111學年度四技二專統測--數學(C)詳解

111 學年度科技校院四年制與專科學校二年制

統一入學測驗-數學(C)

解答：$$x=\log_3 7 \Rightarrow 3^x=7，故選\bbox[red, 2pt]{(B)}$$

解答：$$a_1+a_2+\cdots +a_n = {a_1(1-r^n)\over 1-r} ={2(1-3^n)\over 1-3} =3^{n}-1\gt 2022 \Rightarrow 3^{n} \gt 2023\\ 由於\cases{3^{6}=729\\ 3^{7}=2187}，因此n \ge 7，故選\bbox[red, 2pt]{(B)}$$

解答：$$3.1\le x\le 4.9 \Rightarrow -0.9\le x-4\le 0.9 \Rightarrow |x-4|\le 0.9 \Rightarrow \cases{a=4\\ b=0.9} \Rightarrow ab= 4\times 0.9=3.6\\，故選\bbox[red, 2pt]{(B)}$$

解答：$$(A) \bigcirc:\sin x週期=2\pi \Rightarrow \sin(2x)週期=\pi \Rightarrow |\sin(2x)|的週期=\pi/2\\(B) \bigcirc:\sin x週期=2\pi \Rightarrow 3\sin x週期=2\pi \\(C)\times:\cos x週期=2\pi \Rightarrow \cos(2x)的週期=\pi \\(D)\bigcirc:\cos x週期=2\pi \Rightarrow 4\cos x週期=2\pi \\，故選\bbox[red, 2pt]{(C)}$$

解答：$$A(x+2)(x+3) +B(x-2)(x+3)+ C(x-2)(x+2)\\ =(A+B+C)x^2 +(5A+B)x+ 6A-6B-4C = x^2+2x+7\\ \Rightarrow  A+B+C= 1 ，故選\bbox[red, 2pt]{(A)}$$

解答：$$令f(x,y)=2x-y+12，則A在L右半平面代表f(A)=f(a,-6)\gt 0 \Rightarrow 2a+6+12=2a+18\gt 0\\ \Rightarrow a\gt -9，故選\bbox[red, 2pt]{(D)}$$

解答：

$$\triangle ACP\sim \triangle BDP (AAA) \Rightarrow {\overline{AP}\over \overline{BP}} ={\overline{AC}\over \overline{BD}}={d(A,L)\over d(B,L)} =\left|{1-4+1\over 6-2+1}\right| ={2\over 5}，故選\bbox[red, 2pt]{(A)}$$

解答：$$5x-4\lt x^2 \lt x+2 \Rightarrow 同時滿足\cases{x^2\lt x+2 \Rightarrow x^2-x-2\lt 0 \Rightarrow (x-2)(x+1)\lt 0\\ 5x-4\lt x^2 \Rightarrow x^2-5x+4\gt 0 \Rightarrow (x-4)(x-1)\gt 0} \\ \Rightarrow \cases{-1\lt x\lt 2\\ x\gt 4或x\lt 1} \Rightarrow -1\lt x\lt 1，故選\bbox[red, 2pt]{(A)}$$

解答：$$x^2+y^2+2x+ 4y-3=0 \Rightarrow (x+1)^2 +(y+2)^2 =8 \Rightarrow \cases{圓心P(-1,-2)\\ 半徑r=\sqrt 8= 2\sqrt 2} \\\Rightarrow d(P,L_1) = \left|{ -1-2+1\over \sqrt{2}} \right| = \sqrt 2 ;\\ 又L_1\parallel L_2 \Rightarrow a=b \Rightarrow L_2: ax+ay+10=0 \Rightarrow d(P,L_2)= \left|{ -a-2a+10\over \sqrt{2} a} \right|  = \left|{ 10-3a\over \sqrt{2} a} \right|  =\sqrt 2\\ \Rightarrow (3a-10)^2 =4a^2 \Rightarrow (a-10)(5a-10)=0 \Rightarrow a=2(若a=10\Rightarrow L_2=L_1,不合) \\ \Rightarrow a+b= 2+2=4，故選\bbox[red, 2pt]{(B)}$$

解答：$$利用長除法可得:ax^4+bx^3 +6x^2+5x+2 \\=(x+1)^2(ax^2+(b-2a)x+(6+3a-2b)) +(-7+3b-4a)x-4-3a+2b\\ \Rightarrow \cases{-7+3b-4a=3\\ -4-3a+2b=4 } \Rightarrow \cases{a=-4\\ b=-2} \Rightarrow a+b=-6，故選\bbox[red, 2pt]{(B)}$$

解答：$$依題意\cases{3x+3y-4z= 10\\ 4x+3y-3z=21\\ 2x+6y-2z=26} \Rightarrow \cases{x=6\\ y=4\\ z=5} \Rightarrow 3x+5y-2z=18+20-10=28，故選\bbox[red, 2pt]{(A)}$$

解答：$$假設塔底為D，並假設\overline{CD}=x 則\cases{\triangle ADE: \tan 30^\circ = 塔高/(a+b+x)  \\ \triangle BDE: \tan 45^\circ =塔高/(b+x)  \\ \triangle CDE :\tan 60^\circ =塔高/x  } \\ \Rightarrow 塔高=\sqrt 3x = b+x = (a+b+x)/\sqrt 3 \Rightarrow \cases{b=(\sqrt 3-1)x \\a=(3-\sqrt 3)x}\\ \Rightarrow {a\over b} ={3-\sqrt 3\over \sqrt 3-1} = \sqrt 3，故選\bbox[red, 2pt]{(A)}$$

解答：$$\overline{AB}= \sqrt{2^2+0+0}=2 \Rightarrow \overline{DA} =\overline{DB}=\overline{DC}=2 \Rightarrow \cases{x^2+y^2+z^2 =4 \cdots(1)\\ (x-2)^2 +y^2 +z^2 =4 \cdots(2) \\ (x-1)^2 +(y-\sqrt 3)^2+z^2 =4 \cdots(3)}\\ (1)-(2) \Rightarrow x^2-(x-2)^2 =0 \Rightarrow 2(2x-2)=0 \Rightarrow x=1代入(1)及(3) \Rightarrow \cases{y^2+z^2 =3 \cdots(4)\\ (y-\sqrt 3)^2+z^2 =4 \cdots(5)}\\ (4)-(5) \Rightarrow y^2-(y-\sqrt 3)^2=-1 \Rightarrow y=\sqrt 3/3 代回(1) \Rightarrow 1+{1\over 3}+z^2=4 \Rightarrow z^2={8\over 3}\\ \Rightarrow z={2\sqrt 2\over \sqrt 3} ={2\sqrt 6\over 3}，故選\bbox[red, 2pt]{(C)}$$

解答：$$假設船位空間坐標的原點O(0,0,0)，則直升機位於A(140,80,100)，黑盒子位於B(140,80,-x)，\\ 且\overline{OB}=180 \Rightarrow 140^2+80^2+x^2=180^2\Rightarrow x^2=6400 \Rightarrow x=80，故選\bbox[red, 2pt]{(C)}$$

解答：$$6首歌任排有6!=720種排法；3首慢歌連在一起唱有4!\times 3!=144種排法；\\因此最多有二首慢歌連在一起唱的排法=720-144=576，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{\vec a=(2x+1,-3)\\ \vec b=(3,x-2)} \Rightarrow \vec a-\vec b=(2x-2,-x-1) \Rightarrow \cases{|\vec a-\vec b|^2 = 5x^2-6x+5 \\ |\vec a|^2 +|\vec b|^2 = 5x^2+23} \\ \Rightarrow -6x+5=23 \Rightarrow x=-3，故選\bbox[red, 2pt]{(D)}$$

解答：$$\cases{A=\begin{bmatrix} a& -1\\ 2 & 3\end{bmatrix} \\[1ex] B=\begin{bmatrix} 2& c\\ b & d\end{bmatrix}} \Rightarrow \cases{AB=\begin{bmatrix} 2a-b& ac-d\\ 4+3b & 2c+3d\end{bmatrix} \\[1ex] A+B=\begin{bmatrix} a+2& c-1\\ b+2 & d+3\end{bmatrix}}，因此AB=A+B \Rightarrow \cases{2a-b=a+2\\ ac-d=c-1\\ 4+3b=b+2\\ 2c+3d= d+3} \\ \Rightarrow \cases{ a=1\\b=-1 \\c=1/2 \\d=1}，故選\bbox[red, 2pt]{(C)}$$

解答：$$f(x)=x^3+{12\over x} \Rightarrow f'(x)=3x^2 -{12\over x^2} =9 \Rightarrow 3x^4-9x^2-12=0 \Rightarrow 3(x^2-4)(x^2+1)=0\\ \Rightarrow x^2=4 \Rightarrow a=x=2 \Rightarrow f(2)=8+6=14=b \Rightarrow a+b= 2+14=16，故選\bbox[red, 2pt]{(D)}$$

解答：$$\int {x+3\over 2\sqrt x}\,dx =\int {x \over 2\sqrt x} +{3\over 2\sqrt x}\,dx =\int {1\over 2} \sqrt x  +{3\over 2}{1\over \sqrt x}\,dx = {1\over 3}x^{3/2} +3 x^{1/2}+C，故選\bbox[red, 2pt]{(D)}$$

解答：$$距離=\int_0^3 -{3\over 2}t^2+6t+90\,dt = \left.\left[-{1\over 2}t^3+ 3t^2 +90t \right] \right|_0^3 =-{27\over 2} +27+270 = 283.5，故選\bbox[red, 2pt]{(C)}$$

解答：$$r^6\le {1\over 2} \Rightarrow \log r^6 \le \log 0.5=-0.301 \Rightarrow \log r\le -{0.301\over 6} =-0.05=0.95-1\Rightarrow 1+\log r=0.95 \\ \Rightarrow \log 10r=0.95 \Rightarrow 10r=8.91 \Rightarrow r=0.891，故選\bbox[red, 2pt]{(B)}\\ 註:試題參考數值\log 0.5\approx -0.301、\log 8.91 \approx 0.950$$

解答：$$\cos A={5^2+6^2-4^2\over 2\cdot 5\cdot 6} ={3\over 4} \Rightarrow \sin A={\sqrt 7\over 4} \Rightarrow {4\over \sin A}=2R \Rightarrow 2R={16\over \sqrt 7}，故選\bbox[red, 2pt]{(C)}$$

解答：$$\triangle ABC面積={1\over 2}\overline{AB}\cdot \overline{AC}\sin \angle BAC \Rightarrow {3\sqrt 2\over 2} ={1\over 2}\cdot 3\cdot 2\sin \angle BAC \Rightarrow \sin \angle BAC={\sqrt 2\over 2} \\\Rightarrow \cos \angle BAC=-{\sqrt 2\over 2} ={3^2+2^2-a^2\over 12} \Rightarrow 13-a^2=-6\sqrt 2 \Rightarrow a^2=13+6\sqrt 2，故選\bbox[red, 2pt]{(D)}$$

解答：$$4x^2+6y^2-12y-6=0 \Rightarrow 4x^2+6(y-1)^2=12 \Rightarrow {x^2\over 3} +{(y-1)^2 \over 2}=1 \Rightarrow\cases{x=\sqrt 3\cos \theta \\ y=\sqrt 2\sin \theta +1} \\ \Rightarrow x+3y= \sqrt 3\cos\theta +3\sqrt 2\sin \theta +3 =\sqrt{21}\sin(\alpha+\theta)+3 \Rightarrow 最大值=\sqrt{21}+3，故選\bbox[red, 2pt]{(D)}$$

解答：$$f_1(x)={2x+a\over x^2-2x-3} ={2(x+a/2)\over (x-3)(x+1)} 在x=3連續 \Rightarrow {a\over 2}=-3 \Rightarrow a=-6 \\\Rightarrow f_1(x) ={2\over x+1} \Rightarrow f_1(3)={1\over 2};\\ f_2(x)={x-5\over x-b} \Rightarrow f_2(3)={1\over 2}={-2\over 3-b} \Rightarrow b=7 \Rightarrow a+b=-6+7=1，故選\bbox[red, 2pt]{(C)}$$

==================== END =========================

解題僅供參考，其他歷年試題及詳解