臺南市 115 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:$$假設F(x,y,z)=\sqrt x+y^3+\sqrt z-4 \Rightarrow \nabla\cdot F = \left( {1\over 2\sqrt x},3y^2, {1\over 2\sqrt z} \right) \\ \Rightarrow \nabla\cdot F(4,1,1)= \left( {1\over 4},3,{1\over 2} \right) \Rightarrow 切平面:{1\over 4}(x-4)+3(y-1)+{1\over 2}(z-1)=0 \\ \Rightarrow x+12y+2z=18，故選\bbox[red, 2pt]{(B)}$$
解答:$$ A=\begin{pmatrix} 2 & -1 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \Rightarrow A^2 = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & -3 & -1 \\ -3 & 3 & 3 \\ -1 & 3 & 5 \end{pmatrix} \\ \Rightarrow A^3=\begin{pmatrix} 5 & -3 & -1 \\ -3 & 3 & 3 \\ -1 & 3 & 5 \end{pmatrix} \begin{pmatrix} 2 & -1 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 13 & -9 & -5 \\-9 & 9 & 9 \\-5 & 9 & 13 \end{pmatrix} \\ \Rightarrow \alpha_1^3+ \alpha_2^2+ \alpha_3^3 =tr(A^3) =13+9+13=35 ，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設\cases{a= \sqrt[3]{\sqrt 5+2} \\b= \sqrt[3]{\sqrt 5-2}} \Rightarrow x=a-b \Rightarrow \cases{a^3=\sqrt 5+2\\ b^3=\sqrt 5-2 \\ ab=1} \Rightarrow a^3-b^3=4 \\ \Rightarrow x^3=(a-b)^3=a^3-b^3-3ab(a-b)=4-3x \Rightarrow x^3+ 3x-4=0 \Rightarrow x^3+3x+2=6\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$(A) \bigcirc: 24 \equiv 2 \pmod {11} \Rightarrow 24^{100} \equiv 2^{100} =(2^{10})^{10} \equiv 1 \pmod{11} \\ \qquad \Rightarrow 24^{100}-1 \equiv 0 \pmod{11} \Rightarrow 24^{100}是11的倍數\\ (B)\bigcirc: 3^{90}= (3^{10})^9 \equiv 1^9=1 \pmod{11} \Rightarrow 3^{90}-1是11的倍數\\ (C) \bigcirc: 63 \equiv -3 \pmod{11} \Rightarrow 63^{34} \equiv (-3) ^{34} =3^{34} \pmod{11} =(3^{10})^3\cdot 3^4 \equiv 1^3\cdot 81 \pmod{11} \\ \qquad \Rightarrow 81 \equiv 4 \pmod{11} \Rightarrow 63^{64}-4是11的倍數\\ (D) \times: 17 \equiv 6 \pmod{11} \Rightarrow 17^{48} \equiv 6^{48} =(6^{10})^4\cdot 6^8 \equiv 1^4\cdot 6^8 \pmod{11} \\ \qquad 6^2 \equiv 3 \pmod{11} \Rightarrow 6^4 \equiv -2 \pmod{11} \Rightarrow 6^8 \equiv 4 \pmod{11} \Rightarrow 17^{48}+4不是11的倍數\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設本金P \Rightarrow P(1+0.08)^n \gt 3P \Rightarrow 1.08^n \gt 3 \Rightarrow n\log 1.08 \gt \log 3 \\ \Rightarrow n(\log 108-\log 100) =n(\log(2^2\cdot 3^3)-2)\gt \log 3 \Rightarrow n\gt {0.477\over 0.033} \\ \Rightarrow n\gt 14.45 \Rightarrow n=15，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{A(3,0) \\P(x,y)} \Rightarrow \overline{AP}^2 =(x-3)^2+y^2=(x-3)^2+(x^2-1)=2x^2 -6x+8\\ 當x={3\over 2}=1.5時，\overline{AP}^2有極小值:2\cdot 1.5^2-6\cdot 1.6+8=3.5 \Rightarrow \overline{AP} =\sqrt{3.5}，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{x=-3+5\cos \theta\\ y=4+3\sin \theta} \Rightarrow \cases{\cos \theta=(x+3)/5 \\ \sin \theta=(y-4)/3 } \Rightarrow \cos^2\theta +\sin^2\theta =1={(x+3)^2\over 25}+{(y-4)^2\over 9} \\ \Rightarrow \cases{中心點O(-3,4)\\ a=5\\ b=3} \Rightarrow c=4 \Rightarrow 焦點\cases{F_1(-3+4,4) =(1,4) \\F_2(-3-4,4) =(-7,4)}，故選\bbox[red, 2pt]{(A)}$$
解答:$$\sin C= \sin(\pi-(A+B)) = \sin(A+B) =\sin A\cos B+ \sin B\cos A =2\cos A\sin B \\ \Rightarrow \cos A\sin B= \sin A\cos B \Rightarrow \sin(A-B)=0 \Rightarrow A=B \Rightarrow 等腰三角形，故選\bbox[red, 2pt]{(D)}$$
解答:$$相當於過P點斜率最小的直線，顯然是\overleftrightarrow{BP}，故選\bbox[red, 2pt]{(B)}$$
解答:$$a(x^2-4xy+5y^2-y) + b( 3x^2-3y^2+x)+ 12xy-38x+12y+k=0 \\ \Rightarrow (a+3b)x^2+(-4a+12)xy+ (5a-3b)y^2+(b-38)x+ (-a+12)y+k=0 \\ 符合圓的條件:\cases{-4a+12=0\\ a+3b=5a-3b} \Rightarrow \cases{a=3\\ b=2} \Rightarrow 9x^2+9y^2-36x+9y+k=0 \\ \Rightarrow (x-2)^2 + \left( y+{1\over 2} \right)^2={17\over 4}-{k\over 9} =2^2 \Rightarrow k={9\over 4} \Rightarrow 乙和丁正確，故選\bbox[red, 2pt]{(C)}$$
解答:$$假設符合條件的正整數為 N \Rightarrow \cases{N /(5/2)=2N/5為整數\Rightarrow N為5的倍數\\ N/(7/4)=4N/7為整數 \Rightarrow N為7的倍數\\ N/(9/6)=2N/3為整數\Rightarrow N為3的倍數} \\ \Rightarrow N必定是3\times 5\times 7=105的倍數 \Rightarrow \left \lfloor {2026\over 105} \right\rfloor=19，故選\bbox[red, 2pt]{(C)}$$
解答:$$y={2+x\over 2-x} \Rightarrow x={2y-2\over y+1} \Rightarrow f(y)={2y-2\over y+1}\Rightarrow f'(y)={4\over (y+1)^2} \Rightarrow f'(3)={1\over 4}，故選\bbox[red, 2pt]{(D)}$$
解答:$$y=3\tan x-2\csc x\Rightarrow y'= 3\sec^2 x+2 \csc x\cot x \\\Rightarrow y'(\pi/3)=3\cdot 2^2+2 \cdot {2\over \sqrt 3}\cdot {1\over \sqrt 3}=12+ {4\over 3}={40\over 3}=13.333 \approx 13，故選\bbox[red, 2pt]{(A)}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$ \lim_{n\to \infty} \left( {1\over n+1}+{1\over n+2}+ \cdots +{1\over n+n} \right) =\lim_{n\to \infty} \sum_{k=1}^n {1\over n+k} =\lim_{n\to \infty} \sum_{k=1}^n {1\over n(1+k/n)} \\ =\int_0^1 {1\over 1+x}\,dx = \left. \left[ \ln(1+x) \right] \right|_0^1 =\ln 2，故選\bbox[red, 2pt]{(B)}$$
解答:$$\lim_{n\to \infty} \left( {1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+ \cdots +{1\over \sqrt{n^2+n}} \right) =\lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+k}} \\ \Rightarrow \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+n}} \lt \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+k}} \lt \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+1}} \\ \Rightarrow \lim_{n\to \infty} {n\over \sqrt{n^2+n}} \lt \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+k}} \lt \lim_{n\to \infty} \sum_{k=1}^n {n\over \sqrt{n^2+1}} \\ \Rightarrow \lim_{n\to \infty} {1\over \sqrt{1+1/n}} \lt \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+k}} \lt \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{1+1/n^2}} \\ \Rightarrow 1\lt \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+k}} \lt 1 \Rightarrow \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+k}} =1，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{|\vec a-2\vec b|^2=(\vec a-2\vec b) \cdot (\vec a-2\vec b)=1\\ |2\vec a+ \vec b|^2 =(2\vec a+ \vec b) \cdot (2\vec a+ \vec b)=4} \Rightarrow \cases{|\vec a|^2-4\vec a\cdot \vec b+4|\vec b|^2=1 \cdots(1)\\ 4|\vec a|^2+4\vec a\cdot \vec b+|\vec b|^2 =4 \cdots(2)} \Rightarrow 5 \left( |\vec a|^2+ |\vec b|^2 \right)=5 \\ 又由(1)可得-4\vec a\cdot \vec b=1-|\vec a|^2-4|\vec b|^2 \Rightarrow 欲求6|\vec a|^2-4\vec a\cdot \vec b+9|\vec b|^2=6|\vec a|^2 +(1-|\vec a|^2-4|\vec b|^2)+9|\vec b|^2 \\=1+5 \left( |\vec a|^2+ |\vec b|^2 \right)=1+5=6，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)={2-x\over 1-2x} \Rightarrow f_2=f(f(x)) ={2-{2-x\over 1-2x} \over 1-2\cdot {2-x\over 1-2x}} =x \Rightarrow f_n=\cases{x, n為偶數\\ {2-x\over 1-2x}, n為奇數} \\ \Rightarrow f_{10}(x)=x，故選\bbox[red, 2pt]{(D)}$$
解答:$$\sqrt x \Rightarrow x\ge 0 \Rightarrow 取x=y^6 \Rightarrow \sqrt{1+\sqrt{y^6}} =\sqrt[3]{y^6}-1 \Rightarrow \sqrt{1+y^3}=y^2-1 \\ \Rightarrow 1+ y^3=y^4-2y^2+1 \Rightarrow y^4-y^3-2y^2=0 \Rightarrow y^2 (y-2)(y+1)=0 \\ \Rightarrow \cases{y=0 \Rightarrow x=0 \Rightarrow \sqrt[3]x -1\lt0不合 \\ y=-1 \Rightarrow x=1 \Rightarrow \sqrt[3]x-1=0 不合 } \Rightarrow y=2 \Rightarrow x=64，故選\bbox[red, 2pt]{(A)}$$
解答:$$(A)\times :{d\over dx}f({1\over x})=f'({1\over x}) \cdot (-{1\over x^2}) \ne{-1\over (f(x))^2} \\(B)\times :f(x)=xe^2 \Rightarrow f'(x)=e^2 \\(C)\times: x=-1 \Rightarrow f(x)=-x^3-2x \Rightarrow f'(x)=-3x^2-2 \ne |3x^2+2| \\(D) \bigcirc: g(x)=f(x)+1 \Rightarrow g'(x)=f'(x)\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$(C)\bigcirc: 平均值={1\over 4-1} \int_1^4 f(t)\,dt={1\over 3}\cdot 12=4，故選\bbox[red, 2pt]{(C)}$$
解答:$$S={3(5^{13}-1) \over 5-1}={3\over 4}(5^{13}-1) \Rightarrow \log S \approx \log \left( {3\over 4}\cdot 5^{13} \right)=\log 3-2\log 2+13(1-\log 2) \\=0.4771-2\cdot 0.301-13(1-0.301) =8.962 \Rightarrow S為9位數\\ 假設f(n)={5^{n}-1\over 4} \Rightarrow \cases{f(1)=1 \Rightarrow 個位數為1 \\f(2)=6 \Rightarrow 個位數為6 \\f(3)=31 \Rightarrow 個位數為1 \\f(4)=156 \Rightarrow 個位數為6 } \Rightarrow f(13)的個位數為1 \\ \Rightarrow S=3f(13)的個位數為3，故選\bbox[red, 2pt]{(D)}$$
解答:$$P(1,-1)在圓上\Rightarrow 1+1+4-k+m=0 \Rightarrow m=k-6\\ 圓:x^2 + y^2 + 4x + ky + m = 0 \Rightarrow (x+2)^2+ \left( y+{k\over 2} \right)^2 =4+{k^2\over 4}-m \Rightarrow 圓心O(-2,-k/2) \\ \Rightarrow \overline{OP}斜率\times切線斜率({2\over 3})=-1 \Rightarrow \overline{OP}斜率=-{3\over 2} ={-1-(-k/2)\over 1-(-2)} \Rightarrow k=-7\\ \Rightarrow m=-7-6=-13 \Rightarrow 2k+m=-27，故選\bbox[red, 2pt]{(A)}$$
解答:$$誤診率= 健康被檢查為患病+患病被檢查為健康\\= 98\% \times 10\%+ 2\%\times 30\%= 10.4\%，故選\bbox[red, 2pt]{(A)}$$
解答:$$\det(-3A^T B^{-1})= (-3)^3 \det(A^TB^{-1}) =-27 \det(A^T)\cdot \det(B^{-1}) =-27\det(A)\cdot {1\over \det(B)} \\=-27\cdot (-7)\cdot 2= 378，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{x_0 = -3, y_0 = 120 \\x_1 = -2, y_1 = 51 \\x_2 = 3, y_2 = -24 \\x_3 = 4, y_3 = -69} \Rightarrow \cases{f[-3,-2]={51-120\over -2-(-3)} =-69\\ f[-2,3]={-24-51\over 3-(-2)} =-15\\f[3,4]={-69-(-24)\over 4-(-2)} =-45} \Rightarrow \cases{f[-3,-2,3]={-5-(-69)\over 3-(-3)}=9\\ f[-2,3,4]={-45-(-15)\over 4-(-2)} =-5} \\ \Rightarrow f[-3,-2,3,4]={-5-9\over 4-(-3)}=-2，故選\bbox[red, 2pt]{(A)}$$
解答:$$已知 \sum_{k=1}^{10} (a_k + 3k) = 295 \Rightarrow \sum_{k=1}^{10} a_k + 3 \sum_{k=1}^{10} k = 295 \\ \Rightarrow \sum_{k=1}^{10} a_k + 3 \cdot 55=295 \Rightarrow \sum_{k=1}^{10} a_k =130 \Rightarrow \sum_{k=1}^{10} (a_k + 4k) = \sum_{k=1}^{10} a_k + 4 \sum_{k=1}^{10} k =130+4\cdot 55\\= 350，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{A(-1,2,0) \\B( 1,1,1)} \Rightarrow \vec v= \overrightarrow{AB}=(2,-1,1) \Rightarrow \overleftrightarrow{AB}=L: {x+1\over 2}={y-2\over -1}=z=t\\甲:\cases{2=2t-1\\ -1=-t+2\\1=t} \Rightarrow 無解\Rightarrow 甲\not \in L\\ 乙:\cases{3=2t-1 \\0=-t+2 \\2=t} \Rightarrow t=2 \Rightarrow 乙\in L\\ 丙:\cases{-3=2t-1\\ 3=-t+2\\1=t} \Rightarrow 無解\Rightarrow 丙\not \in L \\丁:\cases{21=2t-1\\ -9=-t+2\\11=t} \Rightarrow t=11 \Rightarrow 丁\in L\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$圓心O=\overline{BC}中點\Rightarrow 取\cases{O(0,0) \\C(a,0)\\ D(a,b) \\A(-a,b)\\B(-a,0)} \Rightarrow 矩形面積A =\overline{AD}\times \overline{CD}=2a\times b\\ \overline{OD}=圓半徑\Rightarrow a^2+b^2=2^2=4 \Rightarrow 算幾不等式: a^2+ b^2 \ge 2\sqrt{a^2b^2}=2ab \\ \Rightarrow A=2ab \le 4，故選\bbox[red, 2pt]{(B)}$$
解答:
$$f(x)=x^2+3x+2 = \left( x+{3\over 2} \right)^2-{1\over 4} \Rightarrow 頂點A \left( -{3\over 2},-{1\over 4} \right) \Rightarrow \begin{cases}f(x)\ge 0,& x\ge -1或 x\le -2\\f(x)\le 0,& -2\le x\le -1 \end{cases} \\y=g(x)= |f(x)| 與水平線y=k有兩個交點\Rightarrow k=0或k\gt {1\over 4}，故選\bbox[red, 2pt]{(C)}$$
解答:$$h(x)={9^x\over 9^x+3} \Rightarrow h(1-x)={9^{1-x} \over 9^{1-x}+3}={9\over 9+3\cdot 9^x} ={3\over 3+9^x} \Rightarrow h(x)+h(1-x)=1\\ S= h \left( {1\over 2027} \right)+ h \left( {2\over 2027} \right)+ \cdots+ h \left( {2025\over 2027} \right)+ h \left( {2026\over 2027} \right) \\== h \left( {1\over 2027} \right)+ h \left( {2\over 2027} \right)+ \cdots+ h \left(1- {2\over 2027} \right)+ h \left( 1-{1\over 2027} \right)\\={2026\over 2}=1013，故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)= \sin^2x \cos x =(1-\cos^2x) \cos x= \cos x-\cos^3 x \\\Rightarrow f'(x)=-\sin x+3\cos^2x \sin x=0 \Rightarrow \sin x(3\cos^2x-1)=0 \\ \Rightarrow \cases{\sin x=0 \Rightarrow f(x)=0\\ \cos^2x=1/3 \Rightarrow \sin ^2x=1-\cos^2x=2/3 \Rightarrow f(x)={2\over 3}\cdot \sqrt{1\over 3} ={2\sqrt 3\over 9}}\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$y=f(x) ={3^x-3^{-x} \over 3^x+3^{-x}} ={3^{2x}-1\over 3^{2x}+1} \Rightarrow y\cdot (3^{2x}+1)= 3^{2x}-1 \Rightarrow 3^{2x}={1+y\over 1-y} ={1+f(x) \over 1-f(x)} \\ \Rightarrow \cases{3^{2m} =(1+f(m))/(1-f(m)) =(1+1/5) /(1-1/5)= 3/2\\ 3^{2n} =(1+f(n)) /(1-f(n)) =(1+1/6)/(1-1/6) =7/5} \\ \Rightarrow 3^{2m}\cdot 3^{2n}={3\over 2}\cdot {7\over 5} \Rightarrow 3^{2(m+n)}={21\over 10} \Rightarrow f(m+n)={3^{2(m+n)}-1\over 3^{2(m+n)}+1} ={(21/10)-1\over (21/10)+1}={11\over 31}\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$17^2=8^2+15^2 \Rightarrow \tan \left( \sin^{-1} {8\over 17} \right) ={8\over 15}，故選\bbox[red, 2pt]{(C)}$$
解答:$$\lim_{x\to 0}{\sqrt{x+2}- \sqrt 2\over \sin x} =\lim_{x\to 0}{{d\over dx}(\sqrt{x+2}- \sqrt 2)\over {d\over dx}\sin x} = \lim_{x\to 0}{1/2\sqrt{x+2} \over \cos x} ={1\over 2\sqrt 2}={\sqrt 2\over 4}，故選\bbox[red, 2pt]{(A)}$$
解答:$$(4\cos x\sin x+1)(\sin^2 x+\cos x+1) = (2\sin2x+1)(1-\cos^2x+\cos x+1) \\=(2\sin2x+1) \left[ (\cos x-2)(\cos x+1) \right]=0 \\ \Rightarrow \cases{\sin 2x=-1/2 \Rightarrow x=7\pi/12, 11\pi/12, 19\pi/12, 23\pi/12 \\ \cos x=2 無解\\ \cos x=-1 \Rightarrow x=\pi} \Rightarrow 共5個解，故選\bbox[red, 2pt]{(C)}$$
解答:$$(A) \bigcirc: x^2+x-1=0 \Rightarrow \Delta=1+4=5\gt 0有實根\\ (B)\bigcirc: x^3+2x^2+3x+4=0 三次式一定有實根\\(C) \times: x^4+x^3+3x^2+2x+2 =(x^4+x^3+x^2)+ (2x^2+2x+2)= (x^2+2)(x^2+x+1)=0 \\ \qquad \Rightarrow \cases{x^2+2=0無實根\\ x^2+x+1=0 \Rightarrow \Delta=1-4\lt 0無實根}\\ (D)\bigcirc: x^5+x+1=0 五次式一定有實根\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)= f^{(0)}(x)= \cos x+\sin x= \sqrt 2\sin(x+\pi/4) \Rightarrow \cases{f^{(1)}=\sqrt 2\cos(x+\pi/4) \\f^{(2)}= -\sqrt 2\sin(x+\pi/4) \\f^{(3)}= -\sqrt 2\cos(x+\pi/4) \\ f^{(4)}=\sqrt 2\sin(x+\pi/4) =f^{(0)}} \\ \Rightarrow 循環數4 \Rightarrow 1051=4\times 262+3 \Rightarrow f^{(1051)}=f^{(3)}=-\sqrt 2\cos(x+\pi/4)=-\cos x+\sin x\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$(A) 10^{100}= 100^{50} \\ (B) 5^{150} =125^{50} \\(C)\pi^{200}=(\pi^4)^{50} \approx97^{50} \\ (D)2^{300} =64^{50}\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$f(x)= \sin(x^3-x) \Rightarrow f(-x)=\sin(-x^3+x)= \sin(-(x^3-x))=-\sin(x^3-x)=-f(x)\\ \Rightarrow f(x)為奇函數,對稱原點，故選\bbox[red, 2pt]{(A)}$$
解題僅供參考，其他教甄試題及詳解
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