教育部受託辦理107學年度
公立高級中等學校教師甄選
數學科試題
公立高級中等學校教師甄選
數學科試題
第一部分:選擇題
一、單選題
解:
{A(0,0,0)B(2,0,0)C(2,2,0)D(0,2,0)E(1,0,0)F(0,1,0)G(2,2,1)⇒{→GE=(−1,−2,−1)→GF=(−2,−1,−1)⇒→n=→GE×→GF=(1,1,−3)平面EFG:(x−1)+y−3z=0⇒x+y−3z−1=0⇒D至平面EFG的距離=0+2+0−1√12+12+32=1√11,故選(A)
{log8a+log4b=3log8b+log4a=7⇒{13log2a+12log2b=313log2b+12log2a=7,兩式相加⇒56(log2a+log2b)=10⇒log2ab=12⇒ab=212=4096,故選(D)
解:
b為偶數⇒(a,b,c)共有6×3×6=108種組合;|abbc|>0⇒ac>b2⇒b(偶數)ac數量215−6223−643−42−6105−61−612436145−6254−6363−64共38⇒機率為38108≈0.352,故選(C)
解:
令{f(a,b)=√2a+1+√3b+2g(a,b)=a+b−2,由Lagrange 算子可得{∂∂af=λ∂∂ag∂∂bf=λ∂∂bgg=0⇒{1√2a+1=λ⋯(1)32√3b+2=λ⋯(2)a+b=2⋯(3),由(1)及(2)⇒1√2a+1=32√3b+2⇒12a+1=912b+8⇒18a−12b+1=0⋯(4)由(3)及(4)⇒{a=23/30b=37/30⇒f(23/30,37/30)=√4630+1+√11130+2=√7630+√17130=2√19√30+3√19√30=5√19√30=5√57030=√5706,故選(B)
解:
紅球R,藍球B,黃球Y,現有RRRBBY分給三人a,b,c;⇒abcRRRBBYRYBBBBRYBYRBRBRRBYRBRYRYRBBYRRRYRRBBRBRBBBRRBBRRRYRYRRBYRRRBRBRR⇒共15種分法,故選(B)
解:
{a+b+c+d+e=10a2+b2+c2+d2+e2=205/4⇒{a+b+d+e=10−ca2+b2+d2+e2=205/4−c2柯西不等式:(a2+b2+d2+e2)(12+12+12+12)≥(a+b+d+e)2⇒(2054−c2)×4≥(10−c)2⇒205−4c2≥c2−20c+100⇒5c2−20c−105≤0⇒c2−4c−21≤0⇒(c−7)(c+3)≤0⇒−3≤c≤7⇒{k=7t=−3,故選(A)
解:
{6≤x+y≤8−2≤2x+y≤0⇒封閉區域頂點{P(16,−8)Q(18,−10)R(14,−8)S(12,−6);→A⋅→B=|→A||→B|cos60∘=1×2×12=1⇒→u⋅→v=(→A+→B)⋅(x→A+y→B)=x|→A|2+(x+y)→A⋅→B+y|→B|2=x+(x+y)+4y=2x+5y;令f((x,y))=2x+5y⇒{f(P)=32−40=−8f(Q)=36−50=−14f(R)=28−40=−12f(S)=24−30=−6⇒→u⋅→v最大值為−6,故選(C)
解:
正弦定理:¯ACsin∠B=¯ABsin∠C=2R⇒25/13sin∠B=3sin∠C=2×52=5⇒{sin∠B=5/13sin∠C=3/5⇒{cos∠B=12/13cos∠C=−4/5⇒¯BC=¯AB⋅cos∠B+¯AC⋅cos∠C=3613−2013=1613再由餘弦定理:cos∠BAC=32+(25/13)2−(16/13)22⋅3⋅(25/13)=9+36916915013=1890169×13150=6365,故選(D)
二、複選題

解:
a−c2b=a−b2c⇒ac−c22bc=ab−b22bc⇒ac−c2=ab−b2⇒b2−c2=ab−ac⇒{b=c⋯(1)b+c=a⇒aa+b+c=a2a=12同理,b+c2a=a−c2b⇒b2+bc=a2−ac⇒c(a+b)=(a+b)(a−b)⇒{a=−b⋯(2)b+c=a,由(1)及(2)⇒{a=−bc=b⇒aa+b+c=−bb=−1因此aa+b+c=−1,12,故選(BC)
解:
f(x)=x3−kx=x(x2−k)=0⇒x=−√k,0,√k⇒∫10|f(x)|dx=|∫√k0f(x)dx|+|∫1√kf(x)dx|=|[14x4−k2x2]|√k0|+|[14x4−k2x2]|1√k|=14k2+14(k−1)2=14(2(k−12)2+12)⇒k=12時,∫10|f(x)|dx有最小值:14⋅12=18⇒{a=1/2b=1/8,故選(AC)
解:
(A)◯:bnbn−1=2an2an−1=2an−an−1=2d⇒<bn>為公比2d的等比數列(B)×:d>0⇒r=2d>1(C)◯:{b1=2a1=21=2r=2d=√2⇒bn>4096⇒b1rn−1=2⋅2(n−1)/2=2(n+1)/2>4096=212⇒n+12>12⇒n>23⇒n=24(D)◯:n∑i=1ai=2a1+(n−1)d2⋅n=2+(n−1)/22⋅n=22⇒(n+11)(n−8)=0⇒n=8⇒8∑i=1bi=b1−b1r81−r=2−2⋅241−√2=(25−2)(√2+1)=30(1+√2),故選(ACD)
解:
(A)◯:迴歸直線必經(50,M)⇒M=34×50+20=57.5(B)◯:相關係數r⇒34=r×σyσx⇒r=34×812=0.5(C)×:迴歸直線只是推估,不能保證(D)◯:σy>σx⇒物理成績較分散,故選(ABD)
第二部份:綜合題
一、填充題
(sin263∘−3sin227∘)(sin29∘−3cos2171∘)=(cos227∘−3(1−cos227∘))(sin29∘−3cos29∘)=(−3+4cos227∘)(sin29∘−3(1−sin29∘))=(−3+4cos227∘)(−3+4sin29∘)=cos27∘(−3+4cos227∘)sin9∘(−3+4sin29∘)cos27∘sin9∘=−cos81∘sin27∘cos27∘sin9∘(∵{cos(3x)=4cos3x−3cosxsin(3x)=−4sin3x+3sinx)=−sin9∘sin27∘cos27∘sin9∘=−tan27∘=tan(360∘−27∘)=tan333∘⇒θ=333∘
解:
a2=2+1=3=a1+3a3=3+2√4+1=8=a2+5a4=8+2√9+1=15=a3+7⋯⋯+an=an−1+(2n−1)an=3+5+7+⋯+2n−1=∑nk=2(2k−1)⇒a30=30∑k=2(2k−1)=32×29−29=899
解:
令g(x)=xf(x)⇒{g(0)=0g(t)=tf(t)=12,t=1−2019⇒g(t)=a(t−1)(t−2)⋯(t−2019)+12⇒g(0)=a×−2019!+12=0⇒a=12×2019!⇒g(2020)=a×2019!+12=12+12=1⇒f(2020)=12020×g(2020)=12020
解:
12×3×4×5+13×4×5×6+14×5×6×7+15×6×7×8+⋯=13(12×3×4−13×4×5)+13(13×4×5−14×5×6)+13(14×5×6−15×6×7)+13(15×6×7−16×7×8)+⋯=13×12×3×4=172
解:
A=[10−12]⇒A2=[10−12][10−12]=[10−34]⇒A4=[10−34][10−34]=[10−1516]⇒A8=[10−1516][10−1516]=[10255256]=[1001]+255[00−11]⇒(a,b)=(1,255)
解:
{P(x,−x2+5x+1)Q(x,−x+6)⇒¯PQ=−x2+5x+1+x−6=−(x2−6x+9)+4=−(x−3)2+4≤4⇒最大值為4
解:
n12log2n−1[12log2n−1]小計1−3[−1,0)−1−34−15[0,1)0016−63[1,2)14864−100[2,3)274119⇒119
解:
(√m+√m2−n+√m−√m2−n)2=62⇒2m+2√n=36⇒n=(m−18)2由m2≥n⇒n=(m−18)2=m2−36m+182⇒−36m+182≤0⇒m≥9⇒∑n=92+82+⋯+12=9×10×196=285
解:
a_n=a_{n-1}-a_{n-2},n\ge 3\\ \Rightarrow \begin{array}{l} a_1=a_1\\ a_2=a_2\\a_3= a_2-a_1\\ a_4=a_3-a_2= a_2-a_1-a_2=-a_1 \\ a_5=a_4-a_3 =-a_1-(a_2-a_1) =-a_2 \\a_6=a_5-a_4=-a_2-(-a_1) =-a_2+a_1 \\\hdashline a_7 =a_6-a_5=-a_2+a_1-(-a_2)=a_1 \\ a_8=a_7-a_6= a_1-(-a_2+a_1) =a_2 \\ a_9=a_8-a_7=a_2-a_1\\ \cdots\end{array} \Rightarrow \cases{a_m=a_n,\text{if }(m \mod 6)=(n \mod 6)\\ \sum_{k=n}^{n+5} a_k=0,k\ge 1}\\ \cases{\sum_{n=1}^{40}a_n=30 \\ \sum_{n=1}^{80}a_n= 78} \Rightarrow \cases{a_{37}+ a_{38}+a_{39}+a_{40} =30\\ a_{79}+a_{80}=78} \Rightarrow \cases{a_{1}+ a_{2}+a_{3}+a_{4} =30\\ a_{1}+a_{2}=78} \\ \Rightarrow \cases{a_{1}+ a_{2}+(a_{2}-a_1)+(-a_{1}) =30\\ a_{1}+a_{2}=78} \Rightarrow \cases{2a_{2} -a_{1} =30\\ a_{1}+a_{2}=78} \Rightarrow \cases{a_1=42 \\ a_2=36}\\ \sum_{n=1}^{123}a_n= a_{121}+ a_{122}+a_{123}= a_1+a_2+a_3= 2a_2 = \bbox[red, 2pt]{72}
二、計算證明題
f(x)=x^3-4x^2+2x+4 = (x-2)(x^2-2x-2),因此f(x)=0 \Rightarrow x=0,1\pm \sqrt{3} \\ \Rightarrow \cases{\alpha =1-\sqrt 3\\ \beta =2 \\ \gamma =1+\sqrt 3} \Rightarrow {1\over x-\alpha}+ {1\over x-\beta} +{1\over x-\gamma} ={ 3x^2-2(\alpha+ \beta +\gamma)x +\alpha\beta+\beta\gamma +\gamma\alpha\over (x-\alpha)(x-\beta)(x-\gamma)} \\= {3x^2-8x+2 \over (x-1+\sqrt 3)(x-2)(x-1-\sqrt 3)} ={(x-{4+\sqrt{10}\over 3})(x-{4-\sqrt{10}\over 3}) \over (x-1+\sqrt 3)(x-2)(x-1-\sqrt 3)};\\因此 {1\over x-\alpha}+ {1\over x-\beta} +{1\over x-\gamma}>0 \\\Rightarrow (x-{4+\sqrt{10}\over 3})(x-{4-\sqrt{10}\over 3})(x-1+\sqrt 3)(x-2)(x-1-\sqrt 3)>0 \\ \Rightarrow \bbox[red, 2pt]{\cases{x > 1+\sqrt 3\\ 2< x < {4+\sqrt{10}\over 3}\\ 1-\sqrt 3< x <{4-\sqrt{10} \over 3} }}
解:
解:
|z-1-i|-|z+1+i|=2 \Rightarrow |(x-1)+(y-1)i|-|(x+1)+(y+1)i|=2 \\ \Rightarrow (x-1)^2+(y-1)^2 =4+4\sqrt{(x+1)^2+(y+1)^2} +(x+1)^2+(y+1)^2 \\ \Rightarrow -x-y-1=\sqrt{(x+1)^2+(y+1)^2} \Rightarrow (x+y+1)^2=(x+1)^2+(y+1)^2 \\ \Rightarrow x^2+y^2 +2xy+2x+2y+1 =x^2+y^2 +2x+2y+2 \Rightarrow \bbox[red, 2pt]{xy=1/2}
\lim_{n\to \infty}\left(1+{1\over n} \right)^n = \lim_{n\to \infty} 1+{n\over 1!}\cdot {1\over n} +{n(n-1)\over 2!}\cdot {1\over n^2} +{n(n-1)(n-2)\over 3!}\cdot {1\over n^3} +\cdots \\ = \lim_{n\to \infty} 1+{1\over 1!}+{1\over 2!}(1-{1\over n})+ {1\over 3!}(1-{1\over n})(1-{2\over n})+ \cdots \\ =1+{1\over 1!} +{1\over 2!}+{1\over 3!}+\cdots =\sum_{n=1}^\infty {1\over n!},故得證
-- END (僅供參考) --
請問計算第二題的最後1行是不是少+1?所以答案是1/2嗎?
回覆刪除另外,填充第1題有算式嗎?謝謝
謝謝提醒,已經補齊及修訂!!
刪除