114年馬祖高中教甄(二)-數學詳解

 國立馬祖高級中學114學年度第2次專任教師甄試

單選題 

解答:$$\cases{a=9 \Rightarrow -3\le x\le 15 \\b=1 \Rightarrow x\lt -3或x\gt -1} \Rightarrow 兩者交集 \Rightarrow 0\le x\le 15 \Rightarrow 整數x共有16個 \Rightarrow a+b=10\\，故選\bbox[red, 2pt]{(2)}$$

解答:$$當底面為正方形時，表面積最小。假設正方形邊長為x \Rightarrow 12x^2=588 \Rightarrow x=7 \\ \Rightarrow 表面積=x^2+2(12x+12x) =385，故選\bbox[red, 2pt]{(3)}$$

解答:$$圓C:(x-1)^2+(y-1)^2=1 \Rightarrow d(L_1,L_2)=2 \Rightarrow {|12-k|\over 5} =2 \Rightarrow k=2，故選\bbox[red, 2pt]{(4)}$$

解答:

$$\triangle ABC三角分別為60^\circ-30^\circ-90^\circ \Rightarrow \cases{ \overline{BC} =\sqrt 3\cdot \overline{AC}=12} \\ 又\cases{\angle AMB= 120^\circ \Rightarrow \angle MAB=\angle MBA=30^\circ\\ \angle ANC= 120^\circ \Rightarrow \angle NAC= \angle NCA =30^\circ} \Rightarrow 取\cases{C(0,0) \\A(0,4\sqrt 3) \\B(12,0) \\M(8,4\sqrt 3) \\N(-2,2\sqrt 3,)} \\ \Rightarrow \overline{MN} = \sqrt{10^2 +( 2\sqrt 3)^2}= \sqrt{112} =4\sqrt{7}，故選\bbox[red, 2pt]{(4)}$$

解答:$$\begin{array}{} a& b& c& 數量\\\hline 1& 4& 8-20& 13\\ 1& 5& 9-20& 12\\ \cdots& \cdots& \cdots\\1& 16& 20&1\\\hdashline 2& 5& 9-20& 12\\ 2&6&10-20&11 \\ \cdots& \cdots& \cdots\\ 2& 16& 20& 1\\ \hdashline \vdots & \vdots &\vdots \\\hdashline 13& 16& 20& 1\\\hline\end{array} \Rightarrow 總計=\sum_{k=1}^{13} {k(k+1)\over 2} = 455，故選\bbox[red, 2pt]{(2)}$$

解答:

$$f(x)=(x+1)^2(x-2)-3 \xrightarrow{向左平移4單位} g(x)=f(x+4) =(x+5)^2(x+2)-3 \\ \xrightarrow {再向上平移3單位} h(x)=g(x+3) =(x+5)^2(x+2) \Rightarrow \cases{k=3\\ \alpha=-5\\ \beta=-2} \Rightarrow k+\alpha-\beta= 0，故選\bbox[red, 2pt]{(5)}$$

解答:$$\cases{C_1與座標軸交於A(2,0), B(0,2), C(-2,0), D(0,-2) \\ C_2與x=y交於E(1,1), F(-1,-1)\\ C_2與x+y=0交於G(1,-1),H(-1,1)} \Rightarrow \cases{A,B,E共線\\ B,C,H共線\\ C,D,F共線\\ A,D,G共線} \\ \Rightarrow可決定 \cases{C^8_2-4(C^3_2-1)=20條直線 \\ C^8_3-4=52個三角形} \Rightarrow 20+52=72，故選\bbox[red, 2pt]{(1)}$$

解答:$$a_n=\sin(n\pi+{\pi\over 6}) \Rightarrow \langle a_n\rangle =-{1\over 2},{1\over 2},-{1\over 2},{1\over 2},\dots \\ \Rightarrow \sum_{n=1}^\infty (a_n)^n=-{1\over 2}+({1\over 2})^2+(-{1\over 2})^3+ ({1\over 2})^4+\dots  =-{1\over 2}\left(1+ (-{1\over 2})^2 +(-{1\over 2})^3 +(-{1\over 2})^4 + \cdots\right) \\=-{1\over 2} \cdot {1\over 1-(-1/2)} =-{1\over 3}，故選\bbox[red, 2pt]{(1)}$$

解答:$$\cases{2白球:機率=1/C^n_2\\ 2綠球:機率=C^{n-3}_2/C^n_2 \\ 1藍1白:機率=C^2_1/C^n_2 \\ 1藍1綠: 機率=C^{n-3}_1/C^n_2 \\ 1白1綠: 機率=C^2_1C^{n-3}_1/C^n_2} \\\Rightarrow E_n={1\over C^n_2} (2n+ 2C^{n-3}_2+ 3nC^2_1+ (2n+1)C^{n-3}_1+ (n+1)C^2_1C^{n-3}_1) =10-{6\over n}為整數 \\ \Rightarrow n=6\Rightarrow E_n=9 \Rightarrow n+E_n=15，故選\bbox[red, 2pt]{(3)}$$

解答:$$\log(2^{2^{10}}) =2^{10}\log 2=1024 \log 2=1024\times 0.301= 308.224  \Rightarrow \cases{n=309\\ m=308} \\ \Rightarrow \log a=0.224 \lt \log 2 \Rightarrow b=1 \Rightarrow m+n+b= 618，故選\bbox[red, 2pt]{(5)}$$

解答:$$|\sin (\pi x)-\cos(\pi x)|=1 \Rightarrow x=0,{1\over 2},1,{3\over 2},2,\dots, 30 \\ \Rightarrow \sum a=(1+2+\cdots+30) +{1\over 2}(1+3+\cdots+{59})=465+450 =915，故選\bbox[red, 2pt]{(5)}$$

解答:$$\cases{O(0,0) \\ P(2\cos\theta+2, 2\sin \theta)\\ Q(4,3)} \Rightarrow \triangle OPQ面積={1\over 2} \begin{vmatrix} 0& 0& 1\\ 2\cos \theta+2& 2\sin \theta& 1\\ 4& 3& 1\end{vmatrix} =|3\cos \theta-4\sin \theta+3|\\ =5\sin(\alpha-\theta)+3 \Rightarrow m=8,此時\cases{\cos \theta=3/5\\ \sin \theta=-4/5} \Rightarrow P=({16\over 5},-{8\over 5}) =(a,b) \\\Rightarrow m+2a-b=8+{32\over 5}+{8\over 5} =16，故選\bbox[red, 2pt]{(2)}$$

解答:$$假設\cases{\vec b=\overrightarrow{AB} \\ \vec c=\overrightarrow{AC} \\\vec d= \overrightarrow{AD}} \Rightarrow  {\vec d\cdot \vec b\over |\vec b|^2}\vec b={5\over 2}\vec b \Rightarrow {\vec d\cdot \vec b\over |\vec b|^2}={5\over 2} \Rightarrow {(3\vec b+2 \vec c)\cdot \vec b\over |\vec b|^2}={3+{2\vec c\cdot \vec b\over |\vec b|^2}} ={5\over 2} \\ \Rightarrow {\vec c\cdot \vec b\over |\vec b|^2}=-{1\over 4} \Rightarrow k=-{1\over 4} \Rightarrow \cases{A(0,0)\\ B(b,0) \\ C(-b/4,y) \\D(5b/2, x)} \Rightarrow \vec d=3\vec b+2\vec c \Rightarrow x=2y \Rightarrow D(5b/2,x) \\ \Rightarrow {ABDC\over \triangle ABC} ={\triangle ABC+ \triangle BCD\over \triangle ABC} ={ \begin{vmatrix} 5b/2& 2y& 1\\ -b/4& y& 1\\ b& 0& 1\end{vmatrix}+\begin{vmatrix} -b/4& y& 1\\ 0& 0& 1\\ b& 0& 1\end{vmatrix}\over \begin{vmatrix} -b/4& y& 1\\ 0& 0& 1\\ b& 0& 1\end{vmatrix} } =5=h \\ \Rightarrow 4k+2h=-1+10=9，故選\bbox[red, 2pt]{(4)}$$

解答:$$ABCD為矩形\Rightarrow \overrightarrow{AD} \bot \overrightarrow{AB} \Rightarrow \overrightarrow{AD} \times \overrightarrow{AB} =0，其它皆不一定正確，故選\bbox[red, 2pt]{(1)}$$

解答:$$\overleftrightarrow{FG} \bot \overleftrightarrow{DH} \Rightarrow (2,3,-1) \cdot(1,a,-4) =0 \Rightarrow  a=-2 \\ \vec n=(2,3,-1) \times(1,a,-4) =-7(2,-1,1) \Rightarrow 包含\overleftrightarrow{FG} 的平面E:2(x-3)-(y-5)+(z-2)=0 \\ \Rightarrow  E:2x-y+z=3 \Rightarrow 邊長=兩歪斜線距離=(1,2,12)與E的距離={2-2+12-3\over \sqrt{6}} ={3\over 2} \sqrt 6\\，故選\bbox[red, 2pt]{(3)}$$

解答:$$1,z,z^3共線\Rightarrow {z^3-z\over z-1} =k \in \mathbb R \Rightarrow z^2+z-k=0 \Rightarrow z={-1\pm \sqrt{1+4k}\over 2}  \Rightarrow 實部=-{1\over 2}\\ -2, z,z^2共線\Rightarrow {z^2-z\over z+2}=t\in \mathbb R \Rightarrow z^2-(t+1)z-2t=0 \Rightarrow z={t+1 \pm \sqrt{(t+1)^2+8t}\over 2} \\ \Rightarrow 實部={t+1\over 2}=-{1\over 2} \Rightarrow t=-2 \Rightarrow z={-1\pm \sqrt{15}i\over 2} \\ 若z={-1+ \sqrt{15}i\over 2} \Rightarrow z^2={-7-\sqrt{15}i\over 2} \Rightarrow z^3={11-3\sqrt{15}i\over 2} \Rightarrow \cases{A(z)=(-1/2,\sqrt{15}/2) \\ B(z^2)=(-7/2,-\sqrt{15}/2) \\C(z^3)=(11/2,-3\sqrt{15}/2) } \\ \Rightarrow \triangle ABC面積={1\over 2}\begin{vmatrix} -1/2& \sqrt{15}/2& 1\\ -7/2& -\sqrt{15}/2& 1 \\11/2& -3\sqrt{15}/2& 1\end{vmatrix} =6\sqrt{15}，故選\bbox[red, 2pt]{(3)}$$

解答:$$\cases{0\le x\le 15\\ x+y\le 24 \\ x\le 3y \\0\le y\le 2x} 所圍區域頂點坐標\cases{A(8,16) \\ B(15,9) \\C(15,5) \\ O(0,0)} \Rightarrow 獲利函數f(x,y)=7x+6y\\ \Rightarrow \cases{f(A)=152\\ f(B) =159\\ f(C)= 135\\ f(O)=0} \Rightarrow 獲利最大159萬元，故選\bbox[red, 2pt]{(4)}$$

解答:$$P(3,0)及Q(1,1)均在直線x+2y=3上\Rightarrow \cases{T(P)=(3,-6) \\T(Q)=(1,-1)} \Rightarrow \cases{3a-6b=3\\ a-b=3} \\ \Rightarrow \cases{a=5\\b=2} \Rightarrow a-b=3，故選\bbox[red, 2pt]{(5)}$$

解答:$$f(x)=2x^3-3ax^2+6 \Rightarrow f'(x)=6x^2-6ax \Rightarrow f''(x)=12x-6a\\ f'(x)=0 \Rightarrow 6x(x-a)=0 \Rightarrow x=0,a \Rightarrow \cases{f''(0)=-6a\lt 0 \Rightarrow f(0)為極大值\\ f''(a)=6a\gt 0 \Rightarrow f(a)為極小值} \\ \Rightarrow S(t)=2t^3-3\cdot 2t^2+6 \Rightarrow \int_0^2 S(t)\,dt = \int_0^2 2t^3-6t^2+6\, dt=4，故選\bbox[red, 2pt]{(1)}$$

解答:$$A\cap B \Rightarrow 1-(x-1)^2=ax \Rightarrow x^2+(a-2)x =0 \Rightarrow x=0,2-a\\ \Rightarrow 旋轉體積=\pi \int_0^{2-a} (1-(x-1)^2-ax)\,dx ={9\over 16}\pi \Rightarrow {(2-a)^3\over 6} ={9\over 16} \Rightarrow a={1\over 2}，故選\bbox[red, 2pt]{(2)}$$

解答:$$假設\cases{\langle a_n\rangle 公差為d_1\\ \langle b_n\rangle 公差為d_2} \Rightarrow \lim_{n\to \infty}{a_n\over b_n} =\lim_{n\to \infty}{a_1+(n-1)d_1\over b_1+(n-1)d_2} = {d_1\over d_2}=5 \\ \Rightarrow \lim_{n\to \infty}{a_1+a_2 +\cdots+a_n\over n\cdot b_{2n}} =\lim_{n\to \infty}{(2a_1+(n-1)d_1)n/2 \over n\cdot (b_{1}+(2n-1)d_2)} ={d_1\over 4d_2}={5\over 4}，故選\bbox[red, 2pt]{(3)}$$

解答:$$\cases{甲袋中取出1白1紅的機率=C^3_1C^2_1/C^5_2=3/5\\ 乙袋中取出1白1紅的機率=C^1_1C^2_1/C^3_2=2/3 }   \Rightarrow {3/5\over 3/5+2/3} ={9\over 19}，故選\bbox[red, 2pt]{(2)}$$

解答:$$\overrightarrow{BC} \cdot \overrightarrow{AC} =(-\overrightarrow{AB}+ \overrightarrow{AC}) \cdot \overrightarrow{AC} =-\overrightarrow{AB}\cdot \overrightarrow{AC}+ \overline{AC}^2=-20+ \overline{AC}^2 =5 \\ \Rightarrow \overline{AC}=5，故選\bbox[red, 2pt]{(1)}$$

解答:$$A=\begin{bmatrix} a& b\\ c&-a \end{bmatrix} \Rightarrow A^{-1}={1\over a^2+bc} A \Rightarrow A-A^{-1} ={a^2+bc-1\over a^2+bc}A \\ 又\det(A)=-a^2-bc={1\over 2} \Rightarrow a^2+bc=-{1\over 2} \\\Rightarrow A-A^{-1} =3A \Rightarrow \det(A-A^{-1}) =3^2\det(A) ={9\over 2}，故選\bbox[red, 2pt]{(4)}$$

解答:$$ax^2+(2a+b)x-12= a(x^2+(2-a)x-2a)+6=ax^2+(2a-a^2)x-2a^2+6 \\ \Rightarrow \cases{2a+b=2a-a^2\\ -12=-2a^2+6} \Rightarrow \cases{a=3\\ b=-9} \Rightarrow a-b=12，故選\bbox[red, 2pt]{(5)}$$

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