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：$$令A、B均為S的非空且互斥的子集合，並令C=S-A-B；\\對任一元素x\in S，x一定在A、B、C其中之一；也就是每個元素有3種選擇，共有3^{10}種；\\但A、B均非空，需扣除A=\varnothing，或B=\varnothing 的情形，即3^{10}-2\times 2^{10}+1 (1代表A=B=\varnothing)\\ =59049-2\times 1024+1=57002 \Rightarrow m=57002\div 2(A、B對調)=28501 \\ \Rightarrow \cases{a=8\\b=5\\ c=0\\d=1} \Rightarrow a+b+c+d = 14，故選\bbox[red,2pt]{(A)}$$
解答：$$(A)\times: 沒有最大的質數\\(B)\times: 7663=79\times 97\\(C)\times: 若p,p+2的最大公因數是k,k\gt 1；由於p是質數，因此k\ge p \Rightarrow k=p或p+1\\ \qquad \Rightarrow (p+2)/k \lt 2\Rightarrow \gcd(k,p+2)=1 \Rightarrow k與p+2互質，矛盾!\\故選\bbox[red,2pt]{(D)}$$
解答：$$兩交點\cases{P(x_1,y_1)\\ Q(x_2,y_2)} 對稱於直線\;y=x \Rightarrow \cases{P(x_1,y_1)\\ Q(y_1,x_1)} ;\\ 又P、Q皆在直線\;y=kx+1\;上\Rightarrow \cases{y_1=kx_1+1\\ x_1=ky_1+1} \Rightarrow y_1-x_1=k(x_1-y_1) \Rightarrow k=-1\\ 將直線\;y=-x+1代入曲線\;x^2+y^2-x-y=4 \Rightarrow x^2+(-x+1)^2-x+x-1=4\\ \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=2,-1 \Rightarrow \cases{P(2,-1)\\ Q(-1,2)} \\ \Rightarrow x_1+y_1+x_2+y_2= 2-1-1+2=2，故選\bbox[red,2pt]{(C)}$$
解答：$$f(x)= 24x^{24} +\sum_{k=1}^{23}(24-k)(x^{24-k}+x^{24+k})\\ =x+ 2x^2+\cdots +23x^{23}+ 24x^{24}+23x^{25}+\cdots +2x^{46}+x^{47} \\ \Rightarrow xf(x)=x^2+ 2x^3+\cdots +23x^{24}+ 24x^{25}+23x^{26}+\cdots +2x^{47}+x^{48} \\ \Rightarrow f(x)-xf(x)=x+x^2+\cdots +x^{24}-(x^{25}+x^{26}+\cdots +x^{47}+x^{48}) \\ \qquad\qquad=x+x^2+\cdots +x^{24}-x^{24}(x+x^2+\cdots +x^{24}) \\ \qquad\qquad =(1-x^{24})(x+x^2+\cdots +x^{24})\\ \Rightarrow f(x)={x(1-x^{24})(1+x+\cdots +x^{23}) \over 1-x} = {x(1-x) (1+x+\cdots +x^{23})(1+x+\cdots +x^{23}) \over 1-x} \\=x(1-x^{23})^2 \Rightarrow f(x)=0的相異根為 0及z_k=e^{ik\pi/12},k=1-23 \\ \Rightarrow z_k^2= e^{ik\pi/6},k=1-23 \Rightarrow Im(z_k^2)= \sin{k\pi \over 6},k=1-23 \\ \Rightarrow \sum_{k=1}^{23} |\sin{k\pi \over 6}| = 4\sum_{k=1}^{5} \sin{k\pi \over 6} =4({1\over 2}+{\sqrt 3\over 2 }+1 +{\sqrt 3\over 2}+{1\over 2})=8+4\sqrt 3 \\ \Rightarrow m+n+p =8+4+3=15，故選\bbox[red,2pt]{(B)}$$
解答：
$$A為切點\Rightarrow \angle OAB=90^\circ \Rightarrow \tan \theta={\overline{AB}\over \overline{OA}} =\overline{AB}\\ 令\cases{\overline{OC}=a \\ \angle OBC=\angle CBA =\alpha} \Rightarrow \cases{\tan 2\alpha= 1/\overline{AB} =1/\tan \theta\\ \tan \alpha =(1-a)/\overline{AB} =(1-a)/\tan \theta}\\ \Rightarrow \tan 2\alpha ={2\tan \alpha\over 1-\tan^2 \alpha} \Rightarrow {1\over \tan \theta}={2(1-a)/\tan \theta \over 1-(1-a)^2/\tan^2\theta} \Rightarrow 1-a=-\tan^2\theta+\tan \theta \sec\theta\\ \Rightarrow a=1+\tan^2\theta-\tan\theta\sec \theta =1+{\sin^2\theta \over \cos^2\theta} -{\sin\theta \over \cos^2\theta} ={\cos^2\theta +\sin^\theta-\sin\theta\over \cos^2 \theta} \\ ={1-\sin\theta \over 1-\sin^2\theta} ={1-\sin \theta \over (1+\sin\theta)(1-\sin \theta)} ={1\over 1+\sin \theta},故選\bbox[red,2pt]{(C)}$$
二、填充題
解答：$$令正方形邊長為a \Rightarrow \cases{\cos \angle PBA={a^2+5^2-7^2 \over 2\cdot 5\cdot a} ={a^2-24\over 10a}\\ \cos \angle PBC ={a^2+5^2-1^2 \over 2\cdot 5\cdot a}={a^2 +24\over 10a}} \\ 由於\angle PBA+\angle PBC=90^\circ，\cos \angle PBC=\sin \angle PBA \Rightarrow \left( {a^2-24\over 10a} \right)^2 +\left({a^2 +24\over 10a} \right)^2=1 \\ \Rightarrow (a^2-24)^2 +(a^2+24)^2 =100a^2 \Rightarrow a^4-50a^2+24^2=0 \Rightarrow a^2= 25\pm 7\\ \Rightarrow 正方形面積=a^2=\bbox[red,2pt]{32} (a^2=25-7=18 \Rightarrow a=3\sqrt 2\Rightarrow 對角線\overline{AC}=6 \not \gt 7=\overline{PA})$$
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：$$A=\begin{bmatrix} 5& -3 \\ 4& -2\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow (\lambda-1)(\lambda -2)=0\\ \lambda_1 =1 \Rightarrow (A-\lambda_1 I)X=0 \Rightarrow \begin{bmatrix} 4& -3 \\ 4& -3\end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}=0 \Rightarrow 4x_1=3x_2，取\vec u=\begin{bmatrix} 3 \\ 4 \end{bmatrix} 及\vec v=\begin{bmatrix} 4 \\ -3 \end{bmatrix}\\ \Rightarrow U=(\vec n_1={\vec u\over |\vec u|},\vec n_2={\vec v\over |\vec v|}) =\begin{bmatrix} 3/5& 4/5 \\ 4/5 & -3/5\end{bmatrix}\Rightarrow A\vec n_1=\vec n_1 及 \\ A\vec n_2 =\begin{bmatrix} 5& -3 \\ 4& -2\end{bmatrix} \begin{bmatrix} 4/5 \\ -3/5 \end{bmatrix} = \begin{bmatrix} 29/5 \\ 22/5 \end{bmatrix}=7\vec n_1+2\vec n_2 \Rightarrow AU=A(\vec n_1 \;\vec n_2) =\begin{bmatrix} \vec n_1 \\ \vec n_2 \end{bmatrix}\begin{bmatrix} 1 & 7 \\ 0 & 2 \end{bmatrix} \\ \Rightarrow \bbox[red,2pt]{\cases{U=\begin{bmatrix} 3/5& 4/5 \\ 4/5 & -3/5\end{bmatrix} \\ T=\begin{bmatrix} 1 & 7 \\ 0 & 2 \end{bmatrix} }}$$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：$$x^5+x^4+ 4x^3+7x^3 +9x+18 = (x^3+ax^2+bx+c)(x^2+dx+e),a,b,c,d,e\in \mathbb{Z}\\ \Rightarrow \cases{a+d=1\\ e+ad+b=4\\ ae+bd+c=7 \\ be+cd=9\\ ec=18}，取a=0 \Rightarrow \cases{d=1\\ b+e=4\\ b+c=7\\ be+c=9\\ ec=18}，由\cases{c-e=3\\ ce=18} \Rightarrow \cases{c=6\\e=3} \Rightarrow \cases{a=0\\ b=1\\ c=6\\d=1\\ e=3} \\ \Rightarrow \bbox[red,2pt]{(x^3+x+6)(x^2 +x+3 )}$$
================================ END ============================
沒有留言:
張貼留言