國立竹東高級中學 113 學年度第一次教師甄試數學科試卷
一、填充題(每題 7 分,共 56 分)
解答:x2−3x+1=0⇒x−3+1x=0⇒x+1x=3⇒{(x+1x)2=32(x+1x)3=33⇒{x2+1x2+2=9x3+1x3+3(x+1x)=27⇒{x2+1x2=7x3+1x3=18⇒(x2+1x2)2=72⇒x4+1x4=47⇒(x4+1x4)(x3+1x3)=47×18⇒x7+1x7+x+1x=846⇒x7+1x7=746−3=843解答:f(x)=x+√k−x2⇒k−x2≥0⇒k≥x2≥0f′(x)=0⇒1=x√k−x2⇒x2=k2⇒x=±√k2⇒最大值f(√k2)=√k2+√k2=2√13⇒k=26f(x)最小值發生在k−x2=0⇒x2=k⇒x=−√26⇒f(−√26)=−√26
解答:{F1:x+y+z=0F2:x+y+z=k⇒d(F1,F2)=|k|√3{L1=E1∩E2:x=y=zL2=E2∩E3:x−4=y+4=zL3=E1∩E3:x−8=y−4=z⇒{A=F1∩L1=(0,0,0)B=F1∩L2=(4,−4,0)C=F1∩L3=(4,0,−4)⇒△ABC面積=12‖→AB×→AC‖=12⋅16√3=8√3⇒三角柱體積=8√3⋅|k|√3=30⇒k=±154
解答:[x′y′]=[0110][1301][xy]=[yx+3y]⇒[xy]=[y′−3x′x′]⇒x2+y2=1⇒(y′−3x′)2+x′2=1⇒Γ:(y−3x)2+x2=1
解答:bn=2n∑k=11n⋅k√n2+k2⇒an=12bn, as n→∞⇒lim
解答:\bbox[cyan, 2pt]{此題不計分}
解答:\cases{{a\over 3}+{b\over 4}+{c\over 6}=1 \\{a\over 4}+{b\over 5}+{c\over 7}=1 \\{a\over 6}+{b\over 7}+{c\over 9}=1 } \Rightarrow \cases{4a+3b+2c=12 \\ 35a+28b +20c=140\\ 21a+18b+14c=126} \Rightarrow \begin{bmatrix}4 & 3 & 2 \\35 & 28 & 20 \\21 & 18 & 14\end{bmatrix} \begin{bmatrix}a\\ b \\c \end{bmatrix}= \begin{bmatrix}12 \\140 \\ 126 \end{bmatrix} \\\Rightarrow A= \begin{bmatrix}4 & 3 & 2 \\35 & 28 & 20 \\21 & 18 & 14\end{bmatrix} \Rightarrow A^{-1}=\begin{bmatrix}16 & -3 & 2 \\-35 & 7 & -5 \\21 & \frac{-9}{2} & \frac{7}{2}\end{bmatrix} \Rightarrow \begin{bmatrix}a\\ b \\c \end{bmatrix}= A^{-1} \begin{bmatrix}12 \\140 \\ 126 \end{bmatrix} = \begin{bmatrix}24 \\-70\\ 63 \end{bmatrix} \\ \Rightarrow a+b+c= 24-70+63= \bbox[red, 2pt]{17}
解答:a_2=a_1 \Rightarrow a_3=a_2+a_1 =2a_2 \Rightarrow a_4=a_3+a_2+a_1=4a_2 \Rightarrow a_n=2^{n-2}a_2 \\ \Rightarrow {1\over a_1}+ {1\over a_2}+ \cdots+ {1\over a_n}+\cdots= {1\over 7}+ {1\over 7}+{1\over 7\cdot 2}+{ 1\over 7\cdot 2^2}+\cdots \\={1\over 7} +{1\over 7} (1+{1\over 2}+ {1\over2^2}+\cdots)={1\over 7} +{1\over 7}\cdot 2= \bbox[red, 2pt]{3\over 7}
二、計算證明題(共 44 分)
解答:由於\sum_{k=i}^n C^n_kC^k_i= C^n_i 2^{n-i}, 因此\sum_{k=3}^n C^n_kC^k_3= C^n_3 2^{n-3} =\bbox[red, 2pt]{{1\over 3}2^{n-4}(n-2)(n-1)n}\\可以想成有n個物品分成三群,各有3個,k-3個,及n-k個的方法數\\ 先將n個取3個有C^n_3,剩下n-3個分成2群有2^{n-3}種分法,因此有C^n_32^{n-3}種分法解答:\textbf{(1)}\; \Gamma_3:y=\sqrt{(1-x^2)^3} \Rightarrow V_3= \int_{-1}^1 (1-x^2)^3\pi\,dx =2 \pi \int_{0}^1 (1-x^2)^3 \,dx \\\quad = 2\pi \left. \left[-{1\over 7}x^{7}+{3\over 5}x^{5}-x^3+x \right] \right|_0^1 =2\pi \cdot {16\over 35}=\bbox[red, 2pt]{32\pi\over 35}\\ \textbf{(2)}\; 假設I_n= \int_0^1 (1-x^2)^n \,dx, 取\cases{u=(1-x^2)^n \\ dv=dx} \Rightarrow \cases{u=-2nx(1-x^2)^{n-1} \\ v=x} \\\quad \Rightarrow I_n=\left. \left[ x(1-x^2)^n\right] \right|_0^1+2n \int_0^1 x^2(1-x^2)^{n-1}\,dx =0+2n \int_0^1 (1-(1-x^2))(1-x^2)^{n-1}\,dx \\\quad = 2n \int_0^1 (1-x^2)^{n-1}\,dx -2n \int_0^1 (1-x^2)^n\,dx =2nI_{n-1}-2nI_n \Rightarrow I_n={2n \over 2n+1} I_{n-1} \\ \Rightarrow {V_{n+1} \over V_n} ={{2n+2\over 2n+3}I_n\over I_n} ={2n+2\over 2n+3} \quad \bbox[red, 2pt]{QED}
解答:y={4-3\sin x\over 2+\cos x} \Rightarrow 3\sin x+y\cos x=4-2y \Rightarrow \sqrt{y^2+9} \sin(x+\alpha) =4-2y \\ \Rightarrow \sin(x+\alpha)={4-2y\over \sqrt{y^2+9}} \Rightarrow \left| {4-2y\over \sqrt{y^2+9}}\right|\le 1 \Rightarrow (4-2y)^2 \le y^2+9\\ \Rightarrow 3y^2-16y+7\le 0 \Rightarrow (y-{8+\sqrt{43}\over 3}) (y-{8-\sqrt{43}\over 3}) \le 0 \Rightarrow \bbox[red, 2pt]{{8-\sqrt{43}\over 3}\le y\le {8+\sqrt{43}\over 3}}
解答:假設A(a,b), 由於{1\over 3}(2A+B)=P(1,2) \Rightarrow B(3-2a,6-2b) \Rightarrow \cases{a^2+b^2=37\\ (3-2a)^2+(6-2b)^2=37} \\ \Rightarrow a+2b=13 \Rightarrow a=13-2b \Rightarrow (13-2b)^2+b^2=37 \Rightarrow 5b^2-52b+132=0 \\ \Rightarrow (b-6)(5b-22)=0 \Rightarrow \cases{b=6 \Rightarrow a=1\\ b=22/5 \Rightarrow a=21/5} \Rightarrow \cases{A(1,6), B(1,-6)\\ A(21/5,22/5), B(-27/5,-14/5)} \\\Rightarrow \overleftrightarrow{AB} : \bbox[red, 2pt]{x=1,3x-4y+5=0}
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計算第二題的(1):V_3是對(1-x^2)^3積分,不是(1-x^2)^6,故答案有誤.
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