115年嘉科實中教甄-數學詳解

國立嘉科實驗高級中學115學年度高中部教師甄選

一、 填充題（每題 7 分，共 56 分）

解答:$$\sum_{k=1}^n k(k+1) = \sum_{k=1}^n( k^2+k) ={n(n+1)(2n+1) \over 6}+{n(n+1)\over 2} ={n(n+1)(n+2) \over 3} \\ \Rightarrow {1\over 1\times 2}+ {1\over 1\times 2+2\times 3}+ \cdots+{1\over 1\times 2+ 2\times 3+ \cdots+n(n+1)}+\cdots \\= \lim_{n\to \infty}\sum_{m=1}^n {1\over \sum_{k=1}^m k(k+1)} = \lim_{n\to \infty}\sum_{m=1}^n{3\over m(m+1)(m+2)}\\ = {3\over 2}\lim_{n\to \infty} \sum_{m=1}^n  \left( {1\over m(m+1)} -{1\over (m+1)(m+2)} \right) ={3\over 2}\cdot {1\over 1\cdot 2} = \bbox[red, 2pt]{3\over 4}$$

解答:$$A= \sum_{n=0}^6 C^6_n \cdot 69^{6-n}= (1+69)^6= 70^6 =(2\cdot 5\cdot 7)^6 =2^6\cdot 5^6 \cdot 7^6 \\ \Rightarrow A之正因數個數=(6+1) \cdot (6+1) \cdot (6+1) =7^3= \bbox[red, 2pt]{343}$$

解答:$$取\cases{\sqrt{2025}=m\\ \sqrt{2026}=n} \Rightarrow \cases{m^2=2025\\ n^2=2026} \Rightarrow 2025a^2=2026b^2 \Rightarrow m^2a^2=n^2b^2 \Rightarrow a={nb\over m} \\ \Rightarrow {1\over nb/m}+{1\over b}=1 \Rightarrow b={m+n\over n} \Rightarrow a={n\over m}\cdot {m+n\over n} ={m+n\over m} \\ \Rightarrow \sqrt{2025a+2026b} = \sqrt{m^2\cdot {m+n\over m}+n^2\cdot {m+n\over n}} =\sqrt{(m+n)^2}= m+n=\bbox[red, 2pt]{ \sqrt{2025}+ \sqrt{2026}}$$

解答:$$取{a\over x+3^2}+{b\over x+5^2}+{c\over x+7^2} =1 \\\Rightarrow a(x+25)( x+49)+ b(x+9)(x+49)+ c(x+9)(x+25) =(x+9)(x+25)(x+49)\\ 令f(x)=(x+9)(x+25)(x+49)- [a(x+25)( x+49)+ b(x+9)(x+49)+ c(x+9)(x+25)] \\ \Rightarrow f(2^2)=f(4^2)=f( 6^2)=0 \Rightarrow f(x)=(x-4)(x-16)(x-36) \Rightarrow x^2係數=-56 \\ \Rightarrow (x+9)(x+25)(x+49)- [a(x+25)( x+49)+ b(x+9)(x+49)+ c(x+9)(x+25)] 的x^2係數\\=83-(a+b+c) \Rightarrow 83-(a+b+c)=-56 \Rightarrow a+b+c=83+56= \bbox[red, 2pt]{139{}}$$

解答:$$N=2026 \Rightarrow P(X=x) ={{x-1\choose k-1} \over {N\choose k}} \Rightarrow E(X) = \sum_{x=k}^N x\cdot {{x-1\choose k-1} \over {N\choose k}} ={k\over {N\choose k}} \sum_{x=k}^N {x\choose k} ={k\over {N\choose k}} {N+1\choose k+1} \\=k\cdot {k!(N-k)!\over N!} \cdot {(N+1)!\over (k+1)! (N-k)!} ={k(N+1) \over k+1} ={2027k\over k+1} \ge 2000 \Rightarrow k\ge {2000\over 27} \approx74.1 \\ \Rightarrow k= \bbox[red, 2pt]{75}$$

解答:$$f(x)=0的四根為z_i, i=1-4 \Rightarrow 四根乘積=z_1 \cdot z_2\cdot z_3\cdot  z_4=1 \Rightarrow |z_1| \cdot |z_2|\cdot |z_3|\cdot  |z_4|=1 \\依題意|z_i|\le 1 \Rightarrow |z_1| = |z_2|= |z_3| =  |z_4|=1 \Rightarrow 四根可寫成e^{\pm i\theta}, e^{\pm i\phi} \\ \Rightarrow f(x) =(x-e^{i\theta})(x-e^{-i\theta}) (x-e^{i\phi})(x-e^{-i\phi}) =(x^2-2\cos \theta x+1)(x^2-2\cos \phi x+1) \\=x^4-2(\cos \theta +\cos \phi) x^3+ (2+ 4\cos \theta \cos \phi)x^2-2(\cos \theta+ \cos \phi) x+1 \\ \Rightarrow \cases{a=-2(\cos \theta+\cos \phi) \\b=2+4\cos \theta \cos \phi\\ c=-2(\cos \theta+\cos \phi)} \Rightarrow a^2+c^2-8b =8(\cos \theta+ \cos \phi)^2-8(2+4\cos \theta\cos \phi) \\= 8(\cos \theta-\cos \phi)^2-16 \Rightarrow 最大值=8(1-(-1))^2-16=32-16= \bbox[red, 2pt]{16}$$

解答:$$\textbf{Case I: AAAA}(四個數字都相同) \Rightarrow 乘積為a^4=(a^2)^2為完全平方數，共6個\\ \textbf{Case II: AABB}(兩對相同的數字) \Rightarrow 乘積為a^2b^2=(ab)^2為完全平方數，共C^6_2\cdot C^4_2=90個\\ \textbf{Case III: AAAB}(三個數字相同、一個不同) \Rightarrow 乘積為a^3b=a^2(ab) \Rightarrow ab需為完全平方數\\ \qquad 只有(1,1,1,4)或(4,4,4,1)，共2\times {4!\over 3!}=8個\\ \textbf{Case IV: AABC}(一對相同、兩個不同) \Rightarrow 乘積為a^2bc \Rightarrow bc為完全平方數 \Rightarrow \cases{b,c\in \{1,4\} \\a\in \{2,3,5,6\}} \\\qquad 共C^4_1\cdot {4!\over 2!}=48個\\ \textbf{Case V:ABCD} (四數字均不同),只能是(2,3,4,6)或(1,2,3,6)兩種，共有2\cdot 4!=48個 \\ 總共有6+90+8+48+48= 200 \Rightarrow 機率={200\over 6^4} ={200\over 1296} = \bbox[red, 2pt]{25\over 162}$$

解答:$$\cases{u=3x+2y\\ v=3x-2y}\Rightarrow \left|{\partial(u,v) \over \partial(x,y)} \right| = \begin{Vmatrix} 3&2\\ 3&-2 \end{Vmatrix} =12 \\\Rightarrow |u-2|+ |v+5|+|u+2|+|v-4|\le 14 為一矩形，面積為4\times 10=40 \\\Rightarrow 原式所圍面積=40\times {1\over 12} = \bbox[red, 2pt]{10\over 3}$$

二、 計算證明題（每題 10 分，共 20 分）

解答:

$$假設\cases{\overline{AC}=b\\ \overline{AB}=c} \Rightarrow \cases{\overline{BC} =b+c/2\\ \overline{PA}=3c/4\\ \overline{PB}=c/4} \Rightarrow \triangle ABC:\cos A={b^2+c^2-(b+c/2)^2\over 2bc} ={3c-4b\over 8b} \\ \Rightarrow \triangle APC: \cos A= {b^2+(3c/4)^2-\overline{PC}^2 \over 3bc/2} \Rightarrow \overline{PC}^2 =b^2+{9\over 16}c^2-{3\over 2}bc \cos A\\ =b^2+{9\over 16}c^2-{3\over 2}bc \cdot {3c-4b\over 8b} =b^2+{3\over 4}bc \\ \cos \angle CPA= {\overline{PC}^2+ (3c/4)^2-b^2 \over (3c/2)\cdot \overline{PC}} ={3c+4b\over 8\overline{PC}} \Rightarrow \cos 2\angle CPA=2\cos^2 \angle CPA-1 \\=2\cdot {(3c+4b)^2\over 64\overline{PC}^2}-1 =2\cdot {(3c+4b)^2\over 64(b^2+3bc/4)^2}-1 ={3c-4b\over 8b}= \cos A\\ \Rightarrow \cos 2\angle CPA= \cos A加上\cos \angle CPA ={3c+4b\over 8\overline{PC}} \gt 0 \Rightarrow \angle CPA是銳角 \\ \Rightarrow \angle A=2\angle CPA \;\bbox[red, 2pt]{故得證}$$

 

解答:$$利用柯西不等式(\text{Titu's Lemma}): {a_1^2\over b_1} +{a_2^2 \over b_2} +{a_3^2\over b_3} \ge {(a_1+a_2 +a_3)^2 \over b_1+b_2+b_2}\\ 因此{x^3\over 2+x}+ {y^3\over 2+y}+ {z^3\over 2+z} ={(x^2)^2\over 2x+x^2}+ {(y^2)^2\over 2y+y^2}+ {(z^2)^2\over 2z+z^2} \ge{(x^2+y^2+z^2)^2 \over x^2+y^2+z^2+2(x+y+z)} \\={9\over 3+2(x+y+z)}  \Rightarrow {x^3\over 2+x}+ {y^3\over 2+y}+ {z^3\over 2+z} \ge {9\over 3+2(x+y+z)}\\ 再一次柯西不等式: (x^2+y^2+z^2) (1^2+1^2+1^2) \ge (x+y+z)^2 \Rightarrow 9\ge (x+y+z)^2 \\ \Rightarrow x+y+z\le 3 \Rightarrow {9\over 3+2(x+y+z)} \ge {9\over 3+2\cdot 3} =1 \Rightarrow {x^3\over 2+x}+ {y^3\over 2+y}+ {z^3\over 2+z}\ge 1\; \bbox[red, 2pt]{故得證}$$

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解題僅供參考，其他教甄試題及詳解