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2021年3月23日 星期二

110年身障生升四技二專-數學(A)詳解

110 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)

解答$$由f(x)=2x^3+ax^2 +bx+6 ,因此 \cases{f(1)=-6\\ f(-1)=12} \Rightarrow \cases{2+a+b+6=-6\\ -2+a-b+6=12} \\ \Rightarrow \cases{a+b =-14\\ a-b=8} \Rightarrow \cases{a=-3 \\ b=-11} \Rightarrow 2a-b=-6+11=5,故選\bbox[red,2pt]{(D)}$$
解答$$\cases{a=2^7 \times 5^6\\ b=2^4\times 5^5} \Rightarrow \log_{10}a +\log_{10}b =\log_{10}(2^7 \times 5^6) +\log_{10} (2^4\times 5^5 )\\= 7\log_{10}2+ 6\log_{10}5 +4\log_{10}2+ 5\log+{10}5 =11\log_{10} 2 +11\log_{10} 5 =11\log_{10}(2\times 5) \\ =11\log_{10}(10)=11,故選\bbox[red,2pt]{(B)}$$
解答$$\cases{A(2,3)\\ B(-1,2)\\ C(3,0)} \Rightarrow \cases{\overrightarrow{AB} =(-3,-1) \\ \overrightarrow{BC}=(4,-2)} \Rightarrow 2\overrightarrow{AB} -\overrightarrow{BC} =(-6,-2)-(4,-2)=(-10,0) \\ \Rightarrow |2\overrightarrow{AB} -\overrightarrow{BC}| =\sqrt{(-10)^2+0^2}=10,故選\bbox[red,2pt]{(C)}$$
解答$$(ab,a-b)在第二象限\Rightarrow \cases{ab \lt 0\\ a-b \gt 0} \Rightarrow \cases{a\gt 0\\ b\lt 0}\\ 直線ax+by=3過兩點\cases{P(0,3/b)\\ Q(3/a,0)} \Rightarrow 不過第二象限,故選\bbox[red,2pt]{(B)}$$
解答$$\cases{A(x,y)\\ P(1,5)\\ B(3,7)\\ \overline{AP}: \overline{BP}=3:2} \Rightarrow P={2\over 5}A+ {3\over 5}B \Rightarrow \cases{1={2\over 5}x+ {3\over 5}\cdot 3 \\ 5={2\over 5}y+ {3\over 5}\cdot 7} \Rightarrow \cases{x=-2\\ y=2},故選\bbox[red, 2pt]{(B)}$$
解答$$利用長除法可得f(x)=(2x-3)(x^2+3x+4)+a+12,由於2x-3為其因式\\,因此x^2+3x+4也是因式(a=-12),故選\bbox[red, 2pt]{(A)}$$
解答$$三角形外心即為外接圓圓心,故選\bbox[red, 2pt]{(B)}$$
解答$$f(x,y)=2y-6 \Rightarrow f(3,0)=0-6=-6\ne 0,故選\bbox[red, 2pt]{(A)}$$
解答$$(A)\times: (20-10)\ne (25-20),B不是等差數列\\(B)\times:(15-0) \ne (25-15),A也不是等差數列 \\(C)\times: A、B都不是等差數列\\故選\bbox[red, 2pt]{(D)}$$
解答$${1-(1/2)^5\over 1-1/2} =2-{1\over 2^4},故選\bbox[red, 2pt]{(B)}$$
解答$$-x^2-2x-1 \gt 0 \Rightarrow x^2+2x+1 \lt 0 \Rightarrow (x+1)^2 \lt 0 \Rightarrow 無解,故選\bbox[red, 2pt]{(D)}$$
解答$$9^{x+1}-({1\over 3})^{x^2-4x-5}=0 \Rightarrow 3^{2x+2} =3^{-x^2+4x+5} \Rightarrow 2x+2= -x^2+4x+5 \Rightarrow x^2-2x-3=0 \\\Rightarrow (x-3)(x+1)=0 \Rightarrow x=3,-1 \Rightarrow \alpha+\beta = 3-1=2,故選\bbox[red, 2pt]{(D)}$$
解答$$\tan \alpha=-{4\over 3} \Rightarrow \cases{\cos \alpha = \pm {3\over 5} \\ {\sin \alpha \over \cos \alpha}=-{4\over 3} \Rightarrow \sin \alpha=-{4\over 3}\cos \alpha}\\ \Rightarrow \sin 2\alpha = 2\sin \alpha\cos \alpha =2 \cdot (-{4\over 3}\cos \alpha)\cdot \cos \alpha =-{8\over 3}\cos ^2\alpha = -{8\over 3}(\pm {3\over 5})^2 =-{24\over 25},故選\bbox[red, 2pt]{(A)}$$
解答$$ 3\tan 945^\circ +4\sin 690^\circ -\cos 600^\circ \\=3\tan (360^\circ \times 2+225^\circ) +4\sin (360^\circ\times 2-30^\circ) -\cos (360^\circ\times 2-120^\circ)\\ =3\tan 45^\circ-4\sin 30^\circ -\cos 120^\circ =3\cdot 1-4\cdot {1\over 2} +{1\over 2} = {3\over 2},故選\bbox[red, 2pt]{(D)}$$
解答:兩個方法都可以求出交點數,故選\(\bbox[red,2pt]{(A)}\)

解答:就是3的排列數,即3!=6,故選\(\bbox[red,2pt]{(D)}\)

解答$$C^8_6=28,故選\bbox[red,2pt]{(C)}$$
解答$${1000\over 300000} ={1\over 300},故選\bbox[red,2pt]{(C)}$$
解答擲骰子是獨立事件,且骰子是公正的,甲乙機率相同,故選\(\bbox[red, 2pt]{(A)}\)
解答


$$\cases{x\ge 0\\ y\ge 0\\ x+y \le 10\\ 2x+3y\le 28}所圍區域頂點坐標為\cases{A(0,0)\\ B(10,0)\\ C(2,8)\\ D(0,28/3)} \Rightarrow  \cases{ f(A)=0\\ f(B)=30\\ f(C)=38\\ f(D)=37{1\over 3}} \Rightarrow f(x,y)有極大值38\\,故選\bbox[red,2pt]{(C)}$$

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