115年新竹新科國中教甄-數學詳解

 新竹市立新科國民中學 115 學年度正式教師甄選

一、填充題 (共 25 題)

解答:$$假設\cases{B(0,0) \\A(0,3) \\C(4,0)\\D(4,3) }\Rightarrow P(4t,3t) \Rightarrow \cases{E(0,3t) \\F(4t,3)} \Rightarrow \overline{EF}^2=16t^2+(3t-3)^2=25t^2-18t+9 \\ \Rightarrow t={18\over 50}時,\overline{EF}有最小值=\sqrt{3600\over 625} ={60\over 25} = \bbox[red, 2pt]{12\over 5}$$

解答:$$\overline{DF} \parallel \overline{BG} \Rightarrow \overline{AF}:\overline{FG} = \overline{AD} : \overline{DB} = \triangle ADF: \triangle BDF= 20:30=2:3 \Rightarrow \cases{\overline{AF}=2k\\\ \overline{FG} =3k} \\ \Rightarrow \triangle ABF: \triangle BFG= \overline{AF}: \overline{FG}=2:3 \Rightarrow \triangle BFG={3\over 2}\cdot (20+30)=75 \\ \Rightarrow \overline{BE} : \overline{EC} =\overline{FG}: \overline{GC}= \triangle BFG: \triangle BCG=75:60= \bbox[red, 2pt]{5:4}$$

解答:$$假設\cases{A的x坐標為x_A\\ C的x坐標為x_C}，圖形對稱軸為x=-7 \Rightarrow {x_A+ x_C\over 2}=-7 \Rightarrow x_A+ x_C=-14 \\ \cases{\overline{AB}=-x_A\\ \overline{BC}=x_C} \Rightarrow -x_A:x_C=5:1 \Rightarrow 5x_C=-x_A \Rightarrow -5x_C+x_C=-14 \Rightarrow x_C={7\over 2} \\ \Rightarrow x_A=-14-{7\over 2}=-{35\over 2} \Rightarrow \overline{AC}=x_C-x_A={7\over 2}+{35\over 2}= \bbox[red, 2pt]{21}$$

解答:$$k=2026x \Rightarrow \cases{a=k-1\\ b=k\\ c=k+1} \Rightarrow \cases{a+b+c=3k\\ ab+bc+ca=3k^2-1} \\ \Rightarrow a^2+b^2+c^2-ab-bc -ca=(a+b+c)^2-3(ab+bc +ca) =9k^2-3(3k^2-1) =\bbox[red, 2pt] 3$$

解答:$${(2026^2-2025^2)+(2024^2-2023^2)+ \cdots+(4^2-3^2)+(2^2-1^2) \over 2027} \\= {4051+4047+\cdots+7+3 \over 2027} ={(4051+3)\cdot 1013\over 2\cdot 2027} =\bbox[red, 2pt]{1013}$$

解答:$$\cases{a+b=2026\\ ab=1 \\ c+d=2025\\ cd=1} \Rightarrow (a+c)(b+c)(a-d)(b-d)= (ab+c(a+b)+c^2) (ab-d(a+b)+d^2) \\=(1+2026c+c^2)(1-2026d+d^2) \\=1-2026d+d^2+2026c-2026^2cd+ 2026cd^2+ c^2-2026c^2d+ c^2d^2 \\=1-2026d+d^2+2026c-2026^2+2026d+c^2-2026c+1 \\ =2+c^2+d^2-2026^2=2+(c+d)^2-2cd-2026^2=2+2025^2-2-2026^2= \bbox[red, 2pt]{-4051}$$

解答:$$義勇選到A車廂的機率為{1\over 8}，伊之助選到A-1或A+1車廂的機率為{1\over 4} \\\Rightarrow 兩人車廂相鄰的機率={1\over 8}\times{1\over 4}={1\over 32}, \\又義勇有8個車廂可供選擇，因此機率為{1\over 32}\times 8= \bbox[red, 2pt]{1\over 4}$$

解答:

$$假設\cases{S_1的邊長為a\\ S_3的邊長為b} \Rightarrow \cases{S_2的上側邊長=3366-(a+b)\\ S_2的左側邊長=a+b-2026} \Rightarrow 3366-(a+b)=a+b-2026 \\ \Rightarrow a+b={1\over 2}(3366+2026)=2696 \Rightarrow S_2邊長=3366-2696= \bbox[red, 2pt]{670}$$

解答:$$假設\cases{甲材料買了a包\\ 乙材料買了b包} \Rightarrow 在3a+5b=58的條件下，求180a+260b的最小值\\ 3a+5b=58 \Rightarrow b={58-3a\over 5}為整數  \Rightarrow f(a)=180a+260\cdot {58-3a\over 5} =24a+3016 \\ \Rightarrow  a=1 \Rightarrow b=11 \Rightarrow  f(a)= \bbox[red, 2pt]{3040}$$

解答:

$$\cases{正五邊形內角=3\cdot 180/5=108^\circ\\ 正三角形內角=60^\circ} \Rightarrow \angle ABF=60^\circ+108^\circ=168^\circ \\ 又\overline{AB} =\overline{BC} =\overline{BF} \Rightarrow \angle BAF=\angle BFA={180^\circ-168^\circ\over 2}=6^\circ \Rightarrow \angle AFE=108^\circ-6^\circ=102^\circ\\ A,F,E,P共圓\Rightarrow \angle APE= 180^\circ-\angle AFE= \bbox[red, 2pt]{78^\circ}$$

解答:$$B(4,3)對稱x軸的對稱點B'(4,-3) \Rightarrow 直線\overline{AB'}:4x+3y=7 與x軸交於(7/4,0)\\ \Rightarrow P=({7\over 4},0) \Rightarrow x坐標為\bbox[red, 2pt]{7\over 4}$$

解答:

$$取\cases{P= \overleftrightarrow{AF} \cap \overleftrightarrow{BC} \\Q= \overleftrightarrow{DE} \cap \overleftrightarrow{BC} \\R= \overleftrightarrow{AF} \cap \overleftrightarrow{DE} } \Rightarrow \cases{\triangle ABP為正三角形且邊長為2 \\\triangle CDQ為正三角形且邊長為3 \\ \triangle EFR為正三角形 } \\\Rightarrow \triangle PQR為正三角形且邊長=\overline{PB} +\overline{BC}+ \overline{CQ}=2+4+3=9 \Rightarrow \overline{ER}=9-3-4=2 \\ \Rightarrow \cases{\triangle ABP面積= {\sqrt 3\over 4}2^2 = \sqrt 3 \\\triangle CDQ面積= {\sqrt 3\over 4}3^2 = 9\sqrt 3/4 \\\triangle EFR面積= {\sqrt 3\over 4}2^2 = \sqrt 3 \\\triangle PQR面積= {\sqrt 3\over 4}9^2 = 81\sqrt 3/4} \Rightarrow 六邊形面積= 大三角形-3個小三角形\\ ={81\over 4}\sqrt 3- \left( \sqrt 3+{9\over 4}\sqrt 3+ \sqrt 3 \right) = \bbox[red, 2pt]{16\sqrt 3}$$

解答:$$兩邊之和大於第三邊 \Rightarrow \cases{10+15\gt k\\ 10+k\gt 15} \Rightarrow 5\lt k\lt 25\\ \textbf{Case I }:15是最長邊 \Rightarrow 10^2+k^2 \lt 15^2 \Rightarrow k^2\lt 125 \Rightarrow k\le 11 \Rightarrow 5\lt k\le 11\\ \qquad \Rightarrow k=6,7,8, 9,10,11，共6個\\ \textbf{Case II }:k是最長邊\Rightarrow 10^2 +15^2\lt k^2 \Rightarrow 325\lt k^2 \Rightarrow k\ge 19 \Rightarrow 19\le k\lt 25\\ \qquad \Rightarrow k=19,20,21,22, 23, 24，共6個 \\ 合計:6+6=\bbox[red, 2pt]{12}個$$

解答:$$九個數字皆是2的次方,即2^a, 其中a_i=2,3,4,5,6,7,8,9,10 \Rightarrow \sum a_i=54 \\ \Rightarrow 任一行、列或對角線相乘的數字為2的{54\over 3}=18 次方\Rightarrow 中心點B=2^{18/3} =64\\ 用次方的角度看九宮格 \Rightarrow \begin{array}{|c|c|c|}\hline 7&a&p\\\hline q & 6& c\\\hline r& 10& s\\\hline \end{array} \Rightarrow a+6+10=18 \Rightarrow a=2 \\ \Rightarrow p=18-7-a=9 \Rightarrow r=3\Rightarrow s=5 \Rightarrow c=18-p-s=4 \\ \Rightarrow \cases{A=2^2\\ B=2^6\\ C=2^4} \Rightarrow A+B+C =4+64+16= \bbox[red, 2pt]{84}$$

解答:$$假設三角形三邊長為a,b,c，則三角形面積A={1\over 2}\cdot a\cdot 3= {1\over 2}\cdot b\cdot 4= {1\over 2}\cdot c\cdot x\\ \Rightarrow \cases{a=2A/3\\ b=A/2\\ c=2A/x} \Rightarrow |a-b|\lt c\lt a+b\Rightarrow  \left| {2A\over 3}-{A\over 2}\right|\lt {2A\over x} \lt {2A\over 3}+{A\over 2} \\ \Rightarrow {A\over 6} \lt {2A\over x} \lt {7A\over 6} \Rightarrow {1\over 6} \lt {2\over x} \lt {7 \over 6} \Rightarrow 6 \gt {x\over 2} \gt {6\over 7} \Rightarrow \bbox[red, 2pt]{12\gt x\gt {12\over 7}}$$

解答:$$假設定價為x元\Rightarrow \cases{ 每碗售價P(x)=30+2x\\ 銷量Q(x)=200-10x \\每碗獲利P(x)-18=(30+2x)-18 =12+2x} \\ \Rightarrow 總獲利: (12+2x)(200-10x)=3200 \Rightarrow x^2-14x+40=0 \Rightarrow (x-4)(x-10)=0 \\ \Rightarrow \cases{x=4 \Rightarrow P(4)=38\\ x=10 \Rightarrow P(10)=50} \Rightarrow 定價應設為

\bbox[red, 2pt]{38} 元 \;(定價要儘可能便宜)$$

解答:$$利用公式: C^n_k+ C^n_{k+1}= C^{n+1}_{k+1}\\S=C^{100}_1+C^{101}_2+ \cdots+C^{130}_{31} =C^{100}_0+C^{100}_1+C^{101}_2+ \cdots+C^{130}_{31}-C^{100}_0 \\=C^{101}_1+C^{101}_2+\cdots +C^{130}_{31}-1 =C^{102}_2 +C^{102}_3+ \cdots +C^{130}_{31}-1 \\ =\cdots =C^{130}_{30}+C^{130}_{31}-1=  \bbox[red, 2pt]{C^{131}_{31}-1}$$

解答:$$\sum_{n=1}^\infty {1\over n^2+ n} =\sum_{n=1}^\infty \left( {1\over n}-{1\over n+1} \right) = 1-{1\over 2}+{1\over 2}-{1\over 3}+\cdots = \bbox[red, 2pt]1$$

解答:$$ f(x) = x^3 - x^2 - x + 5 \Rightarrow f'(x)=3x^2-2x-1 \Rightarrow f''(x)=6x-2\\ f''(x)=0 \Rightarrow x={1\over 3} \Rightarrow \cases{f''(x)\lt 0, x\lt 1/3\\ f''(x)\gt 0, x\gt 1/3} \Rightarrow x={1\over 3}為反曲點\Rightarrow f(1/3)={124\over 27} \\ \Rightarrow 反曲點坐標: \bbox[red, 2pt] { \left( {1\over 3},{124\over 27} \right)}$$

解答:$$威爾遜定理 \text{(Wilson's Theorem): } (p-1)! \equiv -1 \pmod p,其中 p為質數\\ p=11 \Rightarrow 10! \equiv -1\pmod {11}\equiv 10\pmod {11} \Rightarrow x=\bbox[red, 2pt]{10}$$

解答:$$1\le k\le n \Rightarrow \frac{1}{\sqrt{n^2+n}} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{\sqrt{n^2+1}} \\ \Rightarrow \sum_{k=1}^n \frac{1}{\sqrt{n^2+n}} \leq \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \leq \sum_{k=1}^n \frac{1}{\sqrt{n^2+1}}   \Rightarrow   \frac{n}{\sqrt{n^2+n}} \leq \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \leq  \frac{n}{\sqrt{n^2+1}} \\ \Rightarrow \lim_{n\to \infty} \frac{n}{\sqrt{n^2+n}} \leq \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \leq \lim_{n\to \infty}  \frac{n}{\sqrt{n^2+1}} \\ \Rightarrow 1\leq \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \leq 1 \Rightarrow \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}=\bbox[red, 2pt] 1$$

解答:$$\int_0^{\pi/2} x\cos x\,dx = \left. \left[ x\sin x+ \cos x \right] \right|_0^{\pi/2} =\bbox[red, 2pt]{{\pi\over 2}-1}$$

解答:$$f(x)=g(x) \Rightarrow {1\over x}=3-2x \Rightarrow 1=3x-2x^2 \Rightarrow 2x^2-3x+1=0 \Rightarrow (2x-1)(x-1)=0 \\ \Rightarrow x={1\over 2}, 1 \Rightarrow \int_{1/2}^1 \left( 3-2x-{1\over x} \right)\,dx = \left. \left[ 3x-x^2-\ln x \right] \right|_{1/2}^1 =2- \left( {5\over 4}+\ln 2 \right)= \bbox[red, 2pt]{{3\over 4}-\ln 2}$$

解答:$$A= \begin{bmatrix}4& 1\\0& 2 \end{bmatrix} \Rightarrow \det(A-\lambda I) =(\lambda-4)(\lambda-2) =0 \Rightarrow \lambda= \bbox[red, 2pt]{2,4}$$

解答:$$y''-5y'+6y=0 \Rightarrow \alpha^2-5\alpha+6=0 \Rightarrow (\alpha-3)(\alpha-2)=0 \Rightarrow \alpha=2,3 \\ \Rightarrow y= \bbox[red, 2pt]{C_1e^{2x}+ C_2e^{3x}}$$

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