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2023年6月7日 星期三

112年花蓮縣國中聯合教甄-數學詳解

花蓮縣 112 學年度國民中小學教師聯合甄選

選擇題【共 25 題,每題 4 分,共 100 分】請以 2B 鉛筆於答案卡上作答,單選題;答錯不倒扣。

1. 某校圖書館有 5000 元購書經費,今有三種圖書每本單價分別為 100,150,200 元。該校欲一共購買三十本,每種書都至少一本,且恰好把 5000 元用完。試問有幾個不同的購書方案?
(A) 9  (B)  10  (C) 11   (D)  12

解答:$$假設\cases{100元的書買x本\\ 150元的書買y本\\ 200元的書買z本} \Rightarrow \cases{100x+150y+200z=5000\\ x+y+z=30} \Rightarrow \cases{2x+3y+4z=100\\ x+y+z=30\cdots(1)} \\ \Rightarrow y+2z=40 \cdots(2),由(1)及(2)\Rightarrow (x,y,z)=(1,18,11),(2,16,12),\dots,(9,2,19) ,共九種買法\\,故選\bbox[red, 2pt]{(A)}$$
解答:$$2x^2+xy+y^2=2 \Rightarrow 4x+y+xy'+2yy'=0 \Rightarrow y'={-4x-y\over x+2y} \Rightarrow y'(1,-1)={-3\over -1}=3 \\\Rightarrow 切線斜率=3  \Rightarrow 切線方程式: y=3(x-1)-1=3x-4 \Rightarrow y=3x-4,故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)=x^x= e^{x\ln x } \Rightarrow f'(x)=(\ln x+1)e^{x\ln x}= (\ln x+1)x^x \\ \Rightarrow f'(2)=(\ln 2+1)2^2= 4(\ln 2+1),故選\bbox[red, 2pt]{(D)}$$
解答:$$(A)\bigcirc:\int_e^\infty{1\over x(\ln x)^2}\,dx = \left. \left[-{1\over \ln x} \right] \right|_e^\infty =0-(-1)=1,故選\bbox[red, 2pt]{(A)}$$
解答:$$a_{n+1}=a_n+\sqrt{a_n}+{1\over 4}= (\sqrt{a_n}+{1\over 2})^2 \\ \Rightarrow a_2=({3\over 2})^2={9\over 4} \Rightarrow a_3=2^2=4 \Rightarrow a_4=({5\over 2})^2={25\over 4} \Rightarrow a_5=3^2 \Rightarrow a_n=\begin{cases}k^2 & n=2k-1\\ (2k+1)^2/4 &n=2k  \end{cases} \\ \Rightarrow a_{99}=a_{2\cdot 50-1}=50^2=2500,故選\bbox[red, 2pt]{(C)}$$
解答:$$a_n={(-1)^{n-1}\over n\cdot 3^n}x^n \Rightarrow \lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{(-1)^{n}\cdot x^{n+1}\over (n+1)\cdot 3^{n+1}}  \cdot {n\cdot 3^n \over (-1)^{n-1}\cdot x^n}\right| \\= \lim_{n\to \infty}\left| { n\over (n+1)\cdot 3}\cdot x \right| ={1\over 3}|x|\lt 1 \Rightarrow |x|\lt 3\\ x=3 \Rightarrow a_n=(-1)^{n-1}/n \Rightarrow \sum_{n=1}^\infty a_n=1-{1\over 2}+{1\over 3}-\cdots \Rightarrow 級數收斂\\ x=-3 \Rightarrow a_n=-{1\over n} \Rightarrow \sum_{n=1}^\infty a_n =-(1+1/2+1/3+\cdots )\Rightarrow 級數發散\\ 因此收斂區間為(-3,3],故選\bbox[red, 2pt]{(B)}$$
解答:$$\iint_D(x+y)\,dA = \int_0^1\int_x^1 (x+y)\,dydx =\int_0^1 \left. \left[ xy+{1\over 2}y^2 \right]\right|_x^1\,dx =\int_0^1 (x+{1\over 2}-{3\over 2}x^2)\,dx \\= \left. \left[ {1\over 2}x^2+{1\over 2}x -{1\over 2}x^3\right]\right|_0^1 ={1\over 2},故選\bbox[red, 2pt]{(A)}$$
解答:$$f可積 \Rightarrow F可微,故選\bbox[red, 2pt]{(B)}$$
解答:$$(A)解釋變數(自變數)為類別變數,不是連續型\\(B) 變異數分析用來檢驗平均數是否相等,假設他們的平均數是相等的\\ (D) F分布,不是卡方\\,故選\bbox[red, 2pt]{(C)}$$
解答:$$(A)相關係數與因果無關\\(B) X是自變項,Y是應變項\\(C) \beta與\rho 同正負\\,故選\bbox[red, 2pt]{(D)}$$
解答:$$本題與第六題相關,A發散,B收斂,故選\bbox[red, 2pt]{(C)}$$
解答:$$甲:常見的反例f(t)=e^{i2\pi } \\ 乙:反例是微分的反函數不得為0\\,故選\bbox[red, 2pt]{(D)}$$
解答:$$y=\cos(e^{-\theta^2}) \Rightarrow y'=-\sin(e^{-\theta^2})\cdot -2\theta e^{-\theta^2} =2\theta e^{-\theta^2} \sin(e^{-\theta^2}),故選\bbox[red, 2pt]{(D)}$$
解答:$$\lim_{x\to 0^+} x^2\ln x =\lim_{x\to 0^+} {\ln x\over {1\over x^2}} = \lim_{x\to 0^+} {1/x\over {-2\over x^3}} =\lim_{x\to 0^+} -{x^2\over 2}=0,故選\bbox[red, 2pt]{(B)}$$
解答:$$算幾不等式:{x^2\over 8}+{y^2\over 2}\ge 2\sqrt{{x^2\over 8}\cdot {y^2\over 2}} \Rightarrow 1\ge 2\cdot {xy\over 4} \Rightarrow xy\le 2,故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{逆時針旋轉45^\circ矩陣A=\begin{bmatrix} \sqrt 2/2 & -\sqrt 2/2\\ \sqrt 2/2& \sqrt 2/2 \end{bmatrix}\\ 對x軸鏡射矩陣B=\begin{bmatrix}1& 0 \\ 0& -1 \end{bmatrix}} \Rightarrow BA=\begin{bmatrix} \sqrt 2/2 & -\sqrt 2/2\\ -\sqrt 2/2& -\sqrt 2/2 \end{bmatrix},故選\bbox[red, 2pt]{(A)}$$
解答:$$(A)\det \left(\begin{bmatrix} 1-\lambda & 2\\ 3 & 4-\lambda\end{bmatrix} \right)=0 \Rightarrow \lambda^2-5\lambda-2=0 \Rightarrow \lambda \not \in \mathbb N \\(B)\det \left(\begin{bmatrix} 1-\lambda & 4\\ 2 & 3-\lambda\end{bmatrix} \right)=0 \Rightarrow \lambda^2-4\lambda-5=0 \Rightarrow \lambda=5,-1 \not \in \mathbb N \\(C)\det \left(\begin{bmatrix} 2-\lambda & 3\\ 1& 4-\lambda\end{bmatrix} \right)=0 \Rightarrow \lambda^2-6\lambda+5=0 \Rightarrow \lambda =1,5 \\ (D)\det \left(\begin{bmatrix} 3-\lambda & 4\\ 1 & 2-\lambda\end{bmatrix} \right)=0 \Rightarrow \lambda^2-5\lambda+2=0 \Rightarrow \lambda \not \in \mathbb N \\,故選\bbox[red, 2pt]{(C)}$$
解答:$$假設\cases{A(\alpha,0)\\ B(\beta,0)\\ C(0,c)} \Rightarrow \cases{\overrightarrow{AC}=(\alpha,-c) \\\overrightarrow{BC}=(\beta,-c) \\ \alpha\beta =c/a} \Rightarrow \overrightarrow{AC} \cdot \overrightarrow{BC}=0 \Rightarrow \alpha\beta +c^2=0 \Rightarrow {c\over a}+c^2=0 \\ \Rightarrow c+ac^2=0 \Rightarrow c(1+ac)=0 \Rightarrow ac=-1,故選\bbox[red, 2pt]{(A)}$$
解答:$$x遞減則y遞減\Rightarrow r\gt 0 \Rightarrow y截距為負值 \Rightarrow 不過第二象限,故選\bbox[red, 2pt]{(B)}$$
解答:$$每個集合可以含1不含1,含2不含2,含3不含3,有2^3種可能\\ 因此組數=2^3\cdot 2^3\cdot 2^3-2^2\cdot 2^2\cdot 2^2\cdot 3+ 2\cdot 2\cdot 2\cdot 3-1=343,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{A(0,2)\\ B(3,1)} \Rightarrow\overline{AB}的中垂線 L:3x-y-3=0 ,由於\overline{AB}為兩圓的一弦,因此圓心在L上\\ 因此假設圓心C(a,3a-3) ,圓與x軸相切,因此圓半徑3a-3\\ \Rightarrow 圓方程式:(x-a)^2+(y-3a+3)^2=(3a-3)^2,\\將A(0,2)代入上式\Rightarrow a^2+(5-3a)^2=(3a-3)^2 \Rightarrow a^2-12a+16=0 \\假設兩根為\alpha,\beta \Rightarrow \alpha+\beta =12 \Rightarrow 兩圓半徑之和=(3\alpha-3) +(3\beta-3) =3(\alpha+\beta)-6\\ =3\times 12-6=30,故選\bbox[red, 2pt]{(B)}$$
解答:$$函數不一定可微,泰勒級數為無窮級數,極限為0才收斂,故選\bbox[red, 2pt]{(C)}$$
解答:$$E在\overline{AB}的中垂線上,又\overline{AB}=2 \Rightarrow \overline{BE}在\overline{AB}的投影長度為1 \Rightarrow \overrightarrow{BE}\cdot \overrightarrow{BA}=2\cdot 1=2 \\ \Rightarrow \overrightarrow{BE}\cdot \overrightarrow{AB}=-2,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cot(2\alpha) -\sqrt 3=\sec \alpha \Rightarrow {\cos 2\alpha \over \sin 2\alpha}-{1\over \cos \alpha}=\sqrt 3 \Rightarrow {1-2\sin^2\alpha\over 2\sin \alpha\cos \alpha}-{1\over \cos \alpha}=\sqrt 3 \\ \Rightarrow 1-2\sin^2\alpha-2\sin\alpha=2 \sqrt 3\sin \alpha\cos \alpha =2\sqrt 3\sin \alpha\cdot \sqrt{1-\sin^2\alpha} \\ \Rightarrow (1-2\sin^2\alpha-2\sin\alpha)^2=12\sin^2\alpha (1-\sin^2\alpha)\\ \Rightarrow 16\sin^4\alpha +8\sin^3\alpha -12\sin ^2\alpha-4\sin \alpha+1=0 \\ \Rightarrow -4\sin\alpha(3\sin\alpha- 4\sin^3\alpha)-(-8\sin^3\alpha+6\sin \alpha)+2\sin \alpha+1=0\\ \Rightarrow -4\sin \alpha\sin(3\alpha)-2\sin(3\alpha)+2\sin \alpha+1=0 \Rightarrow -2\sin(3\alpha)(2\sin\alpha+1)+2\sin \alpha+1=0\\ \Rightarrow (1-2\sin(3\alpha))(2\sin \alpha+1)=0 \Rightarrow \sin(3\alpha)={1\over 2} \Rightarrow 3\alpha=30^\circ \Rightarrow \alpha=10^\circ,故選\bbox[red, 2pt]{(A)}$$
解答:$$|z^2+1|=|z| \Rightarrow (|z^2+1|)^2=(|z|)^2 \Rightarrow (z^2+1)(\bar z^2+1)=z\bar z\\ \Rightarrow z^2\bar z^2+ z^2+\bar z^2+1= z^2\bar z^2+ (z+\bar z)^2-2z\bar z+1=z\bar z \\ \Rightarrow z^2\bar z^2 +(z+\bar z)^2-3z\bar z+1=0 \Rightarrow r^4-3r^2+s^2+1=0,其中\cases{r^2=z\bar z\\ s=z+\bar z},r,s \in \mathbb R\\ \Rightarrow r^2={3+\sqrt{5-4s^2}\over 2} \Rightarrow r^2的最大值={3+\sqrt 5\over 2},其中s=0\\ \Rightarrow r的最大值=\sqrt{3+\sqrt 5\over 2} ={\sqrt {6+2\sqrt 5}\over 2} ={1+\sqrt 5\over 2},故選\bbox[red, 2pt]{(D)}$$

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  1. 第四題 (D) wolfram 積分出來也是0 耶...

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