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2025年5月12日 星期一

114年全國高中教甄聯招-數學詳解

 教育部受託辦理114學年度
公立高級中等學校教師甄選

第一部分:選擇題( 共40分)
一、單選題( 每題2分, 共22分)

解答:{a1+a2++a10=60a2+a4++a10=40{a11r101r=60(1)a1r1r101r2=40(2)(1)(2)=1+rr=32r=2a1(2101)=60a1=6010230.059(A)
解答:={1:(1,1,1,1)=18:{(1,1,2,4)=12(1,2,2,2)=427:(1,3,3,3)=464:{(1,4,4,4)=4(2,2,4,4)=6125:(1,5,5,5)=4216:{(1,6,6,6)=4(2,3,6,6)=12(3,3,4,6)=1263=6364XGeo(p=6364)E(X)=1p=6463=144720.57(C)
解答:f(x)=cos(2x)3sin(2x)=2(sinπ6cos(2x)+cosπ6sin(2x))=2sin(2x+π6)f(x+5π6)=2sin(2x+π)=2sin(2x)g(x)=2sin(2x)|x|y=g(x)x{Γ1:y=2sin(2x)Γ2:y=|x|(0,0)π2<2<3π4,(π4,π2)(B)


解答:22+2=C122/C512×14C122/C512×14+C132/C512×34=11/85011/850+39/850=1150(B)
解答:f(x)=x(x2+1)(x3+x+2)f(x)=6x5+8x3+6x2+2x+2f(x)=30x4+24x2+12x+2>0f(x),{f(x)5f(0)=2,f(x)=0,k,f(k)=0a0f(x)=0a=0,a=m<k(kmf(x)dx=0kf(x)dx)a(B)
解答:f(x)=(x+15x)n4096=f(1)=2nn=12f(x)=(x+15x)12=12k=0C12kx12kxk/5=12k=0C12kx126k/51265kk=0,5,10{31010(10!)113C1133!10!×C1133!13!=1526(D)
解答:|x3+y41145|=160|20x+15y1368|x,yZ20x+15y520x+15y13682,3,8220x+15y1368=220x+15y=13704x+3y=274{x=1y=90|x3+y41145|=1602=130(C)
解答:rref([123142311131211])=[100101020013]{x=1y=2z=3[110α011β101γ]{x+y=αy+z=βx+z=γ{α=3β=5γ=4(B)
解答:an=23+1231×33+1331×43+1431××n3+1n31=3317×47213×513321××(n+1)(n2n+1)(n1)(n2+n+1)=3(n+1)!/2(n1)!(n2+n+1)=32(n+1)nn2+n+1lim
解答:a_n=3a_{n-1}-2(-1)^{n-1} \Rightarrow a_n+ k(-1)^{n} =3(a_{n-1}+k(-1)^{n-1}) \Rightarrow a_n =3a_{n-1} +3k(-1)^{n-1}-k(-1)^n \\ \Rightarrow -2(-1)^{n-1}= 3k(-1)^{n-1} -k(-1)^n =4k(-1)^{n-1} \Rightarrow k=-{1\over 2} \\ \Rightarrow a_n-{1\over 2}(-1)^n =3(a_{n-1}-{1\over 2}(-1)^{n-1}) \Rightarrow 取b_n= a_n-{1\over 2}(-1)^n \Rightarrow b_n=3b_{n-1},b_1=1  \\ \Rightarrow b_n=3^{n-1} \Rightarrow b_{114} =3^{113} =a_{114}-{1\over 2}(-1)^{114} \Rightarrow a_{114} =3^{113}+{1\over 2} \\ \Rightarrow \log(3^{113}) =113\log 3=113\cdot 0.4771= 53.9123 \Rightarrow a_{113} 是54位數,故選\bbox[red, 2pt]{(A)}

解答:

此題圖形簡單,可直接手繪求解,故選\bbox[red, 2pt]{(A)}

二、複選題(每題3分,共18分, 全對才給分)

解答:(A)\bigcirc: 1\gt \tan A\tan B\gt 0\Rightarrow \cases{\tan A\gt 0\\ \tan B\gt 0} \Rightarrow \tan C=\tan(\pi-(A+B))=-\tan(A+B) \\\qquad =-{\tan A+ \tan B\over 1-\tan A\tan B}\lt 0 \Rightarrow C為鈍角\\ (B)\bigcirc: \sin A+\cos A=\sqrt 2(\sin A\cos 45^\circ+\sin 45^\circ \cos A) =\sqrt 2\sin(45^\circ+A) ={1\over 4} \Rightarrow \sin(A+45^\circ) ={1\over 4\sqrt 2} \\\qquad \Rightarrow A+45^\circ \gt 135^\circ \Rightarrow A\gt 90^\circ \Rightarrow A為鈍角\\ (C) \bigcirc: \cases{若C為鈍角\Rightarrow \cos C=-\cos(A+B) \gt 0矛盾 \\ 若A為鈍角 \Rightarrow \sin C= \sin(A+B) \lt 0 矛盾} \Rightarrow \triangle ABC為銳角\triangle\\(D)\times: 若A為鈍角,仍符合\cases{\sin A=5/6\\ \sin B=4/5\\ \cos A=-\sqrt{11}/6\\ \cos B=3/5} 且\cases{\cos C=-\cos(A+B) \gt 0\\ \sin C=\sin(A+B)\gt 0}\\,故選\bbox[red, 2pt]{(ABC)}

解答:(A)\bigcirc: 若\lim_{x\to 0} f(x)=L\lt \infty 存在 \Rightarrow \lim_{x\to 0}\left(f(x)+{|x|\over x} \right) =L+\lim_{x\to 0}{|x|\over x} 不存在,矛盾\\ (B)\times: f(x)=x \Rightarrow \lim_{x\to 0}\left(f(x) \cdot {|x|\over x} \right) =\lim_{x\to 0} |x|=0存在,但 \lim_{x\to 0} f(x) =0亦存在\\ (C)\bigcirc: \lim_{x\to 0}\left(f(x) \cdot {|x|\over x} \right)=L\lt \infty 存在\Rightarrow \lim_{x\to 0}\left(f(x) \cdot {|x|\over x} \right)^2 =L^2 \Rightarrow  \lim_{x\to 0}\left((f(x))^2 \cdot {|x|^2\over x^2} \right) \\\qquad =\lim_{x\to 0} (f(x))^2 =L^2存在 \\ (D)\times: 令f(x)=\begin{cases}0 & x\lt 0\\ x+1& x \gt 0 \end{cases},則\lim_{x\to 0}f(x)不存在, 但\lim_{x\to 0}\left( f(x)\cdot {[x]\over x}\right) =0存在\\,故選\bbox[red, 2pt]{(AC)}


解答:\cases{333\cdots3=3+3\cdot 10+3\cdot 10^2+\cdots+3\cdot 10^{98} ={1\over 3}(10^{99}-1)\\ 666\dots6 = 6+ 6\cdot 10+\cdots+6\cdot 10^{98} ={2\over 3}(10^{99}-1)} \\\Rightarrow \overbrace{333\cdots3}^{99個} \times \overbrace{666\cdots 6}^{99個}={2\over 9}(10^{99}-1)^2 =2\times \overbrace{11\cdots 1}^{99個} \times \overbrace{99\cdots 9}^{99個} =2\times \overbrace{11\cdots 1}^{99個} \times (1\overbrace{00\cdots 0}^{99個} -1) \\=  \overbrace{22\cdots 2}^{99個}  \overbrace{00\cdots 0}^{99個} -\overbrace{22\cdots 2}^{99個} = \overbrace{22\cdots 2}^{98個} 1 \overbrace{77\cdots 7}^{97個}8 \\(A)\times: n=99\times 2=198\\ (B)\bigcirc\\ (C)\times:最高位數字為2\\ (D)\bigcirc\\,故選\bbox[red, 2pt]{(BD)}
解答:y=x^3+ax^2+a \Rightarrow y' =3x^2+2ax\\ 假設切點P(x_0,y_0)=(x_0, x_0^3+ax_0^2+a) \Rightarrow 切線斜率= 3x_0^2+2ax_0 \\ 假設原點O(0,0) \Rightarrow \overline{OP}斜率=切線斜率 \Rightarrow {x_0^3+ax_0^2+a\over x_0} =3x_0^2+2ax_0 \\ \Rightarrow 2x_0^3+ax_0^2-a=0有三相異實數解,因此取f(x)=2x^3+ax^2-a \Rightarrow f'(x)=6x^2+2ax =0 \\\Rightarrow 2x(3x+a)=0 \Rightarrow x=0, -a/3 \Rightarrow f(0)f(-a/3)\lt 0 \Rightarrow (-a)({a^3\over 9}-a) \lt 0 \\ \Rightarrow (-a^2)({a^2\over 9}-1)\lt 0 \Rightarrow {a^2\over 9}-1\gt 0 \Rightarrow a\gt 3,a\lt -3 \\ \Rightarrow \cases{(A)\bigcirc: \pi\approx 3.14\gt 3\\ (B) \bigcirc: \sqrt{2025} =45 \gt 3 \\(C) \times: \log 114\lt \log 1000=3 \\(D) \bigcirc: \displaystyle {2025\over 114} \gt {342\over 114}=3},故選\bbox[red, 2pt]{(ABD)}
解答:(A)\bigcirc:a_n的個位數不可能為5,而b_n的個位數皆是5,因此\langle c_n\rangle 各項均異\\ (B)\times: 若a_{30} =c_{45}代表c_1-c_{45}中有30個a_n,15個b_n,而\cases{b_{15}= 5^{15} \Rightarrow \log b_{15} =15(1-0.301)= 10.485 \\ a_{30} =2^{30} \Rightarrow \log a_{30} =30\times 0.301= 9.03}\\ \qquad  \Rightarrow a_{30} \lt b_{15} \quad 矛盾 \\(C) \times: 若b_{10} =c_{30}代表c_1-c_{30}中有10個b_n,20個a_n,而\cases{b_{10} =5^{10} \Rightarrow \log b_{10} =6.99\\a_{20}=2^{20} \Rightarrow \log a_{20} =6.02 \\a_{21} =2^{21} \Rightarrow \log a_{21}=6.321\\ a_{22}=2^{22} \Rightarrow \log a_{22}=6.622 \\ a_{23}= 2^{23} \Rightarrow \log a_{23} =6.923} \\\qquad \Rightarrow 比b_{10}小的a_n有23個,所以b_{10} 在\langle c_n\rangle排名不是第30個,應該是c_{33} \\(D) \bigcirc: \cases{c_k=a_{20} =2^{20} \\c_{k+h}=a_{30} =2^{30}} \Rightarrow \cases{c_1-c_k 中有k-20個b_n\\ c_1-c_{k+h}中有k+h-30個b_n} \Rightarrow \cases{\log 5^{k-20} \lt \log 2^{20} \\ \log 5^{k+h-30} \lt \log 2^{30}} \\ \qquad \Rightarrow \cases{k-20 \lt 8.612 \\ k+h-30\lt 12.918} \Rightarrow h-10\lt 4.30 \Rightarrow h\lt 14.30 \Rightarrow h=14(可驗算)\\,故選\bbox[red, 2pt]{(AD)}

解答:

(A)\bigcirc: \overleftrightarrow{OB}是\angle B的角平分線 \Rightarrow 對稱點A'\in \overleftrightarrow{OB} \\(B)\times: 過A且與\overleftrightarrow{OB}垂直的直線\overleftrightarrow{AA'}: x-y=6 \Rightarrow \overleftrightarrow{OB}\cap \overleftrightarrow{AA'}=(4,-2)= (A+A')/2 \\ \qquad \Rightarrow A'=(6,0) \\(C) \times:假設A對\overleftrightarrow{OC}的對稱點為 \Rightarrow A''=({2\over 5}, {4\over 5}) \Rightarrow \overleftrightarrow{BC} =\overleftrightarrow{A'A''}: x+7y=6 \\ (D)\bigcirc: 圓心O =\overleftrightarrow{OB} \cap \overleftrightarrow{OC}=(3,-1) \Rightarrow  圓半徑r=d(O, \overleftrightarrow{BC} ) =\sqrt 2 \\ \qquad \Rightarrow 圓方程式: (x-3)^2+ (y+1)^2=2  \Rightarrow x^2+y^2-6x +2y+8=0\\,故選\bbox[red, 2pt]{(AD)}

第二部分: 綜合題( 共60分)
一、 填充題(每題4分,共36分)

解答:平面E的法向量\vec n \parallel(\vec b\times \vec c) =(-10, 5,5) \Rightarrow \vec a在\vec n的正射影\vec u= (4,1,3) \cdot (-2,1,1)\cdot {(-2,1,1)\over (-2)^2+1^2+1^2} \\= -{2\over 3}(-2,1,1)= ({4\over 3},-{2\over 3}, -{2\over 3}) \Rightarrow x\vec b+ y\vec c=\vec a-\vec u \Rightarrow (2x+3y,3x+7y, x-y) =({8\over 3},{5\over 3},{11\over 3}) \\ \Rightarrow (x,y)= \bbox[red,2pt]{ \left({41\over 15},-{14\over 15} \right)}

解答:{\triangle ABC\over \triangle BCD} ={3\over 1} \Rightarrow {\overline{CD} \over \overline{AD}} ={1\over 2} \Rightarrow {\tan A\over \tan C} ={\overline{BD}/\overline{AD} \over \overline{BD}/\overline{CD}} ={1\over 2} \Rightarrow \cases{\tan A=k\\ \tan C=2k} \\ \Rightarrow \tan B=-\tan(A+C)   ={3k\over 2k^2-1} \Rightarrow {2\over \tan A}+{1\over \tan B}+{3\over \tan C} ={2\over k}+{2k^2-1\over 3k}+ {3\over 2k} \\={ 4k^2+19\over 6k} ={2k\over 3}+{19\over 6k} \ge 2\sqrt{{2k\over 3}\times{19\over 6k}} = \bbox[red, 2pt]{{2\over 3}\sqrt{19}}
解答:f(k) = \cos{k\pi \over 11} \Rightarrow f(k)=\begin{cases} 0 & 1\le k\le 5\\ -1 & 6\le k\le 16\\ 0 & 17\le k\le 21\\ 1& k=22\end{cases} \Rightarrow \sum_{k=1}^{11} f(k)= -10 \\ \Rightarrow \sum_{k=1}^{114=22\cdot 5+4} f(k) = -10\cdot 5+0= \bbox[red, 2pt]{-50}
解答:\log_{n+1} a_n= {\log a_n\over \log (n+1)} =1+{1\over (n+1)\log(n+1)} = {(n+1)\log(n+1)+1\over (n+1)\log(n+1)} \\ \Rightarrow \log a_n= {(n+1)\log(n+1)+1\over (n+1) }= \log(n+1)+{1\over n+1} =\log ((n+1)\cdot 10^{1/n+1})\\ \Rightarrow a_n=(n+1)10^{1/n+1} \Rightarrow {a_n\over n+1} =10^{1/n+1}\lt 1.2 \Rightarrow {1\over  n+1}\lt \log 1.2 \approx 0.079 \approx {1\over 12.6} \\ \Rightarrow n= \bbox[red, 2pt]{12}
解答:
4\times {2\pi\over 12}(\cos 0+ \cos{2\pi\over 12}+ \cos{4\pi\over 12}) ={2\pi\over 3}(1+{\sqrt 3\over 2}+{1\over 2}) = \bbox[red, 2pt]{{\pi\over 3}(3+\sqrt 3)}
解答:\cases{\overrightarrow{OA} \times \overrightarrow{OB} =\overrightarrow{OC} \\ \overrightarrow{OA} \times \overrightarrow{OC} =\overrightarrow{OD}} \Rightarrow \cases{\overrightarrow{OA} \bot \overrightarrow{OC} \\\overrightarrow{OA}\bot \overrightarrow{OD} \\\overrightarrow{OC} \bot \overrightarrow{OD} \\ \overrightarrow{OB} \bot \overrightarrow{OC}} \Rightarrow \cases{A(k,0,0)\\ C(0,k,0)\\ D(0,0, m)\\ B(a,0,b)} \Rightarrow \overrightarrow{OA} \times \overrightarrow{OC} =(0,0,k^2) =(0,0,m) \Rightarrow m=k^2 \\ \Rightarrow \overrightarrow{OA} \times \overrightarrow{OB} =(k,0,0)\times (a,0,b) =(0,-bk,0) =(0,k,0) \Rightarrow b=-1 \Rightarrow \overline{BD} =\sqrt{a^2+(b-m)^2} \\=\sqrt{a^2+b^2-2bm+m^2} =\sqrt{k^2+2m+m^2} =\sqrt{k^2+2k^2 +k^4} = \bbox[red, 2pt]{k\sqrt{k^2+3}}
解答:(\sqrt 5+ \sqrt 6+\sqrt 7)( \sqrt 6+\sqrt 7-\sqrt 5) (\sqrt 5+\sqrt 7-\sqrt 6) (\sqrt 5+\sqrt 6-\sqrt 7) \\= ( (\sqrt 6+\sqrt 7)+\sqrt 5)( (\sqrt 6+\sqrt 7)-\sqrt 5) (\sqrt 5+(\sqrt 7-\sqrt 6)) (\sqrt 5 -(\sqrt 7-\sqrt 6)) \\=((\sqrt 6+\sqrt 7)^2-5) (5-(\sqrt 7-\sqrt 6)^2) =(8+2\sqrt{42}) (-8+2\sqrt{42}) =(2\sqrt{42})^2-8^2 \\=168-64= \bbox[red, 2pt]{104}
解答:分組的方法數:C^5_1C^4_2=30\\ 10張椅子5人入座後剩下5張椅子,剩下5張椅子中有6個間隔取3個給各組有C^6_3,\\組內排列數:2!\cdot 2!=4及3組排列數:3\\ 因此共有30\times C^6_3\times 4\times 3=\bbox[red, 2pt]{7200}坐法

解答:\sum_{k=1}^{114} (k!\times k)= \sum_{k=1}^{114} ((k+1)!-k!) =(2!-1!)+(3!-2!) +\cdots (115!-114!) =115!-1\\ 又115!是2025的倍數,因此(115!-1)除以2025的餘數=2025-1= \bbox[red, 2pt]{2024}

二、計算題(每題8分,共24分)

解答:



令\cases{\vec u= \overrightarrow{AB'}=\overrightarrow{AB}/|\overrightarrow{AB}| \\ \vec v=\overrightarrow{AD'}=\overrightarrow{AD}/|\overrightarrow{AD}| \\ \vec w = \overrightarrow{AC'} = \sqrt 3 \overrightarrow{AC}/|\overrightarrow{AC}| } \Rightarrow \cases{\vec u+\vec v=\sqrt 3 \vec w \\ |\vec u| =| \vec v|=  1\\ |\vec w|=\sqrt 3} \Rightarrow \cos \angle AD'C' = {|\vec v|^2+ |\vec u|^2-(\sqrt 3|\vec w|)^2 \over 2\cdot |\vec u| |\vec v|} \\={1+1- 3\over 2} =-{1\over 2} \Rightarrow \angle AD'C'=120^\circ \Rightarrow \angle DAC=\angle CAB=30^\circ \\ \Rightarrow \cases{ \overrightarrow{AC} = \overrightarrow{AB}逆時針旋轉30^\circ \times s = s \begin{bmatrix} \cos 30^\circ & -\sin 30^\circ\\ \sin 30^\circ & \cos 30^\circ \end{bmatrix} \begin{bmatrix} 5\\ 2\sqrt 3 \end{bmatrix} =s({3\over 2}\sqrt 3, {11\over 2} )\\ \overrightarrow{AD} = \overrightarrow{AB} 逆時針旋轉60^\circ \times t =t \begin{bmatrix}\cos 60^\circ & -\sin 60^\circ\\ \sin 60^\circ & \cos 60^\circ \end{bmatrix}\begin{bmatrix} 5\\ 2\sqrt 3 \end{bmatrix} =t(-{1\over 2}, {7\over 2}\sqrt 3)} \\ \Rightarrow \overrightarrow{CD} =\overrightarrow{AD}-\overrightarrow{AC} \Rightarrow (-7,-{\sqrt 3\over 3}) =(-{1\over 2}t-{3\over 2}\sqrt 3s, {7\over 2}\sqrt 3t-{11\over 2}s) \Rightarrow \cases{s =4\sqrt 3/3 \\t=2} \\ \Rightarrow \cases{\overrightarrow{AC} =(6, 22\sqrt 3 /3)\\ \overrightarrow{AD} = (-1,7\sqrt 3)} \Rightarrow \cases{|\overrightarrow{AB}|= \sqrt{37}\\ |\overrightarrow{AC}|= 4 \sqrt{111} /3 \\ |\overrightarrow{AD}|= 2\sqrt{37}} \Rightarrow ABCD面積=\triangle ABC+ \triangle ACD \\= {1\over 2} \sin 30^\circ (|\overrightarrow{AB}|\cdot |\overrightarrow{AC}| + |\overrightarrow{AD}|\cdot |\overrightarrow{AC}|) = \bbox[red, 2pt]{37\sqrt 3}
解答:\lim_{x\to 1}{1\over x-1}\left( {1-x^{100} \over 1-x}-100\right) =\lim_{x\to 1}{1\over x-1}\left( {(1-x)(1+x+x^2+ \cdots+ x^{99}) \over 1-x}-100\right) \\= \lim_{x\to 1}{1\over x-1}\left(  1+x+x^2 +\cdots+x^{99}-100\right) = \lim_{x\to 1} {(x+x^2+ \cdots+ x^{99}-99)'\over (x-1)'} \\=\lim_{x\to 1} (99x^{98} +98x^{97}+ \cdots +2x+ 1) =99+98+ \cdots +1 =50\times 99= \bbox[red, 2pt]{4950}
y={\cos x+2\sin x\over 2+\cos x} \Rightarrow  2y=(1-y)\cos x+2\sin x \\=\sqrt{(1-y)^2+2^2} \left( {1-y\over \sqrt{(1-y)^2+2^2}} \cos x+ {2\over \sqrt{(1-y)^2+2^2}} \sin x\right) =\sqrt{y^2-2y+5} \sin(x+\theta) \\ \Rightarrow \sin(x+\theta) ={2y\over \sqrt{y^2-2y+5}} \Rightarrow \left| {2y\over \sqrt{y^2-2y+5}}\right| \le 1 \Rightarrow 4y^2 \le y^2-2y+5 \\ \Rightarrow 3y^2+2y-5\le 0  \Rightarrow (3y+5) (y-1)\le 0 \Rightarrow -{5\over 3}\le y\le 1 \Rightarrow \bbox[red, 2pt]{\cases{最大值1\\ 最小值-{5\over3}}}

\bbox[cyan, 2pt]{試題疑義申復更正答案}


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解題僅供參考,其他教甄試題及詳解




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