111年台中女中教甄-數學詳解

臺中市立臺中女子高級中等學校 111 學年度第二次教師甄選

壹、 填充題：每格 4 分，全對才給分，共 72 分。

1.某家蛋捲店有五種口味的蛋捲，分別為原味、抹茶、巧克力、芝麻與咖啡五種；購買時以兩包為一組裝成禮盒，這兩包口味可以一樣，也可以不同。 某人想同時買三盒禮盒，試問他有 種組合方式。

解答： $$禮盒內的組合:\cases{兩包口味相同有5種\\ 兩包口味均不同:C^5_2=10} \Rightarrow 有10+5=15種不同的包裝\\三盒禮盒的組合:\cases{三盒都一樣:15種\\ 二盒相同另一盒不同:15\times 14 =210\\ 三盒都不一樣:C^{15}_3 =455} \Rightarrow 共有15+ 210+ 455=\bbox[red, 2pt]{680}組合方式$$

解答： $$A_n(x_n,y_n) 逆時針旋轉90度變為 B_n(-y_n,x_n)\Rightarrow  A_{n+1}=(4A_n+ 3B_n)/7 ={1\over 7}(4x_n-3y_n,3x_n+4y_n) \\ \Rightarrow A_{n+1}={1\over 7}\begin{bmatrix} 4 & -3\\ 3 & 4\end{bmatrix}\begin{bmatrix}x_n\\ y_n \end{bmatrix} \Rightarrow T=\bbox[red, 2pt]{{1\over 7}\begin{bmatrix} 4 & -3\\ 3 & 4\end{bmatrix}}$$

解答：

$$\cases{\lim_{x\to \infty}\cfrac{f'(x)}{x^2-3x+7}=3\\ f(x)在x=-1時有極值} \Rightarrow \cases{f'(x)= 3x^2+ax +b\\ f'(-1)=0} \Rightarrow \cases{f(x)= x^3+{1\over 2}ax^2+bx +c \\a-b=3}  \\ 又\lim_{x\to 0}\cfrac{f(x)}{x }= \lim_{x\to 0}x^2+{1\over 2} x+ b+{c\over x}= 1 \Rightarrow \cases{c=0\\ b=1} \Rightarrow a=4 \Rightarrow f(x)=x^3+2x^2+x\\ f''(x)=0 \Rightarrow 6x+4=0 \Rightarrow x=-2/3 \Rightarrow y=f(-2/3)= -{8\over 27}+ {8\over 9}-{2\over 3}= -{2\over 27} \\y=k與y=f(x)所圍兩面積相等 \Rightarrow y=k經過y=f(x)的對稱中心B(-2/3,-2/27) \\\Rightarrow k=\bbox[red,2pt]{-2\over 27}$$

解答： $$假設\cases{C為原點\\ \overrightarrow{OB}為x軸方向\\ \overrightarrow{OD}為y軸方向 \\ \overrightarrow{OG}為z軸方向} \Rightarrow \cases{A(12,6,0)\\ B(12,0,0)\\ C(0,0,0) \\D(0,6,0)\\ E(12,6,6) \\F(12,0,6)\\ G(0,0,6)\\ H(0,6,6)} \Rightarrow E=ACH平面: x-2y+2z=0\\ P(x,y,z)\in E \Rightarrow \cases{d_1^2= d(P,\overline{AD})^2= (6-y)^2+z^2 \\ d_2^2 = d(P,\overline{DH})^2 = x^2+(6-y)^2 \\ d_3^2 = d(P,\overline{CD})^2 = x^2+z^2} \Rightarrow d_1^2 +d_2^2 +d_3^2 = 2(x^2+ (y-6)^2 +z^2)\\利用Lagrange 算子求極值: 令\cases{f(x,y,z)=x^2+(y-6)^2 +z^2\\ g(x,y,z)=x-2y+2z} \Rightarrow \cases{f_x= \lambda g_z\\ f_y=\lambda g_y\\ f_z=\lambda g_z} \Rightarrow \cases{2x =\lambda \\ 2(y-6)= -2\lambda \\ 2z=2\lambda} \\ \Rightarrow \cases{x=z/2\\ y=6-z} 代入g(z/2,6-z,z)=0 \Rightarrow \cases{x=4/3\\ y=10/3 \\ z=8/3} \Rightarrow f(4/3,10/3,8/3) = 16\\ \Rightarrow d_1^2 +d_2^2 +d_3^2的極值= 2f(4/3,10/3,8/3) =2\times 16=\bbox[red, 2pt]{32}$$

解答：

$$1-\sqrt{4-y^2}=x \Rightarrow (1-x)^2 = 4-y^2 \Rightarrow (x-1)^2+y^2 = 4 \Rightarrow \cases{圓心(1,0)\\ 半徑r=2}\\ 此圓需滿足\cases{x\le 0\\ 1-\sqrt{4-y^2}\le 0} \Rightarrow \cases{x\le 0\\ -\sqrt 3\le y\le \sqrt 3}，因此剩下1\le x\le 0的部份，見上圖著色部分\\ 令3x-y=k，此為斜率3的直線，因此k的極值出現在圓左右兩切線；\\而此圓僅剩左半部，依此經過(0,-\sqrt 3)的直線y=3x-\sqrt 3 \Rightarrow k=\sqrt 3=M\\ 現在來求左切線:y=3x-k代入圓方程式\Rightarrow (x-1)^2+(3x-k)^2=4 \\ \Rightarrow 10x^2-(2+6k)x + k^2-3 判別式為0 \Rightarrow (2+6k)^2-40(k^2-3)=0 \Rightarrow k^2-6k-31=0\\\Rightarrow k=3- 2\sqrt {10}(3+ 2\sqrt{10} 為右切線) \Rightarrow m=3-2\sqrt{10} \Rightarrow M+m= \bbox[red,2pt]{3-2\sqrt{10}+ \sqrt 3}$$

解答：

$$假設O=\overline{AC}\cap \overline{BD} \Rightarrow \cfrac{\triangle ABD}{\triangle CBD} = \cfrac{{1\over 2} \overline{AB}\cdot \overline{AD}\sin A}{{1\over 2}\overline{CB}\cdot \overline{CD}\sin (180^\circ-A)} =\cfrac{  \overline{AB}\cdot \overline{AD} }{ \overline{CB}\cdot \overline{CD} } = {2\cdot 4\over 3\cdot 4}={2\over 3}\\ 又\cfrac{\triangle ABD}{\triangle CBD} =\cfrac{\overline{AO}}{\overline{OC}}，因此\cfrac{\overline{AO}}{\overline{OC}}={2\over 3}; 同理，\cfrac{\overline{BO}}{\overline{OD}} ={3\cdot 2\over 4\cdot 4}={3\over 8};\\ \cases{\cfrac{ \overline{BO}}{\overline{OD}} ={3\over 8} \Rightarrow \overrightarrow{AO} = {3\over 11}\overrightarrow{AD} + {8 \over 11} \overrightarrow{AB} \\\cfrac{\overline{AO}}{\overline{OC}}={2\over 3} \Rightarrow \overrightarrow{AO}= {2\over 5}\overrightarrow{AC}} \Rightarrow {2\over 5}\overrightarrow{AC}= {3\over 11}\overrightarrow{AD} + {8 \over 11} \overrightarrow{AB} \Rightarrow \overrightarrow{AC}= {15\over 22}\overrightarrow{AD} + {20 \over 11} \overrightarrow{AB} \\ \Rightarrow x-y =  {20 \over 11}-{15\over 22} =\bbox[red, 2pt]{25 \over 22}$$

解答： $$\triangle ABC={1\over 2}\cdot \overline{AB}\cdot \overline{DQ} \Rightarrow {1\over 2}\cdot 3\cdot \overline{DQ}=12 \Rightarrow \overline{DQ}=8 \Rightarrow \overline{DO}= \overline{DQ}\sin 30^\circ = 4\\ \Rightarrow 四面體ABCD體積={1\over 3}\cdot \triangle ABC\cdot \overline{DO} ={1\over 3}\cdot 15\cdot 4= \bbox[red, 2pt]{20}$$

解答：

$$令\cases{P(3,4)\\ Q(z_2)\\ R(z_1) =(\cos\theta,\sin \theta)\\ O(0,0)}，由\cases{|z_2|=2 \Rightarrow Q在半徑為2的圓上\\ |z_2-(3+4i)|=3 \Rightarrow \overline{PQ}=3 \\\overline{OP}=5} \Rightarrow Q({6\over 5},{8\over 5}) \\ 又|z_1-z_2|= \overline{QR} =\sqrt 3 \Rightarrow ({6\over 5}-\cos\theta)^2 +({8\over 5}-\sin \theta)^2 = (\sqrt 3)^2 \Rightarrow {12\over 5}\cos \theta +{16\over 5} \sin \theta =2\\ \Rightarrow {12\over 5}\cos \theta +{16\over 5} \sqrt{1-\cos^2 \theta} =2 \Rightarrow 100\cos^2\theta-60\cos \theta -39=0 \\\Rightarrow \cos\theta ={3+4\sqrt 3\over 10}({3- 4\sqrt 3\over 10}\lt 0，負值不合) \Rightarrow \sin \theta = {4-3\sqrt 3\over 10} \Rightarrow z_1= \bbox[red, 2pt]{{3+4\sqrt 3\over 10}+ {4-3\sqrt 3\over 10}i}$$

解答：

$$\Gamma: y^2=4x \Rightarrow \cases{準線 M:x=-1 \\ 焦點F(1,0)}，又直線L斜率為2 \Rightarrow L: y=2x+a\\ 將L代入\Gamma \Rightarrow (2x+a)^2 = 4x \Rightarrow 4x^2 +(4a-4)x+a^2 =0 \Rightarrow x=\cfrac{(1-a)\pm \sqrt{1-2a}}{2}\\ \overline{AF} +\overline{BF} =  d(A,M)+ d(B,M) = \left( \cfrac{(1-a)+ \sqrt{1-2a}}{2}+1\right)+\left( \cfrac{(1-a)- \sqrt{1-2a}}{2}+1\right) \\ =1-a+2 = 4\Rightarrow a=-1 \Rightarrow \cases{x={2\pm \sqrt 3\over 2}\\ L:y=2x-1} \Rightarrow \cases{A((2+\sqrt 3)/2, 1+\sqrt 3)\\ B((2-\sqrt 3)/2, 1-\sqrt 3)} \\ \Rightarrow \overline{AB}= \sqrt{(\sqrt 3)^2 +(2\sqrt 3)^2} =\bbox[red,2pt]{\sqrt{15}}$$

解答： $$f(x)= (x-\sqrt 3)^{50}+ (x+1)^{50} =\sum_{k=0}^{50} a_kx^k \Rightarrow a_0-a_2+a_4-\cdots -a_{50}={1\over 2} (f(i)+f(-i))\\ =Re(f(i)) =Re((i-\sqrt 3)^{50}+(i+1)^{50}) = Re((2e^{5\pi i/6})^{50} +(\sqrt 2 e^{\pi i/4})^{50}) =2^{49} =k\\ \Rightarrow \log_4 k= \log_4 2^{49}= \log_4 4^{49/2} =\bbox[red,2pt]{49\over 2}$$

解答： $$剛好繞兩圈且不會在第一圈結束的情形:\\(1)12奇排列:在第一圈(6奇)就結束\\ (2)1偶10奇:只有"奇奇奇奇奇偶+奇奇奇奇奇奇"1種情況\Rightarrow 機率= ({1\over 2})^{11} \\ (3)2偶8奇排列數=C^{10}_2= 45 需扣除\cases{1偶4奇+1偶4奇:5\times 5=25\\ 6奇+ 2偶2奇: C^4_2= 6\\ 2偶2奇+ 6奇:C^4_2= 6}剩下45-25-6-6= 8種情況\\ \quad \Rightarrow 機率=8\times ({1\over 2})^{10} \\(4) 3偶6奇排列數=C^9_3 =84 需扣除\cases{6奇+3偶:1\\ 4奇1偶+2奇2偶:5\times 6=30\\ 2奇2偶+4奇1偶=30\\ 3偶+6奇:1}剩下84-62=22 \\ \quad \Rightarrow 機率=22\times ({1\over 2})^{9} \\ (5)4偶4奇排列數=C^8_4 =70 需扣除\cases{1偶4奇+3偶:5\\ 2偶2奇+ 2偶2奇: 36\\ 3 偶+ 1偶4奇: 5}剩下70-46= 24 \\ \quad \Rightarrow 機率=24\times ({1\over 2})^{8} \\ (6)5偶2奇排列數=C^7_2 =21 需扣除\cases{2偶2奇+ 3偶:6\\ 3偶+2偶2奇:6}剩下21-12=9 \\\quad \Rightarrow 機率= 9\times ({1\over 2})^{7} \\(7)6偶 排列:3偶繞一圈提前結束\\ 因此機率=({1\over 2})^{11} + 8\times ({1\over 2})^{10} +22\times ({1\over 2})^{9}+  24\times ({1\over 2})^{8} +9\times ({1\over 2})^{7}= \bbox[red, 2pt]{441\over 2048}$$

解答：

$$兩圖形\cases{\Gamma_1:y=x(x-3)/3\\ \Gamma_2: x=y(y-3)/3} 對稱於直線L:y=x，兩圖形\cases{y=x(x-3)/3\\ y=x}交於\cases{O(0,0)\\ P(6,6)} \\ \Rightarrow \Gamma_1與\Gamma_2所圍面積=2倍(\Gamma_1與L所圍面積)= 2\int_0^6x-{1\over 3}x(x-3)\,dx =\int_0^6 -{2\over 3}x^2+4x \,dx \\ =\left.\left[ -{2\over 9} x^3 +2x^2\right] \right|_0^6 =\bbox[red,2pt]{24}$$

解答： $$邊長6\sqrt 2 \Rightarrow \cases{O(0,0,0)\\ A(6,0,6)\\ B(0,6,6)\\ C(6,6,0)} \Rightarrow \cases{\overline{OD} =\overline{OA}/6 \Rightarrow D=(A+5O)/6= (1,0,1)\\ \overline{OE}= \overline{OB}/3 \Rightarrow B=(B+2O)/3= (0,2,2)\\ \overline{OF}= \overline{OC} /2 \Rightarrow F=(O+C)/2= (3,3,0) \\G=(B+C)/2= (3,6,3)} \\\Rightarrow \cases{\overrightarrow{GD} =(-2,-6,-2) \\ \overrightarrow{GE} =(-3,-4,-1) \\ \overrightarrow{GF} =(0,-3,-3) } \Rightarrow DEFG體積= {1\over 6}\begin{Vmatrix}-2 & -6 &-2 \\ -3 & -4 & -1\\ 0 & -3 & -3\end{Vmatrix} ={1\over 6}\times 18=\bbox[red, 2pt]{3}$$

解答： $$假設通過P(-2,5)之直線L: y=m(x+2)+5，\\又\overline{PA} :\overline{PB} =1:2 \Rightarrow P=(2A+B)/3，因此假設A(a,b) \Rightarrow B(-6-2a,15-2b)\\ A、B皆在圓上\Rightarrow \cases{a^2+b^2+2a-6b-3=0 \Rightarrow (a+1)^2(b-3)^2=13 \cdots(1)\\ (-6-2a)^2+(15-2b)^2+2(-6-2a)-6(15-2b)-3=0} \\ \Rightarrow a=2b-14 代回(1) \Rightarrow (2b-13)^2 +(b-3)^2=13 \Rightarrow 5b^2-58b+165=0 \Rightarrow (b-5)(5b-33)=9\\ \Rightarrow \cases{b=5 \Rightarrow a=-4\\ b=33/5 \Rightarrow a=-4/5}，P(a,b)=(-4,5)、(-4/5, 33/5)皆在L上\\ \Rightarrow \cases{5=-2m+5\\ 33/5=m(6/5)+5} \Rightarrow \bbox[red,2pt]{\cases{m=0\\ m=4/3}}$$

解答： $$找(a,b,c)滿足a(x+y+2z) + b(y+2z)+c (x+2y+5z)= 2x-3y-8z\\ \Rightarrow \cases{a +c=2\\ a+b+2c=-3 \\ 2a+2b+5z= -8} \Rightarrow \cases{a=4\\ b=-3\\ c=-2} \Rightarrow \cases{-8\le 4(x+y+2z) \le 12\\ -3\le -3(y+2z)\le 9\\ -6\le -2(x+2y+ 5z) \le 8}\\ 三式相加 \Rightarrow 2x-3y-8z\le 29 =m，此時 \cases{x+ y+2z = 3\\ y+2z=-3\\ x+2y+5z=-4} \Rightarrow \cases{x=6\\ y=5\\ z=-4} \\ \Rightarrow (p,q,r,m)= \bbox[red,2pt]{(6,5,-4,29)}$$

解答： $$\lim_{n\to \infty} {1\over n^6} \sum_{k=1}^n [(n^2+nk+k^2)(n +k)^3] =\lim_{n\to \infty}  \sum_{k=1}^n {1\over n} \left[(1+ {k \over n}+({k\over n})^2)(1 +{k\over n})^3 \right] \\ =\int_0^1 (1+x+x^2)(1+x)^3 \,dx = \int_0^1 x^5+4x^4 +7x^3 +7x^2+4x +1\,dx\\ ={1\over 6}+ {4\over 5}+{7\over 4} +{7\over 3}+ 2+ 1 = \bbox[red,2pt]{161\over 20}$$

解答： $$a+b^2+ c^4=\underbrace{{a\over 8}+\cdots +{a\over 8}}_{8個} +\underbrace{{b^2\over 4}+\cdots +{b^2\over 4}}_{4個} + {c^4\over 2}+ {c^4\over 2} =28\\ \Rightarrow {28\over 14}\ge \sqrt[14]{({a\over 8})^8\cdot ({b^2\over 4})^4\cdot ({c^4\over 2})^2} =\sqrt[14]{(abc)^8\over 2^{34}} = \sqrt[14]{64^8\over 2^{34}} = \sqrt[14]{ 2^{14}} = 2\\ \Rightarrow 剛好等號成立，因此{a\over 8}= {b^2\over 4}= {c^4\over 2}=k\Rightarrow \cases{a=8k \\ b^2=4k \\ c^4=2k } \Rightarrow a+b^2+c^4 = 14k = 28 \Rightarrow k=2 \\ \Rightarrow \cases{a=16\\ b=2\sqrt 2\\ c=\sqrt 2} \Rightarrow a+b+c = \bbox[red, 2pt]{ 16+3\sqrt 2}$$

解答： $$a_1,a_2,a_3成等差 \Rightarrow \cases{a_1=a_2-d\\ a_3=a_2+d}，又a_1為a_2及a_3的等比中項\Rightarrow (a_2-d)^2 = a_2(a_2+d)\\ \Rightarrow d^2 = 3a_2d \Rightarrow d=3a_2 \Rightarrow a_1=a_2-d=-2a_2 \Rightarrow a_1=-{2\over 3}d \Rightarrow a_n = -{2\over 3}d +(n-1)d = (n-{5\over 3})d\\ \Rightarrow \lim_{n\to \infty} S_n = \lim_{n\to \infty} \sum_{k=1}^n(n-{5\over 3})d\cdot ({1\over 2})^n =d\left(\lim_{n\to \infty} \sum_{k=1}^n {n\over 2^n} -{5\over 3} \cdot ({1\over 2})^n\right) \\ =d(2-{5\over 3}) ={1\over 3}d = 5 \Rightarrow d= 15  \Rightarrow a_1 = -{2\over 3}\cdot 15= \bbox[red, 2pt]{-10}$$

 貳、計算證明題：請寫出詳細計算與證明過程，否則不予給分， 共 28 分。

解答：

$$作\overline{OD}\parallel \overline{AB}，其中O\in\overline{AC} \Rightarrow \cases{\angle DOC=\angle A=2\angle CED (圓心角=2 圓周角)\\ \angle ODC= \angle B =\angle C\Rightarrow \overline{OD} =\overline{OC}=圓半徑r} \Rightarrow O為\triangle CDE 外心 \\ 令\triangle CDE 外接圓交\overline{AC} 於F \Rightarrow \triangle ACE\sim \triangle ADF (AAA) \Rightarrow \cfrac{\overline{AD}}{ \overline{AC}} =\cfrac{\overline{AF}}{ \overline{AE}} \cdots(1)\\ 又\cases{\overline{AB} \parallel \overline{DO} \Rightarrow \angle BAD=\angle ADO\\ \angle ADO=180^\circ-\angle DOC= 180^\circ-\angle A=180^\circ -\angle BED= \angle AED} \Rightarrow \triangle ADO \sim \triangle ABE (AAA) \\ \Rightarrow \cfrac{\overline{OD}}{ \overline{AE}} =\cfrac{\overline{AD}}{ \overline{AB}} =\cfrac{\overline{AD}}{ \overline{AC}}\stackrel{(1)}{=}\cfrac{\overline{AF}}{ \overline{AE}} \Rightarrow \overline{AF}= \overline{OD}= r\\ \overline{OD}\parallel \overline{AB} \Rightarrow \cfrac{\overline{CD}}{ \overline{BD}} =\cfrac{\overline{OC}}{ \overline{AO}} =\cfrac{r}{ 2r} \Rightarrow \overline{BD}= 2\overline{CD}，\bbox[red,2pt]{故得證}$$

解答： $$a,b,c成等差\Rightarrow a+c=2b，再由正弦定理:{a\over \sin A}={b\over \sin C}={c\over \sin C}=2R \Rightarrow \cases{\sin A= a/2R\\ \sin B= b/2R\\ \sin C=c/2R} \\\Rightarrow  \sin A+\sin C =(a+c)/2R =2b/2R =2\sin B \Rightarrow \sin A+\sin C=2\sin B\cdots(1)\\ 而A+B+C=180^\circ \Rightarrow \sin B=\sin (A+C) \stackrel{(1)}{\Rightarrow} 2\sin B= 2\sin(A+C)= \sin A+\sin C \\ \cases{2\sin(A+C)= 4\sin{A+C\over 2}\cos {A+C\over 2} \\ \sin A+\sin C=2 \sin{A+C\over 2}\cos {A-C\over 2}},兩式相等\Rightarrow 2\cos{A+C\over 2} =\cos{A-C\over 2} \\ \Rightarrow 2\left(\cos{A\over 2}\cos {C\over 2}-\sin{A\over 2}\sin {A\over 2} \right) = \cos{A\over 2}\cos {C\over 2}+ \sin{A\over 2}\sin {C\over 2} \Rightarrow 3\sin{A\over 2}\sin {C\over 2} = \cos{A\over 2}\cos{C\over 2}\\ \Rightarrow \cfrac{\sin{A\over 2}}{\cos{A\over 2}} \cdot \cfrac{\sin{C\over 2}}{\cos{C\over 2}} =\cfrac{1}{3} \Rightarrow \tan{A\over 2}\tan{C\over 2}= \bbox[red,2pt]{1\over 3}$$

解答： $$x^k-x^{k-1}-x^{k-2}-\cdots -x-1 = x^k-(x^{k-1}+x^{k-2}+\cdots +1) = x^k-\cfrac{x^k-1}{x-1} \\ =\cfrac{x^{k+1}- 2k^k+1}{x-1} \Rightarrow \alpha_k 為f_k(x)=x^{k+1}- 2x^k+1 =0正實根 (\alpha_k\ne 1,k\gt 1) \\ \Rightarrow \cases{f(0)= 1\\ f(1)= 0 \\ f(2)=1 \\ f'_k(x)= (k+1)x^k-2kx^{k-1}}，因此f'_k(x)=(k+1)x^{k-1}(x-{2k\over k+1})=0 \Rightarrow \cases{x=0\\ x=2-{2\over k+1}} \\ \Rightarrow \cases{f(x)遞減,x\in (0,2-{2\over k+1}) \\ f(x)遞增,x\gt 2-{2\over k+1}} \Rightarrow 2-{2\over k+1}\lt \alpha_k \lt 2,k\gt 1\\ 由夾擠定理: \lim_{k\to \infty} \alpha_k =2，因此\alpha_k收斂且收斂至2，\bbox[red, 2pt]{故得證}$$

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解題僅供參考，其他教甄歷年試題及詳解