高雄市 115 學年度市立國民中學教師聯合甄選
科目:學科專業科目-數學
說明:本試卷共計 50 題，為四選一單選題(每題 2 分，共 100 分)
解答:$$直角\triangle 斜邊上的中線長=外接圓半徑 \Rightarrow 斜邊長=直徑=2\\ 假設直角\triangle兩股長分別為a,b \Rightarrow \cases{a^2+b^2=2^2=4 \\面積={1\over 2}ab} \Rightarrow 算幾不等式:a^2+b^2 \ge 2\sqrt{a^2b^2}\\ \Rightarrow ab\le {a^2+b^2\over 2}=2\Rightarrow {1\over 2}ab\le 1，故選\bbox[red, 2pt]{(D)}$$
解答:$$取g(x)=f(x)-2 \Rightarrow 2,-1為g(x)=0的兩根 \Rightarrow g(x)=f(x)-2=a(x-2)(x+1)(x-b) \\ \Rightarrow \cases{g(1)=f(1)-2=-2= -2a(1-b) \\ g(-2)=f(-2)-2 =-8=4a(-2-b)} \Rightarrow \cases{a=1\\ b=0} \Rightarrow g(x)=f(x)-2=x(x-2)(x+1) \\ \Rightarrow g(3)=f(3)-2=12 \Rightarrow f(3)=14，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{\overline{BC} = \sqrt{\overline{AB}^2 - \overline{AC}^2} = \sqrt{10^2 - 6^2} = 8 \\ \overline{BD} = \sqrt{\overline{AB}^2 - \overline{AD}^2} = \sqrt{10^2 - 8^2} = 6 =\overline{AC}} \Rightarrow \overparen{AC} = \overparen{BD} \Rightarrow \overparen{AC} + \overparen{AD} = \overparen{BD} + \overparen{AD} = \overparen{AB}\\ \overline{BE} = \sqrt{\overline{AB}^2 - \overline{AE}^2} = \sqrt{10^2 - 5^2} = \sqrt{75} \Rightarrow \overparen{AF} > \overparen{BE} \Rightarrow \overparen{AE} + \overparen{AF} > \overparen{AE} + \overparen{BE} = \overparen{AB} \\ 綜合以上:\cases{ \overparen{AC} + \overparen{AD} = \overparen{AB} \\ \overparen{AE} + \overparen{AF} > \overparen{AB} }，故選\bbox[red, 2pt]{(B)}$$
解答:$$u={6x-2\over x^2+5} \Rightarrow u+{1\over u}=2 \Rightarrow u^2-2u+1=0 \Rightarrow(u-1)^2=0 \\ \Rightarrow u= {6x-2\over x^2+5}=1 \Rightarrow x^2-6x+7=0 \Rightarrow 兩根之和=6，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{3^{0} \equiv 1 \text{ (mod 5) }\\ 3^{1} \equiv 3 \text{ (mod 5) }\\ 3^{2} \equiv 4 \text{ (mod 5) }\\ 3^{3} \equiv 2 \text{ (mod 5) }\\ 3^{4} \equiv 1 \text{ (mod 5) } } \Rightarrow 週期為4 \Rightarrow \cases{ a \equiv 0\text{ (mod 4)} \Rightarrow 3a+1 \equiv 1\text{ (mod 4)} \Rightarrow 3^{3a+1} \equiv 3^1 \equiv 3 \text{ (mod 5)} \\ a \equiv 1\text{ (mod 4)} \Rightarrow 3a+1 \equiv 0\text{ (mod 4)} \Rightarrow 3^{3a+1} \equiv 3^0 \equiv 1 \text{ (mod 5)} \\ a \equiv 2\text{ (mod 4)} \Rightarrow 3a+1 \equiv 3\text{ (mod 4)} \Rightarrow 3^{3a+1} \equiv 3^3 \equiv 2 \text{ (mod 5)} \\ a \equiv 3\text{ (mod 4)} \Rightarrow 3a+1 \equiv 2\text{ (mod 4)} \Rightarrow 3^{3a+1} \equiv 3^2 \equiv 4 \text{ (mod 5)} } \\ \Rightarrow \cases{a\equiv 0 \text{ (mod 5)} \Rightarrow a^3 \equiv 0 \text{ (mod 5)} \\ a\equiv 1 \text{ (mod 5)} \Rightarrow a^3 \equiv 1\text{ (mod 5)} \\a\equiv 2 \text{ (mod 5)} \Rightarrow a^3 \equiv 3 \text{ (mod 5)} \\a\equiv 3 \text{ (mod 5)} \Rightarrow a^3 \equiv 2 \text{ (mod 5)} \\a\equiv 4 \text{ (mod 5)} \Rightarrow a^3 \equiv 4 \text{ (mod 5)} }\;; 考慮 3^{3a+1}與 a^3 同餘\text{ (mod 5)} 的情形 \\ \cases{餘數皆為3:同時滿足a \equiv 0 \text{ (mod 4)} 與 a\equiv 2 \text{ (mod 5)} \Rightarrow a \equiv 12 \text{ (mod 20)} \\ 餘數皆為1:同時滿足a \equiv 1 \text{ (mod 4)} 與 a\equiv 1 \text{ (mod 5)} \Rightarrow a \equiv 1 \text{ (mod 20)} \\餘數皆為2:同時滿足a \equiv 2 \text{ (mod 4)} 與 a\equiv 3 \text{ (mod 5)} \Rightarrow a \equiv 18 \text{ (mod 20)} \\ 餘數皆為4:同時滿足a \equiv 3 \text{ (mod 4)} 與 a\equiv 4 \text{ (mod 5)} \Rightarrow a \equiv 19 \text{ (mod 20)} }\\ \Rightarrow 每 20 個連續整數中恰好有 4 個 \Rightarrow 2026=20\times 101+6 \Rightarrow 共有101\times 4+1=405個數同餘\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{A(-3,3) \\B(2,-4)\\ C=(x,0)或(0,y) \\ \triangle ABC為直角三角形} \Rightarrow \overleftrightarrow{AB}斜率m_{AB} ={-4-3\over 2-(-3)} =-{7\over 5} \\ \textbf{Case I }\angle A=90^\circ \Rightarrow \overline{AC} \bot \overline{AB} \Rightarrow m_{AC}={5\over 7} \Rightarrow \overleftrightarrow{AC}: y-3={5\over 7}(x+3) \Rightarrow 5x-7y+36=0 \\\qquad \Rightarrow \cases{C_1= \overleftrightarrow{AC} \cap x軸=(-36/5,0)\\ C_2=\overleftrightarrow{AC} \cap y軸=(0,36/7)} \\ \textbf{Case II }\angle B=90^\circ \Rightarrow \overline{BC} \bot \overline{AB} \Rightarrow m_{BC}={5\over 7} \Rightarrow \overleftrightarrow{BC}: 5x-7y=38 \\ \qquad \Rightarrow \cases{C_3= \overleftrightarrow{BC} \cap x軸= (38/5,0)\\ C_4=\overleftrightarrow{BC} \cap y軸=(0,-38/7)} \\ \textbf{Case III }\angle C=90^\circ \Rightarrow \overline{AC} \bot \overline{BC} \Rightarrow \cases{C=(x,0) \Rightarrow ({0-3\over x+3})\cdot ({0+4\over x-2}) =-1 \\C=(0,y) \Rightarrow ({y-3\over 0+3})\cdot ({y+4\over 0-2})=-1}\\\qquad \Rightarrow \cases{x^2+x-18=0 \Rightarrow x\not \in \mathbb Q \\ y^2+y-18=0 \Rightarrow y\not \in \mathbb Q} \\ 因此C點共有4種可能，故選\bbox[red, 2pt]{(B)}$$
解答:$$假設\cases{x=P(A) \\y=P(B)} \Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B) =P(A)+P(B)-P(A ) P(B) \\ \Rightarrow 0.84=x+y-xy \Rightarrow 0.84-1=x+y-xy-1 \Rightarrow 0.16 =(1-x)(1-y) \\ \Rightarrow xy =(1-(1-x))(1-(1-y))=1-[(1-x)+(1-y)]+0.16 \\ \Rightarrow xy要最大相當於(1-x)+(1-y)要最小 \Rightarrow (1-x)+(1-y) \ge 2\sqrt{(1-x)(1-y)} =2\sqrt{0.16}=0.8\\ 此時1-x=1-y=0.4 \Rightarrow x=y=0.6 \Rightarrow P(A\cap B)=xy= 0.36，故選\bbox[red, 2pt]{(A)}$$
解答:$$出現至少 2 次正面或 2 次反面 \Rightarrow \cases{X=2 \Rightarrow P(X=2)=p^2+(1-p)^2\\ X=3 \Rightarrow P(X=3)= 1-P(X=3) =1-p^2-(1-p)^2} \\ \Rightarrow E[X]=2\cdot (p^2+(1-p)^2)+ 3(1-p^2-(1-p)^2) =3-p^2-(1-p)^2，故選\bbox[red, 2pt]{(B)}$$
解答:$$利用三角不等式: |x+y|+|x-y-1|\ge |(x+y)-(x-y-1)|=|2y+1| \\ \Rightarrow |x+y|+|x-y-1|+|y| \ge |2y+1|+|y| \Rightarrow 欲求g(y)= |2y+1|+|y| 的最小值 \\ \cases{g(-1/2)=1/2\\g(0)=1 \\-1/2\le y\le 0 \Rightarrow g(y)=2y+1-y=y+1最小值=1/2} \Rightarrow 最小值={1\over 2}，故選\bbox[red, 2pt]{(A)}$$
解答:$$取\cases{A(0,0) \\B(3,0) \\C(0,-3) \\D(0,-1)} \Rightarrow L=\overleftrightarrow{BC}: y=x-3 \Rightarrow E(t,t-3) \Rightarrow \cases{\overrightarrow{DB}=(3,1) \\ \overrightarrow{DE}=(t,t-2)} \\ \Rightarrow \overrightarrow{DB} \cdot \overrightarrow{DE}=4t-2=0 \Rightarrow t={1\over 2} \Rightarrow E({1\over 2},-{5\over 2}) \Rightarrow \overline{CE}= {1\over \sqrt 2} \\ \Rightarrow \triangle CDE面積={1\over 2} \cdot \overline{CD} \cdot \overline{CE}\sin \angle C ={1\over 2}\cdot 2\cdot {1\over \sqrt 2}\cdot {1\over \sqrt 2}={1\over 2}，故選\bbox[red, 2pt]{(A)}$$
$$\overline{AC}平分\angle BCD \Rightarrow \angle BCA= \angle ACD=\theta \Rightarrow \angle CAD=\angle BCA =\theta(內錯角相等) \\ \Rightarrow \overline{AD}=\overline{CD} =\overline{AE} \Rightarrow \angle AED= \angle ADE=\alpha \Rightarrow \cases{\angle BEC=\angle AED=\alpha(對頂角)\\ \angle CBD= \angle ADE=\alpha (內錯角)} \\ \Rightarrow \overline{CB}=\overline{CE} =\overline{DE} \Rightarrow \angle BDC=\angle ACD=\theta \Rightarrow \alpha=2\theta \\ \triangle ADE三內角之和=\theta+2\alpha=\theta+4\theta=5\theta=180^\circ \Rightarrow \theta=36^\circ \Rightarrow \alpha=72^\circ$$
$$\cases{中垂線L:y=-x+2 \Rightarrow 斜率為-1\\ \overleftrightarrow{AB}: x+ay+b=0} \Rightarrow L\bot \overleftrightarrow{AB} \Rightarrow a=-1 \Rightarrow \overleftrightarrow{AB}:y=x+b \\ 求\overleftrightarrow{AB}與L交點P: -x+2=x+b \Rightarrow P的x坐標={2-b \over 2} ={x_a+x_b\over 2},\;其中\cases{A(x_a,y_a) \\B(x_b,y_b)} \\ 將y=x+b代入y=x^2 \Rightarrow x^2-x-b=0 \Rightarrow 兩根之和=x_a+ x_b=1 \Rightarrow {2-b\over 2} ={1\over 2} \\\Rightarrow b=1 \Rightarrow a+2b= -1+2=1，故選\bbox[red, 2pt]{(B)}$$
解答:$$900=2^2\times 3^2\times 5^2 \Rightarrow 正因數個數=(2+1)^3=27 , 其中一組為30\times 30=900\\ 另外26組成對出現，有13組，因此共有13+1=14組，故選\bbox[red, 2pt]{(B)}$$
解答:$$x+y+z=5 \Rightarrow x+y=5-z代入 xy+yz+zx=3 \Rightarrow xy+z(5-z)=3 \\ \Rightarrow xy=z^2-5z+3 \Rightarrow \cases{兩根之和:x+y=5-z\\ 兩根之積:xy=z^2-5z+3} \\\Rightarrow x,y 為t^2-(5-z)t+z^2-5z+3=0 的兩根\Rightarrow \Delta=(5-z)^2-4(z^2-5z+3)\ge 0 \\ \Rightarrow 3z^2-10z-13\le 0 \Rightarrow (3z-13)(z+1)\le 0 \Rightarrow -1\le z\le {13\over 3} \Rightarrow z的最大值為c={13\over 3} \\ 此時\Delta=0 \Rightarrow x=y ={5-z\over 2} ={5-13/3\over 2} ={1\over 3} \Rightarrow a=b={1\over 3} \\ \Rightarrow a+2b+3c={1\over 3}+{2\over 3}+13=14，故選\bbox[red, 2pt]{(B)}$$
解答:$$C=(a,b) \Rightarrow A的x坐標為a-2 代入 y=-3(a-2-a)^2+b=-12+b \Rightarrow \cases{A(a-2,b-12) \\B(a+2,b-12)} \\ \Rightarrow \triangle ABC面積={1\over 2} \cdot 4\cdot (b-(b-12))=24，故選\bbox[red, 2pt]{(D)}$$
解答:$$若\sin C ={1\over 2} \Rightarrow C=30^\circ 或150^\circ \Rightarrow 無法決定唯一的三角形，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{2x^2+(a+3)x+b-3=0 \Rightarrow 兩根之和:-{a+3\over 2}為整數\Rightarrow a是奇數\\ x^2+ax+b=0 \Rightarrow 兩根之和:-{a\over 2}為整數\Rightarrow a是偶數} \Rightarrow 交集為空集合\\，故選\bbox[red, 2pt]{(A)}$$
解答:
解答:$$取\cases{b=ar\\ c=ar^2\\d=ar^3} \Rightarrow a^{80} =a^mr^m= a^nr^{2n} = a^{120}r^{360} \Rightarrow a^{80}=a^{120}r^{360} \Rightarrow a^{40}r^{360}=1 \Rightarrow a=r^{-9} \\ \Rightarrow r^{-720}=r^{-9m}\cdot r^m=r^{-9n}\cdot r^{2n} \Rightarrow \cases{-8m=-720\\ -7n=-720} \Rightarrow \cases{m=90\\ n=720/7} \Rightarrow n-m={90\over 7}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cos A= {1\over 2} ={5^2+4^2-\overline{BC}^2\over 2\cdot 5\cdot 4} \Rightarrow \overline{BC} = \sqrt{21} \Rightarrow \overline{CD} ={4\over 9}\cdot \overline{BC} ={4\over 9}\sqrt{21} \\ \Rightarrow \cos \angle CAD=\cos 30^\circ={\sqrt 3\over 2}={4^2+\overline{AD}^2-112/27\over 8\overline{AD}} \Rightarrow \overline{AD}^2- 4\sqrt 3 \cdot\overline{AD} +{320\over 27}=0 \\ \Rightarrow \overline{AD} ={4\sqrt 3+ 4/3\sqrt 3\over 2} ={20\over 9}\sqrt 3，故選\bbox[red, 2pt]{(D)}$$
解答:
$$由題意可知: B,C,E在同一圓上，其中圓心A, 半徑r=5\\ 依圓冪性質:\overline{BD} \times \overline{DE} =r^2-\overline{AD}^2=5^2-3^2=16，故選\bbox[red, 2pt]{(D)}\\ \bbox[cyan, 2pt]{題目有誤}:應該是求\overline{BD} \times \overline{DE}，而不是求\overline{BD} \times \overline{BE}$$
解答:
$$P在\overline{BD}上\Rightarrow \overline{PC} =\overline{PA} \Rightarrow \overline{PC}+\overline{PM} =\overline{PA} +\overline{PM}要最小 \Rightarrow A,P,M在一直線上 \Rightarrow P=\overline{BD} \cap \overline{AM} \\ \overline{AD} \parallel \overline{BM} \Rightarrow \triangle ADP \sim \triangle MBP \Rightarrow {\overline{DP} \over \overline{PB}} ={\overline{AD} \over \overline{BM}}={10\over 4}={5\over 2} \Rightarrow \triangle ABP ={2\over 7}\triangle ADB ={1\over 7} ABCD \\ \Rightarrow 菱形ABCD面積=7\triangle ABP，故選\bbox[red, 2pt]{(D)}$$
解答:$$取u= {\pi\over 3}+x \Rightarrow x= u-{\pi\over 3} \Rightarrow \sin 3x = \sin(3u-\pi)= -\sin 3u \Rightarrow y= \sin u+\sin 3u \\=\sin u+ \left( 3\sin u-4\sin^3 u \right) =4\sin u-4\sin^3 u\\ 取t=\sin u \Rightarrow y=4t-4t^3 \Rightarrow y'=4-12t^2 =0 \Rightarrow t=\pm {1\over \sqrt 3} \Rightarrow \cases{y(1/\sqrt 3) =8/3\sqrt 3\\ y(-1/\sqrt 3)=-8/3\sqrt 3} \\ \Rightarrow 最小值k=-{8\over 3\sqrt 3} \Rightarrow k^2= {64\over 27}，故選\bbox[red, 2pt]{(C)}$$
解答:
$$y=\log_b x與y=\log_c x的交點為(1,0) \Rightarrow \log_b x\gt \log_c x, \text{ for }x\gt 1 \Rightarrow b\lt c \\假設P為 垂直線:x=a與y=\log_b x的交點 \Rightarrow P的y坐標\lt 1 \Rightarrow a\lt b\\ 因此a\lt b\lt c，故選\bbox[red, 2pt]{(B)}$$
解答:$$B'= \begin{bmatrix}\cos \theta& -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} \begin{bmatrix}b\\a \end{bmatrix} = \begin{bmatrix}b\cos \theta-a\sin \theta\\ b\sin \theta+a\cos \theta \end{bmatrix} \Rightarrow B'的y坐標= b\sin \theta+a\cos \theta，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設\cases{\overline{BC}=a \\\overline{AB} =\overline{AC}=b} \Rightarrow \overline{AM} =\overline{MB}={b\over 2} \Rightarrow \cases{p=\overline{AC}+ \overline{AM}=b+b/2=3b/2\\ q=\overline{BC}+ \overline{BM}= a+b/2} \\ \Rightarrow {p\over q}={5\over 2}={3b/2\over a+b/2} \Rightarrow b=10a \Rightarrow \triangle ABC 三邊長為a,10a,10a \\ 邊長為整數且面積要最小\Rightarrow a=1 \Rightarrow 三邊長為1,10,10 \Rightarrow 底邊上的高h=\sqrt{10^2-({1\over 2})^2} ={\sqrt{399}\over 2} \\ \Rightarrow面積={1\over 2} \times 1\times {\sqrt{399} \over 2}={\sqrt{399}\over 4} ={1\over 2}(1+10+10)\cdot r \Rightarrow r={\sqrt{399} \over 42}，故選\bbox[red, 2pt]{(B)}$$
解答:$$a_{n+1} =a_n+{1\over a_n} \Rightarrow a_{n+1}^2=a_n^2+2+{1\over a_n^2} \Rightarrow a_{n+1}^2-a_n^2 =2+{1\over a_n^2} \\ \Rightarrow \sum_{k=1}^{29} (a_{k+1}^2-a_k^2) = \sum_{k=1}^{29} \left( 2+{1\over a_k^2} \right) \Rightarrow a_{30}^2-1 =2\times 29+ \sum_{k=1}^{29} {1\over a_k^2} =58+a_1^2 +\sum_{k=2}^{29} {1\over a_k^2} \\ \Rightarrow a_{30}^2 =60+\sum_{k=2}^{29} {1\over a_k^2} \gt 60 \Rightarrow a_{30}\gt \sqrt{60} \approx 7.7\\ 又a_n^2 =2n-1+ \sum_{k=1}^{n-1} {1\over a_k^2} \Rightarrow a_k^2 \gt 2k-1 \Rightarrow {1\over a_k^2} \lt {1\over 2k} \Rightarrow \sum_{k=2}^{29}{1\over a_k^2} \lt \sum_{k=2}^{29}{1\over 2k} ={1\over 2} \left( {1\over 2}+{1\over 3}+\cdots+{1\over 29} \right)\\ 由於\sum_{k=2}^{29}{1\over k} \lt \int_1^{29}{1\over x}\,dx = \ln 29 \approx 3.4 \Rightarrow a_{30 }^2 \lt 60+{3.4\over 2}=61.7 \Rightarrow a_{30} \lt \sqrt{61.7} \approx 7.85 \\ \Rightarrow 7.7\lt a_{30} \lt 7.85 \Rightarrow \lfloor a_{30}+0.6 \rfloor =8，故選\bbox[red, 2pt]{(C)}$$
解答:$$B^2-AB=(B-A)B= I \Rightarrow B-A=B^{-1} \Rightarrow \begin{bmatrix}5-a& -2-b\\ -2-c& 1-d \end{bmatrix}= \begin{bmatrix}1& 2\\ 2& 5 \end{bmatrix} \\ \Rightarrow \cases{5-a=1\\ 1-d=5} \Rightarrow 兩式相加:a+d=0，故選\bbox[red, 2pt]{(A)}$$
解答:$$將\vec u與\vec v當成基底 \Rightarrow \vec w=\alpha \vec u+\beta \vec v \Rightarrow \begin{bmatrix}u_1& v_1\\u_2& v_2 \end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}= \begin{bmatrix}w_1\\ w_2 \end{bmatrix} \Rightarrow \alpha={ \begin{vmatrix} w_1& v_1\\ w_2& v_2 \end{vmatrix} \over \begin{vmatrix} u_1& v_1\\ u_2& v_2 \end{vmatrix}}, \beta = { \begin{vmatrix} u_1& w_1\\ u_2& w_2 \end{vmatrix} \over \begin{vmatrix} u_1& v_1\\ u_2& v_2 \end{vmatrix}}\\ 假設D=\begin{vmatrix} u_1& v_1\\ u_2& v_2 \end{vmatrix} \lt 0\quad (因\vec u在第三象限, \vec v在第二象限), 同理可得\cases{\alpha\lt 0\\ \beta\gt 0}，故選\bbox[red, 2pt]{(B)}$$
解答:$$x-1 \lt \lfloor x\rfloor \le x \Rightarrow {n\over 5}-1 \lt \left\lfloor {n\over 5} \right \rfloor \le {n\over 5} \Rightarrow {{n\over 5}-1 \over n}\lt {\left\lfloor {n\over 5} \right \rfloor \over n}\le {{n\over 5}\over n} \\ \Rightarrow {1\over 5}-{1\over n} \lt {\left\lfloor {n\over 5} \right \rfloor \over n}\le{1\over 5 } \Rightarrow \cases{左極限:\lim_{n\to \infty} \left( {1\over 5}-{1\over n} \right) ={1\over 5} \\ 右極限: \lim_{n\to \infty} {1\over 5}={1\over 5}} \Rightarrow 夾擠定理:\lim_{n\to \infty} {\left\lfloor {n\over 5} \right \rfloor \over n} ={1\over 5}\\，故選\bbox[red, 2pt]{(A)}$$
$$E在\overline{AC}的投影點為C,假設D在\overline{AC}的投影點為D' \Rightarrow \overline{BC} \parallel \overline{DD'} \Rightarrow {\overline{CD'} \over \overline{AD'}} ={\overline{BD} \over \overline{AD}}={1\over 1} \\ \Rightarrow \overline{CD'} =\overline{AD'}={8\over 2}=4 \Rightarrow \overrightarrow{ED} \cdot \overrightarrow{AC} = -\overrightarrow{CD'}\cdot \overrightarrow{CA} =-4\cdot 8=-32，故選\bbox[red, 2pt]{(C)}$$
解答:
解答:$$Av= \lambda v \Rightarrow \begin{bmatrix}2& a& 2\\5& b& 3\\-1& 0&-2 \end{bmatrix} \begin{bmatrix}1\\ 1\\ -1\end{bmatrix} = \begin{bmatrix}a\\ b+2\\1 \end{bmatrix} = \lambda \begin{bmatrix}1\\1\\ -1 \end{bmatrix} \Rightarrow \cases{\lambda=-1\\ a=-1\\ b+2=-1 \Rightarrow b=-3} \\ \Rightarrow a+b=-1-3=-4，故選\bbox[red, 2pt]{(D)}$$
$$\cases{A(\cos 20^\circ, \sin 20^\circ) \\ B(\sin 50^\circ, \cos 50^\circ) =(\cos 40^\circ, \sin 40^\circ)} \Rightarrow A,B皆在單位圓上 \Rightarrow \angle AOB=40^\circ-20^\circ=20^\circ \\ \Rightarrow \angle BAO= {180^\circ-20^\circ \over 2}=80^\circ \Rightarrow \overline{AB}與x軸夾角120^\circ \Rightarrow 斜率=\tan 120^\circ =-\sqrt 3，故選\bbox[red, 2pt]{(B)}$$
解答:$$\bbox[cyan, 2pt]{題目有誤}，應該是\lim_{n\to 0} a_n=\lim_{n\to 0} \sum_{k=1}^{2n} {1\over k+n} =\lim_{n\to 0} \sum_{k=1}^{2n} {1\over n(1+k/n)} = \int_0^2 {1\over 1+x}\,dx \\= \left. \left[ \ln(1+x) \right] \right|_0^2 =\ln 3，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(\theta) =8\sin^3\theta-6\sin \theta+2 \Rightarrow f'(\theta)=24\sin^2\theta \cos \theta-6\cos \theta =0 \Rightarrow 6\cos \theta(4\sin^2\theta-1)=0 \\ \Rightarrow \cases{\cos \theta =0 \Rightarrow \cases{\sin \theta= 1 \Rightarrow f(\theta)=8-6+2=4\\ \sin \theta=-1 \Rightarrow f(\theta)=-8+6+2=0} \\ \sin^2\theta=1/4 \Rightarrow \cases{\sin \theta=1/2 \Rightarrow f(\theta)=1-3+2=0\\ \sin \theta=-1/2 \Rightarrow f(\theta)=-1+3+2=4}} \Rightarrow f(\theta)最大值為4，故選\bbox[red, 2pt]{(B)}$$
解答:$$\int_0^1 {2\over (x+1)^2} \,dx = \left. \left[ -{2\over x+1} \right] \right|_0^1= -1-(-2)=1，故選\bbox[red, 2pt]{(B)}$$
解答:$$\int_0^{\pi/2} \sin 2\theta\,d\theta = \left. \left[ -{1\over 2}\cos 2\theta \right] \right|_0^{\pi/2}={1\over 2}-(-{1\over 2})=1，故選\bbox[red, 2pt]{(A)}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$三角形的三邊長均為正數 \Rightarrow a^2-1\gt 0 且2a+1\gt 0\Rightarrow a\gt 1 \\ \Rightarrow \cases{(a^2+a+1)-(a^2-1)=a+2 \gt 0\\ (a^2+a+1)-(2a+1)=a^2-a= a(a-1)\gt 0} \Rightarrow a^2+a+1為最大邊 \\ \Rightarrow \cos A={(a^2-1)^2+ (2a+1)^2-(a^2+a+1)^2 \over 2(a^2-1)(2a+1)} ={-(a^2-1)(2a+1) \over 2(a^2-1)(2a+1)}=-{1\over 2} \Rightarrow \angle A=120^\circ \\ \Rightarrow \cos^2A +\csc^2A= \left( -{1\over 2} \right)^2+ \left( {2\over \sqrt 3} \right)^2={19\over 12}，故選\bbox[red, 2pt]{(D)}$$
解答:$$xy'+2y=4x^2 \Rightarrow y'+{2\over x}y=4x \Rightarrow 積分因子I(x)=e^{\int(2/x)dx} =x^2 \Rightarrow x^2y'+2xy=4x^3 \\ \Rightarrow (x^2y)'=4x^3 \Rightarrow x^2y= \int 4x^3\,dx =x^4+C \Rightarrow y=x^2+{C\over x^2} \Rightarrow y(1)=1+C=4\Rightarrow C=3\\ \Rightarrow y=x^2+{3\over x^2}，故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)={x\over 1+x^4} \Rightarrow f(-x)=-f(x) \Rightarrow f(x)為奇函數\Rightarrow \int_{-\infty}^\infty f(x)\,dx =0，故選\bbox[red, 2pt]{(A)}$$
解答:$$亮點的直線行進斜率的絕對值為{1\over \tan \theta},即每往下走 1 單位，往右走 \tan\theta 單位 \\ 亮點的垂直距離為2+9+9+9=29 \Rightarrow 水平距離為29\tan \theta \\ \Rightarrow \cases{F的x坐標為32-29\tan \theta\\ C的x坐標為16} \Rightarrow \overline{CF}=16-(32-29\tan \theta)=29\tan \theta-16，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{甲班標準化後:人數n_A=50, 平均數\mu_A= 0, 標準差\sigma_A=1 \\乙班標準化後:人數n_B=50, 平均數\mu_B= 0, 標準差\sigma_B=1 } \\ \Rightarrow 合併後平均值\mu= {50\cdot 0+50\cdot 0\over 50+50} =0 \Rightarrow 合併後變異數\sigma^2={n_A(\sigma_A^2+(\mu_A-\mu)^2) +n_B(\sigma_B^2-\mu)^2 \over n_A+n_B} \\={50+50\over 100}=1 \Rightarrow 標準差\sigma =\sqrt 1=1，故選\bbox[red, 2pt]{(D)}$$
解答:$$若y=f(x) 滿足 f(a+x) + f(a-x) = 2b 或f(x)+f(2a-x)=2b，則該函數對稱點為(a, b) \\ 現在f(x)+f(2026-x)=115 \Rightarrow 對稱點為(1013,{115\over 2}) \Rightarrow f({x\over 2}) 對稱點為(2026,{115\over 2})，故選\bbox[red, 2pt]{(C)}$$
解答:$$將 \triangle ABC面積S={a^2+ b^2-c^2\over 2}代入餘弦定理: a^2+b^2-c^2 =2ab\cos C \\ \Rightarrow S={2ab\cos C\over 2} =ab\cos C ={1\over 2}ab\sin C \Rightarrow \tan C=2，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cos z=2 \Rightarrow {e^{iz}+e^{-iz} \over 2}=2 \Rightarrow e^{iz}+e^{-iz}=4 \Rightarrow e^{2iz}-4e^{iz}+1=0 \\ \Rightarrow e^{iz}= 2\pm \sqrt 3 \Rightarrow iz= \ln(2\pm \sqrt 3)+i(2n\pi), n\in \mathbb Z \Rightarrow z=2n\pi+i\ln(2\pm \sqrt 3) \\ \Rightarrow \cases{(A)\bigcirc: n=1 \Rightarrow z=2\pi+i\ln (2+\sqrt 3)為其一解\\ (B) \bigcirc: n=0 \Rightarrow z=i\ln(2+\sqrt 3)為其一解 \\(C)\bigcirc: n=0 \Rightarrow z=i\ln(2-\sqrt 3) 為其一解 \\(D)\times: n\ne {1\over 2}}，故選\bbox[red, 2pt]{(D)}$$
解題僅供參考，其他教甄試題及詳解
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