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:令A、B均為S的非空且互斥的子集合,並令C=S−A−B;對任一元素x∈S,x一定在A、B、C其中之一;也就是每個元素有3種選擇,共有310種;但A、B均非空,需扣除A=∅,或B=∅的情形,即310−2×210+1(1代表A=B=∅)=59049−2×1024+1=57002⇒m=57002÷2(A、B對調)=28501⇒{a=8b=5c=0d=1⇒a+b+c+d=14,故選(A)解答:(A)×:沒有最大的質數(B)×:7663=79×97(C)×:若p,p+2的最大公因數是k,k>1;由於p是質數,因此k≥p⇒k=p或p+1⇒(p+2)/k<2⇒gcd(k,p+2)=1⇒k與p+2互質,矛盾!故選(D)
解答:兩交點{P(x1,y1)Q(x2,y2)對稱於直線y=x⇒{P(x1,y1)Q(y1,x1);又P、Q皆在直線y=kx+1上⇒{y1=kx1+1x1=ky1+1⇒y1−x1=k(x1−y1)⇒k=−1將直線y=−x+1代入曲線x2+y2−x−y=4⇒x2+(−x+1)2−x+x−1=4⇒x2−x−2=0⇒(x−2)(x+1)=0⇒x=2,−1⇒{P(2,−1)Q(−1,2)⇒x1+y1+x2+y2=2−1−1+2=2,故選(C)
解答:f(x)=24x24+23∑k=1(24−k)(x24−k+x24+k)=x+2x2+⋯+23x23+24x24+23x25+⋯+2x46+x47⇒xf(x)=x2+2x3+⋯+23x24+24x25+23x26+⋯+2x47+x48⇒f(x)−xf(x)=x+x2+⋯+x24−(x25+x26+⋯+x47+x48)=x+x2+⋯+x24−x24(x+x2+⋯+x24)=(1−x24)(x+x2+⋯+x24)⇒f(x)=x(1−x24)(1+x+⋯+x23)1−x=x(1−x)(1+x+⋯+x23)(1+x+⋯+x23)1−x=x(1−x23)2⇒f(x)=0的相異根為0及zk=eikπ/12,k=1−23⇒z2k=eikπ/6,k=1−23⇒Im(z2k)=sinkπ6,k=1−23⇒23∑k=1|sinkπ6|=45∑k=1sinkπ6=4(12+√32+1+√32+12)=8+4√3⇒m+n+p=8+4+3=15,故選(B)
解答:
A為切點⇒∠OAB=90∘⇒tanθ=¯AB¯OA=¯AB令{¯OC=a∠OBC=∠CBA=α⇒{tan2α=1/¯AB=1/tanθtanα=(1−a)/¯AB=(1−a)/tanθ⇒tan2α=2tanα1−tan2α⇒1tanθ=2(1−a)/tanθ1−(1−a)2/tan2θ⇒1−a=−tan2θ+tanθsecθ⇒a=1+tan2θ−tanθsecθ=1+sin2θcos2θ−sinθcos2θ=cos2θ+sinθ−sinθcos2θ=1−sinθ1−sin2θ=1−sinθ(1+sinθ)(1−sinθ)=11+sinθ,故選(C)
二、填充題
解答:令正方形邊長為a⇒{cos∠PBA=a2+52−722⋅5⋅a=a2−2410acos∠PBC=a2+52−122⋅5⋅a=a2+2410a由於∠PBA+∠PBC=90∘,cos∠PBC=sin∠PBA⇒(a2−2410a)2+(a2+2410a)2=1⇒(a2−24)2+(a2+24)2=100a2⇒a4−50a2+242=0⇒a2=25±7⇒正方形面積=a2=32(a2=25−7=18⇒a=3√2⇒對角線¯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:
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} }}或\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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