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2023年7月19日 星期三

112年台北市聯合轉學考-升高二(技高)-數學詳解

臺北市高級中等學校 112 學年度聯合轉學考
升高二數學科試題(技高)

一、單選題:

解答:$$已知\cases{P_1(k,-1) \\P_2(4,k)\\ P_3(5,-3)},且\overline{P_1P_2} =\overline{P_1P_3} \Rightarrow (k-4)^2+(k+1)^2=(k-5)^2+2^2 \\ \Rightarrow 2k^2-6k+17= k^2-10k+29 \Rightarrow k^2+4k-12=0 \Rightarrow (k+6)(k-2)=0\\ \Rightarrow k=2或-6,故選 \bbox[red, 2pt]{(B)}$$
解答:$$\cases{有二相異實根\Rightarrow 判別式b^2-4ac \gt 9\\ 圖形凹向下\Rightarrow a\lt 0\\ x=0 \Rightarrow c\gt 0\\ 極大值位於第二象限\Rightarrow -{b\over 2a} \lt  \Rightarrow b\lt 0} \Rightarrow P(b^2-4ac \gt 0,abc \gt 0)在第一象限,故選 \bbox[red, 2pt]{(A)}$$
解答:$$(x-5)^2 \gt (2x-4)^2 \Rightarrow x^2-10x+25 \gt 4x^2-16x+16 \Rightarrow x^2-2x-3\lt 0 \\ \Rightarrow (x-3)(x+1)\lt \Rightarrow -1\lt x\lt 3,故選 \bbox[red, 2pt]{(D)}$$
解答:$$\cases{無實數解\Rightarrow 判別式4-4k^2 \lt 0 \Rightarrow 1\lt k^2  \Rightarrow k\gt 1或k\lt -1\\ 又圖形\le 0無實數解\Rightarrow 圖形凹向上 \Rightarrow k\gt 0} \Rightarrow k\gt 1,故選 \bbox[red, 2pt]{(B)}$$
解答:$$\cases{A(-2,3)\\ B(6,2)\\ P(x,0)} \Rightarrow \overline{PA}^2+\overline{PB}^2 =f(x)=(x+2)^2+9 +(x-6)^2+4 = 2x^2-8x+53\\ \Rightarrow 最小值=f(2) =8-16+53=45,故選 \bbox[red, 2pt]{(C)}$$
解答:$$\cases{A(2,5)\\ B(-1,1) \\ C(10,-1)} \Rightarrow \cases{\overline{AB}=\sqrt{3^2+4^2} =5\\ \overline{AC}= \sqrt{8^2+6^2} =10}\\ \overline{AC}為\angle A的角平分線 \Rightarrow {\overline{BD}\over \overline{DC}} = {\overline{AB} \over \overline{AC}} ={5 \over 10} ={1\over 2} \Rightarrow D=(2B+C)/3 =\left({8\over 3},{1\over 3}\right),故選 \bbox[red, 2pt]{(A)}$$
解答:$$\cases{\overline{AB}斜率= {5-a\over -2} \\ L:y=3ax-5 斜率=3a} \Rightarrow -{5-a\over 2}=3a \Rightarrow a=-1,故選 \bbox[red, 2pt]{(C)}$$
解答:$$\cases{y截距-4\\ L斜率-2} \Rightarrow x截距= -4/-2=2 \Rightarrow 面積={1 \over 2}  \cdot 4\cdot 2=4,故選 \bbox[red, 2pt]{(D)}$$
解答:$$\cases{a= \cos 20^\circ =\sin 70^\circ \\ b=\sin 50^\circ \\ c=\cos 10^\circ =\sin 80^\circ } \Rightarrow c\gt a\gt b,故選 \bbox[red, 2pt]{(D)}$$
解答:$$\cases{f(x)= \sin(-2x)的週期為A=\pi \\ g(x)=2\tan(x/2)的週期為B=2\pi \\ h(x)=-{1\over 3}|\cos x|的週期為C=\pi} \Rightarrow a+b+c=4\pi,故選 \bbox[red, 2pt]{(D)}$$
解答:$$假設\overline{AD}=a,\cos \angle ADB=-\cos \angle ADC \Rightarrow 4^2+a^2-5^2=-(4^2+a^2-7^2) \\ \Rightarrow 2a^2=42 \Rightarrow a=\sqrt{21},故選 \bbox[red, 2pt]{(A)}$$

解答:$$(A) (x+4)^2+(y-3)^2=12  \Rightarrow 圓心在第二象限\\(B) (x+4)^2+(y+3)^2=12 \Rightarrow 圓心在第三象限\\ (C)(x-4)^2+(y+3)^2=12 \Rightarrow 圓心在第四象限\\ (D) (x-4)^2+(y-3)^2=12 \Rightarrow 圓心在第一象限\\,故選 \bbox[red, 2pt]{(A)}$$
解答:$$\angle A:\angle B:\angle C=1:3:2 \Rightarrow \cases{\angle A=k\\ \angle B=3k\\ \angle C=2k} \Rightarrow \angle A+\angle B+\angle C=6k=180^\circ\\ \Rightarrow \cases{\angle A-30^\circ\\ \angle B=90^\circ \\ \angle C=60^\circ} \Rightarrow {\overline{AC}+ \overline{BC} \over \overline{AB} + \overline{BC}} ={2+ 1\over \sqrt 3+1} ={3 \over \sqrt 3+1},故選 \bbox[red, 2pt]{(B)}$$
解答:$$\cos \theta ={5^2+8^2-7^2 \over 2\cdot 5\cdot 8} ={1\over 2} \Rightarrow \theta=60^\circ \Rightarrow \vec a與\vec b的夾角=180^\circ -60^\circ =120^\circ,故選 \bbox[red, 2pt]{(C)}$$
解答:$$f(-1)=f(2)= f(3)=0 \Rightarrow -1,2,3為f(x)=0的三根 \Rightarrow f(x)=a(x+1)(x-2)(x-3)\\ 又f(1)=12 \Rightarrow a\cdot 2\cdot (-1)\cdot (-2)=12 \Rightarrow a=3 \Rightarrow f(x) = 3(x+1)(x-2)(x-3) \\ \Rightarrow f(-2)=3\cdot (-1)\cdot (-4) \cdot (-5)=-60,故選 \bbox[red, 2pt]{(A)}$$
解答:$$x=\sqrt 2+1 \Rightarrow (x-1)^2=2 \Rightarrow x^2-2x-1=0\\ \Rightarrow 2x^3-3x^2-x+4 = (x^2-2x-1)(2x+1)+ 3x+5 =3x+5=3\sqrt 2+3+5=3\sqrt 2+8\\,故選 \bbox[red, 2pt]{(A)}$$
解答:$$(3x^5-4x^3+3x^2+1)(-2x^2-3x+1) \\ x^5的係數=3\cdot 1+(-4)\cdot (-2)=3+8=11,故選 \bbox[red, 2pt]{(D)}$$
解答:$$利用長除法:x^3+3x^2+6x+2 = (x^2+x+4)(x+2)-6 \\ \Rightarrow {x^3 +3x^2 +6x+ 2\over x^2+x+4} = (x+2)+{-6 \over x^2+x+4} \Rightarrow \cases{a=1\\ b=2\\ c=-6} \Rightarrow a-b-c=1-2+6=5\\,故選 \bbox[red, 2pt]{(C)}$$
解答:$$f(x)=x^3-5x^2-3=a(x-1)^3+b(x-1)^2+ c(x-1)+d\\ \Rightarrow \cases{f(1)=-7=d\\ f(0)=-3=-a+b-c+d} \Rightarrow a-b+c=-4 \Rightarrow a-b+c-d=-4+7=3,故選 \bbox[red, 2pt]{(C)}$$
解答:$$與3x+4y=2垂直的直線L: 4x-3y=k \Rightarrow 圓心(0,1)至L的距離=圓半徑1\\ \Rightarrow {|-3-k| \over 5}=1 \Rightarrow k=2,-8 \Rightarrow L:\cases{4x-3y+8=0\\ 4x-3y-2=0},故選 \bbox[red, 2pt]{(B)}$$
解答:$$假設公比=r \Rightarrow 等比數列:2,2r,2r^2,2r^3,2r^4,2r^5=486 \Rightarrow r=3\\ \Rightarrow 6個數字總和=2+6+18+54+162+486 =728,故選 \bbox[red, 2pt]{(B)}$$
解答:$${9\over 5}x+32=77 \Rightarrow x=25,故選 \bbox[red, 2pt]{(D)}$$
解答:$${\sum_{n=1}^{15} (4n-2) \over \sum_{n=1}^{15} (3n+2)} ={450\over 390} ={15\over 13},故選 \bbox[red, 2pt]{(B)}$$
解答:$${5\cdot (1/2)^{180/45} \over 9\cdot (1/2)^{180/60}} ={5/16\over 9/8}={5\over 18},故選 \bbox[red, 2pt]{(C)}$$
解答:$$圓C: x^2+y^2-2x-8=0  \Rightarrow (x-1)^2+y^2= 3^2 \Rightarrow 圓心O(1,0) \\ \Rightarrow 外接圓直徑\overline{AO}= \sqrt{4^2+3^2}=5 \Rightarrow 半徑={5\over 2},故選 \bbox[red, 2pt]{(C)}$$
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