114年高中學力鑑定-數學詳解

 114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

一、選擇題：（12 題，每題 5 分，共 60 分）

解答:$$由圖形可知\cases{m_1,m_2\lt 0\\ m_3\gt 0\\ m_4=0}, 又L_2較L_1陡，故選\bbox[red, 2pt]{(B)}$$

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:$$a_1=4 \Rightarrow a_2={1\over 1-4}=-{1\over 3} \Rightarrow a_3={1\over 1+1/3} ={3\over 4} \Rightarrow a_4={1\over 1-3/4} =4 \\ \Rightarrow 循環數3 \Rightarrow a_{1+3n}=4, n\in \mathbb N \Rightarrow {2023}=1+3\cdot 674 \Rightarrow a_{2023}=4，故選\bbox[red, 2pt]{(C)}$$

解答:$$x越大,反而y越小,則相關係數越小，故選\bbox[red, 2pt]{(D)}$$

解答:$$(B) \times:\overline{BC} \parallel \overline{AD} \\ (C) \times: \overline{CD}與\overline{AD}相交 \\(D) \times: \overline{DE}與\overline{AD}相交\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$甲的對面可能為乙,丙,丁, 每人機會相等, 機率為{1\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答:$${\sin(-\theta) \over \sin(\pi+\theta)} -{\tan(\pi +\theta) \over \tan(\pi-\theta)} -{\cos(-\theta) \over \cos(\pi-\theta)} ={-\sin(\theta) \over -\sin \theta} -{\tan \theta \over -\tan \theta} -{\cos(\theta) \over -\cos(\theta)} \\=1+1+1=3，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cos \angle A={\overline{AB}^2+ \overline{AC}^2- \overline{BC}^2 \over 2\cdot \overline{AB}\cdot \overline{AC}} ={25+64-49\over 80} ={1\over 2} \Rightarrow \angle A=60^\circ，故選\bbox[red, 2pt]{(C)}$$

解答:$$\overleftrightarrow{AB}斜率={b-a\over 45-5} ={\log_3 45-\log_3 5\over 40} ={\log_3 9\over 40} ={2\over 40}={1\over 20}，故選\bbox[red, 2pt]{(B)}$$

解答:$$x^2+y^2-2x-2y-3=0 \Rightarrow (x-1)^2+ (y-1)^2=5 \Rightarrow \cases{圓心P(1,1)\\ 圓半徑r=\sqrt 5} \\假設A=(-1,2),則A在圓上且\overleftrightarrow{PA}斜率={1-2\over 1-(-1)} =-{1\over 2} \Rightarrow 切線斜率=2 \\ \Rightarrow 經過A且斜率為2的直線: y=2(x+1)+2 \Rightarrow 2x-y+4=0，故選\bbox[red, 2pt]{(C)}$$

解答:$$\sum_{k=1}^{10} \left( a_k+2k\right) =\sum_{k=1}^{10} a_k +2\sum_{k=1}^{10} k =\sum_{k=1}^{10} a_k +2\cdot 55  =\sum_{k=1}^{10} a_k +110 =240 \\ \Rightarrow  =\sum_{k=1}^{10} a_k =130，故選\bbox[red, 2pt]{(A)}$$

解答:$$50\cdot {3\over 5} +100\cdot {2\over 5} =30+40=70，故選\bbox[red, 2pt]{(A)}$$

二、填充題：（10 題，每題 4 分，共 40 分）

解答:$$r^2\pi \cdot {144\over 360} =40\pi \Rightarrow r^2={40\cdot 360\over 144} =100 \Rightarrow r=\bbox[red, 2pt]{10}$$

解答:$${a+2b\over 2}\ge \sqrt{2ab} \Rightarrow a+2b \ge 2\cdot \sqrt{64} = \bbox[red, 2pt]{16}$$

解答:

$$假設\cases{\overline{EF}=a \\ \overline{AE}=b} \Rightarrow \cases{\overline{AF}^2 =\overline{AE}^2+ \overline{EF}^2\\ \overline{AG}^2= \overline{AB}^2+ \overline{BG}^2 =\overline{AB}^2+ \overline{BC}^2 +\overline{CG}^2} \Rightarrow \cases{36=a^2+b^2\\ 49=2a^2+b^2} \\ \Rightarrow 兩式相減 \Rightarrow a^2= EFGH面積= \bbox[red, 2pt]{13}$$

解答:$$\sqrt{19-8\sqrt 3} =\sqrt{19-2\sqrt {48}} =\sqrt{16}-\sqrt 3=4-\sqrt 3 \Rightarrow 2\lt \sqrt{19-8\sqrt 3}\lt 3 \Rightarrow a=\bbox[red, 2pt]2$$

解答:$$x={3a+2b\over 5} \Rightarrow |x-a|= \left|{2b-2a\over 5} \right|= \left|{2\over 5}(b-a) \right| ={2\over 5}(b-a)=10 \Rightarrow b-a=25 \\ \Rightarrow |x-b|=\left| {{3\over 5}(a-b)}\right| =\left| {{3\over 5}\cdot (-25)}\right| =|-15|=\bbox[red, 2pt]{15}$$

解答:$$前5次出現2次正面及3次反面，且第6次出現正面的機率={C^5_2\over 2^5}\cdot {1\over 2} ={10\over 64} =\bbox[red, 2pt]{5\over 32}$$

解答:$$\begin{bmatrix} a& b\\ c& d\end{bmatrix} = \begin{bmatrix} -1& 3\\ 1&-1\end{bmatrix} \begin{bmatrix} 3& 2\\ 1& 1\end{bmatrix} = \begin{bmatrix} 0& 1\\ 2& 1\end{bmatrix} \Rightarrow abcd= \bbox[red, 2pt]{0121}$$

解答:$$假設f(x)=(x+1)(x-3)p(x)+ ax+b \Rightarrow \cases{f(-1)=-a+b=6\\ f(3)=3a+b=-2} \Rightarrow \cases{a= -2\\ b=4} \\ \Rightarrow 餘式\bbox[red, 2pt]{-2x+4}$$

解答:$${甲生產的不良品\over 所有的不良品} ={50\% \times 5\%   \over 50\% \times 5\% +30\%\times 3\%+ 20\%\times 4\%} = \bbox[red, 2pt]{25\over 42}$$

解答:$$\cases{(x-10)^2 \ge 0\\ x^2+x+1=(x+1/2)^2+3/4 \gt 0 \\ x^2+x-6 =(x+3)(x-2)} \Rightarrow (x-10)^2 (x^2+x-6)(x^2+x+1)\lt 0 \\ \Rightarrow (x+3)(x-2)\lt 0 \Rightarrow -3\lt x\lt 2 \Rightarrow x=-2,-1,0,1 \Rightarrow \bbox[red, 2pt]4個整數解$$

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解題僅供參考，其他學力鑑定試題及詳解