國立基隆女中 113 學年第 2 次教師甄試 筆試試題 數學科
一、 填充題(每題 4 分,共 52 分,題號標示清楚且照順序填答,無須寫出計算過程)
解答:假設等差數列⟨an⟩,首項a,公差d⇒a1+⋯+a113=(2a+112d)⋅1132=2024⇒2a+112d=4048113⇒am+an=4048113, if m+n=114兩式{a1b1+a2b2+⋯+a113b113=4000a1b113+a2b112+⋯+a113b1=k頭尾相加⇒b1(a1+a113)+b2(a2+a112)+⋯+b113(a113+a1)=4000+k⇒4048113(b1+b2+⋯+b113)=4000+k⇒4048113⋅113=4000+k⇒k=48解答:900∑k=22√k+1+√k<900∑k=21√k<900∑k=22√k+√k−1⇒900∑k=22(√k+1−√k)<900∑k=21√k<900∑k=22(√k−√k−1)⇒2(√901−√2)<900∑k=21√k<2(√900−1)⇒2(√901−√2)+1<900∑k=11√k<2(√900−1)+1⇒2(30−√2)+1<900∑k=11√k<59左式=61−2√2=61−2⋅1.414=58.17⇒900∑k=11√k整數部分=58
解答:
面積和=√34x2+3√32(y2+z2)要最小⇒y=z又兩正六邊形不能重疊,因此兩正六邊形夾角最多為120∘,如上圖因此x=√3y⇒x:y:z=√3:1:1
解答:此題相當於求x+y+z=20,且符合1≤x,y,z≤12的整數解個數xy,z個數12y+z=8H2611y+z=7H27⋯⋯⋯7y+z=13H2116y+z=14(12,2),(11,3),…,(2,12),11種5y+z=15(12,3),…,(3,12),10種⋯⋯⋯1y+z=19(12,7),…,(7,12),6種因此共有11∑k=6(H2k+k)=11∑k=6(2k+1)=102+6=108
解答:
解答:此題相當於求x+y+z=20,且符合1≤x,y,z≤12的整數解個數xy,z個數12y+z=8H2611y+z=7H27⋯⋯⋯7y+z=13H2116y+z=14(12,2),(11,3),…,(2,12),11種5y+z=15(12,3),…,(3,12),10種⋯⋯⋯1y+z=19(12,7),…,(7,12),6種因此共有11∑k=6(H2k+k)=11∑k=6(2k+1)=102+6=108
解答:
√(x−3−2siny)2+(x2−2cosy)2=¯PQ,其中{P(x2,x−3)Q(2cosy,2siny)⇒{P∈拋物線:x=(y+3)2Q∈圓:x2+y2=4⇒圓心O(0,0)至P的距離=√x4+x2−6x+9取f(x)=x4+x2−6x+9⇒f′(x)=0⇒2x3+x−3=0⇒x=1⇒P(1,−2)⇒¯PQ最小值=¯OP−圓半徑=√5−2
解答:令{a=6√3−10b=6√3+10k=a1/3−b1/3⇒{a−b=−20ab=8⇒(3√6√3−10−3√6√3+10)3=(a1/3−b1/3)3=a−b−3(ab)1/3(a1/3−b1/3)⇒k3=−20−6k⇒k3+6k+20=0⇒(k+2)(k2−2k+10)=0⇒k=−2
解答:{A(3,2,1)在平面x−y−2z=5的投影點A′(4,1,−1)B(2,−1,5)在平面x−y−2z=5的投影點B′(4,−3,1)⇒P=¯A′B"中點=(4,−1,0)
解答:有異於(0,0,0)的解代表有無限多解\Rightarrow \begin{vmatrix}\sqrt 5(\sin \theta-\cos \theta) & 2 & 1 \\3 & 1& 2\\ 21& 4 & \sqrt 5(\sin \theta-\cos \theta)+14 \end{vmatrix}=0 \\ \Rightarrow (\sin \theta-\cos \theta)^2={9\over 5} \Rightarrow \sin \theta-\cos \theta=-{3\over \sqrt 5} \;(\because -{\pi\over 4}\le \theta\le {\pi\over 4}) \\ \Rightarrow \cases{-3x+2y+z=0\\ 3x+y+2z=0\\ 21x+4y+11z=0} \Rightarrow \begin{bmatrix}-3 & 2& 1\\ 3 & 1& 2 \\21 & 4& 11 \end{bmatrix} \begin{bmatrix}x \\ y\\ z \end{bmatrix} =0 \\ 令A= \begin{bmatrix}-3 & 2& 1\\ 3 & 1& 2 \\21 & 4& 11 \end{bmatrix} \Rightarrow rref(A)= \left[\begin{matrix}1 & 0 & \frac{1}{3}\\0 & 1 & 1\\0 & 0 & 0\end{matrix}\right] \Rightarrow\cases{x+z/3=0\\ y+z=0} \Rightarrow \bbox[red, 2pt]{ \cases{x=t\\ y=3t\\ z=-3t},t\in \mathbb R}
解答:f(x)=(1+3x)^8=\sum_{k=0}^8 a_kx^k \Rightarrow a_1=C^8_1\cdot 3=24\\ f'(x)=24(1+3x)^7 =\sum_{k=1}^8k a_kx^k \Rightarrow f'(1)=24\cdot 4^7= \sum_{k=1}^8 ka_k \Rightarrow \sum_{k=2}^8 ka_k=24\cdot 4^7-a_1 \\=24\cdot 4^7-24=24(4^7-1)=\bbox[red, 2pt]{393192}
解答:令\cases{u=x+2y-z\\ v=4x-7y-5z\\ w=2x-y+3z} \Rightarrow{\partial (u,v,w)\over \partial (x,y,z)} =\begin{Vmatrix}1& 2& -1\\ 4& -7 &-5\\ 2& -1& 3 \end{Vmatrix} =80 \\ \Rightarrow \Omega={\partial (x,y,z)\over \partial (u,v,w)} \int_{-\sqrt 6}^{7\sqrt 6} \int_{-2\sqrt 2}^{6\sqrt 2} \int_{2\sqrt 3}^{5\sqrt 3}1,\,dudvdw ={1\over 80}\cdot 3\sqrt 3\cdot 8\sqrt 2\cdot 8\sqrt 6 =\bbox[red, 2pt]{72\over 5}
解答:S=\sum_{k=0}^\infty {k\over 2^k} =2\sum_{k=0}^\infty {k\over 2^{k+1}} =2\sum_{k=1}^\infty {k-1\over 2^{k}} =2\sum_{k=1}^\infty {k\over 2^{k}}-2 \sum_{k=1}^\infty {1\over 2^{k}} =2S-2 \Rightarrow S=2\\ T=\sum_{k=0}^\infty {k^2\over 2^k} =2\sum_{k=0}^\infty {k^2\over 2^{k+1}} = 2\sum_{k=1}^\infty {(k-1)^2\over 2^{k}} =2\sum_{k=1}^\infty {k^2\over 2^k}-4\sum_{k=1}^\infty{k\over 2^k}+2 \sum_{k=1}^\infty{1\over 2^k} \\\qquad =2T-4S+2 =2T-6 \Rightarrow T=6\\ U=\sum_{k=0}^\infty {k^3\over 2^k} = 2\sum_{k=0}^\infty {k^3\over 2^{k+1}} = 2\sum_{k=1}^\infty {(k-1)^3\over 2^{k}} = 2\sum_{k=1}^\infty {k^3-3k^2+3k-1\over 2^{k}} \\\qquad = 2U-6T+6S-2 =2U-36+12-2 \Rightarrow U=26 \\ \lim_{n\to \infty} \sum_{k=1}^n {a_k\over b_k} =\sum_{k=1}^\infty {k^3+k\over 3\times 2^{k-1}} ={2\over 3}\sum_{k=1}^\infty{k^3+k\over 2^k}= {2\over 3}(U+S) =\bbox[red, 2pt]{56\over 3}
解答:{1\over C^{10}_4}\sum_{y=2}^8 (10-y)(y-1)C^{10-y}_2 ={1176\over 210 } = \bbox[red, 2pt]{28\over 5}
解答:令{a=6√3−10b=6√3+10k=a1/3−b1/3⇒{a−b=−20ab=8⇒(3√6√3−10−3√6√3+10)3=(a1/3−b1/3)3=a−b−3(ab)1/3(a1/3−b1/3)⇒k3=−20−6k⇒k3+6k+20=0⇒(k+2)(k2−2k+10)=0⇒k=−2
解答:{A(3,2,1)在平面x−y−2z=5的投影點A′(4,1,−1)B(2,−1,5)在平面x−y−2z=5的投影點B′(4,−3,1)⇒P=¯A′B"中點=(4,−1,0)
解答:有異於(0,0,0)的解代表有無限多解\Rightarrow \begin{vmatrix}\sqrt 5(\sin \theta-\cos \theta) & 2 & 1 \\3 & 1& 2\\ 21& 4 & \sqrt 5(\sin \theta-\cos \theta)+14 \end{vmatrix}=0 \\ \Rightarrow (\sin \theta-\cos \theta)^2={9\over 5} \Rightarrow \sin \theta-\cos \theta=-{3\over \sqrt 5} \;(\because -{\pi\over 4}\le \theta\le {\pi\over 4}) \\ \Rightarrow \cases{-3x+2y+z=0\\ 3x+y+2z=0\\ 21x+4y+11z=0} \Rightarrow \begin{bmatrix}-3 & 2& 1\\ 3 & 1& 2 \\21 & 4& 11 \end{bmatrix} \begin{bmatrix}x \\ y\\ z \end{bmatrix} =0 \\ 令A= \begin{bmatrix}-3 & 2& 1\\ 3 & 1& 2 \\21 & 4& 11 \end{bmatrix} \Rightarrow rref(A)= \left[\begin{matrix}1 & 0 & \frac{1}{3}\\0 & 1 & 1\\0 & 0 & 0\end{matrix}\right] \Rightarrow\cases{x+z/3=0\\ y+z=0} \Rightarrow \bbox[red, 2pt]{ \cases{x=t\\ y=3t\\ z=-3t},t\in \mathbb R}
解答:f(x)=(1+3x)^8=\sum_{k=0}^8 a_kx^k \Rightarrow a_1=C^8_1\cdot 3=24\\ f'(x)=24(1+3x)^7 =\sum_{k=1}^8k a_kx^k \Rightarrow f'(1)=24\cdot 4^7= \sum_{k=1}^8 ka_k \Rightarrow \sum_{k=2}^8 ka_k=24\cdot 4^7-a_1 \\=24\cdot 4^7-24=24(4^7-1)=\bbox[red, 2pt]{393192}
解答:令\cases{u=x+2y-z\\ v=4x-7y-5z\\ w=2x-y+3z} \Rightarrow{\partial (u,v,w)\over \partial (x,y,z)} =\begin{Vmatrix}1& 2& -1\\ 4& -7 &-5\\ 2& -1& 3 \end{Vmatrix} =80 \\ \Rightarrow \Omega={\partial (x,y,z)\over \partial (u,v,w)} \int_{-\sqrt 6}^{7\sqrt 6} \int_{-2\sqrt 2}^{6\sqrt 2} \int_{2\sqrt 3}^{5\sqrt 3}1,\,dudvdw ={1\over 80}\cdot 3\sqrt 3\cdot 8\sqrt 2\cdot 8\sqrt 6 =\bbox[red, 2pt]{72\over 5}
解答:S=\sum_{k=0}^\infty {k\over 2^k} =2\sum_{k=0}^\infty {k\over 2^{k+1}} =2\sum_{k=1}^\infty {k-1\over 2^{k}} =2\sum_{k=1}^\infty {k\over 2^{k}}-2 \sum_{k=1}^\infty {1\over 2^{k}} =2S-2 \Rightarrow S=2\\ T=\sum_{k=0}^\infty {k^2\over 2^k} =2\sum_{k=0}^\infty {k^2\over 2^{k+1}} = 2\sum_{k=1}^\infty {(k-1)^2\over 2^{k}} =2\sum_{k=1}^\infty {k^2\over 2^k}-4\sum_{k=1}^\infty{k\over 2^k}+2 \sum_{k=1}^\infty{1\over 2^k} \\\qquad =2T-4S+2 =2T-6 \Rightarrow T=6\\ U=\sum_{k=0}^\infty {k^3\over 2^k} = 2\sum_{k=0}^\infty {k^3\over 2^{k+1}} = 2\sum_{k=1}^\infty {(k-1)^3\over 2^{k}} = 2\sum_{k=1}^\infty {k^3-3k^2+3k-1\over 2^{k}} \\\qquad = 2U-6T+6S-2 =2U-36+12-2 \Rightarrow U=26 \\ \lim_{n\to \infty} \sum_{k=1}^n {a_k\over b_k} =\sum_{k=1}^\infty {k^3+k\over 3\times 2^{k-1}} ={2\over 3}\sum_{k=1}^\infty{k^3+k\over 2^k}= {2\over 3}(U+S) =\bbox[red, 2pt]{56\over 3}
解答:{1\over C^{10}_4}\sum_{y=2}^8 (10-y)(y-1)C^{10-y}_2 ={1176\over 210 } = \bbox[red, 2pt]{28\over 5}
解答:假設\cases{\overline{AD}=a\\ \overline{AB}=a+\sqrt{11}\\ \overline{BC}=b\\ \overline{CD}=\sqrt{13}-b} \Rightarrow \overline{AC}^2=\overline{AB}^2+\overline{BC}^2= \overline{AD}^2+ \overline{CD}^2 \\ \Rightarrow 16= a^2 +2\sqrt{11}a+11+b^2= a^2+b^2-2\sqrt{13} b+13 \Rightarrow \sqrt{11}a+ \sqrt{13}b=1 \\ \Rightarrow a={1-\sqrt{13}b\over \sqrt{11}} \Rightarrow ({1-\sqrt{13}b\over \sqrt{11}})^2+b^2-2\sqrt{13} b+13=16 \Rightarrow b={3\sqrt{13}-\sqrt{165} \over 6}\\ 又\overline{AC}\times \overline{BD}= \overline{AB}\cdot \overline{CD}+ \overline{BC}\cdot \overline{AD} \\ \Rightarrow 4\overline{BD}=\sqrt{13}a-\sqrt{11}b+\sqrt{143} \Rightarrow \overline{BD}={1\over 4}(\sqrt{13}a-\sqrt{11}b+\sqrt{143}) \\ ={3\sqrt{13}-6b\over \sqrt{11}} ={\sqrt{165}\over \sqrt{11}} =\bbox[red, 2pt]{\sqrt{15}}
填充第13題是不是有問題啊? 因為b=(3√13−√165)/6<0,但b必須>0.(ps:b=(3√13+√165)/6 也不行,試算過了)
回覆刪除作者已經移除這則留言。
刪除也許四邊形的順序不是ABCD,而是ACBD,這樣結果就符合運算結果!
刪除可是他有說四邊形ABCD喔,所以明確題目有誤(這題其他論壇也有人討論).
刪除