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2021年7月3日 星期六

104年新北市高中教甄聯招-數學詳解

新北市立高級中等學校104學年度教師聯合甄選

一、選擇題

解答ABSC=SABxSxABC3310ABA=B=3102×210+1(1A=B=)=590492×1024+1=57002m=57002÷2(AB調)=28501{a=8b=5c=0d=1a+b+c+d=14(A)
解答(A)×:(B)×:7663=79×97(C)×:p,p+2k,k>1pkpk=pp+1(p+2)/k<2gcd(k,p+2)=1kp+2!(D)
解答{P(x1,y1)Q(x2,y2)y=x{P(x1,y1)Q(y1,x1);PQy=kx+1{y1=kx1+1x1=ky1+1y1x1=k(x1y1)k=1y=x+1x2+y2xy=4x2+(x+1)2x+x1=4x2x2=0(x2)(x+1)=0x=2,1{P(2,1)Q(1,2)x1+y1+x2+y2=211+2=2(C)
解答f(x)=24x24+23k=1(24k)(x24k+x24+k)=x+2x2++23x23+24x24+23x25++2x46+x47xf(x)=x2+2x3++23x24+24x25+23x26++2x47+x48f(x)xf(x)=x+x2++x24(x25+x26++x47+x48)=x+x2++x24x24(x+x2++x24)=(1x24)(x+x2++x24)f(x)=x(1x24)(1+x++x23)1x=x(1x)(1+x++x23)(1+x++x23)1x=x(1x23)2f(x)=00zk=eikπ/12,k=123z2k=eikπ/6,k=123Im(z2k)=sinkπ6,k=12323k=1|sinkπ6|=45k=1sinkπ6=4(12+32+1+32+12)=8+43m+n+p=8+4+3=15(B)
解答

AOAB=90tanθ=¯AB¯OA=¯AB{¯OC=aOBC=CBA=α{tan2α=1/¯AB=1/tanθtanα=(1a)/¯AB=(1a)/tanθtan2α=2tanα1tan2α1tanθ=2(1a)/tanθ1(1a)2/tan2θ1a=tan2θ+tanθsecθa=1+tan2θtanθsecθ=1+sin2θcos2θsinθcos2θ=cos2θ+sinθsinθcos2θ=1sinθ1sin2θ=1sinθ(1+sinθ)(1sinθ)=11+sinθ,(C)

二、填充題

解答
a{cosPBA=a2+527225a=a22410acosPBC=a2+521225a=a2+2410aPBA+PBC=90cosPBC=sinPBA(a22410a)2+(a2+2410a)2=1(a224)2+(a2+24)2=100a2a450a2+242=0a2=25±7=a2=32(a2=257=18a=32¯AC=6
解答拋物線\Gamma:y=x^2+ (a+1)x+b過(3,3) \Rightarrow 3=9+3(a+1)+b \Rightarrow b=-9-3a\\ 又y\ge x \Rightarrow x^2+ (a+1)x+b\ge x \Rightarrow x^2+ax+b \ge 0 \Rightarrow 判別式\;a^2-4b \le 0 \\ \Rightarrow a^2+4(9+3a)\le 0 \Rightarrow a^2+12a+36 \le 0 \Rightarrow (a+6)^2\le 0 \Rightarrow a=-6 \Rightarrow b=-9-3(-6)=9\\ \Rightarrow \Gamma:y=x^2-5x+9 =(x-{5\over 2})^2+{11\over 4}\\\Rightarrow 頂點({5\over 2},{11\over 4})至原點的距離=\sqrt{{25\over 4}+{121\over 16}} =\bbox[red,2pt]{\sqrt{221}\over 4}
解答令g(x,y)=x+2y-2 及f(x,y)=\lambda g(x,y)  \Rightarrow \cases{{\partial\over \partial x}f= \lambda {\partial\over \partial x}g\\ {\partial\over \partial y}f= \lambda {\partial\over \partial y}g} \Rightarrow \cases{2xy=\lambda\\ x^2=2\lambda} \Rightarrow x^2=4xy\\ \Rightarrow x(x-4y)=0\Rightarrow \cases{x=0\\ x=4y} 代入g(x,y)=0 \Rightarrow (x,y)=\cases{(0,1) \\(4/3,1/3)} \\ \Rightarrow \cases{f(0,1)=0 最小值\\ f(4/3,1/3)= 16/27 最大值} \quad \Rightarrow 最大值+最小值=\bbox[red,2pt]{16\over 27}
解答令\cases{a=\sqrt[3]{2+\sqrt \alpha} \\b=\sqrt[3]{2-\sqrt \alpha}\\ x=a+b \in \mathbb{N}} \Rightarrow \cases{a^3+b^3=4\\ ab= \sqrt[3]{4-\alpha} \lt \sqrt[3] 4 (\because \alpha \gt 0)} \\ 又(a+b)^3 =a^3+b^3+3ab(a+b) \Rightarrow x^3=4+3abx \Rightarrow ab={ x^3-4\over 3x} ={x^2\over 3}-{4\over 3x} \lt  \sqrt [3]4 \\ \Rightarrow x=1,2 \Rightarrow \cases{x=1 \Rightarrow ab=-1=\sqrt[3]{4-\alpha} \Rightarrow \alpha=5\\ x=2 \Rightarrow ab=2/3=\sqrt[3]{4-\alpha} \Rightarrow \alpha=100/27} \Rightarrow \alpha=\bbox[red,2pt]{5或{100\over 27}}
解答{\sin (6\pi/16) \over \sin (2\pi/16)} ={\sin (6\pi/16) \cos (2\pi/16)\over \sin (2\pi/16) \cos (2\pi/16)} = { \sin(\pi/2) + \sin(\pi/4)\over \sin(\pi/4)} ={1+1/\sqrt 2 \over 1/\sqrt 2} = \sqrt 2+1\\ \Rightarrow \cases{a= {\sin (6\pi/16) \over \sin (2\pi/16)} \\ b={\sin (7\pi/16) \over \sin (3\pi/16)} \\c={\sin (5\pi/16) \over \sin (1\pi/16)}} \Rightarrow  c-a={\sin(2\pi/16)\sin(5\pi/16)-\sin(6\pi/16)\sin(\pi/16) \over \sin(\pi/16) \sin(2\pi/16)} \\ \Rightarrow c-a的分子={1\over 2}(\cos(3\pi/16)-\cos(7\pi/16))-{1\over 2}(\cos(5\pi/16)-\cos 7\pi/16)) \\={1\over 2}(\cos 3\pi/16)-\cos(5\pi/16)) \gt 0 \Rightarrow c\gt a\cdots(1)\\ 同理,a-b的分子=\sin(6\pi/16)\sin(3\pi/16)-\sin(2\pi/16)\sin (7\pi/16) \\={1\over 2}(\cos(3\pi/16) -\cos(9\pi/16)) -{1\over 2}(\cos(5\pi/15)-\cos(9\pi/16)) \\ ={1\over 2}(\cos(3\pi/16)- \cos(5\pi/16)) \gt 0 \Rightarrow a\gt b\cdots(2)\\ 由(1)及(2)得\bbox[red,2pt]{c\gt a\gt b}
解答由題意可知:\cases{n個根之和=a=2n\\ n個根之積=2^n} \quad \Rightarrow 所有的根皆為2 \\\Rightarrow b=\cases{n\times 2^{n-1},n是奇數\\ -n\times 2^{n-1},n是偶數} \quad= \bbox[red,2pt]{n\times (-2)^{n-1}}
解答令f(x)=\cos x,則\cases{\pi/4 \approx 0.785\\3/5=0.6 \\ \pi/6 \approx 0.524} \Rightarrow f(3/5)={({3\over 5} -{\pi\over 6})f({\pi\over 4})+({\pi\over 4}-{3\over 5})f({\pi\over 6})\over {\pi\over 4}-{\pi\over 6}} \\ \approx ((0.6-0.52)\times 0.707+(0.78-0.6)\times 0.866)\times 3.82 \approx 0.812  \Rightarrow n=\bbox[red,2pt]{8}
解答

令\triangle ABC 邊長分為x,y,z及其邊長上的高分別為\cases{h_x=1/7\\ h_y =1/8\\h_z=1/9}(見上圖),則x,y,z符合題意要求;\\ \triangle ABC 面積={1\over 2}\cdot {x\over 7} ={1\over 2}\cdot {y\over 8} ={1\over 2}\cdot {z\over 9}  \Rightarrow x:y:z=7:8:9 \Rightarrow \cases{x=7k\\ y=8k \\ z=9k}\\ 令s=(x+y+z)\div 2=12k \Rightarrow \triangle ABC 面積=\sqrt{s(s-x) (s-y)(s-z)} =\sqrt{12k\cdot 5k \cdot 4k \cdot 3k} \\ =12\sqrt 5k^2={1\over 2}\cdot {x\over 7} ={k\over 2} \Rightarrow k={1\over 24\sqrt 5} \Rightarrow x+y+z= 24k=\bbox[red,2pt]{1\over \sqrt 5}
解答A=\begin{bmatrix} 5& -3 \\ 4& -2\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow (\lambda-1)(\lambda -2)=0\\ \lambda_1 =1 \Rightarrow (A-\lambda_1 I)X=0 \Rightarrow \begin{bmatrix} 4& -3 \\ 4& -3\end{bmatrix}\begin{bmatrix} x_1\\ x_2  \end{bmatrix}=0 \Rightarrow 4x_1=3x_2,取\vec u=\begin{bmatrix}  3 \\ 4 \end{bmatrix} 及\vec v=\begin{bmatrix}  4 \\ -3 \end{bmatrix}\\ \Rightarrow U=(\vec n_1={\vec u\over |\vec u|},\vec n_2={\vec v\over |\vec v|}) =\begin{bmatrix} 3/5& 4/5 \\ 4/5 & -3/5\end{bmatrix}\Rightarrow A\vec n_1=\vec n_1 及 \\ A\vec n_2 =\begin{bmatrix} 5& -3 \\ 4& -2\end{bmatrix} \begin{bmatrix}  4/5 \\ -3/5 \end{bmatrix} = \begin{bmatrix}  29/5 \\ 22/5 \end{bmatrix}=7\vec n_1+2\vec n_2 \Rightarrow AU=A(\vec n_1 \;\vec n_2) =\begin{bmatrix} \vec n_1 \\ \vec n_2 \end{bmatrix}\begin{bmatrix}  1 & 7 \\ 0 & 2 \end{bmatrix} \\ \Rightarrow \bbox[red,2pt]{\cases{U=\begin{bmatrix} 3/5& 4/5 \\ 4/5 & -3/5\end{bmatrix} \\ T=\begin{bmatrix}  1 & 7 \\ 0 & 2 \end{bmatrix} }}\lambda_2 =2 \Rightarrow (A-\lambda_2 I)X=0 \Rightarrow \begin{bmatrix} 3& -3 \\ 4& -4\end{bmatrix}\begin{bmatrix} x_1\\ x_2  \end{bmatrix}=0 \Rightarrow x_1=x_2,取\vec u=\begin{bmatrix}  1 \\ 1 \end{bmatrix} 及\vec v=\begin{bmatrix}  1\\ -1 \end{bmatrix}\\ \Rightarrow U=(\vec n_1={\vec u\over |\vec u|},\vec n_2={\vec v\over |\vec v|}) =\begin{bmatrix} 1/\sqrt 2& 1/\sqrt 2 \\ 1/\sqrt 2 & -1/\sqrt 2\end{bmatrix}\Rightarrow A\vec n_1=2\vec n_1 及 \\ A\vec n_2 =\begin{bmatrix} 5& -3 \\ 4& -2\end{bmatrix} \begin{bmatrix}  1/\sqrt 2 \\ -1/\sqrt 2 \end{bmatrix} = \begin{bmatrix}  8/\sqrt 2 \\ 6/\sqrt 2 \end{bmatrix}=7\vec n_1+\vec n_2 \Rightarrow AU=A(\vec n_1 \;\vec n_2) =\begin{bmatrix} \vec n_1 \\ \vec n_2 \end{bmatrix}\begin{bmatrix}  2 & 7 \\ 0 & 1 \end{bmatrix} \\ \Rightarrow \bbox[red,2pt]{\cases{U=\begin{bmatrix} 1/\sqrt 2& 1/\sqrt 2 \\ 1/\sqrt 2 & -1/\sqrt 2\end{bmatrix} \\ T=\begin{bmatrix}  2 & 7 \\ 0 & 1 \end{bmatrix} }}答案不是唯一,上述兩組答案都對,學校公布的答案是後者。
解答x^5+x^4+ 4x^3+7x^3 +9x+18 = (x^3+ax^2+bx+c)(x^2+dx+e),a,b,c,d,e\in \mathbb{Z}\\ \Rightarrow \cases{a+d=1\\ e+ad+b=4\\ ae+bd+c=7 \\ be+cd=9\\ ec=18},取a=0 \Rightarrow \cases{d=1\\ b+e=4\\ b+c=7\\ be+c=9\\ ec=18},由\cases{c-e=3\\ ce=18} \Rightarrow \cases{c=6\\e=3} \Rightarrow \cases{a=0\\ b=1\\ c=6\\d=1\\ e=3} \\ \Rightarrow \bbox[red,2pt]{(x^3+x+6)(x^2 +x+3 )}
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解題僅供參考,其他教甄試題及詳解 

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