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2023年9月25日 星期一

112年高中學力鑑定-數學詳解

112 年度自學進修普通型高級中等學校畢業程度學力鑑定考試
數學科試題本

一、選擇題:(12 題,每題 5 分,共 60 分)

解答:$$由圖形可知:\cases{m_2\lt m_1 \lt 0\\ m_3 \gt 0 \\ m_4=0},故選\bbox[red, 2pt]{(B)}$$
解答:$$y=f(x)= 2x^3-4x+3= 2(x-1)^2+1 \Rightarrow 對稱軸為x=1,故選\bbox[red, 2pt]{(D)}$$
解答:$$x越大則y越大,因此r\gt 0,故選\bbox[red, 2pt]{(C)}$$
解答:$$\sin x 週期為2\pi \Rightarrow \sin 4x週期為{2\pi\over 4}={\pi \over 2},故選\bbox[red, 2pt]{(D)}$$


解答:$$x=(\sqrt 2)^\sqrt 2 \Rightarrow x^\sqrt 2 =\left( (\sqrt 2)^\sqrt 2\right)^\sqrt 2 = (\sqrt 2)^2 = 2,故選\bbox[red, 2pt]{(B)}$$
解答:$$x^2+y^2-4x+2y+k=0 \Rightarrow (x^2-4x+4)+(y^2+2y+1)+k-5=0\\ \Rightarrow (x-2)^2+(y+1)^2 =5-k \Rightarrow 5-k \gt 0 \Rightarrow 5\gt k,故選\bbox[red, 2pt]{(A)}$$


解答:$$\log{25\over 6}-\log{3\over 44}+\log{18\over 11}=\log 25-\log 6-\log 3+\log 44+\log 18-\log 11\\ =2\log 5-(\log 2+\log 3)-\log 3+(2\log 2+\log 11)+ (2\log 3+\log 2)-\log 11\\=2(1-\log 2)+2\log 2 = 2,故選\bbox[red, 2pt]{(B)}$$
解答:$$出現偶數的機率=出現奇數的機率={1\over 2} \Rightarrow 擲一次骰子的期望值={1\over 2}(5+1)=3\\ \Rightarrow 擲三次骰子的期望值= 3\times 3=9,故選\bbox[red, 2pt]{(C)}$$
解答:$$\vec a\bot \vec b \Rightarrow \vec a\cdot \vec b=0 \Rightarrow (1,2) \cdot (t+1,-3t+2)=t+1-6t+4=-5t+5=0 \Rightarrow t=1\\,故選\bbox[red, 2pt]{(D)}$$
解答:$$(A)\bigcirc: \sqrt{4^2+5^2}= \sqrt{41} \\(B)\times: 5\ne 7 \\ (C)\times: \sqrt{3^2+4^2+5^2} =5\sqrt 2 \ne 5 \\ (D)\times: (-3,4,5)\ne (3,-4,-5)\\,故選\bbox[red, 2pt]{(A)}$$



解答:$$與\overleftrightarrow{AC}歪斜的直線:\overleftrightarrow{EF}, \overleftrightarrow{DE}, \overleftrightarrow{BE},共三條,故選\bbox[red, 2pt]{(A)}$$
解答:$$(A)\times: 若\cases{B=\begin{bmatrix}0 & 0 \\0 &1 \end{bmatrix} \\[1ex]A= \begin{bmatrix}0 & 0 \\1 &0 \end{bmatrix}} \Rightarrow \cases{AB= \begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} \\[1ex] BA= \begin{bmatrix}0 & 0\\ 1 & 0 \end{bmatrix} } \Rightarrow AB=0但BA\ne 0 \\(B)\times: 若\cases{B=\begin{bmatrix}0 & 0 \\0 &1 \end{bmatrix} \\[1ex]A= \begin{bmatrix}0 & 0 \\1 &0 \end{bmatrix}},則AB=0,但A\ne 0且B\ne 0 \\(C)\bigcirc: AB=I \Rightarrow ABA=A \Rightarrow A^{-1} ABA= A^{-1}A \Rightarrow BA=I \\ (D)\times: (AB)^2 = ABAB \ne A^2B^2\\,故選\bbox[red, 2pt]{(C)}$$

二、填充題:(10 題,每題 4 分,共 40 分)

解答:$${a+2b \over 2} \ge \sqrt{2ab} =\sqrt {2\cdot 32} =8 \Rightarrow a+2b \ge 16 \Rightarrow 最小值=\bbox[red, 2pt]{16}$$
解答:$$\cases{\tan \theta=-4/3\\ \sin \theta \gt 0} \Rightarrow \cases{\sin \theta=4/5\\ \cos =-3/5} \Rightarrow {3+\sin \theta \over 2-3\cos \theta} ={3+4/5 \over 2+9/5} ={19/5\over 19/5}= \bbox[red, 2pt]1$$




解答:$$假設\cases{\overline{EF}= \overline{FG}= \overline{GH}= \overline{HE}=a \\ \overline{AE}= \overline{DH}= \overline{CG}= \overline{BF}=b } \Rightarrow \cases{\overline{AF}=\sqrt{a^2+b^2}=6 \\ \overline{AG} =\sqrt{2a^2+b^2}=7}\\  \Rightarrow EFGH面積=a^2 =7^2-6^2 = \bbox[red, 2pt]{11}$$
解答:$$假設㘣半徑r \Rightarrow r^2\pi \cdot {144\over 360}=40\pi \Rightarrow r= \bbox[red, 2pt]{10}$$
解答:$$假設第六科分數為a \Rightarrow 73+85+85+85+91+a= 6\times 85 \Rightarrow a=91 \\ \Rightarrow 標準差= \sqrt{12^2+6^2+6^2\over 6} =\bbox[red, 2pt]6$$
解答:$$S(n)=n^2-n \Rightarrow a_n = S(n)-S(n-1) =n^2-n-((n-1)^2-(n-1))= 2n-2\\ \Rightarrow \cases{a_{112}= 2\cdot 112-2=222\\ a_{12}= 2\cdot 12-2=22} \Rightarrow a_{112}-a_{12}=  \bbox[red, 2pt]{200} $$
解答:$${0.8\times 0.9\over 0.8\times 0.9+0.2\times 0.4} ={72\over 80}= \bbox[red, 2pt]{0.9}$$
解答:$$1\times 3+2\times 4+\cdots +10\times 12 = \sum_{n=1}^{10}n(n+2) = \sum_{n=1}^{10}n^2 + 2\sum_{n=1}^{10} n\\ ={1\over 6}\cdot 10\cdot 11\cdot 21 +2\cdot {1\over 2}\cdot 10\cdot 11= 385+110= \bbox[red, 2pt]{495}$$
解答:$$7!=\bbox[red, 2pt]{5040}$$
解答:$$\cases{a+b=35\\ c+7=50-35=15\\ a+c=30\\ b+7=50-30=20} \Rightarrow \cases{a=22\\ b=13\\ c=8 } \Rightarrow {b\over b+7} ={13\over 20} =\bbox[red, 2pt]{0.65}$$
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解答僅供參考,其他歷年試題及詳解



 

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