113年高中學力鑑定-數學詳解

113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: $$依題\vec u=(a,b)需同時滿足\cases{\sqrt{a^2+b^2}=1 \\ {3\over a}= {4\over b}}\\(A)\times: \sqrt{4^2+3^2}=5 \ne 1\\(B)\times: {3\over 0}\ne{4\over 1}   \\ (C)\bigcirc: \sqrt{(-3/5)^2+ (-4/5)^2} =1 且{3\over -3/5} ={4\over -4/5} \\ (D)\times: {3\over -1}  \ne {4\over 0}\\ 故選\bbox[red, 2pt]{(C)}$$

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

解答: $$\log a= \log b+2 = \log b+\log 100= \log 100b \Rightarrow a=100b, 故選\bbox[red, 2pt]{(D)}$$

解答: $$擲一次骰子的期望值={1\over 2}(5+1)=3, 擲三次的期望值=3\times 3=9, 故選\bbox[red, 2pt]{(C)}$$

解答: $$(A)\times: 5\ne 7 \\ (B) \bigcirc: \sqrt{4^2+ 5^2}= \sqrt{41} \\(C) \times: \sqrt{3^2+4^2+ 5^2}= 5\sqrt 2 \ne 5\\ (D)\times: (0,4,5) \ne (3,0,0)\\ 故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cases{O(0,0) \\A(2,4)} \Rightarrow \cases{\overline{OA}的中心點P(1,2)\\  \overline{OA}= 2\sqrt 5} \Rightarrow 以P為圓心, 直徑為\overline{OA} 的圓方程式: (x-1)^2 +(y-2)^2 = 5\\ \angle OBA=90^\circ \Rightarrow B在此圓上，因此符合(x-1)^2 +(y-2)^2 = 5的格子點: \\(x,y) =(2,0), (3,1), (3,3), (0,4), (-1,1), (-1,3)，共有六個, 故選\bbox[red, 2pt]{(A)}$$

解答: $$a_1=4 \Rightarrow a_2= {1\over 1-4} =-{1\over 3} \Rightarrow a_3={3\over 4} \Rightarrow a_4= 4 \Rightarrow \langle a_n\rangle =4, -{1\over 3}, {3\over 4} 循環\\  2023=3\times 674+1 \Rightarrow a_{2023}= a_1=4,故選\bbox[red, 2pt]{(B)}$$

解答: $$(A)\times: 圓形為凹向上, 因此a\gt 0\\ (B)\times: 頂點坐標的-{b\over 2a}\gt 0 \Rightarrow ab\lt 0 \Rightarrow b\lt 0 \\(C)\times : y截距\lt 0 \Rightarrow c\lt 0\\ (D)\bigcirc: y=0有兩相異實根 \Rightarrow b^2-4ac \gt 0\\ 故選\bbox[red, 2pt]{(D)}$$

解答: $$-x^2+5x-4 \gt 0 \Rightarrow x^2-5x+4\lt 0 \Rightarrow (x-4)(x-1)\lt 0 \Rightarrow 1\lt x\lt 4 且x-1\ne 1, 故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x)=2x^3-6x^2+6x-2 \Rightarrow f'(x)=6x^2-12x+6 \Rightarrow f''(x)=12x-12\\ f''(x)=0 \Rightarrow 12x-12=0 \Rightarrow x=1 \Rightarrow 對稱中心(1,f(1)) =(1,0), 故選\bbox[red, 2pt]{(B)}$$

解答: $$\begin{vmatrix}x+2& -4\\ -2& x \end{vmatrix}=0 \Rightarrow x^2+2x-8=0 \Rightarrow (x+4)(x-2)=0 \Rightarrow x=2,-4,故選\bbox[red, 2pt]{(A)}$$

解答: $$\vec a\bot \vec b \Rightarrow \vec a \cdot \vec b =0 \Rightarrow (t-3,2t+3)  \cdot (-1,2) =3-t+4t+6=3t+9=0 \Rightarrow t=-3,故選 \bbox[red, 2pt]{(C)}$$

二、填充題：（10 題，每題 4 分，共 40 分）

解答: $${a+2b\over 2} \ge \sqrt{2ab} =\sqrt{64} =8 \Rightarrow a+2b\ge 16 \Rightarrow 最小值\bbox[red, 2pt]{16}$$

解答: $$P\in L:{x\over 2}={y-1\over -1} ={z+1 \over 2} \Rightarrow P=(2t,-t+1,2t-1), t\in \mathbb R 代入E \\ \Rightarrow 6t+2(-t+1)+2t-1=13 \Rightarrow 6t=12 \Rightarrow t=2 \Rightarrow P= \bbox[red, 2pt]{(4,-1,3)}$$

解答: $$C^8_3\cdot C^5_3 \cdot C^2_2 \times 3! =56\cdot 10\cdot 1\cdot 6= \bbox[red, 2pt]{3360},公布的答案是\bbox[cyan,2pt]{560}$$

解答: $$10\times \sin(100\pi \times 0.06-{11\over 6}\pi) = 10\times \sin({25\over 6}\pi)= 10\times \sin({1\over 6}\pi) =10\times {1\over 2} =\bbox[red, 2pt]5$$

解答: $$取出三球都不是紅球的機率={C^6_3\over C^{10}_3} \Rightarrow 至少一紅球機率=1-{C^6_3\over C^{10}_3}= \bbox[red, 2pt]{5\over 6}$$

解答: $$假設A(0,0,0) \Rightarrow G(3,2,1) \Rightarrow \overline{AG} =\sqrt{3^2+2^2+1^2} = \bbox[red, 2pt]{\sqrt{14}}$$

解答: $$3x^2+6x+k \gt 0 \Rightarrow 判別式: 36-12k \lt 0 \Rightarrow \bbox[red, 2pt] {k\gt 3}$$

解答: $$2x+1=0 \Rightarrow x=-{1\over 2} \Rightarrow f(-{1\over 2}) =0 \Rightarrow -{2\over 8}+{a\over 4}+1=0 \Rightarrow a= \bbox[red, 2pt]{-3}$$

解答: $$\cases{\tan \theta=-4/3\\ \sin \theta\gt 0} \Rightarrow \cases{\sin \theta=4/5\\ \cos \theta=-3/5} \Rightarrow {3+\sin \theta \over 2-3\cos \theta} ={3+4/5\over 2+9/5} = \bbox[red, 2pt]1$$

解答: $$\cases{(x-1)^2 \ge 0\\ x^2+x+1 \ge 0} \Rightarrow f(x)\lt 0 \Rightarrow (x-2)^3(x-3) \lt 0 \Rightarrow \bbox[red, 2pt]{2\lt x\lt 3}$$

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解題僅供參考,其他歷年試題及詳解