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2022年2月12日 星期六

104年身心障礙學生四技二專甄試-數學(A)-詳解

104 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:四技二專組-數學(A)

單選題,共 20 題,每題 5 分

解答:$$\cases{L_1:x+y=2\\ L_2:y-x+8=0} \Rightarrow \cases{A=L_1\cap L_2=(5,-3)\\ B=L_1\cap x軸= (2,0)\\ C=L_2\cap x軸=(8,0)} \Rightarrow \triangle ABC={1\over 2}\cdot (8-2)\cdot |-3| =9\\,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{A(a,0)\\ B(0,b)\\ \overline{AC}:\overline{CB}=2:1} \Rightarrow C=(2B+A)/3 = (a/3,2b/3) =(-1,2) \Rightarrow \cases{ a=-3\\ b=3} \\ \Rightarrow B(0,3),故選\bbox[red,2pt]{(D)}$$
解答:$$(\sin 35^\circ +\cos 35^\circ)^2 +(\sin 35^\circ-\cos 35^\circ)^2 = (1+2\sin 35^\circ\cos 35^\circ) +(1-2\sin 35^\circ\cos 35^\circ)\\ =2,故選\bbox[red,2pt]{(A)}$$
解答:$$(A)-\sin 335^\circ = -\sin(-25^\circ) = \sin 25^\circ\\ (B) \cos 155^\circ =-\cos 25^\circ\\ (C)\sin 155^\circ =\sin 25^\circ \\ (D)\cos 65^\circ =\sin 25^\circ\\,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{\vec a=(-1,2)\\ \vec b=(x,2)} \Rightarrow \vec a-2\vec b=(-1-2x,-2),再由\vec a\cdot (\vec a-2\vec b)=0 \Rightarrow (-1,2)\cdot (-1-2x,-2)=0\\ \Rightarrow 1+2x-4=0 \Rightarrow x=3/2,故選\bbox[red,2pt]{(C)}$$
解答:$$曲線2y=x^2上的點P(t,t^2/2)與直線x+y+2=0的距離 ={|t+t^2/2+2|\over \sqrt 2} ={\lvert{1\over 2}(t+1)^2+{3\over 2} \rvert \over \sqrt 2} \\ \Rightarrow 當t=-1時,距離的最小值為{3\over 2\sqrt 2} ={3\sqrt 2\over 4},故選\bbox[red,2pt]{(D)}$$
解答:$$2x^2-3x-2=0 \Rightarrow (2x+1)(x-2)=0 \Rightarrow |\alpha-\beta| =|-{1\over 2}-2|={5\over 2},故選\bbox[red,2pt]{(A)}$$
解答:$$f(x)=ax^3+bx^2-2x+1 = (x^2-4)p(x) \Rightarrow \cases{f(2)=8a+4b-3=0\\ f(-2)=-8a+4b+5 =0} \Rightarrow \cases{a=1/2\\ b=-1/4} \\ \Rightarrow a+b={1\over 4},故選\bbox[red,2pt]{(B)}$$
解答:$$\log_6 14= {\log_3 14\over \log_3 6} ={\log_3 7+\log_3 2\over 1+\log_3 2} ={a+b\over 1+b},故選\bbox[red,2pt]{(C)}$$
解答:$$5^{2x+1}-6\cdot 5^x+1=0 \Rightarrow 5\cdot (5^x)^2-6\cdot 5^x+1=0 \Rightarrow (5\cdot 5^x-1)(5^x-1)=0 \\ \Rightarrow \cases{5^x=1/5\\ 5^x=1} \Rightarrow \cases{x=-1\\ x=0} \Rightarrow a+b=-1+0=-1,故選\bbox[red,2pt]{(D)}$$
解答:$$(3x+1)(x-1)\le 2x^2-7x+13 \Rightarrow 3x^2-2x-1\le 2x^2-7x+13 \Rightarrow x^2+5x-14\le 0\\ \Rightarrow (x+7)(x-2)\le 0 \Rightarrow -7\le x\le 2 \equiv a\le x\le b \Rightarrow a+b=-7+2=-5,故選\bbox[red,2pt]{(B)}$$
解答

$$由上圖可知:圖形經過第三、四象限,故選\bbox[red,2pt]{(B)}$$
解答:$$2x^2+2y^2+4x-6y+3=0 \Rightarrow 2(x^2+2x+1)+2(y^2-3y+{9\over 4})={7\over 2}\\ \Rightarrow (x+1)^2+(y-{3\over 2})^2= {7\over 4} \Rightarrow 圓半徑r={\sqrt 7\over 2} \Rightarrow 圓面積=r^2\pi ={7\over 4}\pi,故選\bbox[red,2pt]{(C)}$$
解答:$$假設過(2,4)的直線L:y=m(x-2)+4與圓相切,也就是d(L,O(1,2))=\sqrt 5 \\ \Rightarrow \lvert {2-m\over \sqrt{m^2+1}}\rvert =\sqrt 5 \Rightarrow (m-2)^2 = 5(m^2+1) \Rightarrow (2m+1)^2=0 \Rightarrow m=-1/2\\ \Rightarrow L:y=-{1\over 2}(x-2)+4 \Rightarrow x+2y=10,故選\bbox[red,2pt]{(A)}$$
解答:$${1\over 6},-{1\over 2},{3\over 2},-{9\over 2}為一等比數列,其中\cases{首項a_1=1/6\\ 公比r=-3} \Rightarrow S(9)=a_1+a_2+ \cdots +a_9\\ =a_1(1+r+r^2+\cdots +r^8)=a_1\cdot {1-r^9\over 1-r} ={1\over 6}\cdot {1+3^9\over 4} ={19684\over 24} ={4921\over 6},故選\bbox[red,2pt]{(D)}$$
解答:$$三種動物排列有3!=6種排法,其中\cases{2隻貓有2!排法\\ 3隻兔有3!排法\\4隻狗有4!排法},因此共有6\cdot 2!\cdot 3!\cdot 4!= 1728種排法\\,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{3的倍數有\lfloor 100/3 \rfloor =33個\\ 5的倍數有\lfloor 100/5 \rfloor =20個\\ 15的倍數有\lfloor 100/15 \rfloor =6個} \Rightarrow 3或5的倍數有33+20-6= 47個 \Rightarrow 機率為{47\over 100},故選\bbox[red,2pt]{(C)}$$
解答:$$ \cases{9開頭:{5!\over 2!2!}=30 \\7開頭:{5!\over 2!2!} =30\\6開頭:{5!\over 2!}  =60 \\5開頭:{5!\over 2!}=60} \Rightarrow 30+30+60+60 = 180個不同的六位數,故選\bbox[red,2pt]{(A)}$$
解答:$$10個數字遞增排序: \begin{array} {|c|c|c|c|c|c|c|c|c|c|c|}\hline i&1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\hdashline x_i& 2 & 3 & 4 & 6 & 6 & 7 & 8 & 8 & 10 & 12\\ \hline\end{array} \\ \Rightarrow \cases{平均數a={1\over 10}\sum x_i=66/10=6.6 \\ 中位數b=(x_5+x_6)/2=(6+7)/2=6.5} \Rightarrow a+b=13.1,故選\bbox[red,2pt]{(D)}$$
解答:$$46-31=15,故選\bbox[red,2pt]{(C)}$$
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