115年身障升大學-數學A詳解

115 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別：大學組-數學 A

解答:$$2^{a+b} =2\times 4^{a-b} =2\times 2^{2a-2b} =2^{2a-2b+1} \Rightarrow a+b=2a-2b+1 \Rightarrow a-3b=-1，故選\bbox[red, 2pt]{(B)}$$

解答:$$y=-x^2+a(x-1)^2 =-x^2+ax^2-2ax+a =(a-1)x^2-2a+a \\ \Rightarrow -1={2a\over 2(a-1)} \Rightarrow 1-a=a\Rightarrow a={1\over 2}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{a,b,c成等比 \Rightarrow b^2=ac\\ a-b,a+c,a-2b+c成等差\Rightarrow 2(a+c)=(a-b)+(a-2b+c) \Rightarrow c=-3b } \\ 將c=-3b代入b^2=ac=a(-3b) \Rightarrow b^2=-3ab \Rightarrow b=-3a \Rightarrow {b\over a}=-3,故選\bbox[red, 2pt]{(C)}$$

解答:$$ \begin{bmatrix}a& 1&2& 1\\b& -1& 1& 2\\ a-b& 2& 1& c \end{bmatrix} \xrightarrow{R_1-R_2\to R_1} \begin{bmatrix}a-b& 2&1& -1\\b& -1& 1& 2\\ a-b& 2& 1& c \end{bmatrix} \\ 第一列與第三列需相同,故c=-1,故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cos \angle ABC={\overline{AB}^2+ \overline{BC}^2-\overline{AC}^2 \over 2\cdot \overline{AB}\cdot \overline{BC}} \Rightarrow {1\over 4}={16+36-\overline{AC}^2\over 48} \Rightarrow \overline{AC} =2\sqrt{10} \\ 假設\angle ACD=\angle ACB=\theta,由於\overline{AD} \parallel \overline{BC}, 所以內錯角\angle DAC=\angle ACB=\theta \Rightarrow \triangle DAC為等腰三角形\\\Rightarrow 可假設\overline{DA} =\overline{DC}=a\\ \cases{\cos \angle ACB=\cos \theta=\displaystyle {40+36-16\over 24\sqrt{10}} ={\sqrt{10}\over 4} \\ \cos \angle DAC= \cos \theta=\displaystyle {a^2+40-a^2 \over 4\sqrt{10}a} ={10\over \sqrt{10}a} ={\sqrt{10} \over a} } \Rightarrow {\sqrt{10} \over 4} ={\sqrt{10}\over a} \Rightarrow a=4, 故選\bbox[red, 2pt]{(A)}$$

解答:

$$P、Q對稱於直線L:x-2y+6=0 \Rightarrow 直線L是\overline{PQ}的中垂線\Rightarrow 圓心O在L上\\ d(P,L)={|0+4+6| \over \sqrt{1+4}} ={10\over \sqrt 5} =2\sqrt 5 \\\Rightarrow 半徑R=\overline{OP} \Rightarrow R^2= d(O,\overline{PQ})^2+ d(P,L)^2 =4^2+(2\sqrt 5))^2 =36 \Rightarrow R=\sqrt{36}=6,故選\bbox[red, 2pt]{(D)}$$

解答:$$假設取出三球的號碼依序為a,b,c \Rightarrow abc=8\\ \Rightarrow (a,b,c)=\cases{(1,2,4) \Rightarrow 排列數3!=6, 每種排列機率皆為\displaystyle {4\times 3\times 1\over 8\times 7\times 6} = {1\over 28}\\(2,2,2)\Rightarrow 排列數1, 機率為\displaystyle {3\times 2\times 1 \over 8\times 7\times 6}={1\over 56}} \\ 第1顆是1號有2種:(a,b,c)=(1,2,4),(1,4,2),機率為 {2\times (1/28) \over 6\times (1/28)+1\times (1/56)} ={4\over 13},故選\bbox[red, 2pt]{(C)}$$

解答:

$$正弦定理:{\overline{AD} \over \sin B} = {\overline{AD} \over \sin 45^\circ} =2\cdot 6 \Rightarrow \overline{AD}=12\cdot {\sqrt 2\over 2}=6\sqrt 2 \\ 再一次正弦定理:{\overline{AD} \over \sin C} =2 R \Rightarrow {6\sqrt 2\over \sqrt 3/2} =2R \Rightarrow R=2\sqrt 6,故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{\vec u=(0,b) \\ \vec v=(x,y)} \Rightarrow \vec u+\vec v=(x,b+y) =(2\sqrt 2,2\sqrt 2) \Rightarrow b+y=2\sqrt 2 \Rightarrow y=2\sqrt 2-b \\ 又\vec u\cdot \vec v=by\ge 1 \Rightarrow b(2\sqrt 2-b)\ge 1 \Rightarrow b^2-2\sqrt 2b +1\le 0 \Rightarrow (b-\sqrt 2)^2 \le 1 \\ \Rightarrow -1\le b-\sqrt 2\le 1 \Rightarrow \sqrt 2-1\le b\le 1+\sqrt 2 \Rightarrow b的最大值為1+\sqrt 2,故選\bbox[red, 2pt]{(A)}$$

解答:$$ g(x)=f(x)+f(-x)= 3\cos(\pi x)+ 4\sin(\pi x)+ 3\cos(-\pi x)+ 4\sin(-\pi x)  \\\qquad = 3\cos(\pi x)+ 4\sin(\pi x)+ 3\cos(\pi x)- 4\sin(\pi x) =6\cos(\pi x),,故選\bbox[red, 2pt]{(C)}$$

解答:$$取a=-{1\over 2} \Rightarrow \cases{\log b=-{1\over \sqrt 2} \\ \log \displaystyle {1\over b} =-\log b={1\over \sqrt 2} \\ (\log b)^2=\displaystyle {1\over 2} \\ \log \sqrt b=\displaystyle {1\over 2} \log b=-{1\over 2\sqrt 2}}  \Rightarrow \log{1\over b} 最大,故選\bbox[red, 2pt]{(B)}$$

解答:$$P在平面E:x+y+z=1的投影點為Q \Rightarrow \overrightarrow{PQ} 平行E的法向量(1,1,1) \Rightarrow \overrightarrow{PQ}=(k,k,k) \\ 假設Q(x_0,y_0,z_0) \Rightarrow \overrightarrow{OQ} =(x_0,y_0,z_0) \Rightarrow \overrightarrow{PQ} \cdot \overrightarrow{OQ} =k(x_0+ y_0+z_0)=3 \\由於Q在E上\Rightarrow x_0+ y_0+z_0=1 \Rightarrow k\cdot 1=3\Rightarrow k=3\Rightarrow  \overrightarrow{PQ} =(3,3,3) \Rightarrow |\overrightarrow{PQ}|= 3\sqrt 3,故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)=(x^2+1)g(x)+x 為三次式且首項係數為1 \Rightarrow g(x)=x+a \\ \Rightarrow f(x)=(x^2+1)(x+a)+x \Rightarrow f(1)=2(1+a)+1=-1\Rightarrow a=-2 \\ \Rightarrow f(x)=(x^2+1)(x-2)+x =x^3-2x^2+2x-1\Rightarrow f'(x)= 3x^2-4x+2 \Rightarrow f''(x)=6x-4 \\ f''(x)=0 \Rightarrow 6x-4=0 \Rightarrow x={2\over 3},故選\bbox[red, 2pt]{(A)}$$

解答:$$公差為{\pi\over 2} \Rightarrow \cases{\theta_1=\theta_2-\pi/2\\ \theta_3=\theta_2+\pi/2} \Rightarrow {\pi\over 2}\le \theta_1\lt \pi \Rightarrow {\pi\over 2}\le \theta_2-{\pi\over 2}\lt \pi  \Rightarrow \pi\le \theta_2\lt  {3\pi\over 2} \\ \Rightarrow \theta_2在第三象限\Rightarrow \sin \theta_2,\cos\theta_2均為負值, 又\sin^2\theta_2 ={9\over 25}\Rightarrow  \sin \theta_2= -{3\over 5} \Rightarrow \cos \theta_2=-{4\over 5} \\ \Rightarrow \sin(\theta_1+ \theta_3) =\sin (2\theta_2) =2\sin \theta\cos \theta =2\cdot (-{3\over 5})\cdot (-{4\over 5}) ={24\over 25},故選\bbox[red, 2pt]{(D)}$$

解答:$$擲三次可能的結果\cases{反:機率{1\over 2},得分0\\ 正反:機率{1\over 4},得分1\\正正反:機率{1\over 8},得分3\\ 正正正:機率{1\over 8},得分8} \\ \Rightarrow 期望值:{1\over 2}\cdot 0+{1\over 4}\cdot 1+{1\over 8}(3+8)={13\over 8},故選\bbox[red, 2pt]{(C)}$$

解答:$$A= \begin{bmatrix}\cos 30^\circ& -\sin 30^\circ\\ \sin 30^\circ& \cos 30^\circ \end{bmatrix} = \begin{bmatrix}\sqrt 3/2& -1/2\\ 1/2& \sqrt 3/2 \end{bmatrix} \Rightarrow A \begin{bmatrix}a\\ a \end{bmatrix} = \begin{bmatrix}{(\sqrt 3-1)a\over 2} \\ {(\sqrt 3+1)a\over 2} \end{bmatrix} = \begin{bmatrix}b\\ c \end{bmatrix} \\ \Rightarrow bc={a^2\over 4}(\sqrt 3-1) (\sqrt 3+1) ={a^2\over 2}=2\Rightarrow a=\pm 2 \\ \Rightarrow \cases{a=2 \Rightarrow \cases{b=\sqrt 3-1\\ c=\sqrt 3+1} \Rightarrow b\lt c 不合\\ a=-2 \Rightarrow \cases{b=-1\sqrt 3\\ c=-1-\sqrt 3} \Rightarrow b\gt c} \Rightarrow a=-2,故選\bbox[red, 2pt]{(A)}$$

解答:$$(A) f(x)=x+2 \Rightarrow y_i=0,1,2,3,4 \\ (B)f(x)=x^2+2 \Rightarrow y_i=6,3,2,3,6\\ (C)f(x)=x^3+2\Rightarrow y_i=-6,1,2,3,10\\ (D)f(x)=(x+2)^4 \Rightarrow y_i=0,1,16,81,256 \\ \Rightarrow 只有(B)的y_i不是遞增數列，比較接近水平線,故選\bbox[red, 2pt]{(B)}$$

解答:$$假設福袋有\cases{餅乾a款\\ 飲料b款\\玩具c款}，需同時滿足\cases{1\le a,b,c\le 3\\ 100a+200b+300c=1000 } \\ \Rightarrow (a,b,c)=\cases{(1,3,1) 有C^3_1C^3_3C^3_1 =9種組合\\(2,1,2)  有C^3_2C^3_1C^3_2 =27種組合\\(3,2,1)  有C^3_3C^3_2C^3_1 =9種組合} \Rightarrow 共9+27+9=45種,故選\bbox[red, 2pt]{(D)}$$

解答:$$(A)\cases{甲:45 \times 4=180\\ 乙:60 \times 2=120} \Rightarrow 第一回合結束:\cases{甲:180-120=60 \times 4=240\\ 乙:120 \times 2=240} \\\qquad \Rightarrow 第二回合結束:\cases{甲:240-240=0\\ 乙:240} \Rightarrow 沒有第三回合\\ (B)\cases{甲:50 \times 4=200\\ 乙:60 \times 2=120} \Rightarrow 第一回合結束:\cases{甲:200-120=80 \times 4=320\\ 乙:120 \times 2=240} \\\qquad \Rightarrow 第二回合結束:\cases{甲:320-240=80 \times 4=320\\ 乙:240\times 2=480} \Rightarrow 第三回合結束
:甲:320-480\lt 0 \\(C)\cases{甲:55 \times 4=220\\ 乙:60 \times 2=120} \Rightarrow 第一回合結束:\cases{甲:220-120=100 \times 4=400\\ 乙:120 \times 2=240} \\\qquad \Rightarrow 第二回合結束:\cases{甲:400-240=160 \times 4=640\\ 乙:240\times 2=480} \\\Rightarrow 第三回合結束甲:640\gt 480 \Rightarrow 還有第四回合\\故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{L_1的方向向量\vec u_1=(1,1,-2) \\ L_2的方向向量\vec u_2=(2,-4,-1) } \Rightarrow  \vec u_1\times \vec u_2 =(-9,-3,-6)    \Rightarrow 取L 的方向向量\vec u_3=(3,1,2) \\ 假設\cases{L_1與L的交點為P \\L_2與L的交點為Q} \Rightarrow \cases{P(t,t,-2t) \\Q(2s+2,-4s+6,-s+1)}\\ \Rightarrow \overrightarrow{PQ}=(2s-t+2,-4s-t+6,-s+2t+1) 與\vec u_3平行 \Rightarrow \cases{2s-t+2=3k\\ -4s-t+6=k\\ -s+2t+1=2k} \\ \Rightarrow \cases{s=1\\ t=1} \Rightarrow \cases{P(1,1,-2) \\ Q(4,2,0)} \Rightarrow (a,b,c)={1\over 2}(P+Q) =({5\over 2},{3\over 2},-1) \Rightarrow a+b+c=3,故選\bbox[red, 2pt]{(A)}$$

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