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2020年11月17日 星期二

108年基隆女中教甄-數學詳解

國立基隆女子高級中學108學年度教師甄選數學科考試試題

一、填充題

1. 三次實係數多項式\(f(x)\),已知\(\lim_{x\to 1} {f(x)\over x-1}=10\),且曲線\( \Gamma: y=f(x)\)圖形的反曲點為(-1,-4),若曲線\(\Gamma\)在點\((1,f(1))\)的切線方程式為\(L:y=g(x)\),請計算由曲線\(\Gamma\)與切線\(L\)所圍成區域的面積為_____。

$$\lim_{x\to 1}{f(x)\over x-1}=10 \Rightarrow f(x)=(x-1)h(x),其中\cases{h(x)=ax^2+bx+c\\ h(1)=10 \Rightarrow a+b+c=10\cdots (1)} \\ 因此f(x)=(x-1)(ax^2+bx+c) = ax^3+(b-a)x^2+(c-b)x-c\\ \Rightarrow f'(x)=3ax^2+2(b-a)x+(c-b) \Rightarrow f''(x)=6ax+2(b-a)\\ y=f(x)的反曲點為(-1,-4) \Rightarrow \cases{f(-1)=-4\\ f''(-1)=0} \Rightarrow \cases{-a+(b-a)+b-c-c=-4\\ -6a+2(b-a)=0}\\ \Rightarrow \cases{a-b+c=2 \cdots(2)\\ b=4a \cdots(3)},由(1),(2),(3)可得\cases{a=1\\b=4\\ c=5} \Rightarrow f(x)=(x-1)(x^2+4x+5)\\ 切線L斜率=f'(1)=10 \Rightarrow L: y=10(x-1)\\\Rightarrow L與\Gamma的交點: 10(x-1)=(x-1)(x^2+4x+5) \Rightarrow x=1,-5\\ \Rightarrow 所圍面積=\int_{-5}^1 (x-1)(x^2+4x+5)-10(x-1)\;dx = \left. {1\over 4}x^4 +x^3-{9\over 2}x^2+5x\right|_{-5}^1 \\ = ({1\over 4}+1-{9\over 2}+5)-({625\over 4}-125-{225\over 2}-25) =\bbox[red,2pt]{108}$$

2. 設數列\(\langle a_n\rangle\)滿足\(a_0=10,a_n=\cfrac{10a_{n-1}-77}{a_{n-1}-8}, n=1,2,3,\dots\),求\(\langle a_n\rangle\)的一般項為___

$$先求中心點: x=\cfrac{10x-77}{x-8} \Rightarrow x^2-18x+77=0 \Rightarrow (x-7)(x-11)=0 \Rightarrow x=7,11\\ 將遞迴平移至中心點: \cfrac{a_n-7}{a_n-11} =\cfrac{{10a_{n-1}-77\over a_{n-1}-8}-7}{{10a_{n-1}-77\over a_{n-1}-8}-11} =\cfrac{3a_{n-1}-21}{-a_{n-1}+11} =-3\cdot \cfrac{a_{n-1}-7}{a_{n-1}-11}\\ 因此令b_n={a_n-7\over a_n-11} \Rightarrow \cases{b_0={10-7\over 10-11}=-3\\b_n=-3b_{n-1}} \\\Rightarrow b_n=-3b_{n-1}=(-3)^2b_{n-2}=\cdots =(-3)^nb_0 =(-3)^{n+1} \\ \Rightarrow b_n ={a_n-7\over a_n-11}=(-3)^{n+1} \Rightarrow 1+{4\over a_n-11}=(-3)^{n+1} \Rightarrow {4\over a_n-11}=(-3)^{n+1}-1 \\ \Rightarrow a_n=\bbox[red,2pt]{\cfrac{4}{(-3)^{n+1}-1}+11} =\cfrac{4}{(-1)^{n+1}\cdot 3^{n+1}+(-1)^{2n+1}}+11 =\cfrac{4}{(-1)^{n+1}( 3^{n+1}+(-1)^{n})}+11 \\ =\cfrac{4\cdot (-1)^{n+1}}{ 3^{n+1}+(-1)^{n}}+11 =11-\cfrac{4\cdot (-1)^{n}}{ 3^{n+1}+(-1)^{n}} =\bbox[red,2pt]{\cfrac{11\cdot 3^{n+1}+7\cdot (-1)^n}{ 3^{n+1}+ (-1)^n}}$$

3. 已知\(x,y\in R\),設\(f(x)=4\times (4^x+4^{-x})-21\times (2^x+2^{-x})+25\),將\(f(x)\)對稱於原點後,再垂直平移\(p\)格得\(g(x)\),且\(g(x)\)有最大值\({121\over16 }\),求\(p\)

$$g(x)=-f(-x)+p =-4(4^x+4^{-x})+ 21(2^x+2^{-x})-25+p \\ =-4(2^x+2^{-x})^2 +21(2^x+2^{-x})-17+p \\\Rightarrow  g(u)=-4u^2+21u-17+p,u=2^x+2^{-x} \Rightarrow g'(u)=0 \Rightarrow u=21/8\\ \Rightarrow g(21/8)={121\over 16} \Rightarrow -4\times {441\over 64}+21\times {21\over 8}-17+p = {121\over 16} \\ \Rightarrow p={121\over 16}+17-{441\over 16} =\bbox[red,2pt]{-3}$$

4. 正數\(x,y,z\)滿足方程組\(\cases{x^2+xy +{y^2\over 3}=25\\ {y^2\over 3}+x^2=9 \\ z^2+xz+x^2=16}\),求\(xy+ 2yz+ 3xz\)為_____

 

$$本題\bbox[red,2pt]{無解}$$

5. 求\(x^{30}\)除以\((x+1)^2(x^2+1)\)的餘式為____


$$P(x)=(x+1)^2(x^2+1) = x^4+2x^3+2x^2+2x+1 \\\Rightarrow x^5=(x-2)P(x)+(2x^2+2x^2+3x+2) =a(x)P(x)+b(x)\\ \Rightarrow x^{30}=(x^5)^6=(a(x)P(x)+b(x))^6 =\sum_{n=0}^6C^6_n a^n(x)P^n(x)b^{6-n}(x) \\ \Rightarrow x^{30}除以P(x)的餘數=b^6(x)除以P(x)的餘數;\\ b^6(x)= (2x^2+2x^2+3x+2)^6 =(2(x+1)(x^2+1)+x)^6 \\=\sum_{n=0}^6C^6_n2^n(x+1)^n(x^2+1)^nx^{6-n} \\\Rightarrow b^6(x)除以P(x)的餘數= x^6+C^6_1(2(x+1)(x^2+1)+x)=c(x)除以P(x)的餘數\\ c(x)= (12x^8 +12x^7+13x^6+12x^5\\ =12x^4-12x^3+13x^2-14x+14)P(x)-14x^3-13x^2-14x-14 \\ \Rightarrow x^{30}除以(x+1)^2(x^2+1)的餘式為\bbox[red,2pt]{-14x^3-13x^2-14x-14}$$

6. \(m\)個互不相同的正偶數和\(n\)個不同的正奇數總和為1987,對於所有這樣的\(m,n\),求\(3m+4n\)的最大值為 ____。 

$$\sum_{i=1}^m a_i+\sum_{j=1}^n b_j= 1987,其中a_i為偶數,b_j是奇數,且a_m\ne a_n,b_s\ne b_t ,m\ne n,s\ne t;\\ \cases{a_1+a_2+\cdots a_m \ge 2+4+\cdots + 2m = m(m+1)\\ b_1+b_2+\cdots b_n \ge 1+ 3+\cdots +2n-1 =n^2} \Rightarrow m^2+m+n^2 \le 1987 \\ \Rightarrow (m+{1\over 2})^2+n^2 \le 1987{1\over 4}\\ 3m+4n= 3(m+{1\over 2})+4n-{3\over 2} \Rightarrow \left((m+{1\over 2})^2+ n^2\right)(3^2+4^2) \ge (3(m+{1\over 2})+4n)^2 \\ \Rightarrow 3(m+{1\over 2})+4n \le \sqrt{((m+{1\over 2})^2+ n^2) \times 25} \le \sqrt{1987{1\over 4}}\times 5\approx 222.89 \\ \Rightarrow 3m+4n =3(m+{1\over 2})+4n-{3\over 2} \le 222.89-{3\over 2}=221.39 \Rightarrow 3m+4n的最大值為\bbox[red,2pt]{221}$$

7. 如圖,已知圓內接正\(\triangle ABC\),在劣弧BC上有一點P。若\(\overline{AP}\)與\(\overline{BC}\)交於D點,且\(\overline{PB}=6, \overline{PC}=10\),則\(\overline{PD}\)之長為____ 。



 

$$\cases{\triangle ABD\sim \triangle CPD  \Rightarrow {a\over 10}={a-b \over \overline{PD}} ={\overline{AD} \over b} \Rightarrow \overline{AD}={ab\over 10}\\ \triangle DBP \sim \triangle DAC \Rightarrow {a-b\over \overline{AD}}={6\over a} ={\overline{PD}\over b} \Rightarrow \overline{AD}={a(a-b)\over 6}} \\ \Rightarrow {ab\over 10} ={a(a-b)\over 6} \Rightarrow a(5a-8b)=0 \Rightarrow b={5\over 8}a \\ \Rightarrow \overline{PD} ={6b\over a} ={6\over a}\times {5a\over 8} = \bbox[red,2pt]{15\over 4}$$

8. 將六個不同球全部放入三個相同箱子中,每個箱子的球數不限,則方法數有_________種。

$$\begin{array}{} 箱A & 箱B & 箱C & 組合數\\\hline 6 & 0 & 0 & 1\\ 5 & 1 & 0 &C^6_6=6 \\ 4 & 1 & 1 & C^6_4=15\\ 4 & 2 & 0 & C^6_4=15\\ 3 & 2 & 1 & C^6_3C^3_2 =60\\ 3 & 3 & 0 & C^6_3\div 2=10\\ 2 & 2 & 2 & C^6_2C^4_2 \div 3!=15\\\hline \end{array} \\\Rightarrow 共有1+6+15+15+60+10+15= \bbox[red,2pt]{122}種分法$$

9. 自點\(P(1,3)\)向拋物線\(\Gamma:y=-x^2\)作切線,則兩切線與\(\Gamma\)所圍封閉區域的面積為=____

$$令切點為Q,Q在y=-x^2上 \Rightarrow Q(a,-a^2) \Rightarrow 切點斜率m=y'=-2x=-2a \\ \Rightarrow 過P(1,3)且斜率為-2a的切線方程式: y=-2a(x-1)+3 \\ 因此切線與拋物線的交點 \Rightarrow -2a(x-1)+3=-x^2 \Rightarrow x^2-2ax+2a+3=0 \\\Rightarrow 判別式=0 \Rightarrow 4a^2-4(2a+3)=0 \Rightarrow (a-3)(a+1)=0 \Rightarrow a=3,-1 \\ \Rightarrow 切點A(3,-9),B(-1,-1) \Rightarrow \overleftrightarrow{AB}:y=-2x-3 \\ 所求面積=\triangle PAB-拋物線與\overleftrightarrow{AB}所圍面積(著色面積)\\ \triangle ABC= {1\over 2}\sqrt{ |\overrightarrow{PA}|^2 |\overrightarrow{PB}|^2-(\overrightarrow{PA}\cdot   \overrightarrow{PB})^2} =  {1\over 2}\sqrt{ |(2,-12)|^2 |(-2,-4)|^2-((2,-12)\cdot (-2,-4))^2}\\ = {1\over 2}\sqrt{148\times 20-1936} = {1\over 2}\sqrt{1024} =16 \\ 著色面積= \int_{-1}^3 -x^2-(-2x-3)\;dx =\int_{-1}^3 -x^2+2x+3\;dx = \left. \left[ -{1\over 3}x^3 + x^2+ 3x\right] \right|_{-1}^3 \\ =9-{-5\over 3} = {32\over 3}\\ 因此所求面積=16-{32\over 3} = \bbox[red,2pt]{16\over 3}$$

10. 試求方程式\(x+\cfrac{x}{\sqrt{x^2-1}}=\cfrac{221}{60}\)的所有實數解為 _______

$$x+{x\over \sqrt{x^2-1}} ={221\over 60} \Rightarrow {1\over \cos \angle A} +{1\over \sin \angle A} ={221\over 60}  \Rightarrow {\sin \angle A+\cos \angle A\over \sin \angle A\cos \angle A}={221\over 60}\\ 令a= \sin \angle A+\cos \angle A \Rightarrow \sin\angle A\cos \angle A={a^2-1\over 2} \Rightarrow {\sin \angle A+\cos \angle A\over \sin \angle A\cos \angle A} ={2a\over a^2-1}={221\over 60}  \\ \Rightarrow 221a^2-120a-221=0 \Rightarrow (17a+13)(13a-17)=0 \Rightarrow a=17/13 (\because a>0) \\ \Rightarrow \sin\angle A\cos \angle A={1\over 2}(({17\over 13})^2-1) ={60\over 169} \Rightarrow (\sin\angle A-\cos \angle A)^2 =1-2\times {60\over 169} ={49\over 169} \\ \Rightarrow \cases{\cases{\sin\angle A+\cos \angle A=17/13 \\ \sin \angle A-\cos \angle A=7/13} \Rightarrow \cases{\sin \angle A = 12/13 \\ \cos \angle A=5/13}\Rightarrow x=13/5\\\cases{\sin\angle A+\cos \angle A=17/13 \\ \sin \angle A-\cos \angle A=-7/13} \Rightarrow \cases{\sin \angle A=5/13 \\ \cos \angle A=12/13} \Rightarrow x=13/12}\\ \Rightarrow x=\bbox[red,2pt]{{13\over 5},{13\over 12}}$$
11. 設矩陣\(A=\left[ \matrix{\cos\theta & \sin \theta\\ \sin \theta & -\cos \theta}\right]\),則\(\sum_{n=1}^{100}A^n\)之值為= _______

$$A=\left[ \matrix{\cos\theta & \sin \theta\\ \sin \theta & -\cos \theta}\right] \Rightarrow A^2=\left[ \matrix{1 & 0\\ 0 & 1}\right]=I \\ \Rightarrow \sum_{n=1}^{100}A^n= A+ I + A+\cdots +A =50A+50I \\=\left[ \matrix{50\cos\theta & 50\sin \theta\\ 50\sin \theta & -50\cos \theta}\right] +\left[ \matrix{50 & 0\\ 0 & 50}\right]\\ =\bbox[red,2pt]{\left[ \matrix{50\cos\theta+50 & 50\sin \theta\\ 50\sin \theta & -50\cos \theta +50}\right]}$$

12. 在坐標平面上,設\(A,B,C\)是橢圓\(\cfrac{x^2}{9}+\cfrac{y^2}{4}=1\)上三點,且\(\triangle ABC\)的重心為 (0,0),已知\(A\)的坐標為\(({3\sqrt 3\over 2},1)\),則\(\overline{BC}\)的長度為 _______

$$\cases{A(3\sqrt 3/2,1)\\ B(x_b,y_b)\\ C(x_c,y_c)},重心(0,0)={1\over 3}(A+B+C) \Rightarrow \cases{x_b +x_c=-3\sqrt 3/2 \cdots(1) \\ y_b+y_c=-1\cdots(2) }\\ B,C皆在橢圓上 \Rightarrow \cases{x_b^2/9+ y_b^2/4=1 \cdots(3)\\ x_c^2/9+ y_c^2/4=1 \cdots(4)};\\將(1)及(2)代入(4)可得 \cfrac{(-3\sqrt 3/2-x_b)^2}{9} +\cfrac{(-1-y_b)^2}{4}=1 \\ \Rightarrow \left( \cfrac{x_b^2}{9} +\cfrac{y_b^2}{4}\right) +\left(\cfrac{3\sqrt 3x_b+ 27/4}{9} +\cfrac{2y_b+1}{4} \right)=1 \\ \Rightarrow \cfrac{3\sqrt 3x_b+ 27/4}{9} +\cfrac{2y_b+1}{4} =0 \Rightarrow 12\sqrt 3x_b+18y_b+36=0 \Rightarrow y_b=-2-\cfrac{2\sqrt 3}{3}x_b \cdots(5)\\ 將(5)代入(3)可得 \cfrac{x_b^2}{9} +\cfrac{4+8\sqrt 3/3 x_b +12/9x_b^2}{4}=1 \Rightarrow 2x_b(2x_b+3\sqrt 3)=0 \\ \Rightarrow \cases{x_b=0\\ x_b=-3\sqrt 3/2} \Rightarrow \cases{y_b=-2 \\ y_b=1}\Rightarrow \cases{(x_b,y_b)=(0,-2)\\ (x_b,y_b)=(-3\sqrt 3/2,1)} \Rightarrow \cases{(x_c,y_c) =(-3\sqrt 3/2,1)\\ (x_c,y_c)=(0,-2)}\\ 也就是\cases{B=(0,-2)\\C=(-3\sqrt 3/2,1)} 或\cases{C=(0,-2)\\B=(-3\sqrt 3/2,1)} \Rightarrow \overline{BC}=\sqrt{\cfrac{27}{4} +9} = \bbox[red,2pt]{\cfrac{3\sqrt 7}{2}}$$

13. 朱媽媽要將紅色、黃色、黑色、綠色及白色5個球分給3個小朋友,分別是小娟得1個、小明得2個、小芳得2個,小芳不喜歡黑色,希望不要拿到,如果朱媽媽隨意分,求小芳達成此願望的機率為______ _

$$全部有\cfrac{5!}{2\times 2} =30種分法;其中小芳拿到黑色的分法有\cfrac{4!}{2} =12種;\\因此小芳如願的機率為\cfrac{30-12}{30}=\cfrac{18}{30} =\bbox[red,2pt]{\cfrac{3}{5}}$$

14. 數學老師把上屆6位學長姐的段考平均(x)與學測平級分(y)做統計分析,結果如右表,
(1)若這6位學長姐的段考平均為80分,試計算段考平均與學測級分兩者的相關係數為_____。
(2)老師現在班上有位學生的段考平均是89分,試預測該生的學測級分為______級分。(請採四捨五入到整數位)
$$\begin{array} {|c|c|c|c|c|c|} \hline編 號& 1& 2& 3& 4& 5 & 6\\\hline 段考平均& 68& 80 & 80& 80& 86& a\\\hline 學測平均& 7 & 9 & 9 & 10 & 12 & 13\\\hline\end{array}$$


(1)$$段考平均為80分 \Rightarrow \bar x=80=(68+80\times 3+86+a)\div 6 \Rightarrow a=86\\ \bar y= (7+9 + 9+10+12+13)\div 6=10\\ 相關係數r=\cfrac{\sum_{i=1}^{6}(x_i-\bar x)(y_i-\bar y)}{\sqrt{\sum_{i=1}^{6} (x_i-\bar x)^2 \times \sum_{i=1}^{6}(y_i-\bar y)^2}} =\cfrac{36+0+0 +0 +12+18}{ \sqrt{(144+36+36)(9+1+1+4+9)}}\\ =\cfrac{66}{72}= \bbox[red,2pt]{\cfrac{11}{12}}$$(2)$$\cfrac{\sigma_Y}{\sigma_X} =\sqrt{\cfrac{\sum_{i=1}^{10}(y_i-\bar y)^2}{\sum_{i=1}^{10}(x_i-\bar x)^2}} = \sqrt{ \cfrac{24}{216}} =\cfrac{1}{3} \Rightarrow 迴歸直線斜率m=r\times \cfrac{\sigma_Y}{\sigma_X} =\cfrac{11}{12}\times \cfrac{1}{3} =\cfrac{11}{36}\\ \Rightarrow 迴歸直線方程式: y=m(x-\bar x)+\bar y=\cfrac{11}{36}(x-80)+10\\ x=89代入可得y= \cfrac{11}{36}(89-80)+10 =\cfrac{11}{4}+10 =12.75 \approx \bbox[red,2pt]{13}$$

15. 在座標平面上,到直線\(x+1=0\)之距離是到點\(F(1,0)\)之距離的兩倍的所有點所形成的圖形是一個橢圓。若此橢圓的焦點坐標為\((a_1,b_1),(a_2,b_2)\),且\(a_1> a_2\),則對數\((a_1,a_2)\)= __________

$$|x-(-1)| = 2\sqrt{(x-1)^2 +y^2} \Rightarrow (x+1)^2 = 4((x-1)^2+y^2) \\ \Rightarrow \cfrac{(x-5/3)^2}{16/9} +\cfrac{y^2}{4/3} =1 \Rightarrow \cases{a=4/3\\ b=2/\sqrt 3} \Rightarrow c=2/3 \Rightarrow \cases{a_1=5/3+2/3=7/3 \\ a_2=5/3-2/3=1} \\ \Rightarrow (a_1,a_2)= \bbox[red,2pt]{\left( {7\over 3},1\right)}$$

計算題

1. 若實數\((x,y)\)滿足不等式組\(\cases{y\ge |x-2| \\ x-3y+6\ge 0}\),求\(x^2+y^2\)的最大值及最小值。

$$\cases{y\ge |x-2|\\ x-3y+y \ge 0} 交集區域為\triangle ABC,見上圖;\\ x^2+y^2 = k^2 相當於以原點為圓心,半徑為k的圓;\\\triangle ABC距原點最遠的地方是頂點B(6,4),最接近的地方是直線\overline{AC}的切點D(1,1);\\ 因此x^2+y^2的最大值=\overline{OB}^2 =52,最小值=\overline{OD}^2 =2,即\bbox[red,2pt]{\cases{最大值:52\\ 最小值:2}}$$

證明題

1. 設\(a,b,c\)都是正數,且\(s=a+b+c\)。試證明\(\cfrac{a^2}{s-a} +\cfrac{b^2}{s-b} +\cfrac{c^2}{s-c} \ge \cfrac{s}{2} \)

$$柯西不等式:\left(({a\over \sqrt{b+c}})^2 +({b\over \sqrt{a+c}})^2 +({c\over \sqrt{a+b}})^2 \right) \left((\sqrt{b+c})^2+ (\sqrt{a+c})^2+ (\sqrt{a+b})^2\right)\\\qquad \quad \ge (a+b+c)^2\\ \Rightarrow \left({a^2\over b+c}+ {b^2\over a+c}+ {c^2\over a+b}\right)(2(a+b+c)) \ge (a+b+c)^2 \Rightarrow \left({a^2\over s-a}+ {b^2\over s-b}+ {c^2\over s-c}\right)(2s) \ge s^2\\ \Rightarrow {a^2\over s-a} +{a^2\over s-a} +{a^2\over s-a} \ge {s\over 2},故得證!!$$

解題僅供參考

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