桃園市立陽明高中數學科教甄試題
一、 填充題，每題 5 分，共 60 分。
解答:$$f(x)=2x^3+x^2+2x+3-\int_1^x f(t)\,dt \Rightarrow f(1)=8-0=8\\ \Rightarrow f'(x)=6x^2+2x+2-f(x) \Rightarrow f'(x)+f(x)=6x^2+2x+2\\ \Rightarrow y'+y=6x^2+2x+2 一階微分方程\\ y'+y=0 \Rightarrow y_h=c_1e^{-x}, y_p=Ax^2+Bx+C \Rightarrow y_p'=2Ax+B\\ \Rightarrow y_p'+ y_p=Ax^2+(2A+B)x+ B+C=6x^2+2x+2 \Rightarrow \cases{A=6\\ B=-10\\ C=12} \\ \Rightarrow y_p=6x^2-10x+12 \Rightarrow y=y_h+y_p \Rightarrow y=c_1e^{-x}+6x^2-10x+12 \\ \Rightarrow y(1)=8 \Rightarrow c_1e^{-1}+8=8 \Rightarrow c_1=0 \Rightarrow y=f(x)=\bbox[red, 2pt] {6x^2-10x+12 }$$
解答:$$6個紅球有5個間隔，但有6個其他球要一起排列，因此不能6個紅球的左右最外側都有其他球\\ \textbf{Case I 最左側或最右側不是紅球: }在6個紅球的5個間隔各插入一球，另一球放在最左或最右側；\\ \qquad 6球任排有{6!\over 3!3!} =20排法，再將此6球平移至最左或最右，因此有20\times 2=40種\\\textbf{Case II 最外兩側都是紅球: }在5個間隔中任取1個(C^5_1)插入1黃1白球(排列數2)，其餘4球任排有\\\qquad {4!\over 2!2!}=6種排法，因此共有5\times 2\times 6=60種 \\ 總共有40+60=\bbox[red, 2pt]{100}種排法$$
解答:$$假設\cases{\overline{AC} =m\\ \overline{AB}=2m} \Rightarrow a\triangle ABC={1\over 2}m\cdot 2m\sin A=1 \Rightarrow \sin A={1\over m^2} \\ \Rightarrow \cos A=\sqrt{1-{1\over m^4}} ={\sqrt{m^4-1\over m^2}} ={m^2+4m^2-\overline{BC}^2\over 4m^2} \Rightarrow \overline{BC}^2=5m^2-4\sqrt{m^4-1} \\ 令f(m)=5m^2-4\sqrt{m^4-1} \Rightarrow f'(m)=0 \Rightarrow 10m-{8m^3\over \sqrt{m^4-1}} =0 \Rightarrow m^2={5\over 3} \\ \Rightarrow \overline{BC}^2={25\over 3}-4\cdot {4\over 3}=3 \Rightarrow \overline{BC} =\bbox[red, 2pt]{\sqrt 3}$$
解答:$$z=a+bi且a^2+b^2=1 \Rightarrow z在單位圓上\Rightarrow \cases{a=\cos \theta\\ b=\sin \theta} \\ \Rightarrow z^2+z-6=(\cos \theta +i \sin \theta)^2+(\cos \theta+i\sin \theta)-6 =(\cos 2\theta+ \cos \theta-6)+i(\sin 2\theta+\sin \theta) \\ \Rightarrow |z^2+z-6|^2= f(\theta) = (\cos 2\theta+ \cos \theta-6)^2 +(\sin 2\theta+\sin \theta)^2=-12 \cos 2\theta-10\cos \theta+38 \\ \Rightarrow f'(\theta) =24\sin 2\theta+10\sin \theta= 48\sin \theta\cos \theta+10\sin \theta=2\sin \theta(5+24\cos \theta) =0\\ \Rightarrow \cases{\sin \theta=0 \Rightarrow f(\theta)非最大值\\ \cos \theta=-5/24 \Rightarrow \cos 2\theta=-{263\over 288} \Rightarrow f(\theta)={1225\over 24}} \Rightarrow |z^2+z-6|最大值= \sqrt{1225\over 24} =\bbox[red, 2pt]{{35\sqrt 6\over 12}}$$
解答:$$令f(x)=\sin^8 x +\cos^8 x，找出f(x)的極大值與極小值，藉由f(x)-m上下調整圖形與x軸有交點\\f'(x)=8\sin ^7x\cos x-8\cos^7 x\sin x= 8\sin x\cos x(\sin^6x- \cos^6x) =4\sin(2x)(\sin^6x- \cos^6x)\\f'(x)=0 \Rightarrow \cases{\sin 2x=0 \Rightarrow x=n\pi/2, n\in \mathbb Z \Rightarrow f(x)=1\\ \sin^6 x=\cos^6x \Rightarrow |\cos x|=|\sin x| \Rightarrow x=(2n-1)\pi/4,n\in \mathbb Z \Rightarrow f(x)=1/8} \\ \Rightarrow f(x)極大值=1,極小值={1\over 8} \Rightarrow \bbox[red, 2pt]{{1\over 8}\le m\le 1}$$
解答:$$S_n^2=a_n\left( S_n-{1\over 2}\right) =(S_n-S_{n-1}) \left( S_n-{1\over 2}\right) =S_n^2-{1\over 2}S_n-S_{n-1}S_n+{1\over 2}S_{n-1} \\ \Rightarrow S_n={S_{n-1} \over 2S_{n-1}+1} \Rightarrow {1\over S_n}={2S_{n-1}+1\over S_{n-1}}=2+{1\over S_{n-1}} \\ \Rightarrow {1\over S_n}=2+{1\over S_{n-1}} =2+2+{1\over S_{n-2}}= \cdots = 2(n-1)+ {1\over S_1} =2n-1 \Rightarrow S_n= \bbox[red, 2pt]{1\over 2n-1}$$
解答:$$\Gamma_1:{x^2\over 4}-{y^2\over 5}=1 \Rightarrow 焦點\cases{F_1(-3,0) \\F_2(3,0) \\c=3}\\ P\in L:x-y+9=0 \Rightarrow P(x,x+9) \in \Gamma\Rightarrow 欲使\overline{PF_1}+ \overline{PF_2}=2a最小 \\ 令 \overline{PF_1}+ \overline{PF_2} =f(x)=\sqrt{(x+3)^2+(x+9)^2} +\sqrt{(x-3)^2+(x+9)^2} \\ \Rightarrow f'(x)= {2x+12\over \sqrt{(x+3)^2+(x+9)^2} }+ {2x+6\over \sqrt{(x-3)^2+(x+9)^2} } =0 \\ \Rightarrow (x+6)^2((x-3)^2+ (x+9)^2) =(x+3)^2((x+3)^2+(x+9)^2) \Rightarrow 54x^2+756x+2430=0 \\ \Rightarrow x^2+14x+45=0 \Rightarrow (x+5)(x+9)=0 \Rightarrow x=-5,-9 \Rightarrow \cases{f(-5)=6\sqrt 5 \\f(-9) =18} \Rightarrow f(-5)\lt f(18) \\ \Rightarrow 橢圓\Gamma貫軸長2a=6\sqrt 5 \Rightarrow a=3\sqrt 5,c=3 \Rightarrow b=6 \Rightarrow {x^2\over a^2}+{y^2\over b^2}=1 \Rightarrow \bbox[red, 2pt]{{x^2\over 45}+{y^2\over 36}=1}$$
解答:$$\cases{x^2+y^2-2y\le 0\\ (x+2y-5)/(x-y-1)=k} \Rightarrow \cases{圓\Gamma:x^2+(y-1)^2\le 1\\ 直線L:(1-k)x +(2+k)y+k-5=0} \\ \Rightarrow \Gamma \cap L\ne \varnothing \Rightarrow d(圓心(0,1),L)\le 1 \Rightarrow {2k-3\over \sqrt{2k^2+2k+5}} \le 1 \Rightarrow (2k-3)^2 \le 2k^2+2k+5 \\ \Rightarrow k^2-7k+2\le 0 \Rightarrow {7-\sqrt{41}\over k}\le k \le {7+\sqrt{41}\over 2} \Rightarrow k的最大值 \bbox[red, 2pt]{7+\sqrt{41}\over 2}$$
解答:$$假設切線方程式: y=ax+b \Rightarrow x^4-2x^3+4x=ax+b \\\Rightarrow x^4-2x^3+(4-a)x-b=0 有相異二重根\alpha,\beta \\ \Rightarrow x^4-2x^3+(4-a)x-b=(x-\alpha)^2 (x-\beta)^2\\ = x^4-2(\alpha+\beta)x^3+ (\alpha^2+ 4\alpha\beta+ \beta^2)x^2 -2\alpha\beta(\alpha+\beta) x+\alpha^2\beta^2 \Rightarrow \cases{\alpha+ \beta=1\\ \alpha^2+ 4\alpha\beta+ \beta^2=0 \\ -2\alpha\beta= 4-a\\ \alpha^2 \beta^2=-b} \\ \Rightarrow \cases{a=3\\ b=-1/4} \Rightarrow \bbox[red, 2pt]{y=3x-{1\over 4}}$$
解答:$$分子:1^2+2^2 +\cdots+n^2 ={1\over 6}n(n+1)(2n+1) \Rightarrow n^3的係數為{1\over 3}\\ 分母: \sqrt 1+\sqrt 2+\cdots+\sqrt n \approx \int_0^n \sqrt x\,dx ={2\over 3}n^{3/2} \Rightarrow (\sqrt 1+\cdots +\sqrt n)^2 \approx {4\over 9}n^3 \Rightarrow n^3的係數為{4\over 9}\\ 因此 \lim_{n\to \infty}{ 1^2+\cdots+n^2 \over (\sqrt 1+\cdots+\sqrt n)^2} ={1/3\over 4/9} = \bbox[red, 2pt]{3\over 4}$$
解答:$$\cases{取出兩球皆1號的機率=C^3_2/C^7_2=1/7\\取出兩球皆2號的機率=C^4_2/C^7_2 =2/7 \\取出1號球,2號球各一的機率=C^3_1C^4_1/C^7_2=4/7}\\ \Rightarrow 長期穩定後:\cases{期望值+2的機率為1/7\\ 期望值+4的機率為2/7\\ 期望值成為3的機率為4/7 } \Rightarrow E={1\over 7}(E+2)+{2\over 7}(E+4) +{4\over 7}\cdot3 \\ \Rightarrow {4\over 7}E={22\over 7} \Rightarrow E=\bbox[red, 2pt]{11\over 2}$$
解答:
此題相當於8個甲與6個乙的排列數，但其中不可以出現連續三個甲，或連續三個乙；因此將甲改成向右一步，將乙改成向上一步，題目變成在一個寬為9,高為7的格子中，由左下角走到右上角,但不可經過陰影區(如上圖)的走法,共有729種走法,機率\(為\bbox[red, 2pt]{729\over 2^{14}}\)。$$$$
二、 作圖、計算證明題， 每題 10 分，詳細配分如各題標示，共 40 分。
解答:$$x+y+z=1 \Rightarrow 可假設 \cases{x=\tan(A/2)\tan(B/2) \\ y=\tan(B/2)\tan (C/2) \\ z=\tan (C/2) \tan(A/2)},其中A+B+C=180^\circ \\ \Rightarrow {x-yz\over x+yz}+{y-zx\over y+zx} +{z-xy\over z+xy} ={1-{yz\over x} \over 1+{yz\over x}} +{1-{zx\over y}\over 1+{zx\over y}}+{1-{xy\over z} \over 1+{xy\over z}} \\={1-\tan^2(C/2) \over 1+\tan^2(C/2)} + {1-\tan^2(A/2) \over 1+\tan^2(A/2)} +{1-\tan^2(B/2) \over 1+\tan^2(B/2)}=\cos C+\cos A+\cos B \\假設\cos A+ \cos B+\cos C=k \Rightarrow 1-2\sin^2{A\over 2}+2\cos{B+C\over 2}\cos{B-C\over 2}=k \\ \Rightarrow 1-2\sin^2{A\over 2}+2\sin{A\over 2}\cos{B-C\over 2}=k \Rightarrow 2\sin^2{A \over 2}-2\cos{B-C\over 2} \sin{A\over 2}+k-1=0 \\ \Rightarrow 判別式\ge 0 \Rightarrow 4\cos^2{B-C\over 2}-8(k-1)\ge 0 \Rightarrow k\le {1\over 2}\cos^2{B-C\over 2}+1\le {1\over 2}+1={3\over 2} \\ \Rightarrow {x-yz\over x+yz}+{y-zx\over y+zx} +{z-xy\over z+xy} \le{3\over 2}, \quad \bbox[red, 2pt]{QED}$$
解答:$$\textbf{(1) }假設\cases{X=\log_2 x\\ Y=\log_3 y}, 由於1\le x\le 2 \Rightarrow 0\le\log_2 x \le 1 \equiv 0\le X\le 1\\\Rightarrow P=(X-1)Y^2-6XY+X+1 =X(Y^2-6Y+1)+1-Y^2 \gt 0 \\ \Rightarrow 1\ge X\gt {Y^2-1\over Y^2-6Y+1} \Rightarrow {Y^2-1\over Y^2-6Y+1} \lt 1 \Rightarrow Y^2-1\lt Y^2-6Y+1 \Rightarrow Y\lt {1\over 3} \\\Rightarrow \log_3y \lt{1\over 3} \Rightarrow y\lt \sqrt[3]3$$
解答:$$E[X(X-1)] = \sum_{x=1}^\infty x(x-1)q^{x-1}p =pq \sum_{x=1}^\infty x(x-1)q^{x-2} =pq {d^2\over dq^2} \sum_{x=0}^\infty q^y =pq {d^2\over dq^2} \left( {1\over 1-q} \right) \\\qquad =pq\cdot {2 \over (1-q)^3} ={2(1-p)\over p^2}, 其中q=1-p \\ Var(X)=E[X(X-1)] +E(X)-[E(X)]^2={2(1-p) \over p^2}+{1\over p}-{1\over p^2} = \bbox[red, 2pt]{1-p\over p^2}$$
解答:
$$\textbf{(1) }r=4\sin \theta 為一圓，圓心(0,2), 半徑=2$$
$$\textbf{(2) }f(x)=e^{-x^2/2} \Rightarrow f'(x)=-xe^{-x^2/2} \Rightarrow f''(x)=-e^{-x^2/2}+x^2e^{-x^2/2}\\ f''(x)=0 \Rightarrow (x^2-1)e^{-x^2/2}=0 \Rightarrow x=\pm 1 \Rightarrow \cases{(\lim_{x\to 1^+}f''(x))(\lim_{x\to 1^-}f''(x)) \lt 0 \\(\lim_{x\to -1^+}f''(x))(\lim_{x\to -1^-}f''(x)) \lt 0} \\ \Rightarrow \cases{(1,f(1)) =(1,e^{-1/2}) \\(-1,f(-1)) =(-1,^{-1/2})} 皆為反曲點$$
$$\textbf{(3) }I=\int_{-\infty}^\infty e^{-x^2/2} \Rightarrow I^2= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)/2}\,dydx \\\qquad \cases{x=r\cos \theta\\ y=r\sin \theta }\Rightarrow I^2= \int_0^{2\pi} \int_0^\infty re^{-r^2/2} \,drd\theta =\int_0^{2\pi} \left. \left[ -e^{-r^2/2}\right] \right|_0^\infty \,d\theta =\int_0^{2\pi} 1\,d\theta =2\pi \\ \quad \Rightarrow I= \bbox[red, 2pt]{\sqrt{2\pi}}$$
解題僅供參考，其他教甄試題及詳解
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