新竹縣立六家高中 112 學年度 第 1 次教師甄選
解答:z21=z22=−2−2√3i=4e4πi/3,取{z1=2e2πi/3z2=−2e2πi/3令{A(z1)B(z2)C(z)O=¯AB中點=原點,由|z−z1|=|z−z2|⇒¯AC=¯BC⇒C在¯AB的中垂線上⇒¯AC2=¯OA2+¯OC2⇒42=22+|z|2⇒|z|=2√3解答:an=nn2−1=n(n−1)(n+1)⇒a1⋅a2⋅a3⋯ak=1⋅21⋅3⋅32⋅4⋅43⋅5⋯k(k−1)(k+1)=2(k−1)!(k+1)=2k(k+1)!=2(1k!−1(k+1)!)⇒n∑k=1(a1⋅a2⋯ak)=2n∑k=1(1k!−1(k+1)!)=2(1−1(n+1)!)=2−2(n+1)!
解答:f(x)=∫x3x2(t2+t−3)3dt⇒f′(x)=(x6+x3−3)3(3x2)−(x4+x2−3)3(2x)⇒f′(−1)=(−3)3⋅3−(−2)⋅(−1)=−81−2=−83
解答:令{→u=2→a+→b→v=→a−4→b⇒{→b=(→u−2→v)/9→a=(4→u+→v)/9⇒3→a+5→b=19(17→u−7→v)⇒{|3→a+5→b|≤19(17|→u|+7|→v|)=19(34+7)=419|3→a+5→b|≥19(17|→u|−7|→v|)=19(34−7)=3⇒(M,m)=(419,3)
解答:f(x)=x2+x+16x⇒f′(x)=1−16x2⇒f″(x)=32x3若f′(x)=0⇒x2=16⇒x=±4⇒f″(4)>0⇒f(4)=9為極小值⇒m≥4f(x)=11⇒x+1+16x=11⇒x+16x=10⇒{x=2x=8因此f(x)∈[9,11]且x∈[2,8]⇒4≤m≤8
解答:f(a,b)=(a−41b−33)2+(a−42b−34)2+(a−40b−35)2+(a−39b−32)2+(a−38b−31)2⇒fa=0⇒5a−200b−165=0⇒a=40b+33代回原式⇒f(b)=(−b)2+(−2b−1)2+(−2)2+(b+1)2+(2b+2)2=10b2+14b+10⇒f′(b)=20b+14=0有極值,此時b=−710⇒a=−710⋅40+33=5⇒(a,b)=(5,−710),但公布的答案是(5,710)
解答:A:0∘N,15∘E⇒A(Rsin90∘cos15∘,Rsin90∘sin15∘,Rcos90∘)=(5(√6+√2),5(√6−√2),0)B:45∘S,120∘W⇒B(Rsin135∘cos(−120∘),Rsin135∘sin(−120∘),Rcos135∘)=(−5√2,−5√6,−10√2)⇒¯AB=√(5√6+10√2)2+(10√6−5√2)2+(10√2)2=20√3⇒∠AOB=120∘=2π3⇒⌢AB=20⋅2π3=40π3
解答:f(x)=x3−3x+1⇒f′(x)=3x2−3⇒{f(a0)=3f′(a0)=9⇒在(a0,f(a0))的切線L:y=f′(a0)(x−a0)+f(a0)⇒y=9(x−2)+3⇒L與x軸交於9(x−2)+3=0⇒x=53⇒a1=53
解答:取△BCD在x−y平面上,原點即為重心,即{B(0,2√3,0)C(−3,−√3,0)D(3,−√3,0)A(0,0,a),又¯AB=5⇒a=±√13取A(0,0,√13)⇒{L1=↔AB=(0,−2√3t,√13t+√13),t∈RL2=↔CD=(s+3,−√3,0),s∈R⇒d(L1,L2)=√(s+3)2+(−2√3t+√3)2+(√13t+√13)2當{s=−3t=7/25時,d(L1,L2)有最小值3√395
解答:一路領先的次數=C116−C117=462−330=132⇒機率=132462=27公式來源
解答:an+2=2an+1+an⇒an−2an−1−an−2=0⇒an=C1(1+√2)n+C2(1−√2)n初始值{a1=1a2=2⇒{C1(1+√2)+C2(1−√2)=1C1(1+√2)2+C2(1−√2)=2⇒{C1=1/2√2C2=−1/2√2⇒an=12√2(1+√2)n−12√2(1−√2)n≜C(αn−βn),其中{α=1+√2β=1−√2C=1/2√2因此{α+β=2αβ=−1⇒α2+β2=(α+β)2−2αβ=6現在{a2k+1=C2(α2k+2+β2k+2−2αk+1βk+1)akak+2=C2(α2k+2+β2k+2−αkβk+2−αk+2βk)⇒a2k+1−akak+2=C2(αkβk+2+αk+2βk−2αk+1βk+1)=C2((αβ)k(α2+β2)−2(αβ)k+1)=18(6(−1)k−2(−1)k+1)=18(6(−1)k+2(−1)k)=(−1)k⇒2023∑k=1(−1)k=(−1+1)+(−1+1)+⋯+(−1+1)−1=−1
解答:f(x)=|2−log3x|⇒f(9)=0為最小值⇒{f(x)遞減,x<3f(x)遞增,x>3⇒{a<b<9<c2−log3c<0因此f(b)=f(c)⇒2−log3b=log3c−2⇒log3bc=4⇒bc=34又2−log3a=2(2−log3b)⇒log3b2a=2⇒b2=9a⇒bac=b2abc=9bc=934=19
解答:令{an:第n個人拿到白球的方法數bn:第n個人拿到不是白球的方法數n=1⇒{a1=1:第1個人只能拿到白球b1=4:第1個人可能拿到紅黃藍黑四色之一n=2⇒{a2=5:第2個拿到白球,第1個人可以拿5色任一球b2=8:第2個人拿到四色之一,第1個人是白色或同色,共4⋅2=8依此類推,可得{an=an−1+bn−1bn=4an−1+bn−1⇒[anbn]=[1141][an−1bn−1]⇒[a6b6]=[1141][a5b5]=[1141]5[a1b1]=[12161244121][14]=[365728]⇒六個人的總方法數=365+728=1093
解答:D是¯BC中點⇒→AD=12→AB+12→AC=12x→AM+12y→AN⇒53→AG=12x→AM+12y→AN⇒→AG=310x→AM+310y→ANM,G,N在一直線上⇒310x+310y=1⇒1x+1y=103,並滿足0<x,y≤1柯西不等式:(1x2+(2y)2)(12+(12)2)≥(1x+1y)2,此時1/x1=2/y1/2⇒1x=4y⇒1x+1y=5y=103⇒y=1510>1違反y≤1y最大只能取1⇒1x=73⇒1x2+4y2=499+4=859
解答:lim
解答:y=f(x)=x^3+kx^2-1 \Rightarrow f'(x)=3x^2+2kx\\ 假設切點P(a,f(a)) \Rightarrow 切線斜率=f'(a) \Rightarrow 切線方程式 y=f'(a)(x-a)+f(a)\\ 切線通過(0,0)\Rightarrow 0=f'(a)(0-a)+f(a) \Rightarrow 0=(3a^2+2ka)(-a)+a^3+ka^2-1 \\ \Rightarrow 2a^3+ka^2+1=0\\ 令g(a)=2a^3+ka^2+1 \Rightarrow g'(a)=6a^2+2ka=0 \Rightarrow 2a(3a+k)=0 \Rightarrow a=0,-k/3\\ g(a)=0 有三相異實根\Rightarrow g(0)\times g(-k/3)\lt 0 \Rightarrow 1\cdot (k^3+27)\lt 0 \Rightarrow \bbox[red, 2pt]{k\lt -3}
解答:
令\overline{BF_2}=k \Rightarrow \overline{AF_1}=3k,由雙曲線定義:\cases{\overline{AF_1}-\overline{AF_2}=2a \\ \overline{BF_1}-\overline{BF_2}=2a}\\由於\angle AF_2F_1+ \angle BF_2F_1 =180^\circ \Rightarrow \cos \angle AF_2F_1 =-\cos \angle BF_2F_1\\ \Rightarrow {5a^2+9k^2-(2a+3k)^2 \over 6\sqrt 5ak}=-{k^2+5a^2-(2a+k)^2 \over 2\sqrt 5ak } \Rightarrow {a^2-12ak\over 3} ={-a^2+4ak \over 1} \\ \Rightarrow {a-12k\over 3\sqrt 5} ={a^2-4ak-2k^2 \over 2a+k} \Rightarrow a^2=6ak \Rightarrow a=6k \Rightarrow \cases{\overline{AF_1}= 15k \\ \overline{BF_1}= 13k}\\ 取s=(\overline{AF_1}+ \overline{BF_1}+ \overline{AB})\div 2= 16k \Rightarrow \triangle AF_1B面積= \sqrt{16k\cdot k \cdot 3k \cdot 12k} =24k^2={32\over 3} \\ \Rightarrow k={2\over 3} \Rightarrow s={32\over 3} \Rightarrow 內切圓半徑={\triangle AF_1B\over s} ={32/3\over 32/3}=\bbox[red, 2pt]1
解答:\mathbf{(1)}\tan a_{n+1}\cdot \cos a_n=1 \Rightarrow \tan a_{n+1}= \sec a_n \Rightarrow \tan^2 a_n= \sec^2 a_{n-1} =1+\tan^2 a_{n-1} \\ \Rightarrow \tan^2 a_n=1+1+\cdots +(1+\tan^2 a_1) =(n-1)+ \tan^2 {\pi\over 6} =n-1+{1\over 3} =\bbox[red, 2pt]{3n-2\over 3} \\\mathbf{(2)} \tan a_{n+1}\cdot \cos a_n= {\sin a_{n+1} \over \cos a_{n+1}} \cdot \cos a_n=1 \Rightarrow \sin a_{n+1}={\cos a_{n+1} \over \cos a_n} \\ \Rightarrow \prod_{k=1}^{m} \sin a_k = \sin a_1 \cdot {\cos a_2\over \cos a_1}\cdot {\cos a_3\over \cos a_2} \cdot {\cos a_4\over \cos a_3} \cdots {\cos a_m\over \cos a_{m-1}} ={\sin a_1\over \cos a_1}\cdot \cos a_m ={1\over 10} \\ \Rightarrow \sec a_m =10\tan a_1={10\over \sqrt 3 } \Rightarrow \sec^2 a_m =1+ \tan^2 a_m ={100\over 3} \\ \Rightarrow 1+{3m-2\over 3}={100\over 3} \Rightarrow m= \bbox[red, 2pt]{33}
解答:\cases{a_n= 3a_{n-1}+ 5b_{n-1} \cdots(1)\\ b_n= a_{n-1}+7b_{n-2}\cdots(2)} \Rightarrow \cases{7a_n= 21 a_{n-1}+ 35b_{n-1}\\ 5b_n= 5a_{n-1}+35b_{n-2}},\;兩式相減 \Rightarrow 7a_n-5b_n= 16a_{n-1} \\ \Rightarrow b_n={7\over 5}a_n-{16\over 5}a_{n-1} \Rightarrow b_{n-1}={7\over 5}a_{n-1}-{16\over 5}a_{n-2} 代入(1) \Rightarrow a_n= 10a_{n-1}-16a_{n-2} \\ \Rightarrow a_n-10a_{n-1} +16a_{n-2}=0 \Rightarrow \alpha^2-10\alpha+16=0 \Rightarrow (\alpha-8)(\alpha-2)=0 \\ \Rightarrow \alpha=8,2 \Rightarrow a_n =C_18^n +C_22^n \\ 由(1) 可得 a_1=3a_0+5b_0 =6+5=11 \Rightarrow \cases{a_1=8C_1+2C_2=11\\ a_0=C_1+C_2 = 2} \Rightarrow \cases{C_1=7/6\\ C_2=5/6} \\ \Rightarrow a_n= \bbox[red, 2pt]{{7\over 6}8^n +{5\over 6}2^n,n\ge 1}
解答:\mathbf{(1)}\;p(X=n)= pq^{n-1} \Rightarrow E(X)=\sum_{n=1}^\infty npq^{n-1} =p{d\over dq} \sum_{n=0}^\infty q^n =p{d\over dq} \left({1\over 1-q}\right) =p\cdot {1\over (1-q)^2} \\ =p\cdot {1\over p^2}=\bbox[red, 2pt]{1\over p} \\\mathbf{(2)}\; E(X(X-1)) =\sum_{n=1}^\infty n(n-1)pq^{n-1} =pq\sum_{n=1}^\infty n(n-1)q^{n-2} =pq\cdot {d^2\over dq^2}\sum_{n=0}^\infty q^n \\=pq \cdot {d^2\over dq^2}\left({1\over 1-q} \right)=pq \cdot {d\over dq} {1\over (1-q)^2} =pq \cdot {2\over (1-q)^3} ={2(1-p)\over p^2}\\ \Rightarrow E(X^2)=E(X(X-1))+E(X)={2(1-p) \over p^2}+{1\over p } ={2-p\over p^2} \\ \Rightarrow Var(X)=E(X^2)-(E(X))^2 ={2-p\over p^2} -{1\over p^2}= \bbox[red,2pt]{1-p\over p^2}
解答:\cases{a_n= 3a_{n-1}+ 5b_{n-1} \cdots(1)\\ b_n= a_{n-1}+7b_{n-2}\cdots(2)} \Rightarrow \cases{7a_n= 21 a_{n-1}+ 35b_{n-1}\\ 5b_n= 5a_{n-1}+35b_{n-2}},\;兩式相減 \Rightarrow 7a_n-5b_n= 16a_{n-1} \\ \Rightarrow b_n={7\over 5}a_n-{16\over 5}a_{n-1} \Rightarrow b_{n-1}={7\over 5}a_{n-1}-{16\over 5}a_{n-2} 代入(1) \Rightarrow a_n= 10a_{n-1}-16a_{n-2} \\ \Rightarrow a_n-10a_{n-1} +16a_{n-2}=0 \Rightarrow \alpha^2-10\alpha+16=0 \Rightarrow (\alpha-8)(\alpha-2)=0 \\ \Rightarrow \alpha=8,2 \Rightarrow a_n =C_18^n +C_22^n \\ 由(1) 可得 a_1=3a_0+5b_0 =6+5=11 \Rightarrow \cases{a_1=8C_1+2C_2=11\\ a_0=C_1+C_2 = 2} \Rightarrow \cases{C_1=7/6\\ C_2=5/6} \\ \Rightarrow a_n= \bbox[red, 2pt]{{7\over 6}8^n +{5\over 6}2^n,n\ge 1}
解答:\mathbf{(1)}\;p(X=n)= pq^{n-1} \Rightarrow E(X)=\sum_{n=1}^\infty npq^{n-1} =p{d\over dq} \sum_{n=0}^\infty q^n =p{d\over dq} \left({1\over 1-q}\right) =p\cdot {1\over (1-q)^2} \\ =p\cdot {1\over p^2}=\bbox[red, 2pt]{1\over p} \\\mathbf{(2)}\; E(X(X-1)) =\sum_{n=1}^\infty n(n-1)pq^{n-1} =pq\sum_{n=1}^\infty n(n-1)q^{n-2} =pq\cdot {d^2\over dq^2}\sum_{n=0}^\infty q^n \\=pq \cdot {d^2\over dq^2}\left({1\over 1-q} \right)=pq \cdot {d\over dq} {1\over (1-q)^2} =pq \cdot {2\over (1-q)^3} ={2(1-p)\over p^2}\\ \Rightarrow E(X^2)=E(X(X-1))+E(X)={2(1-p) \over p^2}+{1\over p } ={2-p\over p^2} \\ \Rightarrow Var(X)=E(X^2)-(E(X))^2 ={2-p\over p^2} -{1\over p^2}= \bbox[red,2pt]{1-p\over p^2}
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解題僅供參考,其他試題及詳解
老師你好,填充第4題,
回覆刪除a向量及b向量與u向量v向量的關係,寫反了
不過後面的3a+5b得到的式子是正確的
謝謝提醒,已修訂
刪除第2題第二行 第一個等號後面的分子地方2 應該是2k! 嗎?
回覆刪除