110學年度四技二專統測--數學(B)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(B)

解答：$$f(x)=({\pi\over 4})^x \Rightarrow \cases{f(0)=1\\ {\pi\over 4} \lt 1 \Rightarrow x越大則f(x)越小}\qquad，故選\bbox[red, 2pt]{(D)}$$

解答：$$\cases{\cases{x+2y=3\\ ax+6y=9} \Rightarrow {1\over a}={2\over 6}={3\over 9} \Rightarrow a=3\\ \cases{2x+y=5\\ 4x+by=10} \Rightarrow {2\over 4} ={1\over b}= {5\over 10} \Rightarrow b=2} \Rightarrow 2a-b=6-2=4，故選\bbox[red, 2pt]{(B)}$$

解答：$$(A)\times: (0,0)代入100x+2y-100 = -100 \not \ge 0\\ (B)\times: (1,1)代入2x+100y+100 = 202 \not \le 0\\ (C)\times: (2,1)代入2x+100y+100 =204 \not \le 0\\ (D)\bigcirc: \cases{(2,-2)代入100x+2y-100 =96 \ge 0\\ (2,-2) 代入2x+100y+100 =-96 \le 0}\\故選\bbox[red, 2pt]{(D)}$$

解答：$$(A)-{9\pi \over 5} +2\pi ={\pi \over 5}=36^\circ\\ (B)-36^\circ\\(C){\pi \over 5}=36^\circ \\ (D)1116^\circ-360^\circ \times 3=36^\circ\\ \Rightarrow (A)=(C)=(D) = 36^\circ，故選\bbox[red, 2pt]{(B)}$$

解答：$$只有9出現2次，其它均出現1次，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{a=\log 2\\ b=\log 3} \Rightarrow \cases{10^a=2\\ 10^b=3 } \Rightarrow 10^{2a+b}=(10^a)^2\cdot 10^b = 2^2\cdot 3=12，故選\bbox[red, 2pt]{(C)}$$

解答：$$x^2+y^2-8x+6y=0 \Rightarrow (x-4)^2 +(y+3)^2 = 4^2+3^2=5^2 \Rightarrow 半徑=5 \Rightarrow 直徑=10\\，故選\bbox[red, 2pt]{(C)}$$

解答：$$3\times 4\times 5=60，故選\bbox[red, 2pt]{(D)}$$

解答：$$抽中三紅球機率={C^5_3\over C^8_3}={10\over 56} ={5\over 28} \Rightarrow 甲獲得金額的期望值= -420\times {5\over 28} +140\times (1-{5\over 28}) \\ ={1120\over 28 } =40，故選\bbox[red, 2pt]{(A)}$$

解答：$$找數字間變化較低的，故選\bbox[red, 2pt]{(C)}$$

解答：$$h(x)=4g(x)-7x+9 \Rightarrow h'(x)=4g'(x)-7 \Rightarrow h'(0)=4g'(0)-7 = 4\times 3-7=5\\，故選\bbox[red, 2pt]{(A)}$$

解答：$$L:2x+3y=1 \Rightarrow y=-{2\over 3}x+{1\over 3} \Rightarrow L的斜率=-{2\over 3} = L_1斜率\Rightarrow y=-{2\over 3}x+b\\ 又L經過(2,0) \Rightarrow 0=-{4\over 3}+b \Rightarrow b={4\over 3} \Rightarrow \cases{m+b= -{2\over 3}+ {4\over 3} ={2\over 3} \\ m\times b= -{2\over 3} \times {4\over 3}=-{8\over 9}}，故選\bbox[red, 2pt]{(A)}$$

解答：

$$C_1:x^2+y^2-2x+2y=0 \Rightarrow (x-1)^2+(y+1)^2 = (\sqrt 2)^2 \Rightarrow  圓心O_1(1,-1),半徑r_1=\sqrt 2\\ C_2: x^2+y^2-4x+4y=0 \Rightarrow (x-2)^2+(y+2)^2=2^2 \Rightarrow  圓心O_2(2,-2),半徑r_2=  2 \\ 又\cases{\overline{O_1O_2}= \sqrt 2=r_1 \\ \text{dist}(O_1,L) =\sqrt 2 =r_1\\ \text{dist}(O_2,L)=0} \Rightarrow O_2在C_1上,且L經過O_2 \Rightarrow \cases{L與C_1相切一點\\ L與C_2交於兩點} \\\Rightarrow 共交3點，故選\bbox[red, 2pt]{(C)}$$

解答：$$\tan \theta={7\over 25} \Rightarrow \cases{\sin\theta,\cos \theta同號\\ \sin\theta = {7\over \sqrt{674}} \\ \cos \theta = {25 \over \sqrt{674}}} \Rightarrow a=\sin\theta \cos\theta ={7\times 25\over 674} ={175\over 674 }\lt {1\over 2}，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{\begin{vmatrix}2 & 3& 4\\ 0 & a & -1\\ 0 & 6 & 2 \end{vmatrix}= 4a+12 \\\begin{vmatrix}0 & b& 2\\ 4 & 6 & 5\\ 0 & 3 & 1 \end{vmatrix}= 24-4b} \Rightarrow 4a+12=24-4b \Rightarrow 4a+4b=12 \Rightarrow a+b=3，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{0\lt \theta \lt {\pi \over 2}\\ \sin \theta ={3\over 5}} \Rightarrow \cos \theta ={4\over 5} \Rightarrow \sin (2\theta)= 2\sin \theta\cos \theta = 2\cdot {3\over 5} \cdot {4\over 5} ={24\over 25} \Rightarrow \cos(2\theta)={7\over 25} \\ \Rightarrow a=\sin(4\theta) = 2\sin (2\theta)\cos (2\theta)= 2\cdot {24\over 25}\cdot {7\over 25} ={336\over 625} \Rightarrow {1\over 2} \lt a \lt {3\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答：$$由題意知:\cases{F_1(2,-3)\\ F_2(-4,-3)\\ 2a=10} \Rightarrow \cases{c=\overline{F_1F_2}\div 2=3\\ a=5} \Rightarrow b=4 \Rightarrow 短軸長=2b=8，故選\bbox[red, 2pt]{(D)}$$

解答：$$股票+債券最多600萬元，即x+y \le 600；又股票不會低於債券的2倍，即x\ge 2y，故選\bbox[red, 2pt]{(A)}$$

解答：$$A(0,0) \Rightarrow B(-240,120) \Rightarrow C(-240+180,120-40)=(-60,80) \\\Rightarrow \overline{AC}= \sqrt{60^2+80^2} =100，故選\bbox[red, 2pt]{(A)}$$

解答：$$\sum_{n=1}^\infty a_n =\sum_{n=1}^\infty {n(n+1)\over 2} \Rightarrow a_n={n(n+1)\over 2} \Rightarrow \cases{a_1=1\\ a_2=3 \\a_3=6 \\a_4=10 \\ a_5=15} \Rightarrow \sum_{n=1}^5 a_n = 35，故選\bbox[red, 2pt]{(A)}$$

解答：$$(11+12+\cdots+17) +(21+\cdots + 27) +\cdots +(71+\cdots +77)-(11+22+\cdots + 77) \\ = (28\times 7\div 2) +(48\times 7\div 2)+\cdots (148\times 7\div 2)- 11(1+2+\cdots+ 7)\\ ={7\over 2}(28+48+68+\cdots +148)-11\times {8\times 7\over 2} =44\times 49-11\times 28=1848，故選\bbox[red, 2pt]{(C)}$$

解答：$$365天取不同的30天有C^{365}_{30}取法，30位同學再排列，有P^{365}_{30}可能，每人在某天出生的機率皆為{1\over 365}\\，因此機率為P^{365}_{30}\times {1\over 365^{30}}，故選\bbox[red, 2pt]{(D)}$$

解答：$$\int_0^1 [2x+f(x)]\;dx +\int_1^3f(x)\;dx +\int_3^0f(x)\;dx \\=\int_0^1 2x\;dx +\int_0^1f(x)\;dx+\int_1^3f(x)\;dx -\int_0^3f(x)\;dx \\ =\int_0^12x\;dx+\int_0^3f(x)\;dx-\int_0^3f(x)\;dx = \int_0^1 2x\;dx = x^2|_0^1= 1，故選\bbox[red, 2pt]{(B)}$$

解答：$$f(x)=x^2+bx+c = (x+1)^2 +(b-2)x+(c-1)\\ \Rightarrow f(x)被(x+1)^2除的餘式(b-2)x+(c-1)能被x-1整除 \Rightarrow b-2+c-1=0 \Rightarrow b+c=3;\\同理,f(x)=(x-1)^2+(b+2)x+(c-1) \Rightarrow (b+2)x+(c-1)能被x+1整除\\ \Rightarrow -b-2+c-1=0 \Rightarrow -b+c=3\\ 因此\cases{b+c=3\\ -b+c=3} \Rightarrow \cases{c=3\\b=0}，故選\bbox[red, 2pt]{(D)}$$

解答：

$$由試題所附資料可知\tan 37^\circ \approx {3\over 4} \Rightarrow \tan 53^\circ ={4\over 3}\\ 現在\cases{\tan 37^\circ =h/(a+31)=3/4\\ \tan 53^\circ =h/a=4/3} \Rightarrow a={4\over 3}h-31 = {3\over 4}h \Rightarrow {7\over 12}h=31 \\ \Rightarrow h={372\over 7} =53.1，故選\bbox[red, 2pt]{(B)}$$

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