108年臺中清水高中教甄-數學詳解

臺中市立清水高級中學108學年度教師甄選

第壹部分、填充題： (每題 5 分，共 60 分)

解：$$令\cases{甲_n:第n天在甲鎮的機率\\ 乙_n:第n天在乙鎮的機率\\ 丙_n:第n天在丙鎮的機率}\qquad\qquad，其中\cases{甲_1=1\\ 乙_1=0 \\丙_1=0} \Rightarrow \cases{甲_2=0\\ 乙_2=1/2 \\丙_2=1/2} \\ \Rightarrow 甲_8={1\over 2}(乙_7+丙_7) = {1\over 2}({1\over 2}(甲_6+丙_6) +{1\over 2}(甲_6+乙_6)) ={1\over 2}甲_6 +{1\over 4}乙_6 +{1\over 4}丙_6\\ ={1\over 4}甲_5 +{3\over 8}乙_5 +{3\over 8}丙_5  ={3\over 8}甲_4 +{5\over 16}乙_4+{5\over 16}丙_4 ={5\over 16}甲_3 +{11\over 32}乙_3+{11\over 32}丙_3 \\={11\over 32}甲_2 +{21\over 64}乙_2 +{21\over 64}乙_2 = \bbox[red,2pt]{21\over 64}$$

解：$$\cases{6p^2 +2019p-14=0 \Rightarrow \cases{p_1=(-2019+\sqrt{2019^2+24\times 14})/12 \\p_2=(-2019-\sqrt{2019^2+24\times 14})/12} \\ 14q^2-2019-6=0 \Rightarrow \cases{q_1=(2019 +\sqrt{2019^2+24\times 14})/28 \\ q_2=(2019 -\sqrt{2019^2+24\times 14})/28}} \\ \Rightarrow p_1q_1 =p_2q_2 = 1，因此{p\over q} ={p_1\over q_2}={p_2\over q_1} = \bbox[red,2pt]{-{7\over 3}}$$

解：$$0\le x\le \pi \Rightarrow -\pi \le -x \le 0 \Rightarrow -{2\pi\over 3} \le {\pi\over 3}-x \le {\pi \over 3};\\現在 y=2\sin\left( {\pi \over 3}+x\right) -2\sin x = 2(\sin{\pi \over 3}\cos x+ \sin x\cos {\pi \over 3})-2\sin x  \\= 2({\sqrt 3 \over 2}\cos x+ {1\over 2}\sin x )-2\sin x = \sqrt 3\cos x-\sin x = 2({\sqrt 3\over 2}\cos x-{1\over 2}\sin x) \\ = 2\sin({\pi \over 3}-x) =\begin{cases} M=\sqrt 3 & 若x=0\\ m=-2 & 若x={5\pi\over 6}\end{cases} \Rightarrow (M,m)= \bbox[red,2pt]{(\sqrt 3,-2)}$$

解：$$令\cases{A:三奇數\\ B:一奇二偶}；\\以直行來看:有ABB,BAB,BBA三種排法；以橫列來看:也是有三種排法；\\1-9共有5個奇數，4個偶數，因此有  4!\times 5!\times 3\times 3排法，機率為{4!\times 5!\times 3\times 3\over 9!} = \bbox[red,2pt]{1\over 14}$$

解：$$\int_{-1}^3\left( \sqrt{16-(x+1)^2}+2\right)dx =\int_{-1}^3\sqrt{16-(x+1)^2} dx+ \int_{-1}^3 2\;dx = 8+\int_{-1}^3\sqrt{16-(x+1)^2} dx\\ 而\int_{-1}^3\sqrt{16-(x+1)^2} dx= \int_0^4 \sqrt{16-u^2}\;du(u=x+1) \\=\int_0^{\pi/2} \sqrt{16-16\sin^2\theta}\cdot 4\cos \theta \;d\theta (\because u=4\sin\theta \Rightarrow du = 4\cos\theta d\theta)\\ =\int_0^{\pi/2}16\cos^2\theta d\theta =\int_0^{\pi/2}16\cdot {\cos 2\theta +1 \over 2}d\theta =\int_0^{\pi/2}16\cdot {\cos 2\theta +1 \over 2}d\theta = 8\int_0^{\pi/2} \left(\cos 2\theta +1  \right) d\theta \\ =8\left. \left[ {1\over 2}\sin 2\theta + \theta \right] \right|_0^{\pi/2} = 4\pi \Rightarrow \int_{-1}^3\left( \sqrt{16-(x+1)^2}+2\right)dx = \bbox[red,2pt]{4\pi + 8}$$

$$令\cases{O為\overline{BC}的中點\\ P為\overleftrightarrow{BA}與\overleftrightarrow{DC}的交點 \\ Q為 \overline{AD}的中點}，見上圖，則{\overline{PQ} \over \overline{PO}} ={\overline{QD}\over \overline{OC}} \Rightarrow {\overline{PQ} \over \overline{PQ} +6\sqrt 2} ={1\over 4} \Rightarrow \overline{PQ} =2\sqrt 6\\ 直角\triangle POC \Rightarrow \overline{PC}^2 =\overline{OP}^2 +\overline{OC}^2 = (6\sqrt 2+2\sqrt 2)^2+ 4^2 =144 \Rightarrow \overline{PC}=12\\ 又{\overline{PD} \over \overline{PC}} ={\overline{PQ} \over \overline{PO}}={1\over 4} \Rightarrow \overline{PD}=3 \Rightarrow \overline{AB}= \overline{CD} = 9-3=6$$

$$沿著\overline{PB}將角錐剪開，如上圖。弧長\stackrel{\frown}{AA'} = \overline{PA}\times \angle APA' \Rightarrow 2\pi=3 \angle APA' \Rightarrow \angle APA'={2\over 3}\pi\\ \triangle PBE': \cos \angle APA' = {\overline{PA}^2 +\overline{PE'}^2- \overline{BE}^2 \over 2\times \overline{PB}\times \overline{PE'}} \Rightarrow -{1\over 2}={12^2+6^2-\overline{BE}^2 \over 2\times 12\times 6}\\ \Rightarrow \overline{BE}=\bbox[red,2pt]{6\sqrt 7}$$

解：$$過(0,0)的直線方程式為\;y=mx，依題意兩圖形\cases{y=x^3+ kx^2+1 \\y=mx} 有相異三交點，\\ 即x^3+ kx^2+1=mx 有相異三實根；斜率m=y'=3x^2+2kx \Rightarrow x^3+ kx^2+1=(3x^2+2kx)x \\ \Rightarrow f(x)=2x^3+kx^2-1=0有相異三實根；f'(x)=0 \Rightarrow 6x^2+2kx=0 \Rightarrow 2x(3x+k)=0\\ \Rightarrow x=0,-k/3；f(x)=0有相異三實根 \Rightarrow f(0)f(-k/3) < 0 \Rightarrow (-1)(-{2\over 27}k^3+{1\over 9}k^3-1) \lt 0 \\ \Rightarrow {1\over 27}k^3-1 \gt 0 \Rightarrow \bbox[red, 2pt]{k \gt 3}$$

解：$$a_1,a_2,a_3成等比 \Rightarrow \cases{a_1=a_2/r\\ a_3=a_2r },r為公比；\\\log_6 a_1 + \log_6 a_2 +\log_6 a_3 = 3 \Rightarrow a_1a_2a_3= {a_2\over r}\cdot a_2 \cdot a_2r =a_2^3 = 6^3 \Rightarrow  a_2=6；\\又a_1+a_2= {a_2\over r}+a_2= 15 \Rightarrow {6\over r}+6=15 \Rightarrow r={6\over 9} \Rightarrow a_1= a_2/r= 9 \\\Rightarrow a_n=a_1r^{n-1} =9 \cdot ({6\over 9})^{n-1} \lt {1\over 10000} \Rightarrow ({6\over 9})^{n-1}\lt {1\over 9}\times 10^{-4} \Rightarrow \log({6\over 9})^{n-1}\lt \log({1\over 9}\times 10^{-4}) \\ \Rightarrow (n-1)(\log 2-\log 3) \lt -4-2\log 3 \Rightarrow n-1 \gt {-4-2\log 3\over \log 2-\log 3} ={-4-2\times 0.4771\over 0.301-0.4771} \approx 28.13 \\ \Rightarrow n \gt 29.13 \Rightarrow n=\bbox[red,2pt]{30}$$

解：

$$依題意:\cases{\sqrt x= \sqrt {y-1323} -\sqrt{z-1323} \cdots(1)\\ \sqrt y=\sqrt{x-675} +\sqrt{z-675}\cdots(2) \\ \sqrt z= \sqrt{y-3675}- \sqrt{x-3675}\cdots(3)}\\ 依此建構一鈍角\triangle ，三邊長分別為\sqrt x,\sqrt y,\sqrt z及其邊上的高分別為\sqrt{1323},\sqrt{675},\sqrt{3675}，見上圖；\\ \cases{直角\triangle AA'B:z=1323+\overline{A'B}^2 \Rightarrow \overline{A'B} =\sqrt{z-1323}\\ 直角\triangle CC'B:x= 3675+\overline{BC'}^2 \Rightarrow \overline{BC'} =\sqrt{x-3675} \\直角\triangle AA'C: y=1323+\overline{A'C}^2 \Rightarrow \overline{A'C} =\sqrt{y-1323} \\直角\triangle ACC': y = 3675 +\overline{AC'}^2 \Rightarrow \overline{AC'} =\sqrt{y-3675} \\直角\triangle ABB':z= 675+\overline{AB'}^2 \Rightarrow \overline{AB'} =\sqrt{z-675} \\直角\triangle BB'C:x =675 +\overline{B'C}^2 \Rightarrow \overline{B'C} =\sqrt{x-675}}\\ \Rightarrow \cases{\sqrt x= \overline{A'C} -\overline{A'B} =\sqrt{y-1323}-\sqrt{z-1323} \cdots(1)\\ \sqrt y=\overline{AB'} +\overline{B'C} =\sqrt{z-675} +\sqrt{x-675} \cdots(2)\\ \sqrt z=\overline{AC'} -\overline{BC'} = \sqrt{y-3675} -\sqrt{x-3675} \cdots(3)}\Rightarrow 該\triangle 符合題意要求。\\現在利用建構的\triangle 來求解:\\ \triangle ABC面積= {1\over 2} \cdot \sqrt x\cdot \sqrt{1323}= {1\over 2} \cdot \sqrt y\cdot \sqrt{675}= {1\over 2} \cdot \sqrt z\cdot \sqrt{3675} \\ \Rightarrow 21\sqrt 3\cdot \sqrt x= 15\sqrt 3\cdot \sqrt y = 35\sqrt 3\cdot \sqrt z \Rightarrow \sqrt x:\sqrt y: \sqrt z=5:7:3 \\ 令\cases{\sqrt x=5t\\ \sqrt y=7t\\ \sqrt z=3t}, t\in R 及S=(\sqrt x+\sqrt y+\sqrt z)\div 2 ={15\over 2}t \\ \Rightarrow \triangle ABC面積=\sqrt{s(s-\sqrt x)(s-\sqrt y)(s-\sqrt z)} ={1\over 2}\sqrt y\cdot \sqrt{675} \\ \Rightarrow \sqrt{{15\over 2}t\cdot {5\over 2}t \cdot {1\over 2}t \cdot {9\over 2}t} ={7\sqrt{675}\over 2}t \Rightarrow t=14 \Rightarrow \sqrt z=3t=42 \Rightarrow z=42^2= \bbox[red,2pt]{1764}$$

解：$$由題意可知：\cases{f(b)=3b\\ f(a)=3a}\cdots(1) 或\cases{f(b)=3a\\ f(a)=3b}\cdots(2)\\ (1) \Rightarrow -{1\over 3}a^2+{73\over 3}=3a \Rightarrow a^2+9a-73=0 \Rightarrow a \not \in Z\\ (2) \Rightarrow \cases{-{1\over 3}a^2+{73\over 3}=3b \cdots(3)\\ -{1\over 3}b^2+{73\over 3}=3a \cdots(4)}，(3)-(4) \Rightarrow -{1\over 3}(a^2-b^2)=3(b-a) \Rightarrow a+b=9 \\ \Rightarrow b=9-a 代入(3) \Rightarrow -{1\over 3}a^2+{73\over 3}=3(9-a) \Rightarrow a^2-9a+ 8=0 \Rightarrow (a-8)(a-1)=0 \\\Rightarrow \cases{a=1\\ a=8} \Rightarrow \cases{b=8\\ b=1 (不合,違反b\gt a)} \Rightarrow (a,b)=\bbox[red,2pt]{(1,8)}$$

解：$$f(1)+f(2)+\cdots +f(9)=1+3+5+7+9 =25\\\cases{f(10)+f(11)+\cdots +f(19)=1\times 10(十位數)+ 25(個位數)\\ f(20)+f(21)+\cdots +f(29)=0(十位數)+ 25(個位數)\\ \cdots \\f(90)+f(91)+\cdots +f(99)=9\times 10(十位數)+ 25(個位數) } \\\Rightarrow f(10)+f(11)+\cdots +f(99)=25\times 10 (十位數)+25\times 9(個位數)=475\\ 同理，f(100)+f(101)+\cdots +f(999)\\= 25\times 100 (百位數)+ 25\times 90(十位數) + 25\times 90(個位數)=7000;\\ 又f(1000)=1，因此f(1)+\cdots +f(1000)= 25+475+7000+1 = \bbox[red,2pt]{7501}$$

解：$$\bar x= (1+2 +5+3+4)\div 5=3 \Rightarrow 迴歸直線L:y={3\over 5}(x-\bar x)+\bar y ={3\over 5}(x-3)+\bar y ={3\over 5}x+ \bar y-{9\over 5} \\ \Rightarrow \bar y-{9\over 5}={11\over 5} \Rightarrow \bar y=4 \Rightarrow (3+a+5+ b+6)\div 5 =4 \Rightarrow a+b=6 \cdots(1)\\ 迴歸直線斜率{3 \over 5}= {\sum x_iy_i -n\bar x\bar y\over \sqrt{\sum x_i^2-n\bar x^2} \cdot \sqrt{\sum y_i^2-n\bar y^2}} ={3+2a +25+3b+24-5\times 3\times 4 \over \sqrt{55-45} \cdot \sqrt{a^2+b^2+70-80}}\\ = {2a+3b-8 \over \sqrt {10}\cdot \sqrt{a^2+b^2-10}} \Rightarrow (3\cdot \sqrt{10}\cdot \sqrt{a^2+b^2-10})^2 =(5 \cdot (2a+3b-8))^2\\ \Rightarrow 90(a^2+b^2-10)=25(2a+3b-8)^2 \Rightarrow 18(a^2+(6-a)^2-10) =5(2a+3(6-a)-8)^2 \\ \Rightarrow 31a^2-116a-32=0 \Rightarrow (a-4)(31a+8)=0 \Rightarrow a=4 \Rightarrow b=6-4=2\\ \Rightarrow (a,b)=\bbox[red,2pt]{(4,2)}$$