103年身心障礙學生四技二專甄試-數學(B)-詳解

103 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(B)

單選題，共 20 題，每題 5 分

解答：$$(A)\times: \overleftrightarrow{PQ}斜率={3-(-2)\over -4-1}=-1\ne 1\\(B)\times: \overline{PQ}中點({1-4\over 2},{-2+3\over 2}) =(-{3\over 2},{1\over 2})\\(C)\bigcirc: \overline{PQ}線段長=\sqrt{(-5)^2+5^2}=5\sqrt 2 \\(D)\times: 由(A)知\overleftrightarrow{PQ}斜率=-1，因此與\overleftrightarrow{PQ}垂直之直線斜率為1\\，故選\bbox[red,2pt]{(C)}$$

解答：$$(A)\times: \sin{\pi\over 7}\sec {\pi\over 7}= \sin{\pi\over 7}\cdot {1\over \cos {\pi\over 7}} \ne 1\\ (B)\times: \sin^2{\pi\over 7} +\csc^2{\pi\over 7}= \sin^2{\pi\over 7}+{1\over \sin^2{\pi\over 7}} \ne 1\\(C) \times: \cos{2\pi\over 3} =-{1\over 2}\ne {1\over 2}\\(D)\bigcirc: \cases{\cos{2\pi\over 3} \lt 0\\ \sin {\pi\over 7}\gt 0} \Rightarrow \cos{2\pi\over 3} \lt \sin {\pi\over 7}\\，故選\bbox[red,2pt]{(D)}$$

解答：$$(A)\times: \vec a\cdot \vec b=(1,-2)\cdot (-4,3)= -4-6=-10\ne 9\\(B)\times: -2(\vec a\cdot \vec b)=-2\cdot (-10)=20 \ne -20\\ (C) \times: |\vec a+\vec b|=|(-3,1)|= \sqrt{10} \\(D)\bigcirc: 3(\vec a\cdot \vec b)=3\times (-10)=-30\\，故選\bbox[red,2pt]{(D)}$$

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

解答：$$(A)\times: \log_3 2={\log_5 2\over \log_5 3}={p\over 1/q}=pq\ne {p\over q} \\(B)\bigcirc: 由(A)知: pq = \log_3 2\\ (C)\times: 1-p=1-\log_5 2=\log_5 5-\log_5 2=\log_5 5/2 \ne \log_5 3\\(D)\times: \log_5 8=\log_5 2^3= 3\log_5 2= 3p\ne p^3\\，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{a_6=14\\ a_{17}= 36} \Rightarrow \cases{a_1+5d=14\\ a_1+16d= 36} \Rightarrow \cases{a_1=4\\ d=2}\\(A) \times: 首項a_1=4 \ne 2\\(B)\times: 金差d=2\ne 4\\ (C) \bigcirc:a_5=a_1+4d = 4+8=12 \\(D)\times: S(10)= {(2a_1+9d)\times 10\over 2} =(8+18)5=130 \ne 120\\，故選\bbox[red,2pt]{(C)}$$

解答：$$\sum_{k=1}^\infty {1-3^k\over 4^k} =\sum_{k=1}^\infty \left(({1\over 4})^k-({3 \over 4})^k \right)= {1/4\over 1-1/4} -{3/4\over 1-3/4} ={1\over 3}-3 =-{8\over 3}，故選\bbox[red,2pt]{(A)}$$

解答：$$(A)f(1)=1-3+4-1+2 =3 \ne 2\\ (B)f(-1)= -1+3+4+1+2 =9\ne 3\\(C)\bigcirc: 利用長除法可得: f(x)=(x^2-2x+3)(x^3+2x^2-2x-6)+(-7x+20)\\(D)\times: 由(C)知: 餘式為-7x+20 \ne 7x-20\\，故選\bbox[red,2pt]{(C)}$$

解答：$$-3x^3+ 4x^2-x+2 = a+b(x-1)+c(x-1)^2 +d(x-1)^2\\ =a+b(x-1)+c(x^2 -2x+1)+d (x^3-3x^2+3x-1)\\ =dx^3 +(c-3d)x^2 +(b-2c+3d)x +(a-b+c-d) \\ \Rightarrow \cases{d=-3\\ c-3d= 4\\ b-2c+3d= -1\\ a-b+c-d= 2} \Rightarrow \cases{d=-3\\ c=-5 \\b=-2\\ a=2} \Rightarrow \cases{(A)\times: a+b=0 \ne -8\\ (B)\times: a+b+c=-5\ne -10 \\(C)\times: c+d=-8\ne 0 \\(D)\bigcirc: b+c+d=-10 }，故選\bbox[red,2pt]{(D)}$$

解答：$$\begin{vmatrix} -x & 2x\\ 1-x & 2+x\end{vmatrix} =-x(x+2)-2x(1-x) = x^2-4x=5 \Rightarrow x^2-4x-5=0 \\ \Rightarrow (x-5)(x+1)= 0 \Rightarrow x=5 (x=-1違反x\gt 0)，故選\bbox[red,2pt]{(B)}$$

解答：$$(a^3+ 2b^3)\left({2\over a^3} +{1\over b^3}\right) =4+ ({a\over b})^3+ 4({b\over a})^3 \ge 4 + 2\sqrt{({a\over b})^3 \cdot 4({b\over a})^3}=8，故選\bbox[red,2pt]{(C)}$$

解答：$$個位數字是0 \Rightarrow 4\times 3=12\\ 個位數字是2\Rightarrow 3\times 3=9 (百位數字不為0)\\ 個位數字是4\Rightarrow 3\times 3=9 (百位數字不為0)\\ \Rightarrow 共有12+9+9 =30個偶數，故選\bbox[red,2pt]{(D)}$$

解答：$$A、B獨立\Rightarrow P(A\cap B)=P(A)P(B) \Rightarrow P(A\cup B)= P(A)+P(B)- P(A)P(B) \\= {1\over 4}+{2\over 3} -{1\over 4} \times {2\over 3} ={3\over 4}，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{3球皆為白球的取法:C^5_3\\ 3球皆為紅球的取法:C^3_3} \Rightarrow 3球皆同色的機率={C^5_3+ C^3_3\over C^{10}_3} ={11\over 120}，故選\bbox[red,2pt]{(A)}$$

解答：$$(x+2y)^6 = \sum_{k=0}^6 C^6_kx^k (2y)^{6-k} \Rightarrow x^2y^4的係數=C^6_2\cdot 2^4 =15\cdot 16=240，故選\bbox[red,2pt]{(A)}$$

解答：$$(A)\times: f(x)={x+1\over 3x+1} \Rightarrow f'(x)={1\over 3x+1}-{3(x+1)\over (3x+1)^2} ={-2\over (3x+1)^2} \ne{6x+2 \over (3x+1)^2} \\(B) \bigcirc: f(x)=(3x+4)^2 \Rightarrow f'(x)=2(3x+4)\cdot 3=6(3x+4) \\(C)\times: \int x^3\,dx={x^4\over 4}+C \ne {x^2\over 2}+C\\ (D)\times: f(x)=\sqrt{x+1} \Rightarrow {1\over 2\sqrt{x+1}} \ne {2\over \sqrt{x+1}}\\，故選\bbox[red,2pt]{(B)}$$

解答：$$25x^2+ 16y^2-32y-384=0 \Rightarrow 25x^2+ 16(y^2-2y+1)=384 +16 \Rightarrow 25x^2+16(y-1)^2 = 400\\ \Rightarrow {x^2\over 16} +{(y-1)^2\over 25}=1\Rightarrow \cases{a=5\\ b=4} \Rightarrow \overline{PF_1} +\overline{PF_2} =2a=10，故選\bbox[red,2pt]{(C)}$$

解答：$$x^2+8y-6x+1=0 \Rightarrow (x-3)^2=4\cdot (-2)(y-1) \Rightarrow \cases{(A)\bigcirc: 4|c|=8\\ (B)\times:對稱軸為x=1\\ (C)\bigcirc: 頂點(3,1)\\ (D)\bigcirc:c為負值，開口向下}\\，故選\bbox[red,2pt]{(B)}$$

解答：$$f(x)={a(x^2-1)\over x-1} ={a(x+1)} \Rightarrow f(1)=2a= 6 \Rightarrow a=3，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{\pi/2 \lt \alpha \lt \pi\\ 0\lt \beta \lt \pi/2} \Rightarrow \cases{\sin \alpha,\sin \beta,\cos \beta \gt 0\\ \cos \alpha \lt 0}，因此\cases{\sin \alpha=4/5\\ \cos \beta=5/13} \Rightarrow \cases{\cos \alpha =-3/5\\ \sin \beta=12/13}\\ \Rightarrow \sin(\alpha+\beta)-\cos(\alpha+\beta) =\sin\alpha\cos\beta +\sin\beta\cos \alpha-\cos\alpha\cos\beta + \sin\alpha\sin \beta \\ ={4\over 5}\cdot {5\over 13} +{12\over 13}\cdot {-3\over 5}-{-3\over 5}\cdot {5\over 13} +{4\over 5}\cdot {12\over 13} ={1\over 65}(20-36+15+48)= {47\over 65}，故選\bbox[red,2pt]{(B)}$$

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