朱式幸福: 110學年度四技二專統測--數學(C)詳解

110 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：$${3x-1\over (x-3)(x-1)} ={A\over x-3}+ {B\over x-1} \Rightarrow 3x-1=A(x-1)+B(x-3) = (A+B)x-A-3B\\ \Rightarrow \cases{A+B= 3\\ -A-3B=-1} \Rightarrow \cases{A=4\\ B=-1}，故選\bbox[red, 2pt]{(D)}$$

解答：$$\cases{\tan \theta +\sec \theta = 5\\ \tan \theta -\sec \theta=a} \Rightarrow (\tan \theta +\sec \theta)(\tan \theta -\sec \theta) =5a \Rightarrow \tan^2\theta -\sec^2\theta =5a \\ 由於1+ \tan^2\theta=\sec^2\theta \Rightarrow \tan^2\theta -\sec^2\theta =-1,因此上式\Rightarrow -1=5a \Rightarrow a=-{1\over 5}，故選\bbox[red, 2pt]{(B)}$$

解答：$$利用兩個公式:\cases{\sin 2\theta =2\sin \theta \cos \theta \\ \sin \alpha\cos \beta = {1\over 2}(\sin(\alpha+\beta) +\sin (\alpha-\beta))}\\\sin 10^\circ \cos 10^\circ \cos 50^\circ -\sin 25^\circ \cos 25^\circ\cos 20^\circ ={1\over 2}\sin 20^\circ \cos 50^\circ -{1\over 2}\sin 50^\circ \cos 20^\circ \\ ={1\over 4}(\sin 70^\circ +\sin(-30))-{1\over 4}(\sin 70^\circ +\sin 30^\circ)= -{1\over 4}\sin 30^\circ-{1\over 4}\sin 30^\circ =-{1\over 2}\sin 30^\circ\\ =-{1\over 2}\times {1\over 2}=-{1\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答：$$固體+液體共有93+3=96個樣本，其中只有9個屬於半金屬，因此機率為{9\over 96} ={3\over 32}\\，故選\bbox[red, 2pt]{(B)}$$

解答：$$令f(x)={1\over 3+x+2} \Rightarrow f'(x)=-{1\over (x+5)^2} \Rightarrow f'(0)=-{1\over 25};\\而\lim_{h\to 0}{{1\over 3+h+2}-{1\over 3+2} \over h} =f'(0)=-{1\over 25}，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{a=\sum_{m=1}^7 {m-2 \over 2m-1} =\sum_{k=1}^7 {k-2 \over 2k-1} \\ b=\sum_{k=0}^6 {k-1 \over 2k+1} =\sum_{k=1}^7{k-1-1\over 2(k-1)+1} =\sum_{k=1}^7{k-2\over 2k-1 } \\c = \sum_{i=3}^8 {i-4 \over 2i-5} =\sum_{k=1}^6{k+2-4\over 2(k+2)-5} =\sum_{k=1}^6{k-2\over 2k-1}} \Rightarrow a=b \gt c，故選\bbox[red, 2pt]{(D)}$$

解答：$$I(t)={1\over 1+49\cdot 7^{-t/3}} \Rightarrow \cases{a=I(0)= {1\over 1+49} ={1\over 50} \\ b=I(3)= {1\over 1+49/7}={1\over 8} \\ c=I(6) ={1\over 1+49/49}={1\over 2}} \Rightarrow c=4b，故選\bbox[red, 2pt]{(C)}$$

解答：$$拋物線y=x^2+4x+5 = (x+2)^2+1 \Rightarrow 頂點坐標為(-2,1) \Rightarrow 圓心O(-2,1) \\ \Rightarrow O至y軸距離=2 = 半徑\Rightarrow 圓方程式(x+2)^2+(y-1)^2 =2^2 \\\Rightarrow x^2+y^2+4x-2y+1=0，故選\bbox[red, 2pt]{(D)}$$

解答：$$x^2+4y^2+4x-16y+4=0 \Rightarrow (x+2)^2+4(y-2)^2=16 \Rightarrow {(x+2)^2 \over 4^2} +{(y-2)^2 \over 2^2}=1\\ \Rightarrow 為一左右型的橢圓，且中心點在(-2,2);\\而 {(x+2)^2 \over 4} -{(y-1)^2 \over 5}=1為左右型的雙曲線，且中心點在(-2,1)，故選\bbox[red, 2pt]{(D)} $$

解答：$${k\over 1} ={3\over 4(k+1)} ={k+1\over 8k^2+1} \Rightarrow \cases{4k^2+4k-3=0 \\24k^2+3=4k^2+8k+4} \Rightarrow \cases{(2k+3)(2k-1)=0 \\ (10k+1)(2k-1)=0} \\ \Rightarrow k={1\over 2}，故選\bbox[red, 2pt]{(C)} $$

解答：$$\begin{vmatrix} x+y & x-y & x \\ y+z & y-z & y\\ z+x & z-x & z\end{vmatrix} =\begin{vmatrix} x+y & y+z & z+x \\ x-y & y-z & z-x\\ x & y & z\end{vmatrix} = \begin{vmatrix} y & z & x \\ -y & -z & -x\\ x & y & z\end{vmatrix} = \begin{vmatrix} y & z & x \\ 0 & 0 & 0\\ x & y & z\end{vmatrix}=0，故選\bbox[red, 2pt]{(A)} $$

解答：$$8人取2人在A場地、剩下6人取2人在B場地、剩下4人取2人在C場地，最後2人在D場地；\\即C^8_2C^6_2C^4_2C^2_2 =28\times 15\times 6\times 1 =2520，故選\bbox[red, 2pt]{(B)}$$

解答：$$L_1 \bot L_2 \Rightarrow ab=-1 \Rightarrow (a+b)^2 = a^2+b^2+2ab = 50-2=48;\\ 又\cases{y=ax+b\\ y=bx+a} \Rightarrow 交點P(1,a+b) \Rightarrow \overline{OP} =\sqrt{1+(a+b)^2} =\sqrt{1+48} =7，故選\bbox[red, 2pt]{(B)}$$

解答：$$ab:bc:ca =3:4:6 \Rightarrow {ab\over abc}:{bc\over abc}:{ca\over abc}=3:4:6 \Rightarrow {1\over c}:{1\over a}:{1\over b} =3:4:6 \\ \Rightarrow a:b:c ={1\over 4}:{1\over 6}:{1\over 3} =18:12:24=3:2:4 = \sin A:\sin B:\sin C，故選\bbox[red, 2pt]{(D)}$$

解答：$$f(1)=f(2)=f(-2)=2 \Rightarrow f(x)-2=0的三根為1,2,-2 \Rightarrow f(x)-2=a(x-1)(x-2)(x+2)\\ 又f(-1)=8 \Rightarrow 8-2=a(-2)\cdot(-3)\cdot 1 \Rightarrow 6=6a \Rightarrow a=1 \Rightarrow f(x)=(x-1)(x-2)(x+2)+2\\ \Rightarrow f(x)=x^3-x^2-4x+6 \Rightarrow \cases{b=-1\\ c=-4\\ d=6}，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{\vec a\cdot \vec c=0\\ \vec b\cdot \vec c=0} \Rightarrow \cases{\vec a\bot \vec c \\ \vec b\bot \vec c} \Rightarrow \vec a\parallel \vec b \Rightarrow \vec a\cdot \vec b=\pm|\vec a||\vec b|=-|\vec a||\vec b|(因為\vec a\cdot \vec b \lt 0) =-60，故選\bbox[red, 2pt]{(B)}$$

解答：$$\int_1^3(3x-2)^{110}\;dx = \left. \left[{1\over 333} (3x-2)^{111}\right] \right|_1^3 ={1\over 333}(7^{111}-1)，故選\bbox[red, 2pt]{(A)}$$

解答：$$(A)\times: \tan \theta的週期為\pi \Rightarrow \tan {\theta\over 3}的週期為3\pi\\(B)\times: 應該是\sec^2 \theta -\tan^2\theta =1\\(C)\bigcirc: \sin \theta +\cos \theta =\sqrt 2(\cos 45^\circ\sin \theta+ \sin 45^\circ \cos \theta) =\sqrt 2\sin(\theta+45^\circ) \\\qquad \Rightarrow -\sqrt 2\le \sin \theta +\cos \theta\le \sqrt 2\\(D)\times: \theta ={\pi \over 4}+n\pi, n\in \mathbb{Z}\\，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{{\sqrt 3-i\over \sqrt 3+i} ={(\sqrt 3-i)^2\over (\sqrt 3+i)(\sqrt 3-i)} ={1-\sqrt 3i\over 2} \\{\sqrt 3+i\over \sqrt 3-i} ={(\sqrt 3+i)^2 \over (\sqrt 3-i)(\sqrt 3+i)} ={1+\sqrt 3i\over 2}} \Rightarrow \left({\sqrt 3-i\over \sqrt 3+i}\right)^2 +\left( {\sqrt 3+i\over \sqrt 3-i}\right)^2 = \left({1-\sqrt 3i\over 2}\right)^2 +\left( {1+\sqrt 3i\over 2}\right)^2 \\ ={-1-\sqrt 3i\over 2} +{-1+\sqrt 3i\over 2} =-1，故選\bbox[red, 2pt]{(B)}$$

解答：$$(x^2+2)+{9\over x^2+2} \ge 2 \sqrt{(x^2+2) \cdot {9\over x^2+2}}=6 \Rightarrow (x^2+2)+{9\over x^2+2}-4 \ge 6-4 \\ \Rightarrow x^2-2+{9\over x^2+2} \ge 2 \Rightarrow 最小值為2，故選\bbox[red, 2pt]{(A)}$$

解答：$$三本皆在同一層: 3!\times 3=18\\兩本在同一層，另一本在別層:P^3_2\times 3\times 2 = 36\\三本各在一層:3!=6\\因此共有18+36+6=60種排法，故選\bbox[red, 2pt]{(A)}$$

解答：$$y=mx=-x^2+4x-1 \Rightarrow x^2+(m-4)x+1=0 \Rightarrow 判別式=0 \Rightarrow (m-4)^2=4 \\ \Rightarrow \cases{m=6\\ m=2} \Rightarrow \cases{x^2+2x+1=0\\ x^2-2x+1=0} \Rightarrow \cases{x=-1,不合，需在第一象限\\ x=1} \\\Rightarrow m=2，故選\bbox[red, 2pt]{(B)}$$

解答：$$\int_1^4 (x+{1\over \sqrt x})(\sqrt x-{1\over x})\;dx = \int_1^4 x\sqrt x-{1\over x\sqrt x}\;dx =\int_1^4 x^{3/2}-x^{-3/2} \;dx =\left .\left[ {2\over 5}x^{5/2} +2x^{-1/2}\right]\right|_1^4 \\ =\left( {2\over 5}\cdot 2^5+1\right)-\left( {2\over 5}+2\right)={2^6-2-5\over 5} ={57\over 5}，故選\bbox[red, 2pt]{(A)}$$