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2023年4月19日 星期三

112年台南女中教甄-數學詳解

國立臺南女中 112 學年度第一次教師甄選數學科試題

一、填充題(每題 4 分,共 76 分)

解答
cosAOB=OAOB|OA||OB|=1/211=12AOB=60OAB¯ABP=(A+B)/2=(34,14,24)¯OP=¯OBsin60=33C(a,b,c)P=336C(34,14,24)=(32a,32b,32c){a=3/2b=3/6c=6/6C=(32,36,66)
解答{f(x)y=f(x)f(x)=ax(xb)t=2f(4+t)=f(2t)f(6)=f(0)6a(6b)=0{b=6a=0,ff(x)=ax(x6)=a(x3)29af(x108)12f(x)129a=12a=43f(x)=43x(x6)f(9)=123=36
解答{B=23C=37A=180BC=120:¯BCsinA=2R5sin120=103=2RR=53A=120BOC=240cosBOC=OBOC|OB||OC|=OBOCR212=OBOC25/3OBOC=256
解答ω=1+3i2=cos2π3+isin2π3ω3=11+ω+ω2=0ω296k=1(k)k+1kωk1=96k=1(ω)k+1k=ω22ω3+3ω44ω5+5ω66ω7+7ω88ω9+9ω10+94ω95+95ω9697ω97=ω22+3ω4ω2+56ω+7ω28+9ω+94ω2+9597ω=(14+794)ω2+(2+58++95)+(36+997)ω=(333)ω2+(3+3++3)+(333)ω()=(3×16)ω2+3×16+(3×16)ω=48ω2+4848ω=48(ω2+ω)+48=48(1)+48=96
解答f(x)=10xx216xx260=52(5x)222(8x)2{|5x|5|8x|2{0x106x10:x[6,10]{52(5x)2x=6,22(8x)=0f(6)=240=26
解答|z|=aNz=aeiθz1z2=aeiθ1a2z2iθ=1aeiθ1a2e2iθR1asin(θ)=1a2sin(2θ)asinθ=sin(2θ)=2sinθcosθsinθ(a2cosθ)=0{sinθ=0,,zcosθ=a/2cosθ={1/22/2=1,,θ=2kπ,θ=π3,5π3z=12±32i
解答¯AC=2R,{A(0,0)C(2R,0)B(Rcosθ,Rsinθ)D=(Rcosϕ,Rsinϕ)¯BD=25R2(cosθcosϕ)2+R2(sinθsinϕ)2=20R2(22cosθcosϕ2sinθsinϕ)=2R2(1cos(θϕ))=20(1)AC=32AB+52AD(2R,0)=32(Rcosθ,Rsinθ)+52(Rcosϕ,Rsinϕ)(32Rcosθ+52Rcosϕ,32Rsinθ+52Rsinϕ){3cosθ+5cosϕ=43sinθ+5sinϕ=034+30(cosθcosϕ+sinθsinϕ)=1630cos(θϕ)=18cos(θϕ)=35(1)165R2=20R2=10016R=104¯AC=2R=5
解答nrCnr+1r9:2,2991:9C91,82,C912892:92C92+12=C82,72,C822793:92C93+13=C73,62,C732694:C642595:C552429+C9128+C8227+C7326+C6425+C5524=512+2304+3584+2240+480+16=9136
解答n2+3n+43=k2,k,nN4n2+12n+172=4k2(2n+3)2+163=(2k)2(2k)2(2n+3)2=163(2k+2n+3)(2k2n3)=163×1{2k+2n+3=1632k2n3=1,4k=164k=41n2+3n+43=412=1681
解答y=ax+bx43x2+2x+3=ax+bx43x2+(2a)x+3b=0α,βx43x2+(2a)x+3b=(xα)2(xβ)2{2α+2β=0α2+β2+4αβ=32αβ(α+β)=a2α2β2=3b{α+β=0αβ=3/2{a=2b=3/4L:y=2x+34
解答T=[5/83/83/85/8],T[10]=[p1p2]p1T,T=PDP1=[1111][1/4001][1/21/21/21/2]T=[1111][1/4001][1/21/21/21/2]=[1111][0001][1/21/21/21/2]=[1/21/21/21/2]T[10]=[1/21/2]p1=12
解答{P(0,0,0)A(1,0,1)B(1,1,0)C(0,1,1)O=(A+B+C)/3=(2/3,2/3,2/3)P=O/2=(1/3,1/3,1/3){u=AP=(2/3,1/3,2/3)v=BC=(1,0,1)n=u×v=13(1,4,1)AnE:(x1)+4y+(z1)=0x+4y+z=2{Q=E¯PB=(2/5,2/5,0)R=E¯PC=(0,2/5,2/5)P,A,Q,R16000110112/52/50102/52/51=16825=475=m=13=13475=2175=nmn=421
解答第二張牌比第一張大的情形:(1,2-6),(2,3-6),(3,4-6),(4,5-6),(5,6),\\\quad 共5+4+\cdots+1=15種情形,機率=15/P^6_2=15/30={1\over 2}\\第三張牌比第1張大,也比第2張大的情形(不管第一,二張的大小):\\ \qquad (1,2,3-6): 1,2有2種排列,3-6有4種情形,此情形有2\times 4=8種;\\ \qquad 同理,(1,3,4-6),(1,4,5-6)...(4,5,6),機率=40/P^6_3 ={1\over 3}\\ 第四張比第1,2,3張都大的機率=90/P^6_4={1\over 4}\\ 第五張比前4張都大的機率={1\over 5},第六張比前5張都大的機率={1\over 6}\\ 因此期望值={1\over 2}+ {1\over 3}+ {1\over 4}+ {1\over 5}+ {1\over 6} ={174\over 120} =\bbox[red,2pt]{29\over 20}
解答
解答\angle C=90^\circ \Rightarrow \angle A=90^\circ \Rightarrow 直徑\overline{BD} = \sqrt{12^2+16^2}=20 \Rightarrow 半徑r=10\\ O為\overline{BD}中心點,也是圓心;圓周角\angle D=75^\circ \Rightarrow 圓心角\angle AOC=2\times 75^\circ =150^\circ\\ \Rightarrow \cos \angle AOC={r^2+r^2-\overline{AC}^2 \over 2r^2} \Rightarrow -{\sqrt 3\over 2} ={200-\overline{AC}^2\over  200} \Rightarrow \overline{AC}^2 =100(2+\sqrt 3) \\ \Rightarrow \overline{AC}= 10\sqrt{2+\sqrt 3} =5\sqrt{8+\sqrt{12}} =\bbox[red,2pt]{5(\sqrt 6+\sqrt 2)}


解答9\cos^2 x-3\sin x-7=0 \Rightarrow 9-9\sin^2x-3\sin x-7=0 \Rightarrow 9\sin^2 x+3\sin x-2=0\\ \Rightarrow (3\sin x-1)(3\sin x+2)=0 \Rightarrow \cases{\sin x={1\over 3}\\ \sin x=-{2\over 3} } \\\Rightarrow \cases{x= \theta_1,\pi-\theta_1,\theta_1+2\pi ,3\pi-\theta_1 \\ x=\theta_2,3\pi-\theta_2 , 2\pi+\theta_2 ,5\pi-\theta_2} \Rightarrow 所有根的和=6\pi+10\pi= \bbox[red,2pt]{16\pi}
解答S_n=1+2+\cdots +n={n(n+1)\over 2} \Rightarrow S_n-1={n^2+n-2\over 2} ={(n+2)(n-1)\over 2} \\ \Rightarrow {S_n\over S_n-1}={n(n+1)\over (n-1)(n+2)} \Rightarrow T_n ={2\cdot 3\over 1\cdot 4} \times {3\cdot 4\over 2\cdot 5} \cdot {4\cdot 5\over 3\cdot 6}\cdots {n(n+1) \over (n-1)(n+2)} \\= {3\over 1}\cdot {n\over n+2} \Rightarrow T_{1998} ={3\cdot 1998\over 2000} =\bbox[red, 2pt]{2997\over 1000}
解答假設5個正整數依序從小至大為x_1,x_2,x_3,x_4,x_5;由\cases{全距18\Rightarrow n_5=n_1+18\\ 中位數=眾數=9 \Rightarrow x_3=9}\\ 因此5個正整數依序從小至大為x_1,x_2,9,x_4,x_1+18,\\再由平均數=13 \Rightarrow {2x_1+x_2+x_4+27\over 5}=13 \Rightarrow 2x_1+x_2+x_4=38\\ \begin{array}{} x_1 & x_2 & x_3 & x_4 & x_5 &\\\hline 1 & 9 & 9 & 19& 19 & \times:x_2+x_4\lt 36\\ 2 & 9 & 9 & 20 & 20& \times: x_2+x_4 \lt 34 \\ 3 & 9 & 9 & 21& 21& \times: x_2+x_4\lt 32 \\ 4& 9 & 9 & 21 & 22 & \bigcirc\\ \hdashline 5 &9 &9 &19 &23 & \bigcirc\\ 5 &8 &9 &20 &23 & \times:眾數不是9\\ 5 & 7 & 9 & 21 & 23 & \times:眾數不是9\\ 5 & 6 & 9 & 22 & 23& \times:眾數不是9\\ \hdashline 6 &9 & 9 &17 &24 & \bigcirc \\ 6 & 8& 9 &18&24 & \times:眾數不是9 \\ 6 & 7 & 9 & 19& 24 & \times:眾數不是9\\\hdashline 7 &9 & 9 &15 &25 & \bigcirc\\ 7 & 8 & 9 & 16 & 25 & \times:眾數不是9\\ \hdashline 8 & 9& 9 &13 & 26 & \bigcirc \\\hdashline 9 & 9 & 9 & 11&27 &\bigcirc\end{array}\\ 第二大的數字x_4=21,19,17,15,13,11,共有\bbox[red,2pt]6個
解答假設L:y=ax+b \Rightarrow \cases{d(L,A(7,3))=6\\ d(L,B(6,6))=3} \Rightarrow \cases{|7a+b-3|=6\sqrt{a^2+1} \\ |6a+b-6| =3\sqrt{a^2+1}} \\   先試7a+b-3=2(6a+b-6) \Rightarrow b=9-5a \Rightarrow |7a+9-5a-3|= |2a+6|=6\sqrt{a^2+1} \\ \Rightarrow (2a+6)^2 = 36(a^2+1) \Rightarrow 8a(4a-3)=0 \Rightarrow \cases{a=0 \Rightarrow b=9\\ a=3/4 \Rightarrow b=21/4} \\ \Rightarrow \cases{L:y=9 \Rightarrow d(L,C)=9\\ L:4y=3x+21 \Rightarrow d(L,C)=3} \Rightarrow d(L,C)=\bbox[ red, 2pt]{3或9}
解答假設f(x)=ax^3+bx^2+cx+d \Rightarrow f'(x)=3ax^2 +2bx+c \Rightarrow f'(x)=6ax+2b \\ \Rightarrow f''(x)-3xf'(x)+9f(x) =3bx^2+(6a+c)x +(2b+9d)=0 \Rightarrow \cases{b=0\\ a=-c\\ d=0} \\ \Rightarrow f(x)=ax^3-ax,再將f(2)=6代入\Rightarrow 8a-2a=6 \Rightarrow a=1 \Rightarrow f(x)=x^3-x \\ 因此 g'(x)=f(x)=x^3-x=0 \Rightarrow \cases{x=0\\x=1} \Rightarrow \cases{g(0)= \int_0^0f(t)\,dt=0\\ g(1) = \int_0^1t^3-t\,dt = {1\over 4} -{1\over 2}=\bbox[red, 2pt]{-{1\over 4}}}

二、計算證明題(每題 6 分,共 24 分)

解答假設有理根為{q\over p}(p,q互質) \Rightarrow a\cdot ({q\over p})^2+ b\cdot {q\over p}+c=0 \Rightarrow aq^2+ bpq+cp^2=0\\利用反證法,假設a,b,c都是奇數,由於p,q互質,因此\\ \text{Case I: }p,q都是奇數 \Rightarrow \cases{aq^2為奇數\\ bpq為奇數\\ cp^2為奇數} \Rightarrow 三奇數之和應為奇數,不可能為0\\ \text{Case II: }\cases{p為奇數\\ q為偶數}\Rightarrow \cases{aq^2為偶數\\ bpq為偶數\\ cp^2為奇數} \Rightarrow 二偶一奇之和應為奇數,不可能為0\\ \text{Case III: }\cases{p為偶數\\ q為奇數} \Rightarrow \cases{aq^2為奇數\\ bpq為偶數\\ cp^2為偶數} \Rightarrow 二偶一奇之和應為奇數,不可能為0\\ 因此a,b,c 三數不可能皆為奇數,至少有一個偶數,\bbox[red,2pt]{故得證}
解答\cases{x^3y^2為六位數\\ \log x^3y^2不是整數} \Rightarrow 10^5 \lt x^3y^2 \lt 10^6 \Rightarrow 5\lt \log x^3y^2 \lt 6 \Rightarrow 5\lt \log x+2(\log x+\log y)\lt 6\cdots(1) \\ 又\cases{\log x\ge 0\\ \log y\ge 0} \Rightarrow 0\le 2(\log x+\log y)再加上(1) \Rightarrow 0\le 2(\log x+\log y) \lt 6\\ \Rightarrow 0\le \log x+\log y\lt 3,又\log x+\log y是整數,因此\log x+\log y=0,1,2\\ \text{Case II }\log x+\log y=0: 由於\cases{\log x\ge 0\\ \log y\ge 0} \Rightarrow \log x=\log y=0違反\log x,\log y都不是整數\\ \text{Case I }\log x+\log y=1: 由(1)得5 \lt \log x+2\lt 6 \Rightarrow 3\lt \log x\lt 4加上\log x+\log y=1\\ \qquad \Rightarrow -3\lt \log y\lt -2違反\log y\ge 0\\ 由\text{Case I & Case II}可知: \log x+\log y=2代入(1) \Rightarrow 1\lt \log x\lt 2 \Rightarrow 10\lt x\lt 100\\ \log x+\log y=\log xy=2 \Rightarrow xy=100 \Rightarrow 1\lt y\lt 10 \Rightarrow y=2,3,\dots,9 代入xy=100\\ \Rightarrow (x,y)=\bbox[red,2pt]{(50,2),(25,4),(20,5)}
解答a_k =\sqrt{1+2+\cdots + k} =\sqrt{k(k+1)\over 2}\\ 由於k^2 \le k(k+1)\le (k+1)^2 \Rightarrow k \le \sqrt{k(k+1)}\le k+1 \Rightarrow {k\over \sqrt 2}\le a_k\le {k+1\over \sqrt 2} \\ \Rightarrow  {1\over n^2}\sum_{k=1}^n {k\over \sqrt 2}\le  {1\over n^2} \sum_{k=1}^n a_k\le {1\over n^2}  \sum_{k=1}^n{k+1\over \sqrt 2} \Rightarrow  {1\over n^2}  {n(n+1)\over 2\sqrt 2}\le  {1\over n^2} \sum_{k=1}^n a_k\le {1\over n^2}   {n(n+3)\over 2\sqrt 2} \\ \Rightarrow \lim_{n\to \infty} \left({1\over n^2}  {n(n+1)\over 2\sqrt 2} \right)\le \lim_{n\to \infty}  \left({1\over n^2} \sum_{k=1}^n a_k \right) \le \lim_{n\to \infty}{1\over n^2}   {n(n+3)\over 2\sqrt 2} \\ \Rightarrow {1\over 2\sqrt 2} \le \lim_{n\to \infty} \left( {1\over n^2} \sum_{k=1}^n a_k\right) \le {1\over 2\sqrt 2} \Rightarrow \lim_{n\to \infty} \left( {1\over n^2} \sum_{k=1}^n a_k\right) ={1\over 2\sqrt 2} =\bbox[red, 2pt]{\sqrt 2\over 4}
解答假設兩焦點為F,F' 且\cases{\overline{FA}=p\\ \overline{FB}=q} \Rightarrow \cases{\overline{F'A}= 2a-p \\ \overline{F'B}=2a- q} \\ \Rightarrow \cases{\cos \angle AFF' = {p^2+ (2c)^2-(2a-p)^2 \over 2p(2c)} \\ \cos \angle BFF' ={q^2+(2c)^2-(2a-q^2) \over 2q(2c)}},\angle AFF'+\angle BFF'=180^\circ \\\Rightarrow {p^2+ (2c)^2-(2a-p)^2 \over 2p(2c)}=-{q^2+(2c)^2-(2a-q^2) \over 2q(2c)} \\ \Rightarrow p^2q+4c^2q-q(2a-p)^2 = -pq^2-4c^2p+p(2a-q)^2 \\ \Rightarrow pq(p+q)+4c^2(p+q)-q(4a^2-4ap+ p^2)-p(4a^2-4aq+q^2)=0\\ \Rightarrow pq(p+q)+4c^2(p+q)- 4a^2(p+q)-pq(p+q)+8apq=0\\ \Rightarrow (4c^2-4a^2)(p+q)+8apq \Rightarrow -4b^2(p+q)+8apq=0\\ \Rightarrow {p+q\over pq}={2a\over b^2} \Rightarrow {1\over p}+{1\over q}={2a\over b^2} \Rightarrow {1\over \overline{FA}} +{1\over \overline{FB}} ={2a\over b^2},\bbox[red,2pt]{故得證} 

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解題僅供參考,其他教甄試題及詳解

1 則留言:

  1. 請問第10題為何可以直接假設恰好兩相異實根呢?不是也有兩相異實根+一組共軛虛根的可能嗎?

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