國立嘉義高級中學111學年度第1次教師甄選
填充題:共 20 題,每題 5 分,合計 100 分
解答:12−22+32−42+⋯+392−402+412=21∑k=1(2k−1)2−20∑k=1(2k)2=21∑k=1(4k2−4k+1)−20∑k=1(4k2)=4⋅212−421∑k=1k+21=4⋅441−4⋅21⋅222+21=1764−924+21=861
解答:{n+11=7kn+7=11m⇒{n+18=7k+7=7k′n+18=11m+11=11m′⇒n+18是7的倍數也是11的倍數⇒n+18是77的倍數⇒n+18=77t⇒n=77t−18,t∈Z
解答:甲寄出{遺失(p=0.2)乙收到(p=0.8){回信(p=0.6){遺失(p=0.2)甲收到(p=0.8)不回信(p=0.4)⇒甲沒收到回信且乙有收到信甲沒收到回信=0.8⋅0.4+0.8⋅0.6⋅0.20.2+0.8⋅0.4+0.8⋅0.6⋅0.2=52/12577/125=5277
解答:x+2y+3=0的方向向量為→n=(2,−1),又→a及→b在→n上有相同的正射影⇒→a⋅→n=→b⋅→n⇒(1,m)⋅(2,−1)=(n,2)⋅(2,−1)⇒m+2n=4⇒(m2+n2)(12+22)≥(m+2n)2⇒m2+n2≥42/5=165⇒|→a|2+|→b|2=m2+n2+5≥165+5=415
解答:{L1:x−a2=y−31=z−2−1L2:x−11=y−21=z−a−2⇒{L1通過P(a,3,2)且方向向量為(2,1,−1)L2通過Q(1,2,a)且方向向量為(1,1,−2)⇒平面的法向量為(2,1,−1)×(1,1,−2)=(−1,3,1)⇒平面方程式:−x+3y+z=k⇒x−3y−z=d⇒{b=−3c=−1再將P,Q代入平面方程式⇒{a−9−2=d1−6−a=d⇒a−11=−5−a⇒a=3⇒d=−8⇒(a,b,c,d)=(3,−3,−1,−8)
解答:A=[−311−3]⇒det(A−λI)=0⇒λ=−2,−4{λ=−2⇒(A−λI)X=0⇒X=(t,t)λ=−4⇒(A−λI)X=0⇒X=(s,−s)⇒P=[tst−s]=−2st=−2由於a,b,c,d∈Z,因此{取s=t=1⇒P=[111−1]取s=t=−1⇒P=[−1−1−11]⇒P=[111−1]或[−1−1−11]
解答:最大角為θ⇒cosθ=5+6−72√30=2√30⇒sinθ=√2630⇒△ABC=12⋅√5⋅√6⋅√2630=12√26⇒△A′B′C′=△ABC⋅det(T)=12√26⋅|12−14|=12√26⋅6=3√26
解答:an=1+3+5+⋯+4n+1=(4n+2)(2n+1)÷2=(2n+1)2⇒limn→∞ann2=limn→∞(2n+1)2n2=4
解答:Γ:9(x2−2x+1)+16(y2−4y+4)=144⇒(x−1)216+(y−2)29=1P∈Γ⇒P(4cosθ+1,3sinθ+2)⇒d(P,L)=|8cosθ−18−15sinθ√4+25|=|17sin(α−θ)−18|√29⇒最大值發生在8cosθ−15sinθ=17,即{cosθ=8/17sinθ=−15/17⇒P(3217+1,−4517+2)=(4917,−1117)解答:{a,b,f,g任排有4!排法在5個間隙插入c,d,e,有C53插法c,d,e任排有3!=6種排法⇒c,d,e不相鄰有4!×C53×3=1440排法;{a,b相鄰,f,g任排有3!×2=12排法在4個間隙插入c,d,e,有C43插法c,d,e任排有3!=6種排法⇒c,d,e不相鄰且a,b相鄰有12×C43×3=288排法;因此a,b不相鄰且c,d,e不相鄰有1440−288=1152種
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過P作垂直線,分別交\overleftrightarrow{AD}與x軸交於R、S,並令\overline{PS}=h,見上圖;\\\overline{PS}=h \Rightarrow \overline{CS}= h/\sqrt 3 \Rightarrow \overline{SO}= 1-{h\over \sqrt 3} =\overline{RD} \Rightarrow \cos \angle RAP = \\ \cases{\triangle ADP \sim \triangle QCP \Rightarrow \overline{DP}: \overline{PC} =\overline{AD}: \overline{QC}\\ \triangle CSP \sim \triangle DRP \Rightarrow \overline{DP}: \overline{PC} = \overline{RP}: \overline{PS}} \Rightarrow \cfrac{\overline{AD}}{ \overline{QC}} = \cfrac{\overline{RP}}{ \overline{PS}} \Rightarrow \cfrac{2}{ \overline{QC}} = \cfrac{\sqrt 3-h}{ h} \\ \Rightarrow \overline{QC} ={2h\over \sqrt 3-h} \Rightarrow \cases{\triangle ADP = \overline{AD}\times \overline{RP}\div 2 = \sqrt 3-h\\ \triangle CQP= \overline{QC}\times \overline{PS}\div 2= h^2/(\sqrt 3-h)} \\ \Rightarrow f(h)=\triangle ADP+\triangle CQP = (\sqrt 3-h)+{h^2 \over \sqrt 3-h},則 f'(h)=0 \Rightarrow -1+{2h\over \sqrt 3-h}+ {h^2\over (\sqrt 3-h)^2}=0\\ \Rightarrow -(\sqrt 3-h)^2 +2h(\sqrt 3-h)+h^2=0 \Rightarrow 2h^2-4\sqrt 3h+3=0 \Rightarrow h=\sqrt 3-\sqrt 6/2\\ \Rightarrow f(\sqrt 3-\sqrt 6/2) = {\sqrt 6\over 2}+ \cfrac{\left({2\sqrt 3-\sqrt 6\over 2} \right)^2}{\sqrt 6/2} ={\sqrt 6\over 2}+ {9-6\sqrt 2\over \sqrt 6} =\bbox[red, 2pt]{2\sqrt 6-2\sqrt 3}
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\cos \angle C={1\over 2}= {6^2+2^2-\overline{BD}^2\over 2 \cdot 2\cdot 6}={40-\overline{BD}^2 \over 24} \Rightarrow \overline{BD}=2\sqrt 7 \Rightarrow {2\over \sin \angle CBD} ={2\sqrt 7\over \sqrt 3/2}\\ \Rightarrow \sin \angle CBD = {\sqrt 3\over 2\sqrt 7} \Rightarrow \overline{CP} = \overline{BC}\cdot \sin \angle CBD = 6\cdot {\sqrt 3\over 2\sqrt 7} =3\sqrt{3\over 7}\\ \overline{DP}\parallel \overline{AQ} \Rightarrow \cfrac{\overline{CD}}{\overline{CA}} =\cfrac{\overline{CP}}{ \overline{CQ}} \Rightarrow {2\over 6} ={3\sqrt{3\over 7} \over \overline{CQ}} \Rightarrow \overline{CQ} =9\sqrt{3\over 7} =\bbox[red, 2pt]{{9\over 7}\sqrt{21}}
假設\overline{CE}與\overline{AB}交於O,並以\cases{O為空間的原點\\ \overleftrightarrow{AB}為y軸\\ \overleftrightarrow{CE}為x軸};\\ 由於\cases{\angle AQB=90^\circ \\ \overline{AB}=4\\ \overline{QB}=2} \Rightarrow \angle ABC=60^\circ \Rightarrow \overline{OE}=\overline{BE}\cdot \sin 60^\circ =3\sqrt 3 \Rightarrow 摺起後E(0,0,3\sqrt 3);\\又\cases{\overline{DP} =\overline{CD}\sin 30^\circ = 2\\ \overline{CP}= \overline{CD}\cos 60^\circ =2\sqrt 3 \Rightarrow \overline{OP} =\sqrt 3} \Rightarrow D(-\sqrt 3,2,0) \Rightarrow \overline{DE}= \sqrt{3+4+27} =\bbox[red, 2pt]{\sqrt{34}}
解答:E_1\parallel E_2 \Rightarrow 角柱高h= d(E_1,E_2) = {9\over \sqrt 3} =3\sqrt 3,又\cases{P= E_1\cap E_3\cap E_4 =(11/3, -1/3,-4/3)\\ Q=E_1\cap E_4\cap E_5 = (3,1,-2)\\ R= E_1\cap E_3\cap E_5 =(5/3,-1/3, 2/3)} \\ \Rightarrow \cases{\vec u= \overrightarrow{PQ} =(-2/3,4/3,-2/3)\\ \vec v= \overrightarrow{PR} =(-2,0,2)} \Rightarrow \triangle PQR ={1\over 2} \sqrt{|\vec u|^2 |\vec v|^2 -(\vec u\cdot \vec v)^2} ={1\over 2}\sqrt {64/3-0} =4/\sqrt 3\\ \Rightarrow 三角柱體積={4 \over \sqrt 3}\times 3\sqrt 3= \bbox[red, 2pt]{12}
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通過Q(2,1)的直線L:y=m(x-2)+1 代入y=f(x)=x^2-x \Rightarrow x^2-x=m(x-2)+1\\ \Rightarrow x^2-(m+1)x+2m-1=0 \Rightarrow 判別式=0,即(m+1)^2-4(2m-1)=0 \Rightarrow m^2-6m+5=0\\ \Rightarrow (m-5)(m-1)=0 \Rightarrow \cases{m=5 \Rightarrow L_1: y=5x-9\\ m=1 \Rightarrow L_2:y=x-1} \Rightarrow \cases{A= \Gamma \cap L_2=(1,0)\\ B=\Gamma \cap L_1=(3,6) \\ C= (x=2)\cap \Gamma =(2,2)} \\ 所圍面積=\int_1^2 (x^2-x)-(x-1)\,dx+\int_2^3 (x^2-x)-(5x-9)\,dx \\=\int_1^2 x^2-2x+1\,dx+\int_2^3 x^2-6x+9\,dx =\left. \left[ {1\over 3}x^3-x^2+x \right] \right|_1^2 +\left. \left[ {1\over 3}x^3-3x^2+ 9x \right] \right|_2^3 \\ = {1\over 3}+{1\over 3}= \bbox[red, 2pt]{2\over 3}
解答:(2)\times: 假設兩圓\cases{圓B:(x-a)^2+(y-b)^2 =r^2 \\圓C:x^2+(y+2)^2=4}相切 \Rightarrow 兩圓心距離=兩半徑之和\\ \qquad \Rightarrow a^2 +(b+2)^2 =(r+2)^2 \cdots(1);又(0,4)在圓B上\Rightarrow a^2+(b-4)^2=r^2\cdots(2)\\ \qquad 由(1)及(2)可知圓心(a,b)軌跡為雙曲線\\ (3)\times: 假設兩圓\cases{圓B:(x-a)^2+(y-b)^2 =r^2 \\圓C:x^2+y^2=49}相切 \Rightarrow 兩圓心距離=兩半徑之和 \\ \qquad \Rightarrow \cases{a^2+b^2=(r+7)^2 \\a^2+(b-4)^2=r^2} \Rightarrow a^2+b^2 =(\sqrt{a^2+(b-4)^2}+7)^2 \Rightarrow a^2+{33\over 49}(b-2)^2={33\over 4}為一橢圓 \\故選\bbox[red, 2pt]{(145)}
解答:a_{n+2}\cdot a_n + 6a_{n+2}\cdot a_{n+1} =a_{n+1}\cdot a_n \Rightarrow a_{n+2}( a_n + 6 a_{n+1})= a_{n+1}\cdot a_n \Rightarrow \cfrac{a_n + 6 a_{n+1} }{ a_{n+1}\cdot a_n} ={1\over a_{n+2}}\\ \Rightarrow {1\over a_{n+1}} +{6\over a_n} ={1\over a_{n+2}} \Rightarrow b_{n+2}= b_{n+1}+ 6b_n, 其中\,b_n={1\over a_n} \Rightarrow b_{n+2}-b_{n+1}-6b_n,初始值\cases{b_1=1\\ b_2=1/2}\\ \Rightarrow 特徵多項式\,\lambda^2-\lambda-6=0 \Rightarrow \lambda=3,-2 \Rightarrow b_n= C_1(-2)^n +C_2\cdot 3^n;\\ 再由初始值\cases{b_1=1 =-2C_1+ 3C_2\\ b_2=1/2 =4C_1+9C_2} \Rightarrow \cases{C_1=-1/4\\ C_2=1/6} \Rightarrow b_n= {1\over a_n}= -{1\over 4}(-2)^n -{1\over 6}\cdot 3^n \\ \Rightarrow a_n =\cfrac{1}{-{1\over 4}(-2)^n -{1\over 6}\cdot 3^n} =\cfrac{1}{{1\over 2}(-2)^{n-1} -{1\over 2}\cdot 3^{n-1}} = \bbox[red, 2pt]{\cfrac{2}{ (-2)^{n-1} - 3^{n-1}} , n\in \mathbb{N}}
解答:\cases{\sqrt x\left(1+{1\over x+y}\right)=2 \\\sqrt y\left(1 -{1\over x+y}\right)= \sqrt 2 } \Rightarrow \cases{ 1+{1\over x+y} =2/\sqrt x \cdots(1) \\ 1 -{1\over x+y} = \sqrt 2 /\sqrt y \cdots(2)}\\ \Rightarrow (1)^2 -(2)^2 =\left(1+{1\over x+y}\right)^2- \left(1 -{1\over x+y}\right)^2={4\over x} -{2\over y} \Rightarrow {4\over x+y}= {4y-2x\over xy}\\ \Rightarrow 4xy =(x+y)(4y-2x) \Rightarrow x^2+xy-2y^2=0 \Rightarrow (x+2y)(x-y)=0\\ \Rightarrow x=y (\because x,y\ge 0 \Rightarrow x\ne -2y)代入(1) \Rightarrow 1+{1\over 2x}={2\over \sqrt x} \Rightarrow {(2x+1)^2\over 4x^2} ={4\over x} \\\Rightarrow 4x^2-12x+1=0 \Rightarrow x={3+2\sqrt 2\over 2}(負值不合) \Rightarrow (x,y)= \bbox[red, 2pt]{({3+2\sqrt 2\over 2}, {3+2\sqrt 2\over 2})}
解答:令q=1-p,則E(n)= \sum_{n=1}^\infty nq^{n-1}p =p\sum_{n=1}^\infty{d\over dq}q^n =p{d\over dq}\sum_{n=1}^\infty q^n =p{d\over dq}{q\over 1-q} ={p\over (1-q)^2 }= \color{blue}{1\over p}\\同理,E(n(n-1)) =\sum_{n=1}^\infty n(n-1) q^{n-1}p =pq \sum_{n=1}^\infty n(n-1) q^{n-2} =pq \sum_{n=1}^\infty {d^2\over dq^2}q^n \\= pq {d^2\over dq^2}\sum_{n=1}^\infty q^n =pq {d^2\over dq^2} {q\over 1-q} =pq \cdot {2\over (1-q)^3} ={2(1-p)\over p^2} \\\Rightarrow E(n^2)= E(n(n-1)+n) =E(n(n-1)) +E(n) ={2(1-p)\over p^2} +{1\over p}= \color{blue}{2-p\over p^2}\\ E(n(n-1)(n-2)) = \sum_{n=1}^\infty n(n-1)(n-2) q^{n-1}p = pq^2\sum_{n=1}^\infty n(n-1)(n-2) q^{n-3 } \\=pq^2\sum_{n=1}^\infty {d^3\over dq^3} q^n =pq^2 {d^3\over dq^3}\sum_{n=1}^\infty q^n =pq^2 {d^3\over dq^3}{q\over 1-q}= pq^2 \cdot {6\over (1-q)^4} =\color{blue} {6(1-p)^2\over p^3}\\ 最後E(n^3) = E(n(n-1)(n-2)) +3E(n^2)-2E(n) ={6(1-p)^2\over p^3}+3\cdot {2-p\over p^2}-2\cdot {1\over p} \\=\bbox[red, 2pt]{p^2-6p+6 \over p^3}\\註:如果你有背幾何分布的期望值及變異數,就可以很快算出答案
解答:a_{n+2}\cdot a_n + 6a_{n+2}\cdot a_{n+1} =a_{n+1}\cdot a_n \Rightarrow a_{n+2}( a_n + 6 a_{n+1})= a_{n+1}\cdot a_n \Rightarrow \cfrac{a_n + 6 a_{n+1} }{ a_{n+1}\cdot a_n} ={1\over a_{n+2}}\\ \Rightarrow {1\over a_{n+1}} +{6\over a_n} ={1\over a_{n+2}} \Rightarrow b_{n+2}= b_{n+1}+ 6b_n, 其中\,b_n={1\over a_n} \Rightarrow b_{n+2}-b_{n+1}-6b_n,初始值\cases{b_1=1\\ b_2=1/2}\\ \Rightarrow 特徵多項式\,\lambda^2-\lambda-6=0 \Rightarrow \lambda=3,-2 \Rightarrow b_n= C_1(-2)^n +C_2\cdot 3^n;\\ 再由初始值\cases{b_1=1 =-2C_1+ 3C_2\\ b_2=1/2 =4C_1+9C_2} \Rightarrow \cases{C_1=-1/4\\ C_2=1/6} \Rightarrow b_n= {1\over a_n}= -{1\over 4}(-2)^n -{1\over 6}\cdot 3^n \\ \Rightarrow a_n =\cfrac{1}{-{1\over 4}(-2)^n -{1\over 6}\cdot 3^n} =\cfrac{1}{{1\over 2}(-2)^{n-1} -{1\over 2}\cdot 3^{n-1}} = \bbox[red, 2pt]{\cfrac{2}{ (-2)^{n-1} - 3^{n-1}} , n\in \mathbb{N}}
解答:\cases{\sqrt x\left(1+{1\over x+y}\right)=2 \\\sqrt y\left(1 -{1\over x+y}\right)= \sqrt 2 } \Rightarrow \cases{ 1+{1\over x+y} =2/\sqrt x \cdots(1) \\ 1 -{1\over x+y} = \sqrt 2 /\sqrt y \cdots(2)}\\ \Rightarrow (1)^2 -(2)^2 =\left(1+{1\over x+y}\right)^2- \left(1 -{1\over x+y}\right)^2={4\over x} -{2\over y} \Rightarrow {4\over x+y}= {4y-2x\over xy}\\ \Rightarrow 4xy =(x+y)(4y-2x) \Rightarrow x^2+xy-2y^2=0 \Rightarrow (x+2y)(x-y)=0\\ \Rightarrow x=y (\because x,y\ge 0 \Rightarrow x\ne -2y)代入(1) \Rightarrow 1+{1\over 2x}={2\over \sqrt x} \Rightarrow {(2x+1)^2\over 4x^2} ={4\over x} \\\Rightarrow 4x^2-12x+1=0 \Rightarrow x={3+2\sqrt 2\over 2}(負值不合) \Rightarrow (x,y)= \bbox[red, 2pt]{({3+2\sqrt 2\over 2}, {3+2\sqrt 2\over 2})}
解答:令q=1-p,則E(n)= \sum_{n=1}^\infty nq^{n-1}p =p\sum_{n=1}^\infty{d\over dq}q^n =p{d\over dq}\sum_{n=1}^\infty q^n =p{d\over dq}{q\over 1-q} ={p\over (1-q)^2 }= \color{blue}{1\over p}\\同理,E(n(n-1)) =\sum_{n=1}^\infty n(n-1) q^{n-1}p =pq \sum_{n=1}^\infty n(n-1) q^{n-2} =pq \sum_{n=1}^\infty {d^2\over dq^2}q^n \\= pq {d^2\over dq^2}\sum_{n=1}^\infty q^n =pq {d^2\over dq^2} {q\over 1-q} =pq \cdot {2\over (1-q)^3} ={2(1-p)\over p^2} \\\Rightarrow E(n^2)= E(n(n-1)+n) =E(n(n-1)) +E(n) ={2(1-p)\over p^2} +{1\over p}= \color{blue}{2-p\over p^2}\\ E(n(n-1)(n-2)) = \sum_{n=1}^\infty n(n-1)(n-2) q^{n-1}p = pq^2\sum_{n=1}^\infty n(n-1)(n-2) q^{n-3 } \\=pq^2\sum_{n=1}^\infty {d^3\over dq^3} q^n =pq^2 {d^3\over dq^3}\sum_{n=1}^\infty q^n =pq^2 {d^3\over dq^3}{q\over 1-q}= pq^2 \cdot {6\over (1-q)^4} =\color{blue} {6(1-p)^2\over p^3}\\ 最後E(n^3) = E(n(n-1)(n-2)) +3E(n^2)-2E(n) ={6(1-p)^2\over p^3}+3\cdot {2-p\over p^2}-2\cdot {1\over p} \\=\bbox[red, 2pt]{p^2-6p+6 \over p^3}\\註:如果你有背幾何分布的期望值及變異數,就可以很快算出答案
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