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2023年7月16日 星期日

112年台中市國中教甄聯招-數學詳解

112 學年度臺中市政府教育局受託辦理本市立國民中學
(含本市立高級中等學校附設國中部)教師甄選

選擇題(共 40 題,每題 2.5分,共 100 分)

解答{x1x+3(x+1)(x1)=0x=33x1x+3(x+1)(1x)=0x=1x3x3(x+1)1+x=0x=5315=3(A)(B)
解答f(x)=x3f(0)(D)
解答limnsin(1/n)(1/n)=limnsin(1/n)(1/n)=limn(1/n2)cos(1/n)1/n2=limncos1n=1(LCT):n=1sin1nn=11nn=11nn=1sin1n(D)
解答A1=[abcd],{A[13]=[10]A[42]=[02]{[13]=A1[10]=[ac][42]=A1[02]=[2b2d]{a=1b=2c=3d=1A1=[1231](D)
解答H3(C)
解答{A:B:{1P(AB)=0.6P(AB)=0.4P(A)=0.2P(B)=0.3P(AB)=0.2+0.30.4=0.1(A)
解答162=256(D)
解答E[(X+1)2]=E[X2+2X+1]=E[X2]+2E[X]+E[1]=E[X2]+2+1=E[X2]+3=8E[X2]=5Var(X)=E[X2](E[X])2=51=4Var(X)=4Var(13x)=9Var(X)=36(D)
解答f(x)=1x(x+1)(x+2)=12(1x(x+1)1(x+1)(x+2))f(1)+f(2)++f(10)=12(12111×12)=65264=aba+b=329(D)
解答
{B(0,0)A(0,4)C(3,0){D(0,3)E(3/4,3){L1=CD:x+y=3L2=BE:y=4xF=L1L2=(3/5,12/5){¯CD=32¯DF=352k=33/5=5(C)
解答

y=x1y2=2x+6x22x+1=2x+6x24x5=0(x5)(x+1)=0{x=1y=2x=5y=4{A(1,2)B(5,4){y=x1y2=2x+6{x=y+1x=(y26)/242(y+1)(y26)/2dy=4212y2+y+4dy(A)
解答1:10exdx=[2ex(x1)]|10=22:10xexdx=[ex(x1)]|10=13:π/20esinxsin(2x)dx=[2esinx(sin(x)1)]|π/20=21=3(B)
解答f=cos(xyz)x2y2z{fx=yzsin(xyz)2xy2fy=xzsin(xyz)2x2yfz=xysin(xyz)1f(1,1,0)=(2,2,1):2(x1)+2(y+1)z=0z=2x+2y+4(B)
解答()
解答f(1+x1x)=f(21x1)=xf(21x1)2(1x)2=1x=13f(2)24/9=1f(2)=29(C)
解答f(x)66f(4)f(1)+(41)×6=30(A)
解答xy1111112111113110104199517711+11+10+9+7=48(B)
解答{2465×4×3=6060×3=1803{3234364×3=12,12×3=3618036=144(D)
解答()
解答{a20a{11=a(20a)/C2022=Ca2/C202=a(20a)C202+2Ca2C202=19a190=a10=45a=8(A)
解答(x22kx+k2)+(y22ky+k2)=k24k+2a(xk)2+(yk)2=(k2)2+2a42=(k2)2+2a4>02a4>0a>2(D)
解答6,7,8,924+14+12+22=72{5a10b{5=a/59=(a+72)/910=(a+b+72)/10{(a+72)/9>a/5(a+b+72)/10>18{a<90a+b>108a89b>10889=19b=20(C)
解答AB=4363=64216AB=164216=152216=1927a+b=19+27=46(D)
解答
{E(0,0)B(0,4)C(0,4)D(8,0)F=(B+D)÷2=(4,2)A=3F2E=(12,6)L=AC:5x+6y+24=0k=ABCACD=d(B,L)d(D,L)=4816=3(B)
解答f(x)=cos2(2x)+6sin2x+3=cos2(2x)+61cos2x2+3=cos2(2x)3cos(2x)+6=(cos(2x)32)2+154{b=f(π/2)=10a=f(0)=42ba=204=16(C)
解答limnnk=11k2+nk+n2=limnnk=11/n(k/n)2+(k/n)+1=101x2+x+1dx=[ln|23x+13+23(x2+x+1)|]|10=ln(2+3)ln3=ln2+33=lnαα=23+1=1+233(B)
解答R={(x,y)0y2xx2}(x1)2+y2=1{x=rcosθy=rsinθy2xx2x2+y22xr22rcosθr2cosθ
解答A=\begin{bmatrix}1 & -1 & 0 \\-1 & 3 & -1 \\0 & -1 &2\end{bmatrix}\Rightarrow  \det(A-\lambda I) = \lambda^3- 6\lambda^2+9\lambda-3 =0\Rightarrow \cases{\alpha_1 +\alpha_2 +\alpha_3=6 \\ \alpha_1 \alpha_2 +\alpha_2\alpha_3 +\alpha_3\alpha_1=9\\\alpha_1 \alpha_2 \alpha_3=3} \\ \Rightarrow (\alpha_1 +\alpha_2 +\alpha_3)^2= \alpha_1^2 +\alpha_2^2 +\alpha_3^2+ 2(\alpha_1 \alpha_2 +\alpha_2\alpha_3 +\alpha_3\alpha_1) \Rightarrow 36=\alpha_1^2 +\alpha_2^2 +\alpha_3^2+18 \\ \Rightarrow \alpha_1^2 +\alpha_2^2 +\alpha_3^2=36-18=18 \\又 \alpha_1^3 +\alpha_2^3 +\alpha_3^3-3\alpha_1 \alpha_2 \alpha_3 =(\alpha_1 +\alpha_2 +\alpha_3)(\alpha_1^2 +\alpha_2^2 +\alpha_3^2-(\alpha_1 \alpha_2 +\alpha_2\alpha_3 +\alpha_3\alpha_1)) \\ \Rightarrow \alpha_1^3 +\alpha_2^3 +\alpha_3^3-9=6(18-9) \Rightarrow \alpha_1^3 +\alpha_2^3 +\alpha_3^3=54+9=63,故選\bbox[red, 2pt]{(C)}
解答{x+y/2+y/2+z/3+z/3+z/3\over 6}\ge \sqrt[6]{xy^2z^3\over 36}\\ xy^2z^3有極有值時,x={y\over 2}={z\over 3} \Rightarrow \cases{z=3x\\ y=2x} \Rightarrow x+y+ z=6x =2 \Rightarrow \cases{x=1/3 =\alpha\\ y=2/3 =\beta\\ z=1= \gamma} \\ \Rightarrow 3\alpha+\gamma =1+1=2,故選\bbox[red, 2pt]{(A)}
解答令\cases{a=\sqrt[3]{\sqrt{17}+3} \\b= \sqrt[3]{\sqrt{17}-3}} \Rightarrow x=a-b \Rightarrow x^3=(a-b)^3 =a^3-b^3 -3ab(a-b) =6-3\sqrt[3]{8}x\\ \Rightarrow x^3=6-6x \Rightarrow x^3+6x+7=6-6x+6x+7=13,故選\bbox[red, 2pt]{(D)}
解答\lim_{k\to \infty}\left| {a_{k+1}\over a_k}\right| =\lim_{k\to \infty}\left| {{(k+1)!\over (k+1)^{k+1}}(x-1)^{k+1} \over {k!\over k^k}(x-1)^k}\right| =\lim_{k\to \infty}\left| {({k\over k+1}})^k (x-1)\right| = \left| {{1\over e}}  (x-1)\right|  \lt 1 \\ \Rightarrow 1-e\lt x\lt 1+e \Rightarrow \cases{a=1-e\\ b=1+e} \Rightarrow a+2b=1-e+2+2e=3+e,故選\bbox[red, 2pt]{(D)}
解答\sin x =x-{x^3\over 3!}+{x^5\over 5!}-\cdots \Rightarrow \sin{1\over x} ={1 \over x}-{1\over 3! x^3} +{1\over 5! x^5}-\cdots \\ \Rightarrow x\sin{1\over x} =1-{1\over 3!x^2}+ {1\over 5!x^4} -\cdots \cdots(1)\\ 而\lim_{x\to \infty} \sqrt{x^2-x}-x =\lim_{x\to \infty} {-x\over \sqrt{x^2-x}+x} =\lim_{x\to \infty} {-1\over \sqrt{1- 1/x}+1} =-{1\over 2}\cdots(2) \\ 由(1)及(2)可得: \lim_{x\to \infty} x\left(\sqrt{x^2-x}-x \right) \sin({1\over x}) =-{1\over 2},故選\bbox[red, 2pt]{(A)}
解答f(t)\in P_2(\mathbb R) \Rightarrow f(t)=at^2+bt +c, a,b,c\in \mathbb R \Rightarrow f'(x)=2ax+b\\ \Rightarrow T(f)(x) =3(2ax+b)+ \int_0^x (2at^2+ 2bt +2c) \,dt ={2\over 3}ax^3+ bx^2 +(6a+2c)x+3b\\ \text{If }T(f)(x) =0, \text{ then we have }a=b=c=0 \Rightarrow Null(T)=\{0\} \Rightarrow \text{nullity}(T)=0,故選\bbox[red, 2pt]{(A)}
解答\det(A-\lambda I)=0 \Rightarrow -\lambda^3+8\lambda^2-4\lambda-48=0 \Rightarrow -(\lambda+2)(\lambda-4) (\lambda-6)=0\\ \Rightarrow 特徵值=-2,4,6,故選\bbox[red, 2pt]{(B)}
解答取\cases{p=a-3\\ q=b-2\\ r=c-1}  \Rightarrow a+b+c\le 15 \Rightarrow p+q+r\le 9 \Rightarrow 共有H^4_9=220組整數解,故選\bbox[red, 2pt]{(B)}
解答\cases{\cases{x+y+1=0\\ x-y+3=0} \Rightarrow (x,y)=(-2,1)\\ \cases{x+2y+4=0\\ x-2y+12=0} \Rightarrow (x,y)=(-8,2)} \Rightarrow r=4,s=2 \Rightarrow r+s=6 ,故選\bbox[red, 2pt]{(B)}
解答A=\begin{bmatrix}4 & 4 \\-1 & -1 \end{bmatrix} =\begin{bmatrix}-1 & -4\\1 & 1 \end{bmatrix} \begin{bmatrix}0 & 0 \\0 & 3 \end{bmatrix} \begin{bmatrix}1/3 & 4/3 \\-1/3 & -1/3 \end{bmatrix} \equiv PDP^{-1} \\ \Rightarrow A^n=PD^nP^{-1} \Rightarrow A+ A^2+\cdots +A^n =P(D+D^2+\cdots +D^n)P^{-1} \\=P\begin{bmatrix}0 & 0 \\0 & 3+3^2+\cdots +3^n \end{bmatrix} P^{-1} =P\begin{bmatrix}0 & 0 \\0 & (3^{n+1}-3)/2 \end{bmatrix} P^{-1} =\begin{bmatrix} 2(3^n-1) & 2(3^n-1) \\{1\over 2}(1-3^n) & {1\over 2}(1-3^n) \end{bmatrix} \\ \Rightarrow b+c={1\over 2}(1-3^n)+ {1\over 2}(1-3^n)=1-3^n,故選\bbox[red, 2pt]{(B)}
解答f(x,y)=x^3-y^3+6x^2+ 2x+y+2 =(x^3+ 3x^2 +2x)+3x^2 -(y^3-y)+2 \\=x(x^2+3x+2)+ 3x^2-y(y^2-1)+2= x(x+1)(x+2)+3x^2-(y-1)y(y+1)+2 \\ 連續三整數之積必為3的倍數 \Rightarrow \cases{x(x+1)(x+2)=3k\\ 3x^2=3m\\ (y-1)y(y+1)= 3n} \\ \Rightarrow f(x,y)=3(k+m-n)+2 \ne 0, \forall k,m,n\in \mathbb Z,故選\bbox[red, 2pt]{(A)}
解答\lim_{n\to \infty}\left[2n^2 \left({r\over n}-\sum_{k=1}^r {1\over n+k}\right) \right] =\lim_{n\to \infty}\left[2nr -\sum_{k=1}^r {2n^2\over n+k} \right]\\ =\lim_{n\to \infty}\left[2nr -\sum_{k=1}^r \left((2n-2k)+{2k^2\over n+k} \right) \right] =\lim_{n\to \infty}\left[2nr - \left(2nr-r(r+1)+\sum_{k=1}^r{2k^2\over n+k} \right) \right] \\=\lim_{n\to \infty}\left[r(r+1)-\sum_{k=1}^r{2k^2\over n+k}  \right] =r(r+1)-\lim_{n\to \infty}\sum_{k=1}^r{2k^2\over n+k}  =r(r+1),故選\bbox[red, 2pt]{(D)}
解答{b\over a}={|b^2+c^2-a^2|\over bc} ={b^2+c^2-a^2\over bc} (c\gt b\gt a \Rightarrow c^2+b^2-\gt a^2 )\\ \Rightarrow {b\over 2a}={b^2+c^2-a^2\over 2bc} \Rightarrow {\sin B\over 2\sin A}=\cos  A \Rightarrow \sin B=\sin(2A) \Rightarrow \angle B=2\angle A\\ 同理,{c\over b}={|c^2+a^2-b^2| \over ca} \Rightarrow \sin C= \sin(2B) \Rightarrow C=2\angle B \\ 因此\angle A+ \angle B+ \angle C=\angle A+2\angle A+4\angle A=7\angle A=\pi \Rightarrow \angle A={\pi \over 7},故選\bbox[red, 2pt]{(A)}


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6 則留言:

  1. 老師您好
    32題有亂碼,38題可幫忙解題嗎 謝謝您

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  2. 老師您好,想請教33題

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    1. 已把它寫得再詳細一點了,請參考...

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