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2021年8月25日 星期三

110年蘭陽女中教甄-數學詳解

國立蘭陽女子高級中學110學年度第一次教師甄試 

一、單選(三題):每題4分,共12分


解答$$2x+2\log(2+10^{-x})-\log({1\over 4}+10^x+10^{2x})= 2x+2\log(2+10^{-x}) -\log({1\over 2}+10^x )^2 \\ =2x+2\log  {2+10^{-x} \over 1/2+10^x}  =2x+2\log  {2\cdot 10^x+1  \over (1/2)10^x+10^{2x}}   =2x+2\log  {2\cdot 10^x+1  \over (1/2)10^x(1+2\cdot 10^{x})}  \\ =2x +2\log {2\over 10^x}    =2x+2(\log 2-x) =2\log 2,故選\bbox[red,2pt]{(4)}$$
解答$$C^{100}_{50} ={100!\over 50! 50!}\Rightarrow 分母介於51至100間的質數不會被分子約掉\\,即53,59,61,67,71,73,79,83,89,97,共十個質因數;\\另外,介於2至50間的質數,若在分子出現的頻率多於分母,也符合所求\\ \begin{array}{} 質數 & 分母頻率 & 分子頻率& OK?\\\hline 47 & 2\times \lfloor {50\over 47} \rfloor =2 &\lfloor {100\over 47} \rfloor =2 & \times\\43 & 2\times \lfloor {50\over 43} \rfloor =2 &\lfloor {100\over 43} \rfloor =2 & \times\\41 & 2\times \lfloor {50\over 41} \rfloor =2 &\lfloor {100\over 41} \rfloor =2 & \times\\37 & 2\times \lfloor {50\over 37} \rfloor =2 &\lfloor {100\over 37} \rfloor =2 & \times\\31 & 2\times \lfloor {50\over 31} \rfloor =2 &\lfloor {100\over 31} \rfloor =3 & \bigcirc\\29 & 2\times \lfloor {50\over 29} \rfloor =2 &\lfloor {100\over 29} \rfloor =3 & \bigcirc \\23 & 2\times \lfloor {50\over 23} \rfloor =4 &\lfloor {100\over 23} \rfloor =4 & \times\\19 & 2\times \lfloor {50\over 19} \rfloor =4 &\lfloor {100\over 19} \rfloor =5 & \bigcirc\\17 & 2\times \lfloor {50\over 17} \rfloor =4 &\lfloor {100\over 17} \rfloor =5 & \bigcirc\\ 13 & 2\times \lfloor {50\over 13} \rfloor =6 &\lfloor {100\over 13} \rfloor =7 & \bigcirc\\ 11 & 2\times \lfloor {50\over 11} \rfloor =8 &\lfloor {100\over 11} \rfloor =9 & \bigcirc\\ 7 & 2\times (\lfloor {50\over 7} \rfloor +\lfloor {50\over 7^2} \rfloor) =16 &\lfloor {100\over 7} \rfloor+{100\over 7^2} \rfloor =16 & \times \\ 5 & 2\times (\lfloor {50\over 5} \rfloor +\lfloor {50\over 5^2} \rfloor) =24 &\lfloor {100\over 5} \rfloor+{100\over 5^2} \rfloor =24 & \times \\3 & 2\times (\lfloor {50\over 3} \rfloor +\lfloor {50\over 3^2} \rfloor+\lfloor {50\over 3^3} \rfloor) =44 &\lfloor {100\over 3} \rfloor+{100\over 3^2} +{100\over 3^3}+{100\over 3^4}\rfloor =48 & \bigcirc \\ 2 & 2\times (\lfloor {50\over 2} \rfloor +\cdots+\lfloor {50\over 2^5}\rfloor) =94 &\lfloor {100\over 2} \rfloor+\cdots+ \lfloor{100\over 2^6} \rfloor =97 & \bigcirc \\\hline \end{array}\\ 有八個符合要求,共有10+8=18個,故選\bbox[red,2pt]{(4)}$$
解答$$在七點時,甲乙平行,兩人距丙a公里,丙距丁b公里;\\因此我們有\cases{2(甲-丙)=a \cdots(1)\\ 5(甲+丁)=a+b \cdots(2)\\ 6(乙+丁)= a+b \cdots(3)\\ 6.5(丙+丁)=b \cdots(4)},其中甲、乙、丙、丁代表四人的時速;欲求{a\over 乙-丙}之值\\ 由(1)及(2)可得甲={a\over 2}+丙={a+b\over 2}-丁\Rightarrow 丙+丁={2b-3a\over 10} 代入(4) \Rightarrow a={b\over 6.5}\\由(3)及(4)可得:乙-丙={a+b\over 6}-{b\over 6.5}代入欲求之{a\over 乙-丙} ={a\over {a+b\over 6}-{b\over 6.5} }\\ 將a={b\over 6.5}代入上式\Rightarrow {a\over 乙-丙} = {{b\over 6.5}\over {7.5\over 39}b-{1\over 6.5}b} ={6\over 1.5}=4 \Rightarrow 七點+四小時=上午十一點\\,故選\bbox[red,2pt]{(4)}$$

二、多重選擇題

解答$$2^{48}-1=(2^{24})^2-1= (2^{24}-1)(2^{24}+1) =(2^{12}-1)(2^{12}+1)(2^{24}+1)\\ =(2^{6}-1)(2^6+1)(2^{12}+1)(2^{24}+1)  =  63 \cdot 65\cdot 4097\cdot (2^{24}+1)\\ \Rightarrow 63,65為其因數,故選\bbox[red,2pt]{(23)}$$
解答$$(4)\times: N為\overline{BC}中點\Rightarrow \overline{BN}=6\div 2=3 \Rightarrow \cases{\overline{AN} =3\sqrt 3 \\\overline{ON}= \sqrt{5^2-3^2}=4} \Rightarrow \triangle OAN三邊長為5,3\sqrt 3,4\\\qquad 最大角之\cos \theta ={5^2+4^2-(3\sqrt 3)^2 \over 2\times 5\times 4} \gt 0 \Rightarrow \triangle OAN為銳角三角形\\,故選\bbox[red, 2pt]{(1235)}$$
解答$$ 2^{x-a}+b=2^x \Rightarrow 2^{-a}+b\cdot 2^{-x}=1 \Rightarrow {b\over 2^x}={2^a-1\over 2^a} \Rightarrow 2^x={b\cdot 2^a\over 2^a-1}\\(1)\bigcirc: \cases{a=1\\ b=2}\Rightarrow x=2 為其一解 \\(2)\times: \cases{a\gt 0\\ b\lt 0} \Rightarrow {2^a\over 2^a-1}\gt 0 \Rightarrow 2^x \lt 0矛盾\\(3) \times:\cases{a\lt 0\\ b\gt 0} \Rightarrow {2^a\over 2^a-1}\lt 0 \Rightarrow 2^x \lt 0矛盾 \\(4)\bigcirc: \cases{a=-1\\ b=-4} \Rightarrow x=2為其一解\\(5) \bigcirc: 2^x= {b\cdot 2^a\over 2^a-1} \Rightarrow x= a+\log_2 b-\log_2 (2^a-1)\\,故選\bbox[red,2pt]{(145)}$$
解答$$(1)\times: \cases{f(x)=x+1,\text{ if }x\le 0\\ f(x)=-x+1,\text{ if }x\ge 0}  \Rightarrow f(0)為極大值,但f'(0)不存在\\(3)\times: 例子同(1)\\(5)\times: f'(3)\ne 0\\,故選\bbox[red,2pt]{(24)}$$
解答$$令N=101\times 102 \Rightarrow (7242409)^{10} =(703\times N+103)^{10} =103^{10} \mod N\\ 又103^{10}=(103\times 103)^5 =(N+307)^5 \Rightarrow 103^{10} \mod N= 307^5 \mod N\\ 又307^2 =9\times N+1531 \Rightarrow 307^5 \mod N =1531^2\times 307 \mod N \\ =5407\times 307  \mod N=\bbox[red, 2pt]{1327}$$
解答$$標準差5=\sqrt{f(10)\over 9} \Rightarrow 最小值f(10)=5^2\times 9= \bbox[red,2pt]{225}$$
 
解答$$f(n)= n^3+2n^2-200n =n((n+1)^2-201) \Rightarrow \begin{cases}f(n)\gt 0 & n\ge 14\\ f(n) \lt 0 & n\le 13\end{cases} \\ \Rightarrow \sum_{n=1}^{20} |f(n)| =\sum_{n=1}^{20} f(n) +2|\sum_{n=1}^{13} f(n)| = 7840+ 2\times 8281= \bbox[red,2pt]{24402}$$
解答
解:
由矩陣可知:$$A=\frac { 1 }{ 5 } \begin{bmatrix} 4 & -3 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} \frac { 4 }{ 5 } & -\frac { 3 }{ 5 } \\ \frac { 3 }{ 5 } & \frac { 4 }{ 5 } \end{bmatrix}=\begin{bmatrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{bmatrix}$$即A為一旋轉矩陣,角度為\(\theta\),旋轉狀態如上圖。
(1)\(\sin{\angle P_1OP_3}=\sin{2\theta}=2\sin{\theta}\cos{\theta}=2\times\frac{3}{5} \times\frac{4}{5} = {\frac{24}{25}}\)
(2)$$\triangle P_{ 1 }P_{ 2 }P_{ 3 }=\triangle OP_{ 1 }P_{ 2 }+\triangle OP_{ 2 }P_{ 3 }-\triangle OP_{ 1 }P_{ 3 }\\ =\frac { 1 }{ 2 } a^{ 2 }\sin { \theta } +\frac { 1 }{ 2 } a^{ 2 }\sin { \theta } -\frac { 1 }{ 2 } a^{ 2 }\sin { 2\theta } \\ =a^{ 2 }\times \frac { 3 }{ 5 } -a^{ 2 }\times \frac { 3 }{ 5 } \times \frac { 4 }{ 5 } ={\frac { 3 }{ 25 } a^{ 2 }}$$(3)$$P_{ 1 }=(m,\frac { 1 }{ 10 } m^{ 2 }-10)\Rightarrow a^{ 2 }={ \overline { OP_{ 1 } } }^{ 2 }=m^{ 2 }+{ \left( \frac { 1 }{ 10 } m^{ 2 }-10 \right) }^{ 2 }\\ =\frac { 1 }{ 100 } m^{ 4 }-m^{ 2 }+100=\frac { 1 }{ 100 } { \left( m^{ 2 }-50 \right) }^{ 2 }+75\\ \Rightarrow \triangle P_{ 1 }P_{ 2 }P_{ 3 }=\frac { 3 }{ 25 } \times \left[ \frac { 1 }{ 100 } { \left( m^{ 2 }-50 \right) }^{ 2 }+75 \right] =\frac { 3 }{ 2500 } { \left( m^{ 2 }-50 \right) }^{ 2 }+9\\ \Rightarrow 最小值為\bbox[red,2pt]{9}$$
解答
$$作\overline{BD},並令\cases{\angle CBD=\angle CDB=\theta\\ \overline{AB} =\overline{BC} =\overline{CD}=a},則\cases{\angle C=180^\circ-2\theta\\ \angle ABD=108^\circ-\theta \\ \angle ADB=18^\circ+\theta}\\ 正弦定理:\cases{\triangle ABD\Rightarrow a:\sin (18^\circ+\theta)= \overline{BD}:\sin 54^\circ \\ \triangle BCD\Rightarrow a:\sin \theta = \overline{BD}:\sin (180^\circ-2\theta)} \\\Rightarrow {\sin 54^\circ \over \sin(18^\circ+\theta) } ={\sin(180^\circ-2\theta) \over \sin \theta}={\sin( 2\theta) \over \sin \theta} =2\cos \theta\\ \Rightarrow \sin 54^\circ =2\cos\theta \sin(18^\circ +\theta) =2\cos\theta (\sin 18^\circ\cos\theta +\sin \theta \cos 18^\circ)\\ = \sin 18^\circ \cos 2\theta + \cos 18^\circ \sin 2\theta +\sin 18^\circ \Rightarrow \sin 54^\circ-\sin 18^\circ = \sin(2\theta +18^\circ)\\ 由於\cases{\sin 54^\circ = (\sqrt 5+1)/4\\ \sin 18^\circ= (\sqrt 5-1)/4} \Rightarrow \sin(2\theta +18^\circ)={1\over 2} \Rightarrow 2\theta +18^\circ=30^\circ \Rightarrow 2\theta =12^\circ \\ \Rightarrow \angle C=180^\circ -2\theta = 180^\circ -12^\circ =\bbox[red,2pt]{168^\circ}$$
解答
$${\triangle ADE\over \triangle ABC}={ab\over 9\cdot 8}={1\over 2} \Rightarrow ab=36\\ \overline{DE}^2= (b\cos 40^\circ-a)^2+ b^2\sin^2 40^\circ =b^2- 2ab\cos 40^\circ+a^2 =a^2+b^2-72\cos 40^\circ\\ 當a=b時,a^2+b^2有最小值,即\overline{DE}有最小值;此時a=b=6(\because ab=36),即\overline{AD}=a=\bbox[red,2pt]{6}$$
解答$$R(x)={x\over \ell(x)} ={x\over \overline{AB}} ={x\over \sqrt{(x-4)^2+9}} ={x\over \sqrt{x^2-8x+25}} \\\Rightarrow R'(x)={1\over \sqrt{x^2-8x+25}}-{x(x-4)\over \sqrt{(x^2-8x+25)^3}}\\ 因此R'(x)=0 \Rightarrow {x^2-8x+25-x(x-4)\over \sqrt{(x^2-8x+25)^3}}=0 \Rightarrow 4x=25\\ \Rightarrow x={25\over 4} \Rightarrow R({25\over 4})={25/4\over \sqrt{(9/4)^2+9}}= \bbox[red,2pt]{5\over 3}\\註:公布的答案是\bbox[blue,2pt]{{5\over 2}}$$
解答$$1-99999的數字可以表示成x_1x_2x_3x_4x_5,因此依題意x_1 +x_2 +x_3+x_4+x_5 \le 10\\也就是x_1 +x_2 +x_3+x_4+x_5=k,k=1,2-10,因此共有\sum_{k=1}^{10}H^5_k =3002;\\但H^5_{10}中有x_i=10的數字,共5個,因此符合條件的共有3002-5=2997,再加上最大數字100000\\,總共是2997+1=\bbox[red, 2pt]{2998}個$$
解答$$把表格視為一矩陣A=\{a_{ij}\},則扣除第1行的部份為對稱矩陣;\\欲求之第20行第20列的數字即對角線上第20個數字;\\對角線b_i=a_{ii}=3,7,13,21,31,...,i\in \mathbb{N} \Rightarrow b_n-b_{n-1}=2n,n=2,3,\dots\\ \Rightarrow (b_{20}-b_{19})+ (b_{19}-b_{18}) +\cdots +(b_2-b_1)=2 (20+19+\cdots +2) \\ \Rightarrow b_{20}-b_1=22\times 19 \Rightarrow b_{20}=22\times 19+3= \bbox[red,2pt]{421}$$


解答
$$圓C:x^2+(y-3)^2=2^2 \Rightarrow \cases{圓心P(0,3)\\ 半徑r=2} \Rightarrow \cases{圓面積=r^2\pi= 4\pi\\ P繞x軸轉一圈周長=6\pi} \\ \Rightarrow 甜甜圈體積=4\pi \times 6\pi = \bbox[red,2pt]{24}\pi^2$$
解題僅供參考,其他教甄試題及詳解

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