178+\cdots+179-(n-1) ={n(359-n) \over 2}= 180(n-2) \\ \Rightarrow n^2+n-720=0 \Rightarrow n={-1+\sqrt{2881}\over 2} \approx 26.3 \Rightarrow n=26，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{a+b=c\\ b+c=d\\ c+d=a} \Rightarrow \cases{a=3c/2\\ b=-c/2\\ d=c/2} \Rightarrow a+b+c+d={5\over 2}c希望最大\\ 但b=-{c \over 2}為正整數 \Rightarrow c\lt 0 \Rightarrow c=-2 \Rightarrow a+b+c+d=-5，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{25^x=2000\\ 80^y=2000} \Rightarrow \cases{x\log 25=\log 2000\\ y\log 80=\log 2000} \Rightarrow \cases{1/x=\log 25/\log 2000\\ 1/y=\log 80/\log 2000} \\ \Rightarrow {1\over x}+{1\over y}={\log 25+ \log 80\over \log 2000} ={\log 2000\over \log 2000} =1，故選\bbox[red, 2pt]{(B)}$$

解答:$$該數列為:10,13 ,23, 36,59，故選\bbox[red, 2pt]{(B)}$$

解答:$$(a,b)在y=\log_2 x上\Rightarrow \log_2 a= b\\ (A)\times: \log_2(a-1-1)=\log_2 (a-2)\ne b\\ (B) \times: \log_2 a=b \Rightarrow 2^b=a \Rightarrow 2^{b+1}\ne a \\(C)\bigcirc: \log_2 a=b \Rightarrow \log_2 (2a)-1=b \Rightarrow (2a,b)在y=(\log_2 x)-1上\\ (D)\times:若(b^{-1},a)在圖形y=2^{-x}上 \Rightarrow a=2^{-b^{-1}} \ne 2^b\\，故選\bbox[red, 2pt]{(C)}$$

解答:

$$假設三條直線以L_1,L_2,L_3表示，如上圖；其中只有L_1的斜率為負值，其餘直線斜率皆為正值\\，因此:L_1:2x+y+c=0;又2x+y+c\le 0 \Rightarrow c\lt 0\\ 又\cases{ax+by+b\ge 0\\ dx+ey+f\ge 0}所圍區域包含原點\Rightarrow \cases{b\gt 0\\ f\gt 0}，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(1+2i)=f(1+\sqrt 2)=0 \Rightarrow f(x)=0的三根為1\pm 2i,1+\sqrt 2\\(A)\bigcirc: 三根之積=5(1+\sqrt 2)\gt 0 \Rightarrow f(0)\lt 0 \\(B)\times: f(x)=0只有一實根1+\sqrt 2\Rightarrow  1-\sqrt 2不是f(x)=0的根 \\(C)\times: g(x)=f(x)-3 為三次式\Rightarrow g(x)=0一定有實數解\\ (D)\times: 只有一個實數解\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$f(x)=\sin^2 x+\sin x\cos x+2\cos^2 x = 1+{1\over 2}\sin 2x+\cos^2 x = 1+{1\over 2}\sin 2x+{1\over 2}\cos 2x+{1\over 2} \\\quad ={3\over 2}+{1\over 2}(\sin 2x+\cos 2x) ={3\over 2}+{\sqrt 2\over 2}\sin (2x+\theta) \\ \Rightarrow \cases{(A)\times: 週期為\pi \\ (B)\times: 振幅為\sqrt 2/2\\ (C)\times: f(x)\gt 0 \Rightarrow 無交點\\ (D)\bigcirc:最大值=\displaystyle {3\over 2}+{\sqrt 2\over 2}={3+\sqrt 2\over 2}}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$${\sqrt 3+i\over 2} =\cos{\pi\over 6}+i\sin {\pi\over 6} =e^{\pi i/6} \Rightarrow \left({\sqrt 3+i\over 2} \right)^{69} =e^{69\pi i/6} =e^{23\pi i/2}=-i \\ \Rightarrow \left({\sqrt 3+i\over 2} \right)^{69}+1=1-i=\sqrt 2(\cos{7\over 4}\pi+i\sin {7\over 4}\pi) \Rightarrow \theta ={7\pi\over 4}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{\displaystyle \left|{x-1\over 7} \right| \lt 1 \Rightarrow x=7,\dots,0,-1,\dots,-5,共13個\\ \displaystyle {x-1\over 7}=1 \Rightarrow x=8\\ x^2-81=0\Rightarrow x=\pm 9,共2個} \Rightarrow 合計16個，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)=e^{3x}+x^{3e} \Rightarrow f'(x)=3e^{3x}+3ex^{3e-1} ，故選\bbox[red, 2pt]{(A)}$$

解答:$$A=\begin{bmatrix} a&b\\ c& d\end{bmatrix} \Rightarrow \det(A)=ad-bc=6 \\B=\begin{bmatrix} 3a+4c &3b+4d \\ 5a& 5b\end{bmatrix} \Rightarrow \det(B)=15ab+20bc-15ab-20ad=20(bc-ad ) \\=-20\det(A)=-120，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{L_1:x+(2-m)y+3=0 \Rightarrow x=(m-2)y-3\\ L_2:x+2y+(3-m)=0 \Rightarrow x=m-3-2y} \Rightarrow (m-2)y-3=m-3-2y \\ \Rightarrow my=m \Rightarrow y=1 \Rightarrow x=m-5 \Rightarrow (m-5,1)代入L_3 \Rightarrow (1-m)(m-5)+ 5=0\\ \Rightarrow m^2-6m=0 \Rightarrow m=6,0，故選\bbox[red, 2pt]{(B)}$$

解答:$$柯西不等式: \left( ({5\over a})^2 +({2\over b})^2\right) \left(a^2+b^2 \right)\ge \left( 5+2\right)^2 \Rightarrow {25\over a^2}+{4\over b^2}\ge 49，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{主食材區有5種\\ 配料區有6種} \Rightarrow 共有{C^5_2\times C^6_3}=200種搭配方法\\ 過過敏的搭配方法: \cases{堅果+巧克力酒(不含蘭姆葡萄乾): C^4_1C^4_2 =24\\ 堅果+蘭姆葡萄乾(不含巧克力酒): C^4_1C^4_2 =24\\ 堅果+巧克力酒及蘭姆葡萄乾:C^4_1C^4_1=16} \\ \Rightarrow 發生過過敏的機率={24+24+16\over 200} ={8\over 35}，故選\bbox[red, 2pt]{(A)}$$

解答:$$0.9999\times 2400-0.0001\times (12000000-2400) =1200，故選\bbox[red, 2pt]{(C)}$$

解答:$$9^6=3^{12} \Rightarrow 正因數:3^0,3^1,3^2, \dots,3^{12} \Rightarrow 所有正因數乘積=3^{0+1+2+\cdots +12} \\=3^{78} =9^{39}，故選\bbox[red, 2pt]{(C)}$$

解答:$$||\vec a\cdot (\vec b\times \vec c) || =||(1,-1,3)\cdot (4,-8,-6)||=6，故選\bbox[red, 2pt]{(A)}$$

解答:$$\lfloor \log_4 n \rfloor =\begin{cases}0, & n=1-3\\ 1, & n=4-(4^2-1=15) \\2,& n=16-(4^3-1=63)\\ 3,& n=64-(4^4-1=255)\\  \end{cases} \\ \Rightarrow \sum_{n=1}^{200}\lfloor \log_4 n \rfloor =12+48\cdot 2+137\cdot 3=519，故選\bbox[red, 2pt]{(D)}$$

解答:$$(x-2)^2+y^2 =4 \Rightarrow \cases{圓心P(2,0)\\ 圓面積=4\pi} \Rightarrow P繞y軸一圈長度為4\pi \Rightarrow 旋轉體積=4\pi\times 4\pi =16\pi^2\\，故選\bbox[red, 2pt]{(C)}$$

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