114年台北市聯合轉學考-升高二(技高)-數學詳解

 臺北市立技術型高中暨進修部114學年度聯合轉學考招生考試

升高二數學科試題(技高)

一、單選題：

解答:$$\overline{AP}: \overline{BP}=3:1 \Rightarrow P={3B+A\over 4} ={1\over 4}\left( ({121\over 5},-12) +({663\over 5},24)\right) ={1\over 4}({784\over 5},12) =({196\over 5},3)\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)= 2x^3+5x^2+x-2 =(x+1)(2x^2 +3x-2) =(x+1)(2x-1)(x+2)，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{x\gt \sqrt 5+1\\ x\lt 1-\sqrt 5} \Rightarrow \cases{x-1\gt \sqrt 5\\ x-1\lt -\sqrt 5} \Rightarrow |x-1|\gt \sqrt 5 \Rightarrow (a,b)=(1,\sqrt 5)，故選\bbox[red, 2pt]{(B)}$$

解答:$$3={y截距\over 4} \Rightarrow y截距=12 \Rightarrow 三角形面積={1\over 2} \cdot 4\cdot 12=24，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{L_1:y=2x\\ L_2:y=x/2\\ L_3:y=-2x\\ L_4:2y=x} \Rightarrow \cases{L_1斜率m_1=2\\ L_2斜率m_2= 1/2\\ L_3斜率m_3= -2\\ L_4斜率m_4=1/2 } \Rightarrow \cases{(A)\times: m_1\cdot m_2=1\ne -1\\ (B) \times: L_4=L_2 \Rightarrow 重疊非平行 \\ (C) \bigcirc:m_2\cdot m_3=-1\\ (D) \times: m_1\ne m_3}，故選\bbox[red, 2pt]{(C)}$$

解答:$$L:ax-y-b=0 \Rightarrow y=ax-b \Rightarrow y截距=-b=-3 \Rightarrow b=3 \Rightarrow L:ax-y-3=0 \\ \Rightarrow d((0,0),L) ={|-3|\over \sqrt{a^2+1}} ={3\sqrt 2\over 2} \Rightarrow a=1 \Rightarrow (a,b)=(1,3)，故選\bbox[red, 2pt]{(A)}$$

解答:$$2x-y=10 \Rightarrow y=2x-10 \Rightarrow f(x,2x-10)=x^2+(2x-10)^2=5x^2-40x+100 \\=5(x^2-8x)+100= 5(x^2-8x+16)+100-80 =5(x-4)^2+20 \\ \Rightarrow f(x,y)最小值=f(4,-2)=20，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{開口向下 \Rightarrow a\lt 0\\ 與x軸交於一點\Rightarrow 判別式-4ab=0 \Rightarrow b=0}，故選\bbox[red, 2pt]{(B)}$$

解答:$${-8\over x^2-4}+{x\over x-2} ={-8\over x^2-4}+{x(x+2)\over x^2-4} ={x^2+2x-8\over x^2-4} =0 \Rightarrow x^2+2x-8=0 \\\Rightarrow (x+4)(x-2)=0  \Rightarrow x=-4 (x\ne 2, 否則分母為0)，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(-2)=2025\cdot 2^4-4000\cdot 2^3-99 \cdot 2^2-102\cdot 2+720 =2^3(4050-4000)-99 \cdot 2^2-102\cdot 2+720 \\=50\cdot 2^3-99 \cdot 2^2-102\cdot 2+720=2^2(100-99) -102\cdot 2+720=2^2-102\cdot 2+720 \\=4-204+720=-200+720=520，故選\bbox[red, 2pt]{(A)}$$

解答:$$f(x)=x^3+ax^2+bx-2= g(x)\cdot h(x) =(x-2)(x-1)h(x) \\ \Rightarrow \cases{f(1)=0\\ f(2)=0} \Rightarrow \cases{a+b-1=0\\ 4a+2b+6=0} \Rightarrow \cases{a=-4\\ b=5}，故選\bbox[red, 2pt]{(C)}$$

解答:$$6^2+k^2=8^2+(-6)^2 \Rightarrow k=- 8 ((6,k)在第四象限 \Rightarrow k\lt 0)，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{A(5,-1) \\ B(4,2) \\ C(0,-1) \\ D(x,y)} \Rightarrow \cases{\overrightarrow{AB} =\overrightarrow{DC} \\ \overrightarrow{CB} =\overrightarrow{DA}} \Rightarrow \cases{(-1,3) =(-x,-1-y) \\ (4,3) =(5-x,-1-y)} \Rightarrow \cases{x=1\\ y=-4} \Rightarrow D(1,-4) \\ \Rightarrow \overline{CD}斜率={-4-(-1)\over 1-0}=-3，故選\bbox[red, 2pt]{(D)}$$

解答:$$\sin B=\sin(90^\circ-A) =\cos A={4\over 5}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{a=\sin(-60^\circ) =-\sin 60^\circ=-\sqrt 3/2\\ b=\tan 210^\circ =\tan 30^\circ  =1/\sqrt 3\\ c= \cos(-225^\circ) =\cos 225^\circ=-\cos 45^\circ=-\sqrt 2/2} \Rightarrow b\gt c\gt a，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cos 39^\circ \tan 39^\circ+ \sin 30^\circ\tan 45^\circ\cos 60^\circ+\sin 129^\circ\tan 141^\circ \\=\cos 39^\circ \cdot {\sin 39^\circ \over \cos 39^\circ}+ {1\over 2} \cdot 1 \cdot {1\over 2}  +\sin 51^\circ (-\tan 39^\circ) =\sin 39^\circ+ {1\over 4}- \cos 39^\circ \cdot \tan 39^\circ \\ =\sin 39^\circ+ {1\over 4}- \sin 39^\circ  ={1\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cos(\angle A+ \angle B) =\cos(180^\circ -\angle C) =-\cos \angle C=-{9^2+10^2-5^2 \over 2\cdot 9\cdot 10}= -{13\over 15}，故選\bbox[red, 2pt]{(B)}$$

解答:

$$假設燈塔位於C，則\cases{\angle A=30^\circ\\ \angle B=60^\circ} \Rightarrow \angle C=90^\circ \Rightarrow \cases{\overline{BC} =\overline{AB} \sin 30^\circ=6\\ \overline{AC}=\overline{AB} \sin 60^\circ=6\sqrt 3} \\ \Rightarrow 6\cdot 6\sqrt 3 =12\cdot h \Rightarrow h=3\sqrt 3，故選\bbox[red, 2pt]{(A)}$$

解答:$$依圖形可知:x={\pi\over 2}時,y=0且最大值為2，故選\bbox[red, 2pt]{(D)}$$

解答:$$({1\over 3})^2+({2\over 3})^2={5\over 9} \ne 1 \Rightarrow ({1\over 3},{2\over 3})不是單位向量，故選\bbox[red, 2pt]{(D)}$$

解答:$$\vec a\bot \vec b \Rightarrow \vec a\cdot \vec b=0 \Rightarrow |\vec a-2\vec b|^2=(\vec a-2\vec b) \cdot (\vec a-2\vec b) =|\vec a|^2+4|\vec b|^2-4\vec a\cdot\vec b=1+16-0=17 \\ \Rightarrow |\vec a-2\vec b|=\sqrt{17}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{A(-1,2) \\B(3,-5) \\C(1,6)} \Rightarrow \cases{G=(A+B+C)/3 =(1,1) \\M=(A+C)/2=(0,4)} \Rightarrow \cases{\overrightarrow{BG}= (-2,6) \\ \overrightarrow{AM}= (1,2)} \\ \Rightarrow \overrightarrow{BG} -\overrightarrow{AM}= (-3,4)，故選\bbox[red, 2pt]{(A)}$$

解答:$$x^2+y^2+6x-10y+30=0 \Rightarrow (x+3)^2+(y-5)^2=2^2 \Rightarrow \cases{圓心O(-3,5)\\ 圓半徑r=2} \\ \Rightarrow \overline{OP} =\sqrt{4^2+3^2} =5 \Rightarrow 切線長=\sqrt{\overline{OP}^2-r^2} =\sqrt{25-4} =\sqrt{21}，故選\bbox[red, 2pt]{(C)}$$

解答:$$x^2+y^2-6x +5=0 \Rightarrow  (x-3)^2+y^2 =2^2 \Rightarrow \cases{(A)\times:圓心(3,0) \ne (0,3) \\(B)\times: 圓半徑=2 \ne 3\\ (C)\times: (-3)^2+0^2=9\gt 2^2 \Rightarrow 在圓外\\ (D) \bigcirc: 0^2+2^2=2^2 \Rightarrow (3,2)在圓上}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$圓心(-2,1)至切線距離={|-6+4-23| \over \sqrt{3^2+4^2}} ={5} \\ \Rightarrow \cases{(A) \bigcirc:(1,5)至圓心距離=\sqrt{3^2+4^2}=5 \\(B) \times: (9,-1)至圓心距離= \sqrt{11^2+2^2} \ne 5\\ (C)\times: (-3,8)至圓心距離= \sqrt{1^2+7^2} \ne 5\\ (D) \times: (5,2)至圓心距離=\sqrt{7^2+1^2} \ne 5}，故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考，其他轉學考試題及詳解