113學年度四技二專統測--數學(C)詳解

 113 學年度科技校院四年制與專科學校二年制

統 一 入 學 測 驗-數學(C)

解答: $${5\over (2x+1)(x-2)} ={A\over 2x+1}+{B\over x-2} \Rightarrow A(x-2)+B(2x+1)=5 \\ \Rightarrow (A+2B)x-2A+B=5 \Rightarrow \cases{A+2B=0\\ -2A+B=5} \Rightarrow \cases{A=-2\\ B=1} \Rightarrow 3A+2B=-6+2=-4\\, 故選\bbox[red, 2pt]{(D)}$$

解答:

$$\theta= 35^\circ+ 90^\circ=125^\circ, 故選\bbox[red, 2pt]{(C)}$$

解答: $$\sin \theta=\sin 2024^\circ = \sin(360^\circ \times 5+224^\circ) = \sin 224^\circ \Rightarrow \theta=224^\circ, 故選\bbox[red, 2pt]{(C)}$$

解答: $$直線L與圓C相切\Rightarrow 圓心至L的距離=半徑\\ 圓心(3,-4)至L的距離 ={|3-5+4| \over \sqrt{1^2+1^2}} ={2\over \sqrt 2} =\sqrt 2, 故選\bbox[red, 2pt]{(A)}$$

解答: $$\left[ \begin{array}{rr|r}1 & -1 & 4\\ 2& 3& 3\end{array} \right] \xrightarrow {R_1+R_2 \to R_2} \left[ \begin{array}{rr|r}1 & -1 & 4\\ 3& 2& 7\end{array} \right], 故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\sin \theta \tan \theta= \sin \theta\cdot {\sin \theta \over \cos \theta} ={\sin^2 \theta\over \cos \theta} \lt 0 \Rightarrow \cos \theta \lt 0\\ \cos \theta \cot \theta = \cos \theta \cdot {\cos \theta\over \sin \theta} ={\cos^2 \theta \over \sin \theta} \gt 0 \Rightarrow \sin \theta \gt 0} \Rightarrow \theta 在第二象限, 故選\bbox[red, 2pt]{(B)}$$

解答: $$相鄰的香瓜與木瓜與三種水果排列數=4!\times 2＝48\\ 除了香瓜與木瓜外,剩下四種水果取3種,有C^4_3=4種取法\\ 因此共有48\times 4=192種排列法, 故選\bbox[red, 2pt]{(C)}$$

解答: $${a^2+b\over 2} \ge \sqrt{a^2b} \Rightarrow a^2b \le \left({ a^2+b\over 2}\right)^2 =\left({ 10-b+b\over 2}\right)^2 =5^2=25, 故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{y_1=2x_1+5x_2 \\ y_2=3x_1+8x_2} \Rightarrow \begin{bmatrix}2 & 5 \\3 & 8 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} =\begin{bmatrix}y_1 \\y_2 \end{bmatrix} \Rightarrow A= \begin{bmatrix}2 & 5 \\3 & 8 \end{bmatrix}\\ \Rightarrow A^{-1}={1\over 2\cdot 8-3\cdot 5}\begin{bmatrix}8 & -5 \\ -3 & 2\end{bmatrix}= \begin{bmatrix}8 & -5 \\-3 & 2 \end{bmatrix} \Rightarrow a+b+c+d=8-5-3+2=2\\, 故選\bbox[red, 2pt]{(A)}$$

解答: $$(A)\times: f(5)=10+4=14 \gt 13\\ (B)\times: f(5)=20-4=16 \gt 13\\ (D) \times: f(2)=4-3=1 \lt 4\\, 故選\bbox[red, 2pt]{(C)}$$

解答: $$f(x)=ax^4+bx^2-2x+c \Rightarrow f'(x)=4ax^3+2bx-2 = 8x^3-6x+d\\ \Rightarrow \cases{4a=8\\ 2b=-6\\ d=-2} \Rightarrow \cases{a=2\\ b=-3\\ d=-2} \Rightarrow f(x)= 2x^4-3x^2-2x+c \Rightarrow f(1)=-3+c=5 \Rightarrow c=8\\ \Rightarrow a+b+c+d = 2-3+8-2=5, 故選\bbox[red, 2pt]{(D)}$$

解答: $${1\over \sqrt 2+1} = {\sqrt 2-1\over (\sqrt 2+1)(\sqrt 2-1)}  = \sqrt 2-1 \\\Rightarrow \left( {1\over \sqrt 2+1}-1\right) \left[ \left( {1\over \sqrt 2+1} \right)^2 +{1\over \sqrt 2+1}+1\right] =(\sqrt 2-1-1)[(\sqrt 2-1)^2+\sqrt 2-1+1] \\=(\sqrt 2-2)(3-\sqrt 2) =(\sqrt 2-2)(-3\sqrt 2+3)= -8+5\sqrt 2, 故選\bbox[red, 2pt]{(D)}$$

解答:

$$y=ax^2  \Rightarrow x^2=4\cdot {1\over 4a}y \Rightarrow c={1\over 4a} \Rightarrow \overline{F_1F_2}=4c={1\over a}=8 \Rightarrow a={1\over 8} \Rightarrow c=2\\ \Rightarrow \triangle VF_1F_2 ={1\over 2} \cdot c\cdot \overline{F_1F_2} =8, 故選\bbox[red, 2pt]{(A)}$$

解答: $$\int_0^2 f(x)\,dx = \int_0^1 (\sqrt x+1)\,dx + \int_1^2 (x^2+x)\,dx = \left. \left[ {2\over 3}x^{3/2}+x \right] \right|_0^1 +\left. \left[{1\over 3}x^3+{1\over 2}x^2 \right] \right|_1^2 \\={5\over 3}+({14\over 3}-{5\over 6})= {33\over 6}={11\over 2}, 故選\bbox[red, 2pt]{(B)}$$

解答: $$\lim_{n\to \infty} \left( {n^2-n\over n+1} -{n^2+3n\over n+2}\right) =\lim_{n\to \infty} \left( {(n^2-n)(n+2) -(n^2+3n)(n+1) \over (n+1)(n-2)}  \right) \\= \lim_{n\to \infty} {-3n^2-5n \over n^2-n-2} =-3, 故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{\log x=-2.24\\ \log y=9.28} \Rightarrow \cases{x=10^{-2.24} \\ y=10^{9.28}} \Rightarrow x^2 y=10^{-2.24\cdot 2+8.28} =10^{4.8} \\ \Rightarrow 10^4\lt x^2y \lt 10^5, 故選\bbox[red, 2pt]{(B)}$$

解答: $$, 故選\bbox[red, 2pt]{()}$$

解答: $$當x＝0時,y＝0,即f(0)=0, 只有(D)符合此條件, 故選\bbox[red, 2pt]{(D)}$$

解答: $$\cos \theta-{\sqrt 3\over 2}i = \overline{-{1\over 2}+(\sin \theta )i} =-{1\over 2}-(\sin \theta )i \Rightarrow \cases{\cos \theta =-1/2\\ \sin \theta =\sqrt 3/2} \\ \Rightarrow \sin(2\theta)= 2\sin \theta \cos \theta =-{\sqrt 3\over 2}, 故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{A(0,6,-1) \\ B(3,3,-1)\\ C(4,1,1)} \Rightarrow \cases{\overrightarrow{AC} =(4,-5,2)\\ \overrightarrow{BC} =(1,-2,2) } \Rightarrow \cases{\overrightarrow{AC} \cdot \overrightarrow{BC} =18 \\ |\overrightarrow{BC}|^2= 9} \\\Rightarrow \left({\overrightarrow{AC} \cdot \overrightarrow{BC} \over |\overrightarrow{BC}|^2} \right) \overrightarrow{BC} = 2\overrightarrow{BC} =(2,-4,4), 故選\bbox[red, 2pt]{(C)}$$

解答: $$a_n代表n層金字所需的火柴棒數量 \Rightarrow \cases{a_1=3\\ a_2=9\\ a_3=18} \\ 最底層(第1層)有n個三角形、第2層有n-1個三角形、第3層有n-2個三角形,\dots\\ 因此a_n=3\times(n+(n-1)+\cdots +1)=3\cdot {n(n+1)\over 2} \Rightarrow a_{50}= 3\cdot {50\cdot 51\over 2}=3825, 故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cases{A(-3,4)\\ B(-1,2)\\ C(3,6)} \Rightarrow \cases{L_1= \overleftrightarrow{AB}: x+y-1=0\\ L_2=\overleftrightarrow{AC}: x-3y+15=0\\ L_3=  \overleftrightarrow{BC}: x-y+3=0} \\ 原點三直線的下方\Rightarrow 所圍區域\cases{x+y-1\ge 0\\ x-3y+15\ge 0\\ x-y+3\le 0}, 故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{T_A=4T_B \\ h_B=100} \Rightarrow T_B=0.085\cdot 100^{3/4} \Rightarrow T_A=4T_B=0.34\cdot 100^{3/4} = 0.085h_A^{3/4} \\ \Rightarrow h_A^{3/4}=4\cdot 100^{3/4} \Rightarrow h_A=4^{4/3} \cdot 100 \approx 1.26^4\cdot 100=252, 故選\bbox[red, 2pt]{(C)}$$

解答: $$(B)\times: 極限不存在\\(C)\times: \cases{\lim_{x\to 1^+} {|x-1|\over x-1} =1 \\\lim_{x\to 1^-} {|x-1|\over x-1} =-1} \Rightarrow 極限不存在 \\(D)\times: 極限存在且連續\\, 故選\bbox[red, 2pt]{(A)}$$

解答: $$B(3,2,4)對稱xy平面的對稱點B'(3,2,-4) \Rightarrow P=\overline{AB'} \cap xy平面 =(2,{5\over 2},0), 故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{t=-1 \Rightarrow \overrightarrow{OP}=(-3,2)-(2,1)=(-5,1) \Rightarrow P_1(-5,1) \\ t=1 \Rightarrow \overrightarrow{OP}=(-3,2)+ (2,1)=(-1,3) \Rightarrow P_2=(-1,3)} \\ \Rightarrow \overline{P_1P_2}= \sqrt{20} =2\sqrt 5, 故選\bbox[red, 2pt]{(B)}$$

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解題僅供參考,統測歷年試題及詳解