111年台南女中第2次教甄-數學詳解

國立臺南女中 111 學年度第 2 次教師甄選

填充題： (共 20 題，每題 5 分，共 100 分，答案請化為最簡分數或最簡根式)

解答：$$假設\cases{圓半徑R\\ \angle BOC=\theta} \Rightarrow \angle AOB=2\theta \Rightarrow \cases{\triangle OAB面積={1\over 2}R^2\sin 2\theta = 100 \\ \triangle OBC面積={1\over 2}R^2\sin \theta =80} \Rightarrow {\sin 2\theta\over \sin \theta }={100\over 80} \\ \Rightarrow {2\sin \theta\cos \theta \over \sin \theta}={5\over 4} \Rightarrow \cos\theta ={5\over 8} \Rightarrow \sin \theta = {\sqrt{39}\over 8} \Rightarrow \sin 3\theta = 3\sin \theta -4\sin^3\theta\\ \Rightarrow {\triangle OAC\over \triangle OBC}={\sin 3\theta \over \sin \theta} =3-4\sin^2\theta \Rightarrow \triangle OAC=(3-4\sin^2\theta)\cdot 80 =\bbox[red,2pt]{45}$$

解答：$$令f(x)=(x+1)(x^2+1)(x^4+1) (x^8+1)(x^{16}+1) \\\Rightarrow (x-1)f(x)= (x^2-1)(x^2+1)(x^4+1) (x^8+1)(x^{16}+1) =x^{32}-1\\ \Rightarrow f(x)={x^{32}-1 \over x-1} \Rightarrow f(4)={4^{32}-1\over 3} \Rightarrow \log f(4)=\log (4^{32}-1)-\log 3 \approx \log 4^{32}-\log 3\\ =64\log 2-\log 3 = 64\times 0.301-0.4771 =18.7869 \Rightarrow f(4)為\bbox[red, 2pt]{19}位數$$

解答：$$\begin{array}{cc} x & P(k=x)\\\hline 2,12 & 1/36\\ 3,11 & 2/36\\ 4,10 & 3/36\\ 5,9 & 4/36\\ 6,8 & 5/36\\ 7 & 6/36 \\\hline \end{array},又投擲k個50元硬幣,出現正面的次數為k/2個;\\ 因此期望值=2000+ {50\over 36\cdot 2}\left(2+12+ 2(3+11) + 3(4+10) +4(5+9) +5(6+8)+ 6\cdot 7  \right) \\ = 2000+{25\over 36}\cdot 252=2175，六個人共6\times 2175= \bbox[red,2pt]{13050} $$

解答：$$圓\Gamma通過\cases{O(0,0)\\ A(8,0)\\ B(0,6)} \Rightarrow \cases{圓心C=(A+B)\div 2= (4,3)\\ 半徑R=5} \Rightarrow \overleftrightarrow{AB}斜率=-6/8= -{3\over 4}  \\ \Rightarrow L: 3x+4y+a=0  \Rightarrow d(C,L)=R \Rightarrow {|12+12+a|\over 5}= 5 \Rightarrow |24+a|=25 \\\Rightarrow \cases{a=1(不合，切點需在第一象限)\\ a=-49}  \Rightarrow 3x+4y=49 \Rightarrow x截距= \bbox[red,2pt]{49\over 3}$$

解答：$$假設P=L\cap \overline{AB}，則{\triangle BCP \over \triangle OAB}={1\over 2} \Rightarrow \cfrac{\overline{BC}\times \overline{BP}}{ \overline{BO} \times \overline{AB}} =\cfrac{3\times \overline{BP}}{ 4 \times \overline{AB}} = \frac{1}{2} \Rightarrow \cfrac{ \overline{BP}}{ \overline{AB}} ={2\over 3} \Rightarrow  \cfrac{ \overline{BP}}{ \overline{PA}} ={2\over 1} \\\Rightarrow P=(B+2A)\div 3=(4/3,4/3) \Rightarrow L=\overleftrightarrow{CP}:y={1\over 4}x+1 \Rightarrow x-4y+4=0\\ \Rightarrow (b,c)= \bbox[red,2pt]{(-4,4)}$$

解答：$$\cases{\overline{AF} \bot \overline{FG} \Rightarrow \overline{FG}^2= \overline{AG}^2-\overline{AF}^2 \\ \cases{\overline{AP}\bot \overline{DE} \\ G在平面CDEF} \Rightarrow \angle APG=90^\circ \Rightarrow \overline{AG}^2=\overline{AP}^2 +\overline{PG}^2} \\ \Rightarrow \overline{FG}^2=\overline{AP}^2 +\overline{PG}^2-\overline{AF}^2 = (\overline{AD}\sin 60^\circ)^2+ (20\sqrt 2)^2 -25^2 = 225+800-625=\bbox[red, 2pt]{20}$$

解答：$$恰交3點\Rightarrow k的最小值=( d(O,L))^2=({5\over 10})^2=\bbox[red,2pt]{1\over 4}$$

解答：$$f(x)=ax^3+bx^2 +cx+d  \Rightarrow (x+1)f(x)= ax^4+(a+b)x^3+ (b+c)x^2 +(c+d)x +d\\ f(x)除以x-2餘-5 \Rightarrow f(2)=-5 \Rightarrow 8a+4b+2c+d=-5 \cdots(1)\\ 再用長除法 (x+1)f(x)=(x^3-3)(ax+(a+b))+3x-1 \Rightarrow \cases{b+c=0 \cdots(2)\\ 3a+3b+d=-1 \cdots(3)\\ 3a+c+d =3 \cdots(4)}\\ 由式(1)-(4)可得\cases{a=-1\\ b=-1\\c=1\\d=5} \Rightarrow a+b+c+d = \bbox[red, 2pt]{4}$$

解答：$$f(x)= \int_0^x (at^3+bt^2-9)\,dt = {1\over 4}ax^4+{1\over 3}bx^3-9x \Rightarrow f'(x)=ax^3+bx^2-9\\ x=3有最小值\Rightarrow f'(3)=0 \Rightarrow 27a+9b=9 \Rightarrow 3a+b=1 \cdots(1)\\ 又\int_{-2}^2 f(x)\,dx =\left.\left[  {1\over 20}ax^5 +{1\over 12}bx^4-{9\over 2}x^2 \right] \right|_{-2}^2={16\over 5}a =32 \Rightarrow a=10 再代入(1)\Rightarrow 30+b=1\\ \Rightarrow b=-29 \Rightarrow 2a-b=20+29=\bbox[red, 2pt]{49}$$

解答：$$f(x)=ax^3+bx^2 +cx+d 在x=0的一次近似為y=x+5 (斜率=1)\Rightarrow \cases{f'(0)=1 \Rightarrow c=1 \\f(0)=5 \Rightarrow d=5} \\對稱中心(1,8)=(-{b\over 3a},f(-{b\over 3a})) \Rightarrow \cases{3a+b=0\\ a+b+c+d=8 \Rightarrow a+b=2} \Rightarrow a=\bbox[red, 2pt]{-1}$$

解答：

$$假設\cases{\overline{CD}=x\\ \overline{AB}=\overline{AD}= R\\ \overline{AC}=a}， \triangle ABC 面積={1\over 2}aR\sin \theta = 4\sqrt 3 \Rightarrow \color{blue}{aR\sin \theta} =8\sqrt 3\\  \cos \angle CAD = \cos (\theta+45^\circ) = {a^2+R^2-x^2\over 2aR} \Rightarrow x^2=a^2+R^2-2aR\cos (\theta+45^\circ)\\ =a^2+R^2-2aR(\cos 45^\circ \cos\theta -\sin 45^\circ \sin \theta) =a^2+R^2-\sqrt 2aR(\cos\theta-\sin\theta) \\ =a^2+R^2-\sqrt 2aR\cos\theta+ \sqrt 2\color{blue}{aR\sin \theta} =a^2+R^2-\sqrt 2aR\cos\theta+8\sqrt 6 \ge 2\sqrt{a^2R^2} -\sqrt 2 aR\cos\theta +8\sqrt 6\\ =2aR- \sqrt 2aR\cos\theta +8\sqrt 6 =2\cdot {8\sqrt 3\over \sin \theta}-{8\sqrt 3\over \sin \theta}\cdot \cos \theta +8\sqrt 6= 8\sqrt 6({\sqrt 2-\cos\theta\over \sin \theta}+1)\\ \Rightarrow x^2 \ge 8\sqrt 6({\sqrt 2-\cos\theta\over \sin \theta}+1) =8\sqrt 6( k+1),其中k= {\sqrt 2-\cos\theta\over \sin \theta} \Rightarrow \cos\theta+k\sin\theta =\sqrt 2 \\ \Rightarrow \sqrt{k^2+1}\sin(\theta+\alpha)=\sqrt 2,其中\cases{\sin\alpha = 1/\sqrt{k^2+1} \\\cos \alpha = k/\sqrt{k^2+1}}  \Rightarrow \sin (\theta+\alpha)= {\sqrt 2\over \sqrt{k^2+1}} \le 1\\ \Rightarrow \sqrt 2\le \sqrt{k^2+1} \Rightarrow k\ge 1 \Rightarrow x^2 \ge 8\sqrt 6(1+1)=16\sqrt 6 \Rightarrow x\ge \sqrt{16\sqrt 6} = \bbox[red,2pt]{4\sqrt[4]{6}}$$

解答：

$$x=\lfloor x\rfloor+\{x\} \Rightarrow \{x\}=x-\lfloor x\rfloor \Rightarrow x-\lfloor x\rfloor-kx+3k=0 \Rightarrow  x-\lfloor x\rfloor=k(x-3)\\ 相當於求兩圖形\cases{\Gamma: y= x-\lfloor x\rfloor\\ L:k(x-3)}有三個相異交點時，k的範圍; \\ 而\cases{\Gamma 為斜率為1且通過(m,0)及(m+1,1),m\in \mathbb{Z}不含頂點(m+1,1)的線段\\ L為通過A(3,0)斜率為k的直線}，如上圖；\\因此\cases{k介於\overleftrightarrow{AB}與\overleftrightarrow{AC}之間\\ k介於\overleftrightarrow{AD}與\overleftrightarrow{AE}之間}，其中\cases{B(1,1)\\ C(2,1)\\ D(6,1)\\ E(7,1) } \Rightarrow \bbox[red, 2pt]{\cases{-1 \lt k\lt -{1\over 2} \\ {1\over 4}\lt k\lt {1\over 3}}}$$

解答：$$\overrightarrow{CD} =-\overrightarrow{AC}+\overrightarrow{AD} =m\overrightarrow{AB} +7\overrightarrow{AC} \Rightarrow \overrightarrow{AD}= m\overrightarrow{AB} +8\overrightarrow{AC} \Rightarrow \cfrac{\overline{BE}}{\overline{EC}} =\cfrac{8}{m} =\cfrac{a\triangle ABE}{a\triangle ACE}\\ \Rightarrow {8\over m}= {a\triangle ABE\over 1}  \Rightarrow a\triangle ABE={8\over m}\\ 假設\overrightarrow{AD}=t\overrightarrow{AE}，則 t\overrightarrow{AE}=m\overrightarrow{AB} + 8\overrightarrow{AC} \Rightarrow \overrightarrow{AE}={m \over t}\overrightarrow{AB} +{8 \over t}\overrightarrow{AC} \Rightarrow  {m\over t}+{ 8 \over t}=1 \Rightarrow t=m+8\\ \Rightarrow \cfrac{a\triangle ACD}{a\triangle ACE}= t=m+8 \Rightarrow a\triangle ACD=m+8 \Rightarrow a\triangle CDE=m+7 \\ \Rightarrow a\triangle ABE+a\triangle CDE = {8\over m}+m+7 \ge 2 \sqrt{{8\over m}\cdot m}+7 = \bbox[red,2pt]{4\sqrt 2+7}$$

解答：

$$將三位明星參加的場次以集合表示，見上圖，則\cases{A+D+G+F=15 \cdots(1)\\ B+D+E+G=16 \cdots(2)\\ C+E+F+G=17 \cdots(3)\\ G=5\\ D+E+F= 15}\\ (1)+(2)+(3) \Rightarrow A+B+C+ 2(D+E+F)+ 3G=48 \Rightarrow A+B+C=48-2\cdot 15-3\cdot 5=3\\ \Rightarrow A+B+C+D+E+F+G = 3+15+5=23 \\\Rightarrow 三球員均未得到\text{Double doubles}的場次=40-23= \bbox[red, 2pt]{17}$$

解答：$$假設從頂點K(K=A,B,C,D,F,G,H)走n步到E的方法數記為K_n;\\由於正八邊形是對稱的，因此F_n=D_n,G_n=C_n,H_n=B_n;\\ 因此我們有\cases{A_n= B_{n-1}+H_{n-1}= 2B_{n-1} \cdots(1)\\ B_n = A_{n-1}+ C_{n-1} \cdots(2)\\ C_{n}= B_{n-1}+D_{n-1} \cdots(3)\\ D_n= C_{n-1} \cdots(4)},將\cases{(2)代入(1) \Rightarrow A_n= 2A_{n-2}+ 2C_{n-2}\\ (2)及(4)代入(3) \Rightarrow C_n=A_{n-2} +C_{n-2}+C_{n-2}}\\ \Rightarrow \cases{A_n= 2A_{n-2}+ 2C_{n-2} \cdots(5)\\ C_n= A_{n-2}+ 2C_{n-2} \cdots(6)}，再將\cases{(5)代入(6) \Rightarrow C_n=A_{n-2}+A_n-2A_{n-2} =A_n-A_{n-2}\\ (6)代入(5) \Rightarrow A_n=2(C_n-2C_{n-2})+ 2C_{n-2}= 2C_{n}-2C_{n-2}}\\ \Rightarrow \cases{A_n=2C_n-2C_{n-2}\cdots (7) \\ C_n =A_n-A_{n-2} \cdots(8)}，最後將(8)代入(7)  A_n= 4A_{n-2}-2A_{n-4}\\ 初始值:\cases{A_1=A_2=A_3 =0\\ A_4=2} \Rightarrow A_{奇數}=0，只需計算A_{偶數};  A_6=8  \Rightarrow A_8=28 \Rightarrow a_{10}=96 \\\Rightarrow a_{12}=328 \Rightarrow a_{14}=1120 \Rightarrow a_{16}= \bbox[red, 2pt]{3824}$$

解答：

$$正六邊形每個內角為(6-2)180\div 6=120^\circ，兩頂點距離可能值為1,\sqrt 3, 2，其中6條長度為1\\、6條長度為\sqrt 3,3條長度為2；因此期望值=1\times {6\over 15}+ \sqrt{3}\times {6\over 15}+2\times {3\over 15} = \bbox[red, 2pt]{4+2\sqrt 3\over 5}$$

解答：

$$將數據(X,Y)畫在平面圖上，可看出將(13,12)去掉後，整體的斜率會變得最小\\\begin{array}{} X & Y & X^2 & Y^2 &XY \\ \hline 8 & 11  &64 &121 & 88\\ 9 & 12 & 81 & 144 & 108\\ 10 & 10 & 100 & 100 & 100\\ 11 & 8 & 121 & 64 & 88\\ 12 & 9 & 144 & 81 & 108\\ \hline 50 & 50 & 510 & 510 & 492\end{array} \Rightarrow 迴歸直線斜率b_1={\sum XY-{\sum X\sum Y\over n} \over \sum X^2-{(\sum x)^2\over n}} ={492-50\cdot 50/5\over 510-50^2/5} =-{4\over 5}\\ 迴歸直線通過(\bar x,\bar y)=(50/5,50/5)=(10,10) \Rightarrow y=-{4\over 5}(x-10)+10=-{4\over 5}x +18\\ \Rightarrow (m,k)=\bbox[red,2pt]{(-{4\over 5},18)}$$

解答：

$${x^2\over 64}+{y^2\over 16} \Rightarrow \cases{a=8\\ b=4} \Rightarrow c=\sqrt{64-16} =4\sqrt 3 \Rightarrow \cases{F_1(4\sqrt 3,0)\\ F_2(-4\sqrt 3,0)} \Rightarrow \overline{F_1F_2}=8\sqrt 3\\ 假設\cases{\overline{AF_2}=m \\ \overline{AF_1}=n} \Rightarrow \cases{m+n= 2a = 16\\ m^2+n^2 =\overline{F_1F_2}^2= 192} \Rightarrow \cases{m=8-4\sqrt 2\\ n=8+4\sqrt 2};\\假設\cases{\overline{BF_2}=s \\ \overline{BF_1}=t} \Rightarrow \cases{s+t= 2a = 16\\ (s+m)^2+n^2 =t^2 \Rightarrow (s+8-4\sqrt 2)^2+(8+4\sqrt 2)^2 =t^2} \\ \Rightarrow \cases{s=(24+4\sqrt 2)/17\\ t=(248-4\sqrt 2)/17} \Rightarrow 內切圓半徑=(\overline{AB}+\overline{AF_1}-\overline{BF_1})\div 2 = (s+m+n-t)\div 2\\ =\left(16+{24\sqrt 2\over 17}-{248-4\sqrt 2 \over 17}\right) \div 2=\bbox[red, 2pt]{24+ 4\sqrt 2\over 17}$$

解答：

$$假設\cases{\overline{AF}=\overline{FB}=a\\ \overline{AE}=\overline{EC}=b} \Rightarrow a^2+b^2=({\overline{BC}\over 2})^2 = 9^2=81\\ \cases{\overline{BE}斜率m_1=-b/2a\\ \overline{CF}斜率m_2=-2b/a} \Rightarrow \tan 150^\circ={m_2-m_1\over 1+ m_1m_2} \Rightarrow -{1\over \sqrt 3}={-3b/2a\over 1+b^2/a^2} =-{3b\cdot a^2\over 2(a^2+b^2)a} \\=-{3ab\over 162} =-{ab\over 54} \Rightarrow ab=18\sqrt 3 \Rightarrow \triangle ABC面積=2ab=36\sqrt 3 \Rightarrow \triangle GBC={1\over 3}\triangle ABC= \bbox[red, 2pt]{12\sqrt 3}$$

解答：$$假設\cases{\overline{BC}=a\\ \overline{AC}=b\\ \overline{AB}=c}，則\tan A=\sqrt{15} \Rightarrow \cases{\sin \sqrt{15}/4\\ \cos A=1/4}，因此由餘弦定理:\overrightarrow{AB} \cdot \overrightarrow{AC}=bc/4\\ \cases{\cfrac{\overline{AD}}{\overline{BD}} = \cfrac{1}{2} \Rightarrow \overline{AD}={1\over 3}\overline{AB}={c\over 3}\\\cfrac{\overline{AE}}{\overline{CE}} = \cfrac{1}{4} \Rightarrow \overline{AE}={1\over 5}\overline{AC}={b\over 5}} \Rightarrow \cases{\overline{AD}=c/3\\ \overline{AE} =b/5}  \\ \overrightarrow{AP}\cdot \overrightarrow{AB}= \cases{ \overline{AD}\times \overline{AB} ={c\over 3}\times c={c^2\over 3} \\  (x\overrightarrow{AB} + y \overrightarrow{AC})\cdot \overrightarrow{AB} =xc^2+ybc/4} \Rightarrow 4c^2=12xc^2+ 3ybc \cdots(1)\\ \overrightarrow{AP}\cdot \overrightarrow{AC}= \cases{ \overline{AE} \times \overline{AC} ={b\over 5}\times b={b^2\over 5} \\  (x \overrightarrow{AB} + y\overrightarrow{AC}) \cdot \overrightarrow{AC} =xbc/4+ yb^2} \Rightarrow 4b^2= 5xbc + 20yb^2 \cdots(2) \\ 由(1)及(2) \Rightarrow \cases{12cx+ 3by = 4c\\ 5cx+ 20by=4b} \Rightarrow \cases{x=(80c-12b) /225c\\ y =(48b-20c)/225b} \\\Rightarrow 3x+5y ={480bc-36b^2-100c^2\over 225bc} ={480\over 225}-\left( {36b\over 225c}+{100c\over 225b}\right) \le {480\over 225}-2\sqrt{{36\over 225}\cdot {100\over 225}} \\ ={480\over 225}-2\cdot {60\over 225} ={360\over 225} =\bbox[red, 2pt]{8\over 5}$$

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