110年臺中一中教甄-數學詳解

 臺中市立臺中第一高級中等學校110學年度第1次教師甄選

壹、填充題第一部分

解答：$$\cases{2^k \cdot 4^m \cdot 8^n =512\\ 4^p \cdot 3^q \cdot 6^r=2^{11}\cdot 6^{16}} \Rightarrow \cases{2^{k+2m +3n} =2^9 \\ 2^{2p+r} \cdot 3^{q+r} =2^{27} \cdot 3^{16}} \Rightarrow \cases{k+2m +3n=9 \\ 2p+r=27\\ q+r=16} \\ \Rightarrow \cases{(k,m,n) =(4,1,1), (2,2,1),(1,1,2)\\ (p,q,r)=(14-k,17-2k,2k-1), k=1-8} \Rightarrow\cases{a=3\\ b=8} \\ \Rightarrow (a,b) =\bbox[red, 2pt]{(3,8)}$$

解答：$$A^{-1}B=C \Rightarrow B=AC \Rightarrow \begin{bmatrix} 1 & 4\\-1 & 2\end{bmatrix} =\begin{bmatrix} x & y\\ z & u\end{bmatrix} \begin{bmatrix} 4 & -26 \\-3 & 18\end{bmatrix} =\begin{bmatrix} 4x-3y & -26x+18y \\4z-3u & -26z +18u \end{bmatrix} \\ \Rightarrow \cases{4x-3y=1\\ -26x+18y=4\\ 4z-3u=-1\\ -26z+18u = 2} \Rightarrow (x,y,z, u)=\bbox[red, 2pt]{(-5,-7,2,3)}$$

解答：$$\log_4 x-\log_x 8+2=0 \Rightarrow {1\over 2}\log_2 x -{3\over \log_2 x}+2=0 \Rightarrow (\log_2 x)^2 + 4(\log_2 x)-6=0\\ \Rightarrow \log_2 x= -2+\sqrt{10} (-2-\sqrt{10}不合,\because x\gt 1 \Rightarrow \log_2 x \gt 0) \\ \Rightarrow 2(\log_2 x)^3 +9(\log_2 x)^2-7(\log_2 x)-3 = 2(-2+\sqrt{10})^3 +9(-2+\sqrt{10})^2-7(-2+\sqrt{10})-3\\ =2(-68+22 \sqrt{10})+9(14-4\sqrt{10}) -7(-2+\sqrt{10})-3 = \bbox[red,2pt]{1+\sqrt{10}}$$

解答：

$$點O為\overline{BC}的中點，並令O為坐標原點，則\cases{B(-3,0,0)\\ C(3,0,0)\\ A(0,3\sqrt 3,0)},又\overline{CD} =\overline{BD}，D在平面x=0上，即D(0,y,z)\\ 四面體體積={1\over 3}\triangle ABC \cdot z=18 \sqrt 3 \Rightarrow {1\over 3}\cdot 9\sqrt 3 \cdot z=18 \Rightarrow z=6;\\此外，平面ABC與平面DBC夾角60^\circ \Rightarrow \tan 60^\circ ={z\over y} \Rightarrow y={\sqrt 3\over 3}z= 2\sqrt 3 \Rightarrow D(0,2\sqrt 3,6)\\ \Rightarrow \overline{AD} = \sqrt{(\sqrt 3)^2 +6^2} = \bbox[red, 2pt]{\sqrt{39}}$$

解答：$$\cases{P在x軸移動\\ Q在y軸移動} \quad \Rightarrow \cases{P(a,0)\\ Q(0,b)}，又\cases{\overline{PQ}=10 \\ \overline{PR}: \overline{RQ}=3:2} \Rightarrow \cases{a^2+b^2=10^2\\ R={2\over 5}P+ {3\over 5}Q} \Rightarrow \cases{b^2=100-a^2\\ R({2\over 5}a, {3\over 5}b) \equiv R(x,y)} \\ \Rightarrow {25\over 9}y^2 =100-{25\over 4}x^2 \Rightarrow {x^2\over 16} +{y^2 \over 36}=1 \Rightarrow R的軌跡為一橢圓 \Rightarrow R(4\cos \theta,6\sin \theta)\\ \Rightarrow 令d=R至(2,0)的距離= \sqrt{(4\cos\theta-2)^2+36\sin^2\theta} = \sqrt{20\sin^2\theta -16\cos \theta +20} \\ =\sqrt{-20\cos^2 \theta-16\cos \theta +40} \Rightarrow \cases{\cos\theta =-2/5 \Rightarrow M=6\sqrt{30}/5\\ \cos \theta =1 \Rightarrow m=2} \\ \Rightarrow (M,m)=\bbox[red, 2pt]{({6\sqrt{30}\over 5},2)}$$

貳、填充題第二部分

解答：$$\langle a_n\rangle =x,y,x+y, x+2y,2x+3y,3x+5y,5x+8y, 8x+13y,13x+21y, 21x+34y, 34x+55y,...\\ 由於x,y\in \mathbb{N}, 從右向左試，找第一個a_n=115，可得13x+21y=115 \Rightarrow \cases{x=4\\y=3} \\ \Rightarrow x+y =\bbox[red, 2pt]{7}$$

解答：$$\cases{\lim_{x\to 1}{f(x)\over x-1}=36 \\\lim_{x\to -1}{f(x)\over x+1}= -36 \\\lim_{x\to 2}{f(x)\over x-2}=0 \\\lim_{x\to -2}{f(x)\over x+2}=0} \Rightarrow \cases{f(x)=p(x)(x-1) 且p(1)=36 \\f(x)=q(x)(x+1) 且q(-1)= -36 \\f(x)=r(x)(x-2) 且r(2)=0 \\f(x)= s(x)(x+2)且s(-2)=0 } \\ \Rightarrow f(x)=(ax+b)(x-1)(x+1)(x-2)^2(x+2)^2代回\cases{\lim_{x\to 1}{f(x)\over x-1}=36 \\\lim_{x\to -1}{f(x)\over x+1}= -36   } \\ \Rightarrow \cases{(a+b)\cdot 2\cdot (-1)^2 \cdot 3^2=36\\ (-a+b)\cdot (-2)\cdot (-3)^2 \cdot 1^2=-36} \Rightarrow \cases{a+b= 2\\ -a+b=2} \Rightarrow \cases{a=0\\ b=2} \\ \Rightarrow f(x)=2(x^2-1)(x-2)^2(x+2)^2 \Rightarrow f(3)=2 \cdot 8\cdot 1\cdot 25= \bbox[red, 2pt]{400}$$

解答：$$\cases{E(X)=1 \\ E(X^2)=3 \\ E(Y)=2 \\ E(Y^2)=5 \\ E(XY)=3} \Rightarrow \cases{Var(X)= E(X^2)-(E(X))^2 =3-1=2\\ Var(Y)= E(Y^2)-(E(Y))^2 =5-4=1 \\Cov(X,Y) = E(XY)-E(X)E(Y) =3-2=1} \\ \Rightarrow Var(3X-2Y+7) = 9Var(X)+4Var(Y)-12Cov(X,Y) =18+4-12= \bbox[red, 2pt]{10}$$

解答：$$假設f(x)=x^3+a_2x^2 +a_1x +a_0，且f(x)=0的三根為b,c,d；\\由於abcd=-5 \Rightarrow bcd=-5/a \Rightarrow a_0=5/a \Rightarrow f(x)=x^3+a_2x^2 +a_1x +{5\over a}\\ 再由f(x)=0的三根為b,c,d \Rightarrow \cases{f(x+1)=0的三根為b-1,c-1,d-1 \\f(x+2)=0的三根為b-2,c-2,d-2 \\f(x+3)=0的三根為b-3,c-3,d-3 } \\ \Rightarrow \cases{f(x+1)的常數項= 1+a_2+a_1+5/a= -(b-1)(c-1)(d-1)= -11/a \\f(x+2)的常數項=  8+4a_2 +2a_1+5/a= -(b-2)(c-2)(d-2) = -33/a\\ f(x+3)的常數項= 27+9a_2 +3a_1+5/a= -(b-3)(c-3)(d-3)= -73/a} \\ \Rightarrow \cases{a_1=17/2\\ a_2= -3/2 \\ a=-2} \Rightarrow f(x)=x^3-{3\over 2} x^2+{17\over 2}x- {5\over 2} \Rightarrow f(x-1)=0的三根為b+1,c+1,d+1\\ \Rightarrow (b+1)(c+1)(d+1)= (-1)\times f(x-1)的常數項 =(-1) \times (-1-{3\over 2}-{17\over 2}-{5\over 2}) ={27\over 2} \\ \Rightarrow a(b+1)(c+1)(d+1)= -2\times{27\over 2}= \bbox[red, 2pt]{-27}$$

解答：$$三次函數f(x)=ax^3+ bx^2 +cx +d 的對稱中心點為(-{b\over 3a},f(-{b\over 3a}))，\\因此\cases{-{b\over 3a}=1 \Rightarrow b=-3a \cdots(1)\\ f(-{b\over 3a})=f(1)= a+b+c+d= 2 \cdots(2)} ；又f'(2)= -4 \Rightarrow 12a+4b+c = -4\cdots(3)\\ 將(1)代入(3)\Rightarrow c=-4代入(2) \Rightarrow a-3a-4+d=2 \Rightarrow d=2a+6 \\ \Rightarrow k=\lvert{b^2+c^2 +d^2 \over a} \rvert = \lvert{9a^2+16 +4a^2+24a+36 \over a} \rvert = \lvert{13a^2+ 24a+52 \over a} \rvert =  \lvert 13a+ 24+{52 \over a} \rvert \\ 若a\gt 0,則 13a + {52\over a} \ge 2\sqrt{13a \cdot {52\over a}} = 2\cdot 26=52;若a \lt 0，則13a + {52\over a} \ge -52;\\ 因此k的最小值發生在 13a+  {52 \over a} =-52，此時k=|-52+24|=|-28|= \bbox[red, 2pt]{28}$$

解答：$$ 甲乙皆在A箱抽中1號球的情形:甲抽中1號球的機率{2\over 6},乙隨後抽中1號球的機率{1\over 5}\\，因此機率為{2\over 6}\times {1\over 5}={1\over 15}；甲乙皆在A箱抽中2號球或3號球的機率都是{1\over 15}\\，也就是甲乙在A箱抽中同號球的機率為{1\over 15}\times 3={1\over 5}；\\ 甲乙兩人在三箱皆抽同號的機率就是({1\over 5})^3 ={1\over 125}\\，因此甲乙兩人取得不同的三位數機率為1-{1\over 125}={124\over 125}，而一半是甲大，另一半是乙大\\，因此乙大的機率為{124\over 125}\times {1\over 2}= \bbox[red, 2pt]{62\over 125}$$

解答：$$假設\cases{a=2^x\\ b=3^x}，則6^{x+1}-3\cdot 8^x+ 2\cdot 27^x-36^x = 6ab-3a^3+2b^3-a^2b^2 \\ =3a(2b-a^2) +b^2(2b-a^2) = (3a+b^2)(2b-a^2) = 0 \Rightarrow a^2=2b \Rightarrow 2^{2x}=2\cdot 3^x \\ \Rightarrow 2x\log 2=\log 2+ x\log 3 \Rightarrow x(2\log 2-\log 3)=\log 2 \Rightarrow x={\log 2\over 2\log 2-\log 3} \\ \Rightarrow {x\over 2x-1} ={\log 2/(2\log 2-\log 3) \over \log 3/(2\log 2-\log 3)} ={\log 2\over \log 3} =\bbox[red, 2pt]{\log_3 2}$$

解答：

$$移動矩形ABCD，將A視為原點，\overline{AC}在x軸上，見上圖；\\ \overline{AC}= \sqrt{\overline{AD}^2 +\overline{DC}^2 } = 5，又\overline{AD} \times \overline{DC}=\overline{AC}\times \overline{DE} \Rightarrow \overline{DE}=12/5；\\同理,直角\triangle CDE \Rightarrow \overline{EC}=9/5；\\矩形ABCD在x-y平面上，D繞x軸旋轉\theta ，因此D({16 \over 5},-{12\over 5}\cos \theta,{12\over 5}\sin \theta) \\ \Rightarrow \overline{BD}=\sqrt{({7\over 5})^2 +({12\over 5})^2(1-\cos \theta)^2 +({12\over 5})^2\sin^2 \theta} =\sqrt{{49\over 25}+{144\over 25} (2-2\cos \theta) }\\ = \bbox[red, 2pt]{\sqrt{337-288\cos \theta}\over 5}$$

參、計算證明題(第一題 9 分，第二題 10 分，合計 19 分)

解答：

$$通過A(-6,9)與L垂直的直線L':4x-3y=-51，假設通過B且與L'平行的直線L'':4x-3y=k\\ 正射長=12 \Rightarrow d(L',L'')=12 \Rightarrow k=9 \Rightarrow L'':4x-3y=9 \Rightarrow 假設B(3a,4a-3)\\ \Rightarrow (3a-1)^2+(4a-7)^2=50 \Rightarrow 25a^2-62a=0 \Rightarrow \cases{a=0\\ a=62/25} \Rightarrow \cases{B(0,-3)\\ B(62/25,173/25)} \\ \Rightarrow \cases{\overrightarrow{AB}=(6,-12) \\ \overrightarrow{AB}=({212\over 25}, -{52\over 25})} \Rightarrow 顯然B=(0,-3)有較大的|\overrightarrow{AB}| =\sqrt{180}=\bbox[red, 2pt]{6\sqrt 5}$$

解答：$$\mathbf{(1)}\; f(x)=x^4+x^3+x^2+x+1 = (x-\alpha) (x-\beta) (x-\gamma) (x-\phi) \\ \Rightarrow f'(x)=4x^3+3x^2+2x+1 \\=  (x-\beta) (x-\gamma) (x-\phi) +(x-\alpha)  (x-\gamma) (x-\phi) +(x-\alpha) (x-\beta)  (x-\phi)  +(x-\alpha) (x-\beta) (x-\gamma)  \\ 因此{1\over 1-\alpha}+{1\over 1-\beta} +{1\over 1-\gamma}+{1\over 1-\phi} \\={ (1-\beta)(1-\gamma )(1-\phi)+ (1-\alpha) (1-\gamma )(1-\phi) +(1-\alpha) (1-\beta)(1-\phi)+ (1-\alpha) (1-\beta)(1-\gamma )　\over (1-\alpha) (1-\beta)(1-\gamma )(1-\phi)}\\ ={f'(1)\over f(1)} ={10\over 5} =\bbox[red, 2pt]2 \\\mathbf{(2)}\; \overline{AP} \times \overline{AQ} \times \overline{AR} \times \overline{AS} =|1+i-\alpha| \times|1+i-\beta| \times|1+i-\gamma| \times|1+i-\phi| \\ =|(1+i-\alpha) \times(1+i-\beta) \times(1+i-\gamma) \times(1+i-\phi)| =|f(1+i)| \\=|-4+(2i-2)+ 2i+(1+i)+1|=|5i-4|= \bbox[red, 2pt]{\sqrt{41}}$$

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