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2022年2月7日 星期一

105年身障生升大學-數學甲詳解

105 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:大學組數學甲

單選題,共 20 題,每題 5 分

解答:$$將y=1+2x代入y=x^2-2x+3,可得x^2-2x+3=1+2x \Rightarrow x^2-4x+2=0\\  \Rightarrow \cases{x={2+\sqrt 2} \Rightarrow y=1+2x=5+2\sqrt 2\\ x=2-\sqrt 2 \Rightarrow y=1+2x= 5-2\sqrt 2} \Rightarrow 兩交點\cases{(2+\sqrt 2,5+2\sqrt 2)\\ (2-\sqrt 2,5-2\sqrt 2)}皆在第一象限\\,故選\bbox[red,2pt]{(C)}$$
解答:$$-a^2\lt x\lt a的整數x共有341個\Rightarrow a^2+(a-1)=341 \Rightarrow a^2+a-342=0 \Rightarrow (a-18)(a+19)=0\\ \Rightarrow a=18 \Rightarrow -3a\lt x\lt (a-1)^2 \Rightarrow -54\lt x\lt 289 \Rightarrow 共有53+288+1=342個,故選\bbox[red,2pt]{(C)}$$
解答:$$取\cases{A(1,0)\\ B(\sqrt 3\cdot \cos (5\pi/6),\sqrt 3\sin(5\pi/6))=(-3/2,\sqrt 3/2)} ,則\cases{\vec u=\overrightarrow{OA} =(1,0)\\ \vec v=\overrightarrow{OB} =(-3/2,\sqrt 3/2)} \\ \Rightarrow \cases{|\vec v|=\sqrt 3\\ |\vec v+\vec u|= |(-1/2,\sqrt 3/2)| =1\\ \lvert\vec v+{3\over 2}\vec u\rvert =|(0,\sqrt 3/2)| =3/4\\ |\vec v+2\vec u| =|(1/2,\sqrt 3/2)| =1} \Rightarrow |\vec v|最大,故選\bbox[red,2pt]{(A)}$$
解答:$$從甲乙丙三箱抽出的號碼依序為(2,3,1)或(3,1,2)兩種情形,每種情形的機率皆為({1\over 2})^3={1\over 8}\\因此機率為{1\over 8}\times 2= {1\over 4},故選\bbox[red,2pt]{(B)}$$
解答:$$出現n次正面的機率為P(n)=C^5_n({1\over 3})^n({2\over 3})^{5-n}\\ P(0)=({2\over 3})^5= {32\over 243}\\ P(1)=C^5_1 ({1\over 3}) ({2\over 3})^{4} ={80\over 243} \\ P(2)=C^5_2 ({1\over 3})^2 ({2\over 3})^{3} ={80\over 243} \\ P(3)=C^5_3 ({1\over 3})^3 ({2\over 3})^{2} ={40\over 243}\\ \Rightarrow 全部出現反面機率最小,故選\bbox[red,2pt]{(A)}$$
解答:$$A=\begin{bmatrix} a& b\\ c& d\end{bmatrix} \Rightarrow \cases{A\begin{bmatrix} 1\\ 2\end{bmatrix} =\begin{bmatrix} 5\\ 6\end{bmatrix} \\[1ex] A\begin{bmatrix} 3\\ 4\end{bmatrix} =\begin{bmatrix} 7\\ 8\end{bmatrix}} \Rightarrow \cases{\cases{a+2b=5\\ c+2d=6} \\ \cases{3a+4b= 7\\ 3c+4d=8}} \Rightarrow \cases{a=-3\\ b=4\\c=-4\\ d=5} \\ A=\begin{bmatrix} -3& 4\\ -4& 5\end{bmatrix} \Rightarrow A\begin{bmatrix} 1& 2\\ 3& 4\end{bmatrix} = \begin{bmatrix} 9& 10\\ 11 & 12 \end{bmatrix} ,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{x+2y-z=-3\\ 2x-3y+az= 8\\ 3x+y-8z=b} 有無限多組解\Rightarrow \begin{vmatrix} 1 & 2 &-1 \\ 2 & -3 & a\\ 3 & 1 & -8\end{vmatrix}=0 \Rightarrow a=-9 \\ \Rightarrow \cases{x+2y-z=-3 \cdots(1)\\ 2x-3y-9z= 8 \cdots(2)\\ 3x+y-8z=b \cdots(3)} \xrightarrow{-2(1)+(2),-3(1)+(3)} \cases{-7y-7z=14\cdots(2')\\ -5y-5z=9+b\cdots(3')} \\若 14\times {5\over 7}=9+b 有無限多組解,即b=1;因此a+b=-9+1=-8,故選\bbox[red,2pt]{(A)}$$
解答:$$過A(4,8,-5)且方向向量為(3,-2,1)的直線L,若P在L上 ,則P(4+3t,8-2t,-5+t),t\ge 0;\\(A) P過xy平面\Rightarrow -5+t=0 \Rightarrow t=5\\ (B)P過xz平面\Rightarrow 8-2t=0 \Rightarrow t=4\\ (C)P過yz平面\Rightarrow 4+3t=0 \Rightarrow t=-4/3 \not \ge 0\\ (D)x+y+z=12 \Rightarrow 7+2t=12 \Rightarrow t=5/2 \\ 由以上可知當t=5/2時會碰到x+y+z=12,然後是xz平面(t=4),最後是xy平面\\,故選\bbox[red,2pt]{(D)}$$
解答:$$f(x)=\cases{(x+1)g(x)+a \\ (x+1)(x-2)h(x)+ bx+c},已知\cases{f(-1)=1 \Rightarrow \cases{a=1\\ -b+c=1}\\ g(2)=-2 \Rightarrow f(2)= 3\cdot (-2)+a=-5 = 2b+c} \\ \Rightarrow \cases{a=1\\ b=-2\\ c=-1} \Rightarrow 餘式bx+c =-2x-1,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{A(-4,-4)\\ B(-6,-5)\\ C(4,4)\\ D(5,6)} \Rightarrow \overline{AB}的中垂線L_1'與\overline{CD}的中垂線L_2'交點即為圓心 \\ 令\cases{\vec u=\overrightarrow{AB}=(-2,-1)\\ \vec v=\overrightarrow{CD}=(1,2)} \Rightarrow \cases{L_1':-2(x+4)-(y+4)=0\\ L_2':(x-4)+2(y-4)=0} \Rightarrow \cases{2x+y=-12 \\x+2y=12} \Rightarrow \cases{x=-12\\ y=12}\\ \Rightarrow (-12,12)位於第二象限,故選\bbox[red,2pt]{(B)}$$
解答:$$(1+p)^{12}=1.44 \Rightarrow 12\log(1+p)=\log 1.44 \Rightarrow 18\log(1+p)= {3\over 2}\log 1.44 \Rightarrow (1+p)^{18} =1.44^{3/2} \\ =1.2^3 =1.728 \Rightarrow 利息/本金=72.8\%,故選\bbox[red,2pt]{(C)}$$
解答:$$假設一開始箱子有\cases{a顆藍球\\ b顆綠球} \Rightarrow \cases{E={a\over a+b} \\ {10\over 11}E={a\over a+b+2}} \Rightarrow {10a\over 11(a+b)}= {a\over a+b+2} \\\Rightarrow 11a^2+11ab=10a^2+10ab+20a \Rightarrow a^2+ab-20a=0 \\\Rightarrow a(a+b-20)=0 \Rightarrow a+b=20 \Rightarrow 一開始共有20顆球,故選\bbox[red,2pt]{(C)}$$
解答:$$(A) \times:{1\over 1000} \lt {1\over 567} \Rightarrow \left({1\over 1000}\right)^4 \lt \left({1\over 567}\right)^5 \Rightarrow \log\left({1\over 567}\right)^5\gt \log\left({1\over 1000}\right)^4=-12\\(B)\bigcirc: 由於\cases{({7\over 100000})^3= 3.42\times 10^{-13}\\ ({7\over 90000})^3 \approx 4.7\times 10^{-13}} 且{7\over 100000}\lt {7\over 98765 } \lt {7\over 90000} \\\qquad \Rightarrow 其對數值介於-12與-13之間\\(C)\times:\log\left( {1\over \sqrt{50}}\times 10^{-6}\right)^2=-12+\log {1\over 50}=-12+(\log 2-2) \lt -13 \\\qquad 因此{1\over 7654321} = {1\over 7.654321} \times 10^{-6} \lt {1\over \sqrt{50}}\times 10^{-6} \Rightarrow \left({1\over 7654321}\right)^2 \lt -13\\(D)\times: 101\times 10^{-14} \gt 100\times 10^{-14}=10^{-12} \Rightarrow   \log(101\times 10^{-14}) \gt -12\\,故選\bbox[red,2pt]{(B)}$$
解答
$$在\overline{QR}上取一點S,使得\overline{PS}\bot \overline{QR},如上圖;\\令\cases{h=\overline{PS}\\ a=\overline{QS} \Rightarrow \overline{RS}=6-a},依題意:2\tan Q=\tan R \Rightarrow 2\cdot {h\over a}={h\over 6-a} \Rightarrow a=4\\ \Rightarrow h=\sqrt{3^2-(6-4)^2} =\sqrt 5 \Rightarrow \cases{\cos R={3^2+ 2^2-5\over 2\cdot 3\cdot 2} ={2\over 3}\\\cos R={3^2+6^2-\overline{PQ}^2\over 2\cdot 3\cdot 6} ={45-\overline{PQ}^2\over 36} } \\ \Rightarrow {2\over 3}={45-\overline{PQ}^2\over 36} \Rightarrow \overline{PQ}^2=21 \Rightarrow \overline{PQ}= \sqrt{21},故選\bbox[red,2pt]{(C)}$$
解答:$$(A)\times: \cases{a=4\\ b=2} \Rightarrow \cases{a^b=16\\ b^a=16 } \Rightarrow a^b \not \gt b^a\\ (B)\times: 反例同上\\(C)\times: \cases{a=4\\ b=2} \Rightarrow \cases{(b^a)^b=16^2=256 \\ b^{(a^b)}= 2^{16} } \Rightarrow (b^a)^b \not \gt b^{(a^b)}\\,故選\bbox[red,2pt]{(D)}$$
解答:$$\angle P+\angle Q+\angle R=180^\circ \Rightarrow \tan P+\tan(Q+R)=0;\\(A)\times:\angle P\lt \angle Q+\angle R  \Rightarrow \tan P\gt 0且\tan(Q+R)\lt 0\Rightarrow \tan P不是最小\\ (B)\bigcirc:\angle P\gt \angle Q+\angle R \Rightarrow \tan P\lt 0且\tan Q\gt 0且\tan R\gt 0 \Rightarrow \tan P最小\\ (C) \times: \cases{\angle P=30^\circ\\ \angle Q=40^\circ \\ \angle R=110^\circ} 滿足\cases{\angle P\lt \angle Q\\ \angle P\lt \angle R},但\cases{\tan R\lt 0\\ \tan P\gt 0}\Rightarrow \tan P不是最小 \\(D)\times: \cases{\angle P=90^\circ\\ \angle Q=60^\circ \\ \angle R=30^\circ} 滿足\cases{\angle P\gt \angle Q\\ \angle P\gt \angle R},但 \tan P=\infty \Rightarrow \tan P不是最小\\,故選\bbox[red,2pt]{(B)}$$
解答:$${p+q\over 2}\ge \sqrt{pq} \Rightarrow 5 \ge \sqrt{pq} \Rightarrow 50\ge 2pq \Rightarrow 2pq的最大值為50;\\ \cos R={p^2+q^2-r^2\over 2pq} ={(p+q)^2-2pq-7^2\over 2pq}={10^2-2pq-7^2\over 2pq}={51-2pq\over 2pq}\\ ={51\over 2pq}-1\ge {51\over 50}-1={1\over 50}=0.02,故選\bbox[red,2pt]{(C)}$$
解答:$$z={3\over 5}+ {4\over 5}i =\cos \theta+ i\sin \theta \Rightarrow \cases{1=\cos 0^\circ +i\sin 0^\circ \\ z=\cos \theta+ i\sin \theta\\  z^2 =\cos 2\theta+ i\sin 2\theta}\\ \Rightarrow {1-z\over z^2-z} =-{1\over z}= -{1\over \cos\theta+i\sin \theta} = -\cos\theta+i\sin \theta =\cos(\pi-\theta)+i\sin(\pi-\theta) \\ 由於\cases{\cos\theta =3/5\\ \sin\theta = 4/5} \Rightarrow 45^\circ\lt \theta \lt 60^\circ \Rightarrow 120^\circ \lt \angle PQR =\pi-\theta \lt 135^\circ  ,故選\bbox[red,2pt]{(B)}$$
解答:$${3\over 2}\pi \lt 5\lt {7\over 4}\pi \Rightarrow \cases{-\sqrt 2/2\lt \sin 5 \lt -1 \\ -\sqrt 2\lt \csc 5\lt -1 \\ 0\lt \cos  5\lt \sqrt 2/2 \\ \sqrt 2\lt \sec 5\lt \infty\\ -\infty \lt \tan 5 \lt -1 \\ -1\lt \cot 5 \lt 0}\\ \Rightarrow \sec 5\gt \cos 5\gt \cot 5 \gt \sin x\gt \csc x\gt \tan x,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{Q(0,4)\\ R(-3,0)} \Rightarrow 直線L=\overleftrightarrow{QR}: 4x-3y+12=0 \Rightarrow \cases{d(P(8,0),L)= 44/5\\ d(S(0,-3),L) = 21/5} \\取較長邊長44/5為正方形邊長,故選\bbox[red,2pt]{(A)}$$

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