112年新竹市國中聯合教甄-數學詳解

112 學年度新竹市國民中學教師聯合甄選試題

第二部分：專業科目-數學(70%)

選擇題，第 41 題至第 80 題，每題 2.5 分，共 100 分。

解答：$$x=-{1+i\over \sqrt 2} \Rightarrow x^2=i \Rightarrow f(x)=i^{30}+i^{15}+1 = (-1)^{15}+(-1)^7\cdot i+1=-i，故選\bbox[red,2pt]{(C)}$$

解答：$$假設\cases{f(2+a)=f(2-a)=0\\f(2+b)=f(2-b)=0},即2\pm a,2\pm b 為f(x)=0之四相異實根\\ \Rightarrow 四根之和=8，故選\bbox[red,2pt]{(D)}$$

解答：$$(A)\bigcirc: X\sim b(n=20,p=0.3) \Rightarrow \cases{EX=np=6\\ Var(X)=np(1-p)=4.2} \\(B)\bigcirc\\ (C)\bigcirc\\ (D)\times: p是固定的\\，故選\bbox[red,2pt]{(D)}$$

解答：$$f(x)= \int_0^x \sqrt{1+t^4}\,dt \Rightarrow f'(x)=\sqrt{1+x^4} \Rightarrow f'(1)=\sqrt 2\\又f^{-1}(f(x))=x \Rightarrow (f^{-1})'(f(x))f'(x)=1, \Rightarrow (f^{-1})'(f(x))={1\over f'(x)}\\ x=1代入上式\Rightarrow (f^{-1})'(f(1))={1\over f'(1)} \Rightarrow (f^{-1})'(2)={1\over\sqrt 2}，故選\bbox[red,2pt]{(B)},但公布的答案是\bbox[blue,2pt]{(A)}$$

解答：$$拋物線皆可平移或旋轉成y=ax^2形式，因此皆相似，故選\bbox[red,2pt]{(B)}$$

解答：$$\log_2 \log_4 x+\log_4 \log_2 x=\log_2 ({1\over 2}\log_2 x)+{1\over 2}\log_2 x =-1+\log_2\log_2 x+{1\over 2}\log_2 x={1\over 2}\\ \Rightarrow {3\over 2}\log_2 \log_2 x={3\over 2} \Rightarrow \log_2 \log_2 x=1 \Rightarrow \log_2 x=2\Rightarrow x=4，故選\bbox[red,2pt]{(C)}$$

解答：

$$正六邊形的內角為120^\circ \Rightarrow \triangle ABC為正三角形\Rightarrow \cases{六角形周長＝12m\\六角形面積=12k\\ 六邊形周長=6m \\ 六邊形面積=6k} \\\Rightarrow \cases{a=2\\b=2} \Rightarrow a+b=4，故選\bbox[red,2pt]{(B)}$$

解答：$$y=8^nx^2-2^n(2^n+1)x+1=0 \Rightarrow x={2^n(2^n+1)\pm \sqrt{2^{2n}(2^n+1)^2-4\cdot 8^n}\over 2\cdot 8^n} \\={(2^n+1)\pm (2^n-1)\over 2^{2n+1}}={1\over 2^n},{1\over 2^{2n}} \Rightarrow \overline{A_nB_n}={1\over 2^n}-{1\over 2^{2n}} \\ \Rightarrow \sum_{n=1}^\infty a_n= \sum_{n= 1}^\infty\left({1\over 2^n}-{1\over 2^{2n}}\right)=1-{1\over 3}={2\over 3}，故選\bbox[red,2pt]{(A)}$$

解答：$$令a={1\over 2023},則原式S=1+2r+3r^2+4r^3+\cdots \Rightarrow rS=r+2r^2+3r^3+4r^4+\cdots\\ \Rightarrow S-rS=1+r+r^2+\cdots \Rightarrow S={1+r+r^2+\cdots \over 1-r} ={{1\over 1-r}\over 1-r} ={1\over (1-r)^2} ={1\over ({2022\over 2023})^2}\\ =({2023\over 2022})^2，故選\bbox[red,2pt]{(B)}$$

解答：$$\lim_{x\to 0}{\int_0^x \sqrt{t^4+4}\,dt\over x} =\lim_{x\to 0}{{d\over dx}\int_0^x \sqrt{t^4+4}\,dt\over {d\over dx}x} =\lim_{x\to 0} \sqrt{x^4+4}=2，故選\bbox[red,2pt]{(C)}$$

解答：$$f(x)=(1+x)^n = \sum_{k=0}^n C^n_k x^k \Rightarrow f(3)=4^n = 1+3C^n_1+ 3^2C^n_2 +3^3C^n_3+ 3^4C^n_4+ \cdots \\ \Rightarrow C^n_1+ 3C^n_2+ 3^2C^n_3+\cdots = {4^n-1\over 3}，故選\bbox[red,2pt]{(A)}$$

解答：$$\int_0^1\int_0^1 (1-{1\over 2}x^2-{1\over 2}y^2)\,dxdy = \int_0^1 \left. \left[ x-{1\over 6}x^3-{1\over 2}xy^2 \right]\right|_0^1 \,dy =\int_0^1 {5\over 6}-{1\over 2}y^2 \,dy\\ =\left. \left[ {5\over 6}y-{1\over 6}y^3 \right] \right|_0^1 ={2\over 3}，故選\bbox[red,2pt]{(D)}$$

解答：$$1\times 2+2\times 3+3\times 4 +4\times 5+\cdots +n\times(n+1)= \sum_{k=1}^nk(k+1) =\sum_{k=1}^n(k^2+k) \\={n(n+1)(2n+1)\over 6}+{n(n+1)\over 2} ={n(n+1)(2n+4)\over 6}={n(n+1) (n+2)\over 3} \\ \Rightarrow {1\over 1\times 2+2\times 3+3\times 4 + \cdots +n\times(n+1)} ={3\over n(n+1)(n+2)} ={3\over 2}({1\over n(n+1)}-{1\over (n+1)(n+2}) \\ \Rightarrow 原式={3\over 2}\sum_{n=1}^\infty\left( {1\over n(n+1)}-{1\over (n+1)(n+2} \right) ={3\over 2}({1\over 1\times 2}-{1\over 2\times 3}+{1\over 2\times 3}-{1\over 3\times 4}+\cdots)\\={3\over 2}\times {1\over 2}={3\over 4}，故選\bbox[red,2pt]{(B)}$$

解答：$$f(x)=x^{2023}(x^2+ax+b)=(x-2)^2p(x)+2^{2023}(x-2) \\ \Rightarrow f'(x)=2023x^{2022}(x^2+ax+b)+ x^{2023}(2x+a)=2(x-2)p(x)+(x-2)^2p'(x)+2^{2023}\\ \Rightarrow \cases{f(2) =2^{2023}(4+2a+b) =0\\ f'(2)=2023\cdot 2^{2022}(4+2a+b)+ 2^{2023}(4+a)=2^{2023}} \\ \Rightarrow \cases{4+2a+b = 0\\ 4+a=1} \Rightarrow \cases{a=-3\\ b=2}，故選\bbox[red,2pt]{(C)}$$

解答：$$\lim_{x\to 1}{x^3f(1)-f(x^2)\over x-1} =\lim_{x\to 1}{{d\over dx}(x^3f(1)-f(x^2))\over {d\over dx}(x-1)} =\lim_{x\to 1} ( {3x^2f(1)-2xf'(x^2)}) = 6-2=4\\，故選\bbox[red,2pt]{(B)}，但公布的答案是\bbox[blue, 2pt]{(C)}$$

解答：$$\lim_{n\to \infty}\left[ {1\over \sqrt{n^2+2n}} + {1\over \sqrt{n^2+4n}}+\cdots +{1\over \sqrt{n^2+2kn}} +\cdots +{1\over \sqrt{n^2+2n^2}}\right] \\=\lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+2kn}} =\lim_{n\to \infty} \sum_{k=1}^n {1\over n\sqrt{1+2k/n}} =\int_0^1 {1\over \sqrt{1＋2x}}\,dx =\left.\left[ \sqrt{1+2x}\right] \right|_0^1 =\sqrt 3-1\\，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{x=r\cos\theta\\ y=r\sin \theta} \Rightarrow \int_0^\infty \int_0^\infty e^{-x^2-y^2}\,dxdy = \int_0^{\pi/2} \int_0^\infty e^{-r^2}r\,dr\;d \theta = \int_0^{\pi/2} \left. \left[  -{1\over 2}e^{-r^2} \right] \right|_0^\infty \,d\theta \\=\int_0^{\pi/2}{1\over 2}\,d\theta={\pi\over 4}，故選\bbox[red,2pt]{(A)}$$

解答：$$\lim_{x\to 0^+} x\cdot \tan({\pi\over 2}-x) =\lim_{x\to 0^+} x\cdot \cot(x) =\lim_{x\to 0^+} {x\over 1/\cot x} =\lim_{x\to 0^+}{x \over \tan x}=\lim_{x\to 0^+}{{d\over dx}x \over {d\over dx}\tan x} \\=\lim_{x\to 0^+}{1\over \sec^2 x} =1，故選\bbox[red,2pt]{(A)}$$

解答：$$\lim_{x\to -\infty}(2x+\sqrt{4x^2+3x-2}) =\lim_{x\to -\infty}{-3x+2\over 2x-\sqrt{4x^2+3x-2}} =\lim_{x\to -\infty}{-3+2/x \over 2+2\sqrt{1+3/x-2x^2}} =\\ =-{3\over 4},\bbox[red,2pt]{無解}，公布的答案是\bbox[blue,2pt]{(C)}$$

解答：$$L=\lim_{x\to 1^+}x^{1/(1-x)} \Rightarrow \ln L=\lim_{x\to 1^+}{\ln x\over 1-x}=\lim_{x\to 1^+}{1/x\over -1} =-1  \Rightarrow L=e^{-1}={1\over e}，故選\bbox[red,2pt]{(B)}$$

解答：$${d\over dx} \tan x= \sec^2 x\Rightarrow y=\tan(\pi x^2) \Rightarrow y'=\sec^2(\pi x^2)\cdot 2\pi x,\bbox[red,2pt]{無解},公布的答案是\bbox[blue,2pt]{(C)}$$

解答：$$(x,y)\in x^2+y^2=1 \Rightarrow \cases{x=cos \theta\\ y=\sin \theta} \Rightarrow f(x,y)=3\cos \theta+4\sin \theta =5({3\over 5}\cos \theta+{4\over 5}\sin \theta)\\ = 5\sin(\alpha+\theta) \Rightarrow 極大值=5，故選\bbox[red,2pt]{(C)}$$

解答：$$\int_{-4}^0 \int_{-2}^2 (3x+9y+9)\,dxdy =\int_{-4}^0 \left. \left[{3\over 2}x^2+9xy+9x \right]\right|_{-2}^2 \,dy =\int_{-4}^0 (36y+36)\,dy =\left. \left[ 18y^2 +36y \right] \right|_{-4}^0\\ =0-(288-144)=-144，故選\bbox[red,2pt]{(D)}$$

解答：$$0 \le |x\sin{1\over x}|=|x||\sin{1\over x}|\le |x| \\ 夾擠定理\lim_{x\to 0}|x\sin{1\over x}|=0 \Rightarrow \lim_{x\to 0}(x\sin{1\over x}) =0 \Rightarrow x=0連續，故選\bbox[red,2pt]{(A)}$$

解答：$$六數之和為奇數\Rightarrow \cases{1奇5偶：C^6_1C^5_5=6\\ 3奇3偶：C^6_3 C^5_3=200\\ 5奇1偶：C^6_5C^5_1=30} \Rightarrow 機率={6+200+30\over C^{11}_6} ={236\over 462}={118\over 231}，故選\bbox[red,2pt]{(D)}$$

解答：$$(A)\cases{\begin{bmatrix}1 & 1 \\0 & 0 \end{bmatrix} \begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix} =\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}\\ \begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix} \begin{bmatrix}1 & 1 \\0 & 0 \end{bmatrix}= \begin{bmatrix}0 & 0 \\1 & 1 \end{bmatrix}} \Rightarrow AB\ne BA \\ (B)\begin{bmatrix}1 & 0 & 0 &-1 \\0 & 1& 0 & -1\\ 0 & 0 & 1&-1\\ 0 & 0 & 0 & 1 \end{bmatrix} \xrightarrow{r_4+r_1\to r_1,r_4+r_2\to r_2,r_4+r_3\to r_3} I_4\Rightarrow 四向量獨立 \\（C） 映成(onto)並非1-1條件\\，故選\bbox[red,2pt]{(D)}$$

解答：$$rank(AB)\le \min(rank(A),rank(B))，故選\bbox[red,2pt]{(C)}$$

解答：$$\tan^{-1}x= \int_0^x {1\over 1+t^2}\,dt = \int_0^x 1-t^2+t^4-t^6+\cdots\,dt =x-{1\over 3}x^3+{1\over 5}x^5-{1\over 7}x^7+\cdots \\ \Rightarrow \tan^{-1} 1= {\pi \over 4}=1-{1\over 3}+{1\over 5}-{1\over 7}+\cdots，故選\bbox[red,2pt]{(A)}$$

解答：$$L=\lim_{n \rightarrow \infty}\left(\cos{\pi \over n} \right)^n \Rightarrow \ln L=\lim_{n \rightarrow \infty}{\cos{\pi \over n}\over {1\over n}}=\lim_{n \rightarrow \infty}{{d\over d n}\cos{\pi \over n}\over {d\over d n}{1\over n}} =\lim_{n \rightarrow \infty}{-\sin{\pi\over n}\cdot (-{\pi\over n^2})\over -{1\over n^2}} \\=\lim_{n \rightarrow \infty} -\pi\cdot \sin{\pi \over n} =0  \Rightarrow L=e^0=1，故選\bbox[red,2pt]{(B)}$$

解答：$$\lim_{x\to \infty}{\ln x\over 2\sqrt x} = \lim_{x\to \infty}{{d\over dx}\ln x\over {d\over dx}2\sqrt x} = \lim_{x\to \infty} {1/x\over 1/\sqrt x}= \lim_{x\to \infty}{\sqrt x\over x}= \lim_{x\to \infty}{{d\over dx}\sqrt x\over {d\over dx}x}= \lim_{x\to \infty}{1\over 2\sqrt x} =0，故選\bbox[red,2pt]{(D)}$$

解答：$$樣本是相同，入院測一次，10日後測一次，故選\bbox[red,2pt]{(D)}$$

解答：$$顯然(C)正確，故選\bbox[red,2pt]{(C)}$$

解答：$$顯然(B)錯誤，故選\bbox[red,2pt]{(B)}$$

解答：$$甲乙皆不一定正確，故選\bbox[red,2pt]{(D)}$$

解答：$$f(x)為離散形，仍為線性，故選\bbox[red,2pt]{(A)}$$

解答：$$Q若為一點，甲理論錯誤；點集合長度並非為0，乙也錯誤，故選\bbox[red,2pt]{(D)}$$

解答：$$若g(x)=x-2 \Rightarrow {1\over g(x)}在x=2不可積，故選\bbox[red,2pt]{(B)}$$

解答：$$甲：\int_0^1 7f(x)\,dx= 7\int_0^1 f(x)\,dx =7 \Rightarrow \int_0^1 f(x)\,dx =1，甲正確\\ 乙：\sqrt{\int f(x)\,dx } \ne \int \sqrt{f(x)}\,dx，乙不正確\\，故選\bbox[red,2pt]{(B)}$$

解答：$$甲：0.\overline{352}-0.\overline{212}={352\over 999}-{212\over 999}={140\over 999}=0.\overline{140}\ne 0.\overline{14}\\ 乙：0.\overline{352}\times 2={352\over 999}\times 2={704\over 999}=0.\overline{704}\\ \Rightarrow 甲不正確，乙正確，故選\bbox[red,2pt]{(C)}$$

解答：$${a\over b+c}+{b\over c+a} +{c\over a+b}=1 \Rightarrow {a(a+b+c)\over b+c}+{b(a+b+c)\over c+a} +{c(a+b+c)\over a+b}=a+b+c \\ \Rightarrow {a^2\over b+c}+ a+{b^2\over c+a}+ b+{c^2\over a+b}+c=a+b+c \Rightarrow {a^2\over b+c}+ {b^2\over c+a}+{c^2\over a+b}=0，故選\bbox[red,2pt]{(A)}$$

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解題僅供參考，其他教甄試題及詳解

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解題僅供參考，其他教甄試題及詳解