國立臺灣師範大學附屬高級中學113學年度第2次專任教師甄選
一、 選填題:(每題5分,共90分。填在答案卡上,分數或根式須以最簡形式回答,否則不予計分)
解答:假設1號同學寫出的數字為a,則此數列為⟨an⟩,並令數列bn=an−an−1,n=2,3,…則⟨bn⟩=循環⏞2,3,−1,2,3,−1,⋯⇒bn=bn+3因此25∑n=1an=a1+(a1+b1)+(a1+b1+b2)+(a1+b1+b2+b3)+⋯(a1+b1+⋯+b24)=25a1+24b1+23b2+22b3+21b1+20b2+19b3+⋯+b3=25a1+b1(24+21+18+⋯+3)+b2(23+20+⋯+2)+b3(22+19+⋯+1)=25a+2⋅27⋅82+3⋅25⋅82−23⋅82=25a+424≤2024⇒a≤64解答:(1)⇒f(x)=a(x−1)2+b⇒f(2)=a+b=5(2)⇒y=g(x)=−2(x−1)+b=−2x+b+2(3)⇒h(x)=(a(x−1)2+b)(−2x+b+2)⇒x3係數=−2a⇒|−2a|=6⇒a=3(f(x)有最小值⇒a>0)⇒b=2⇒{f(x)=3(x−1)2+3g(x)=−2x+4⇒h(x)=(3(x−1)2+3)(−2x+4)⇒y=p(x)=h(x−1)=(3(x−2)2+3)(−2x+6)=−6x3+42x2−102x+90⇒h′(x−1)=−18x2+84x−102⇒h″(h−1)=−36x+84=0⇒x=8436=73
解答:
PQRS為平行四邊形⇒{∠SMR=∠MRQ=α¯SR=¯PQ=6∠PMN=∠MNR=θ又¯MN為摺線⇒{¯MR=¯MP=4√3∠RMN=∠PMN=θ△MSR:6sinα=4√3sinS=sinQ=√5/4⇒sinα=√158⇒cosα=78⇒cosα2=√154又2θ+α=180∘⇒sinθ=sin(90∘−α2)=cosα2=√154△RMN:¯MNsinα=4√3sinθ⇒¯MN=4√3√154⋅√158=2√3
解答:
yx數量5014(−2)→253(−4)→492,1(−5)→5220(−5)→511−1(−6)→613−2(−6)→613−3(−4)→49−4(−2)→25−501∑89

解答:依題意假設Q(x)=2(x−α)(x−β)⇒f(x)=Q(x2+8x−17)=2(x2+8x−17−α)(x2+8x−17−β)f(2)=0⇒2(3−α)(3−β)=0⇒α=3或β=3Case I α=β=3⇒f(x)=2(x2+8x−20)2=2(x−2)2(x+10)2Case II α=3,β≠3⇒f(x)=2(x−2)(x+10)(x2+8x−17−β)⇒x2+8x−17−β=0有重根⇒判別式64+4(17+β)=0⇒β=−33令g(x)=Q(2x2−5x+9)=2(2x2−5x+9−α)(2x2−5x+9−β)⇒{Case I α=β=3⇒g(1/2)=2(7−3)2=32Case II α=3,β=−33⇒g(1/2)=2(7−3)(7+33)=320⇒32+320=352

解答:{A(1,2,3)B(3,4,5)P(x,y,0)⇒¯PA2+¯PB2=(x−1)2+(y−2)2+9+(x−3)2+(y−4)2+25=2(x2+y2−4x−6y+32)=2((x−2)2+(y−3)2+19)當{x=2y=3時,有最小值:2×19=38

解答:C^n_2=36 \Rightarrow n=\bbox[red, 2pt]9, \href{https://www.csie.ntu.edu.tw/~yvchen/doc/unlock.pdf}{參考資料}
解答:n=12代入公式:C^n_0+ C^n_1+ C^n_2 +C^n_3 =C^{12}_0 +C^{12}_1 +C^{12}_2 +C^{12}_3 =1+12+66+ 220 =\bbox[red, 2pt]{299}, \\\href{https://www.youtube.com/watch?v=jImSYbmPPVI}{參考資料}
解答:(1+x)^{16}=C^{16}_0+C^{16}_1x+ C^{16}_2x^2+ \cdots +C^{16}_{16}x^{16} \\ \Rightarrow f(x)=x(1+x)^{16} =C^{16}_0x+C^{16}_1x^2 + C^{16}_2x^3+ \cdots +C^{16}_{16}x^{17} \\ \Rightarrow f'(x)=(1+x)^{15}(1+17x) = C^{16}_0+2C^{16}_1x + 3C^{16}_2x^2+ \cdots + 17C^{16}_{16}x^{16} \\ \Rightarrow f''(x)= (1+x)^{14}(32+272x) =2C^{16}_1 + 2\cdot 3C^{16}_2x+ 3\cdot 4C^{16}_3x^2 +\cdots + 16\cdot 17C^{16}_{16}x^{15} \\ \Rightarrow f''(1)=2^{14}\times 304 =2C^{16}_1 + 2\cdot 3C^{16}_2+ 3\cdot 4C^{16}_3+\cdots + 16\cdot 17C^{16}_{16} \\ \Rightarrow {1\over 2^{16}}(2C^{16}_1 + 2\cdot 3C^{16}_2+ 3\cdot 4C^{16}_3+\cdots + 16 \cdot 17C^{16}_{16}) ={2^{14}\times 304\over 2^{16}} = \bbox[red, 2pt]{76}
解答:I=\lim_{n\to \infty}\sum_{k=1}^{2n}{4n^2\over (2n+5k)^3} =\lim_{n\to \infty}\sum_{k=1}^{2n}{4/n\over (2+5k/n)^3} = \int_0^2 {4\over (2+5x)^3} \,dx \\ u=2+5x \Rightarrow du=5\,dx \Rightarrow I=\int_{2}^{12} {4\over 5u^3}\,du =\bbox[red, 2pt]{7\over 72}
解答:x^7=2^7的7根分別為2,\omega,\omega^2,\dots,\omega^6,其\omega =2e^{2\pi/7}\\ \Rightarrow f(x)=x^7-2^7=(x-2)(x-\omega)(x-\omega^2)\cdots(x-\omega^6) \\ \Rightarrow |f(x)|=|x^7-2^7|=|x-2||x-\omega||x-\omega^2| \cdots |x-\omega^6| \\ \Rightarrow |f(1+i)|=|(1+i)^7-2^7|= \overline{PA}\cdot \overline{PB} \cdot \overline{PC}\cdot\cdots \cdot\overline {PG} \\ 又(1+i)^2=2i \Rightarrow (1+i)^6=-8i \Rightarrow (1+i)^7=8-8i \Rightarrow |f(1+i)|=|8-8i-2^7| \\=|-120-8i|= \sqrt{120^2 +8^2} =\bbox[red,2pt]{ 8\sqrt{226}}

解答:\cases{L_1方向向量\vec u=(3,6,-5\sqrt 3) \\L_2方向向量\vec v=(-2,1,0)} \Rightarrow \vec n=\vec u\times \vec v=(5\sqrt 3,10\sqrt 3,15) \\ \Rightarrow 法向量為\vec n且通過L_2上的點(0,0,0)的平面E:\sqrt 3x+2\sqrt3 y+3z=0\\ L_1上的點(1,2,\sqrt 3)至E的距離=2\sqrt 2 =d(L_1,L_2) \Rightarrow 稜長a=2\sqrt 2\cdot \sqrt 2=4\\ \Rightarrow 體積={1\over 12}\sqrt 2a^3= \bbox[red, 2pt]{{16\over 3}\sqrt 2}

解答:X_i=擲出i種點數所需的次數 \Rightarrow X=X_1+X_2+\cdots+X_6且X_i \sim Geo(p_i={7-i\over 6}) \Rightarrow E(X_i)={6\over 7-i}\\ \Rightarrow E(X) =E(X_1+X_2+\cdots+X_6) =E(X_1) +E(X_2)+ \cdots+E(X_6)\\ =1+{6\over 5} +{6\over 4} +{6\over 3} +{6\over 2} +{6\over 1} =\bbox[red, 2pt]{14.7}, \href{https://blog.csdn.net/Rocky6688/article/details/103335048}{參考資料}

解答:\Gamma_1=\Gamma_2 \Rightarrow {2\over 9}x^3-{8\over 3}x+6k+6 =0 \Rightarrow x^3-12x+27k+27=0 \\ 令f(x)=x^3-12x+27k+27 \Rightarrow f'(x)=0 \Rightarrow 3x^2-12=0 \Rightarrow x=\pm 2\\ f(x)=0有三相異實根\Rightarrow f(2)f(-2) \lt 0 \Rightarrow (27k+11)(27k+43)\lt 0 \Rightarrow -{43\over 27}\lt k\lt -{11\over 27} \\ \Rightarrow k=-1 (\because k是整數) \Rightarrow f(x)=x^3-12x=0 \Rightarrow x=0,\pm 2\sqrt 3\\ \Rightarrow \Gamma_1-\Gamma_2 =g(x)={2\over 9}x^3-{8\over 3}x \Rightarrow g(x)是奇函數 \Rightarrow \int_{-2\sqrt 3}^{2\sqrt 3} |g(x)|\,dx =2\int_{0}^{2\sqrt 3} |g(x)|\,dx \\= 2\left. \left[ {2\over 36}x^4-{4\over 3}x^2 \right] \right|_{2\sqrt 3}^0 = \bbox[red, 2pt]{16}
解答:取\cases{u=19x+13y\\ v=25x+17y} \Rightarrow \left|{\partial (u,v) \over \partial (x,y)} \right| =\begin{Vmatrix} 19& 13\\ 25& 17\end{Vmatrix} =2 \\ {|u|\over 3}+{|v|\over 4}=1所圍面積= 3\times 4\times 2=24 \Rightarrow {|19x+13y|\over 3}+{|25x+17y|\over 4}=1所圍面積= {24\over 2}= \bbox[red, 2pt]{12}
解答:

解答:依題意假設Q(x)=2(x−α)(x−β)⇒f(x)=Q(x2+8x−17)=2(x2+8x−17−α)(x2+8x−17−β)f(2)=0⇒2(3−α)(3−β)=0⇒α=3或β=3Case I α=β=3⇒f(x)=2(x2+8x−20)2=2(x−2)2(x+10)2Case II α=3,β≠3⇒f(x)=2(x−2)(x+10)(x2+8x−17−β)⇒x2+8x−17−β=0有重根⇒判別式64+4(17+β)=0⇒β=−33令g(x)=Q(2x2−5x+9)=2(2x2−5x+9−α)(2x2−5x+9−β)⇒{Case I α=β=3⇒g(1/2)=2(7−3)2=32Case II α=3,β=−33⇒g(1/2)=2(7−3)(7+33)=320⇒32+320=352

解答:{A(1,2,3)B(3,4,5)P(x,y,0)⇒¯PA2+¯PB2=(x−1)2+(y−2)2+9+(x−3)2+(y−4)2+25=2(x2+y2−4x−6y+32)=2((x−2)2+(y−3)2+19)當{x=2y=3時,有最小值:2×19=38

解答:C^n_2=36 \Rightarrow n=\bbox[red, 2pt]9, \href{https://www.csie.ntu.edu.tw/~yvchen/doc/unlock.pdf}{參考資料}
解答:n=12代入公式:C^n_0+ C^n_1+ C^n_2 +C^n_3 =C^{12}_0 +C^{12}_1 +C^{12}_2 +C^{12}_3 =1+12+66+ 220 =\bbox[red, 2pt]{299}, \\\href{https://www.youtube.com/watch?v=jImSYbmPPVI}{參考資料}
解答:(1+x)^{16}=C^{16}_0+C^{16}_1x+ C^{16}_2x^2+ \cdots +C^{16}_{16}x^{16} \\ \Rightarrow f(x)=x(1+x)^{16} =C^{16}_0x+C^{16}_1x^2 + C^{16}_2x^3+ \cdots +C^{16}_{16}x^{17} \\ \Rightarrow f'(x)=(1+x)^{15}(1+17x) = C^{16}_0+2C^{16}_1x + 3C^{16}_2x^2+ \cdots + 17C^{16}_{16}x^{16} \\ \Rightarrow f''(x)= (1+x)^{14}(32+272x) =2C^{16}_1 + 2\cdot 3C^{16}_2x+ 3\cdot 4C^{16}_3x^2 +\cdots + 16\cdot 17C^{16}_{16}x^{15} \\ \Rightarrow f''(1)=2^{14}\times 304 =2C^{16}_1 + 2\cdot 3C^{16}_2+ 3\cdot 4C^{16}_3+\cdots + 16\cdot 17C^{16}_{16} \\ \Rightarrow {1\over 2^{16}}(2C^{16}_1 + 2\cdot 3C^{16}_2+ 3\cdot 4C^{16}_3+\cdots + 16 \cdot 17C^{16}_{16}) ={2^{14}\times 304\over 2^{16}} = \bbox[red, 2pt]{76}
解答:I=\lim_{n\to \infty}\sum_{k=1}^{2n}{4n^2\over (2n+5k)^3} =\lim_{n\to \infty}\sum_{k=1}^{2n}{4/n\over (2+5k/n)^3} = \int_0^2 {4\over (2+5x)^3} \,dx \\ u=2+5x \Rightarrow du=5\,dx \Rightarrow I=\int_{2}^{12} {4\over 5u^3}\,du =\bbox[red, 2pt]{7\over 72}
解答:x^7=2^7的7根分別為2,\omega,\omega^2,\dots,\omega^6,其\omega =2e^{2\pi/7}\\ \Rightarrow f(x)=x^7-2^7=(x-2)(x-\omega)(x-\omega^2)\cdots(x-\omega^6) \\ \Rightarrow |f(x)|=|x^7-2^7|=|x-2||x-\omega||x-\omega^2| \cdots |x-\omega^6| \\ \Rightarrow |f(1+i)|=|(1+i)^7-2^7|= \overline{PA}\cdot \overline{PB} \cdot \overline{PC}\cdot\cdots \cdot\overline {PG} \\ 又(1+i)^2=2i \Rightarrow (1+i)^6=-8i \Rightarrow (1+i)^7=8-8i \Rightarrow |f(1+i)|=|8-8i-2^7| \\=|-120-8i|= \sqrt{120^2 +8^2} =\bbox[red,2pt]{ 8\sqrt{226}}

解答:\cases{L_1方向向量\vec u=(3,6,-5\sqrt 3) \\L_2方向向量\vec v=(-2,1,0)} \Rightarrow \vec n=\vec u\times \vec v=(5\sqrt 3,10\sqrt 3,15) \\ \Rightarrow 法向量為\vec n且通過L_2上的點(0,0,0)的平面E:\sqrt 3x+2\sqrt3 y+3z=0\\ L_1上的點(1,2,\sqrt 3)至E的距離=2\sqrt 2 =d(L_1,L_2) \Rightarrow 稜長a=2\sqrt 2\cdot \sqrt 2=4\\ \Rightarrow 體積={1\over 12}\sqrt 2a^3= \bbox[red, 2pt]{{16\over 3}\sqrt 2}

解答:X_i=擲出i種點數所需的次數 \Rightarrow X=X_1+X_2+\cdots+X_6且X_i \sim Geo(p_i={7-i\over 6}) \Rightarrow E(X_i)={6\over 7-i}\\ \Rightarrow E(X) =E(X_1+X_2+\cdots+X_6) =E(X_1) +E(X_2)+ \cdots+E(X_6)\\ =1+{6\over 5} +{6\over 4} +{6\over 3} +{6\over 2} +{6\over 1} =\bbox[red, 2pt]{14.7}, \href{https://blog.csdn.net/Rocky6688/article/details/103335048}{參考資料}

解答:\Gamma_1=\Gamma_2 \Rightarrow {2\over 9}x^3-{8\over 3}x+6k+6 =0 \Rightarrow x^3-12x+27k+27=0 \\ 令f(x)=x^3-12x+27k+27 \Rightarrow f'(x)=0 \Rightarrow 3x^2-12=0 \Rightarrow x=\pm 2\\ f(x)=0有三相異實根\Rightarrow f(2)f(-2) \lt 0 \Rightarrow (27k+11)(27k+43)\lt 0 \Rightarrow -{43\over 27}\lt k\lt -{11\over 27} \\ \Rightarrow k=-1 (\because k是整數) \Rightarrow f(x)=x^3-12x=0 \Rightarrow x=0,\pm 2\sqrt 3\\ \Rightarrow \Gamma_1-\Gamma_2 =g(x)={2\over 9}x^3-{8\over 3}x \Rightarrow g(x)是奇函數 \Rightarrow \int_{-2\sqrt 3}^{2\sqrt 3} |g(x)|\,dx =2\int_{0}^{2\sqrt 3} |g(x)|\,dx \\= 2\left. \left[ {2\over 36}x^4-{4\over 3}x^2 \right] \right|_{2\sqrt 3}^0 = \bbox[red, 2pt]{16}
解答:取\cases{u=19x+13y\\ v=25x+17y} \Rightarrow \left|{\partial (u,v) \over \partial (x,y)} \right| =\begin{Vmatrix} 19& 13\\ 25& 17\end{Vmatrix} =2 \\ {|u|\over 3}+{|v|\over 4}=1所圍面積= 3\times 4\times 2=24 \Rightarrow {|19x+13y|\over 3}+{|25x+17y|\over 4}=1所圍面積= {24\over 2}= \bbox[red, 2pt]{12}
解答:
\Gamma_1:\overline{BC}=10 \Rightarrow c=5 \Rightarrow b^2=d(B,L)\cdot d(C,L) ={25\over \sqrt 5} \cdot {15\over \sqrt 5} =75 \Rightarrow b=5\sqrt 3 \Rightarrow a=\sqrt{75+25} = 10\\ \Rightarrow \Gamma_1:{(x+1)^2\over 100} +{y^2\over 75}=1 \\ \Gamma_2:c=5 \Rightarrow {(x+1)^2\over k}-{y^2\over 25-k}=1通過D(-6,{9\over 4}) \Rightarrow {25\over k}-{81/16\over 25-k}=1 \Rightarrow 16k^2-881k+10000=0\\ \quad \Rightarrow (16k-625)(k-16) =0 \Rightarrow \cases{k=625/16 \Rightarrow a=25/4 \gt c(5)不合\\ k=16 \Rightarrow a=4} \Rightarrow \Gamma_2:{(x+1)^2\over 16}-{y^2\over 9}=1 \\ 因此 \Gamma_1 \cap \Gamma_2 \Rightarrow (x+1)^2=100\left( 1-{y^2\over 75}\right) =16\left(1+{y^2\over 9} \right) \Rightarrow y^2=27 \Rightarrow |y|=3\sqrt 3 \\ \Rightarrow s_{\triangle ABC}={1\over 2}\cdot \overline{BC}\cdot |y|={1\over 2}\cdot 10\cdot 3\sqrt 3= \bbox[red, 2pt]{15\sqrt 3}\\ 橢圓兩焦點至切線距離的乘積=b^2, \href{https://www.sohu.com/a/575876952_121124316}{參考資料}
解答:x^2+xy+y^2 =[x,y] \begin{bmatrix}1& 1/2\\ 1/2 & 1 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} =[x,y] \left[ \begin{matrix}\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix} \right] \left[ \begin{matrix}\frac{1}{2} & 0 \\0 & \frac{3}{2} \end{matrix} \right] \left[ \begin{matrix}\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix} \right]^T\begin{bmatrix}x\\ y \end{bmatrix} \\ 取[x',y'] =[x,y] \left[ \begin{matrix}\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix} \right] = \left[{\sqrt 2\over 2}(-x+y), {\sqrt 2\over 2}(x+y)\right] \Rightarrow \cases{x={y'-x'\over \sqrt 2} \\y={x'+y'\over \sqrt 2}} \\ \Rightarrow \cases{x^2+xy +y^2 ={x'^2\over 2} +{y'^2\over 2/3}=1 \cdots(1)\\ x^2-3xy-2y^2= x'^2-3x'y'-2y'^2 \cdots(2)} ,由(1) \Rightarrow \cases{x'=\sqrt 2\cos \theta\\ y'=\sqrt{2\over 3} \sin \theta} 代入(2) \\ \Rightarrow 2\cos^2\theta-2\sqrt 3\sin \theta\cos \theta-{4\over 3}\sin^2\theta ={5\over 3} \cos 2\theta-\sqrt 3\sin 2\theta +{1\over 3} \\ \Rightarrow 最大值=\sqrt{{25\over 9}+3}+{1\over 3} = \bbox[red, 2pt]{1+2\sqrt{13} \over 3}
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解答:x^2+xy+y^2 =[x,y] \begin{bmatrix}1& 1/2\\ 1/2 & 1 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} =[x,y] \left[ \begin{matrix}\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix} \right] \left[ \begin{matrix}\frac{1}{2} & 0 \\0 & \frac{3}{2} \end{matrix} \right] \left[ \begin{matrix}\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix} \right]^T\begin{bmatrix}x\\ y \end{bmatrix} \\ 取[x',y'] =[x,y] \left[ \begin{matrix}\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix} \right] = \left[{\sqrt 2\over 2}(-x+y), {\sqrt 2\over 2}(x+y)\right] \Rightarrow \cases{x={y'-x'\over \sqrt 2} \\y={x'+y'\over \sqrt 2}} \\ \Rightarrow \cases{x^2+xy +y^2 ={x'^2\over 2} +{y'^2\over 2/3}=1 \cdots(1)\\ x^2-3xy-2y^2= x'^2-3x'y'-2y'^2 \cdots(2)} ,由(1) \Rightarrow \cases{x'=\sqrt 2\cos \theta\\ y'=\sqrt{2\over 3} \sin \theta} 代入(2) \\ \Rightarrow 2\cos^2\theta-2\sqrt 3\sin \theta\cos \theta-{4\over 3}\sin^2\theta ={5\over 3} \cos 2\theta-\sqrt 3\sin 2\theta +{1\over 3} \\ \Rightarrow 最大值=\sqrt{{25\over 9}+3}+{1\over 3} = \bbox[red, 2pt]{1+2\sqrt{13} \over 3}
解答:請參閱下方學校提供的解答
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解題僅供參考,教甄歷年試題及詳解
選填D y=0時, x=-5~5吧?從圖也可以看出來
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