$$圓C: x^2+y^2+6x+8y+10=0 \Rightarrow (x^2+ 6x+9) +(y^2+8y+16)+10=25 \\ \Rightarrow (x+3)^2+(y+4)^2=15 \Rightarrow \cases{圓心M(-3,-4)\\ 圓半徑r=\sqrt{15}} \Rightarrow \overline{MO}=5 \\ \Rightarrow \overline{PO} \cases{最大值為5+r=5+\sqrt{15} \lt 9\\ 最小值為5-r=5-\sqrt{15} \lt 2} \Rightarrow \sqrt{a^2+b^2}=\overline{PO}=2,3,\dots,8 \Rightarrow 共7個整數\\ \Rightarrow 在\overleftrightarrow{MO}兩側各有7個整數點，共14組，故選\bbox[red, 2pt]{(B)}$$
解答:$$角度在第二、三象限，餘弦值為負數，因此僅考慮(B),(C), (D)\\ 越靠近180^\circ,餘弦值越小，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{ \overline{AB}=3\\ \overline{BC}= 4\\\overline{AC}=5 } \Rightarrow \angle B=90^\circ \Rightarrow \cases{B(0,0)\\ A(0,3) \\C(4,0)} \Rightarrow 直線L= \overleftrightarrow{AC}: 3x+4y=12 \\ 假設P(x,y)在L上 \Rightarrow 矩形面積=x\cdot y = x\left(3 - \frac{3}{4}x\right) = -\frac{3}{4}x^2 + 3x = -{3\over 4}(x-2)^2+3 \\ \Rightarrow 當 x = 2 時 (此時y={3\over 2})，矩形面積有最大值 2\cdot {3\over 2} =3，故選\bbox[red, 2pt]{(B)}$$
解答:$$初始狀態:原正三角形邊長為 4\Rightarrow 面積為 \frac{\sqrt{3}}{4} \times 4^2 = 4\sqrt{3}\\第1次操作:原三角形平分成 四等份\Rightarrow \cases{左下塗色面積L_1= \sqrt 3\\ 右下塗色面積R_1=\sqrt 3} \\ 第2次操作:R_1平分成 四等份\Rightarrow \cases{R_1左下塗色面積L_2= \sqrt 3/4\\ R_1右下塗色面積R_2=\sqrt 3/4} \\第k次操作:R_{k-1}平分成 四等份\Rightarrow \cases{R_{k-1}左下塗色面積L_k= R_{k-1}/4\\ R_{k-1}右下塗色面積R_k= R_{k-1}/4} \\ 因此第5次操作後的面積:\cases{L_1 = \sqrt{3} \\L_2 = \sqrt{3} \times \frac{1}{4} \\L_3 = \sqrt{3} \times \frac{1}{16} \\L_4 = \sqrt{3} \times \frac{1}{64} \\L_5 = \sqrt{3} \times \frac{1}{256}\\ R_5 = \sqrt{3} \times \frac{1}{256}} \\\Rightarrow 總面積= \sqrt{3} \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} \right) + \frac{\sqrt{3}}{256} = \frac{341\sqrt{3}}{256} + \frac{\sqrt{3}}{256} = \frac{342\sqrt{3}}{256} \\= {171\sqrt 3\over 128}，故選\bbox[red, 2pt]{(D)}$$
解答:$$取f(x,y)=2x+y-1 \Rightarrow A, B在\overline{AB}的異側 \Rightarrow f(A)\cdot f(B)\le 0 \Rightarrow(2k+3)(2k) \le 0 \\ \Rightarrow -{3\over 2}\le k\le 0，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{a = \displaystyle \frac{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})}{\sqrt{5} + \sqrt{2}} = \frac{5 - 2}{\sqrt{5} + \sqrt{2}} = \frac{3}{\sqrt{5} + \sqrt{2}} \\b = \displaystyle \frac{(\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{3})}{\sqrt{6} + \sqrt{3}} = \frac{6 - 3}{\sqrt{6} + \sqrt{3}} = \frac{3}{\sqrt{6} + \sqrt{3}} \\c = \displaystyle \frac{(\sqrt{10} - \sqrt{7})(\sqrt{10} + \sqrt{7})}{\sqrt{10} + \sqrt{7}} = \frac{10 - 7}{\sqrt{10} + \sqrt{7}} = \frac{3}{\sqrt{10} + \sqrt{7}}} \Rightarrow a\gt b\gt c，故選\bbox[red, 2pt]{(A)}$$
解答:$$x+2+{1\over x+1}=1+ x+1+{1\over x+1} \ge 1+2\sqrt{(x+1)\cdot {1\over x+1}} =3，故選\bbox[red, 2pt]{(A)}$$
解答:$$若a\gt 0 \Rightarrow |ax+3|\lt b \Rightarrow -b-3\lt ax\lt b-3 \Rightarrow -{b+3\over a}\lt x\lt {b-3\over a} \\ \qquad \Rightarrow \cases{-(b+3)/a=-3\\ (b-3)/a=5} \Rightarrow {-6\over a}=2 \Rightarrow a\lt 0不合 \\ 若a\lt 0 \Rightarrow |ax+3|\lt b\Rightarrow -b-3\lt ax\lt b-3 \Rightarrow {b-3\over a}\lt x\lt -{b+3\over a} \\ \qquad \Rightarrow \cases{(b-3)/a=-3\\ -(b+3)/a=5} \Rightarrow -{6\over a}=2 \Rightarrow a=-3 \Rightarrow {b-3\over -3}=-3 \Rightarrow b=12，故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)=(x^2+2x+ 3) (x^2+2x+5)+2x^2+4x-2 \\ \Rightarrow f'(x)=(2x+2) (x^2+2x+5)+(x^2+2x+3)(2x+2)+4x+4 =(2x+2)(2x^2+4x+10) \\ f'(x)=0 \Rightarrow 4(x+1)(x^2+2x+5)=0 \Rightarrow 4(x+1)((x+1)^2+4)=0 \Rightarrow x=-1\\ \Rightarrow f(-1)=2\cdot 4-4=4，故選\bbox[red, 2pt]{(A)}$$
解答:$$(ax^2+1)^5 = \sum_{k=0}^5 C^5_k(ax^2)^{5-k}\cdot 1^k\\ 2(5-k)=6 \Rightarrow k=2 \Rightarrow x^6係數為C_2^5 \cdot a^{5-2} = \frac{5 \times 4}{2 \times 1} \cdot a^3 = 10a^3 =80 \Rightarrow a=2，故選\bbox[red, 2pt]{(C)}$$
解答:$$ (\sin A + \sin B)^2= \sin^2 A + 2\sin A\sin B + \sin^2 B = \sin^2 C + 3\sin A\sin B \\ \Rightarrow \sin^2 A + \sin^2 B - \sin^2 C = \sin A\sin B\\ 將正弦定理{a\over \sin A}={b\over \sin B}={c\over \sin C} \Rightarrow a:b:c =\sin A:\sin B: \sin C代入上式\\ \Rightarrow a^2+b^2-c^2=ab \Rightarrow \cos C={a^2+b^2-c^2\over 2ab}={ab\over 2ab}={1\over 2} \Rightarrow \angle C=60^\circ，故選\bbox[red, 2pt]{(B)}$$
解答:
$$假設\overline{BC}=a \Rightarrow \triangle ABC= \triangle ABD+ \triangle DBC \\ \Rightarrow {1\over 2}\cdot 6\cdot a\sin60^\circ ={1\over 2}\cdot 6\cdot 2\sqrt 3\sin30^\circ +{1\over 2}\cdot 2\sqrt 3\cdot a \sin 30^\circ \Rightarrow {3\over 2}\sqrt 3a=3\sqrt 3+ {1\over 2}\sqrt 3a \\ \Rightarrow a=3 \Rightarrow \cos B={6^2+a^2-\overline{AC}^2\over 2\cdot 6\cdot a} \Rightarrow {1\over 2}={45-\overline{AC}^2\over 36} \Rightarrow \overline{AC} =3\sqrt 3，故選\bbox[red, 2pt]{(D)}$$
二、 多選題(40 分)：
解答:$$(A)\bigcirc: (1+x)(3-x) \ge 0 \Rightarrow (x+1)(x-3)\le 0 \Rightarrow -1\le x\le 3 \\(B)\bigcirc: (x+1)^3(x-3)\le 0 \Rightarrow (x+1)(x-3)\le 0 \Rightarrow -1\le x\le 3 \\(C) \times: x=0代入\cases{左式:(x+1)^3(x-3)^2=9\\右式: x-3=-3} \Rightarrow 9\not \le -3 \\(D)\bigcirc: |x-1|\le 2 \Rightarrow -2\le x-1\le 2 \Rightarrow -1\le x\le 3 \\(E)\times:(x-1)^2\le 4\Rightarrow |x-1| \le 2 \Rightarrow -1\le x\le 3\\，故選\bbox[red, 2pt]{(ABDE)}$$
解答:$$不能圍成三角形有兩種情況:其中兩直線互相平行，或是三條直線交於同一個點\\ 情況一:直線之間互相平行 \cases{L_1 \parallel L_2 \Rightarrow 斜率-3=-k/2 \Rightarrow k=6\\ L_2 \parallel L_3 \Rightarrow 斜率1=-k/2 \Rightarrow k=-2} \\ 情況二:三直線交於一點 \Rightarrow {L_1與L_2交於(-1,2)}代入L_3 \Rightarrow k\cdot (-1)+ 2\cdot 2+1=0 \Rightarrow k=5 \\ 因此k=-2,5,6，故選\bbox[red, 2pt]{(ADE)}$$
解答:$$(A)\bigcirc: {1\over 2}\gt {1\over 3} \Rightarrow 115^{1/3} \lt 115^{1/2} \\(B) \bigcirc: \cases{a=11^{1/3} \\b=5^{1/2}} \Rightarrow \cases{a^6=11^2=121\\ b^6=5^3=125} \Rightarrow b^6\gt a^6 \Rightarrow 5^{1/2} \gt 11^{1/3} \\(C)\bigcirc: \log 6^{100}=100(\log 3+\log 2) =77.81 \Rightarrow \log 6\lt 0.81\lt \log 7 \Rightarrow 最高位數字為6\\ (D) \bigcirc: \log 6^{100} =77.81 \Rightarrow 6^{100}為77+1=78 位數\\ (E) \bigcirc: \cases{\log 2^{250} =250\times 0.301=75.25\\ \log 6^{100} =77.81} \Rightarrow 6^{100} \gt 2^{250}\\，故選\bbox[red, 2pt]{(ABCDE)}$$
解答:$$(A)\bigcirc: f(3) = 3^5 - 5\cdot (3^4) + 8 \cdot (3^3) - 7\cdot (3^2) + 6\cdot (3) - 4=5\\ (B)\bigcirc: 由(A)可知: f(3) =5 \Rightarrow f(x)除以x-3的餘式為5\\(C)\times: f(x)= (x-3)(x^4 - 2x^3 + 2x^2 - x + 3) + 5 \\\qquad = 2(x-3) \cdot \left[ \frac{1}{2}(x^4 - 2x^3 + 2x^2 - x + 3) \right] + 5 \Rightarrow 商式為 \frac{1}{2}x^4 - x^3 + x^2 - \frac{1}{2}x + \frac{3}{2} \\(D)\times: f(2) = 2^5 - 5(2^4) + 8(2^3) - 7(2^2) + 6(2) - 4=-4 \ne -31 \\(E)\bigcirc: 假設餘式為ax+b \Rightarrow \cases{f(3)=3a+b=5\\ f(1) =a+b=-1} \Rightarrow \cases{a=3\\ b=-4} \Rightarrow 餘式為3x-4\\，故選\bbox[red, 2pt]{(ABE)}$$
解答:$$L:2x-y+k=0 \Rightarrow \cases{x截距:-k/2\\ y截距:k} \Rightarrow 面積= \frac{1}{2} \times \left\vert{} -\frac{k}{2} \right\vert{} \times \vert{}k\vert{} = 9 \Rightarrow {1\over 4}k^2=9 \\ \Rightarrow k=\pm 6，故選\bbox[red, 2pt]{(BC)}$$
解答:$$R+W+N= 100 \Rightarrow Y=R+{N\over 5} = R + \frac{100 - R - W}{5} = \frac{4}{5}R - \frac{1}{5}W + 20 \\ 又X= R - \frac{W}{4} \Rightarrow \frac{4}{5}X = \frac{4}{5}\left(R - \frac{W}{4}\right) = \frac{4}{5}R - \frac{1}{5}W =Y-20 \Rightarrow Y={4\over 5}X+20 \\(A)\bigcirc: Y-X = \left( {4\over 5}X+20 \right) -X=20-0.2X \ge 0, 0\le X\le 100 \Rightarrow Y\ge X \\(B)\bigcirc: 假設X的平均數為\bar X \Rightarrow \bar Y=0.8\bar X+20, 結果與(A)相同: \bar X\le \bar Y \\(C) \times: \vert{}Y_1 - Y_2\vert{} = \vert{}(0.8X_1 + 20) - (0.8X_2 + 20)\vert{} = 0.8\vert{}X_1 - X_2\vert{} \Rightarrow |X_1-X_2|\gt |Y_1-Y_2| \\(D) \bigcirc: Y=0.8X+20 \Rightarrow Y 隨著 X 的增加而嚴格遞增 \\(E)\bigcirc: Y=0.8X+20 \Rightarrow 相關係數為1，故選\bbox[red, 2pt]{(ABDE)}$$
解答:$$(A)\times: m_{AB} ={-4-(-4) \over 3-1}=0 \ne 2 \\(B)\times: m_{AC} ={0-(-4) \over 5-1}=1\ne 0 \\(C) \bigcirc: \overline{AB}$ 的中點坐標為 (\frac{1+3}{2}, \frac{-4+(-4)}{2}) = (2, -4) \Rightarrow 中垂線:x=2 \\(D)\times: \overline{AC}$ 的中點坐標為 $(\frac{1+5}{2}, \frac{-4+0}{2}) = (3, -2) \Rightarrow 中垂線x+y-1=0 \\(E)\bigcirc: 外心即為兩中垂線交點\Rightarrow x = 2 與 x + y - 1 = 0 的交點為(2,-1)\\，故選\bbox[red, 2pt]{(CE)}$$
解答:$$ \theta_1, \theta_2, \theta_3, \theta_4 分別落在第一、二、三、四象限 \Rightarrow \cases{ \cos \theta_1=\cos \theta_4=1/3\\ \cos \theta_2=\cos \theta_3=- 1/3} \\(A)\times \cases{\cos\theta_1=0.333\\ \cos(\pi/4)=0.707} \Rightarrow \theta_1\gt {\pi \over 4} \\(B)\bigcirc: \cos(180^\circ-\theta_1)=-\cos\theta_1=-{1\over 3} =\cos \theta_2 \Rightarrow 180^\circ-\theta_1=\theta_2\Rightarrow \theta_1+ \theta_2=180^\circ \\(C)\bigcirc: \theta_3在第三象限\Rightarrow \cos\theta_3=-{1\over 3} \\(D) \times: \cos \theta_4={1\over 3} \ne {2\sqrt 2\over 3} \\(E) \times: \cases{\cos \theta_4=1/3\\ \cos(\theta_3+90^\circ)=-\sin \theta_3 \ne 1/3} \Rightarrow \theta_4 \ne \theta_3+ 90^\circ\\，故選\bbox[red, 2pt]{(BC)}$$
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解題僅供參考，其他轉學考試題及詳解
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