111年台中市國中聯合教甄-數學詳解

111 學年度臺中市政府教育局受託辦理本市立國民中學

(含本市立高級中等學校附設國中部)教師甄選

選擇題（共 50 題，每題 2分，共 100 分）

解答： $$f(x)= x\sin(x)+e^{-x} \Rightarrow \cases{f(2h)= 2h\sin(2h)+e^{-2h}\\ f(0)=1\\ f(-2h)=2h\sin(2h) +e^{2h}} \\ \Rightarrow \lim_{h\to 0}{f(2h)-2f(0)+f(-2h)\over h^2} =\lim_{h\to 0}{4h\sin(2h)+e^{2h}+ e^{-2h}-2 \over h^2}\\  =\lim_{h\to 0}{4 \sin(2h)+ 8h\cos(2h)+ 2e^{2h}-2 e^{-2h}  \over 2h} \text{ by l'Hospital rule} \\  =\lim_{h\to 0}{16 \cos(2h)- 16h\sin(2h)+ 4e^{2h}+4 e^{-2h}  \over 2} \text{ by l'Hospital rule} \\={24\over 2}=12，故選\bbox[red, 2pt]{(D)}$$

解答： $$x^2 +3xy+y^2 +x-y=10\quad \underrightarrow{對x微分} \quad 2x+3y+ 3xy'+2yy'+1-y'=0\\ \Rightarrow y'=\left. -{2x+3y+1\over 3x+2y-1} \right|_{(1,2)} =-{3\over 2} \Rightarrow 切線斜率為-{3\over 2}且通過(1,2)\\ \Rightarrow 切線方程式:y=-{3\over 2}(x-1)+2 \Rightarrow y=-{3\over 2}x +{7\over 2}，故選\bbox[red, 2pt]{(B)}$$

解答： $$f(x)= {d\over dx} \left(\int_1^{\cos x} \sqrt{1+t^4}\,dt \right) = \sqrt{1+(\cos x)^4}\cdot (-\sin x)\\ \Rightarrow f({\pi \over 4})= \sqrt{1+{1\over 4}}\cdot (-{\sqrt 2\over 2}) ={\sqrt 5\over 2}\cdot (-{\sqrt 2\over 2}) = -{\sqrt{10}\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答： $$\sum_{k=1}^\infty {(x+1)^k\over k\cdot 2^k} \Rightarrow a_n={(x+1)^n\over n\cdot 2^n}  \Rightarrow 收斂區間r =\lim_{n \to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n \to \infty} \left|{x+1\over 2} \cdot {n\over n+1} \right| \lt 1\\ \Rightarrow \left|{x+1\over 2}  \right| \lt 1 \Rightarrow -2\lt x+1\lt 2 \Rightarrow -3\lt x\lt 1 \Rightarrow 收斂區間[-3,1) (x=-3級數收斂)\\ \Rightarrow \cases{a=-3\\ b=1} \Rightarrow a+4b = -3+4=1，故選\bbox[red, 2pt]{(C)}$$

解答： $$直線L通過A(2,1,3)且方向向量為(1,-3,2)(即平面E:x-3y+2z=4的法向量)\\ \Rightarrow L:{x-2\over 1}={y-1\over -3} ={z-3\over 2}; 若P\in L \Rightarrow P(t+2,-3t+1,2t+3),t\in \mathbb{R}\\ \Rightarrow d(P,E)= \left|{t+2+9t-3+ 4t+6-4\over \sqrt{1+9+4}} \right| ={14t+1\over \sqrt{14}} \Rightarrow 當t=-{1\over 14}時，P({27\over 14}, {17\over 14},{20\over 7})在E上\\ 因此P=(A+Q)/2，Q為A的對稱點 \Rightarrow Q= 2P-A= ({13\over 7}, {10\over 7},{19\over 7}) =(\alpha,\beta,\gamma) \\ \Rightarrow \beta +\gamma = {29\over 7}，故選\bbox[red, 2pt]{(B)}$$

解答： $$A=\left[\begin{matrix}0 & -1 \\1 & 2\end{matrix}\right] =\left[\begin{matrix}-1 & 1 \\1 & 0\end{matrix}\right] \left[\begin{matrix}1 & 1 \\0 & 1\end{matrix}\right]\left[\begin{matrix}0 & 1 \\1 & 1\end{matrix}\right] =PDP^{-1} \Rightarrow A^n = P \left[\begin{matrix}1 & n \\0 & 1\end{matrix}\right]P^{-1} \\ \Rightarrow A^{111}=P \left[\begin{matrix}1 & 111 \\0 & 1\end{matrix}\right]P^{-1} =\left[\begin{matrix}-1 & -110 \\1 & 111\end{matrix}\right] \left[\begin{matrix}0 & 1 \\1 & 1\end{matrix}\right]=\left[\begin{matrix}-110 & -111 \\111 & 112\end{matrix}\right] =\left[\begin{matrix}a & b \\c & d\end{matrix}\right]\\ \Rightarrow a+c+d = -110+111+112 =113，故選\bbox[red, 2pt]{(D)}$$

解答： $$個位數必須是2,4,6,8，有四種選擇；百位數有5種選擇、萬位數有5種選擇、千位數有4種選擇\\、十位數有3種選擇，共有4\times 5\times 5\times 4\times 3= 1200個偶數，故選\bbox[red, 2pt]{(A)}$$

解答： $$10^{n-2} \gt 9^n \Rightarrow \log 10^{n-2}\gt \log 9^n \Rightarrow n-2 \gt 2n\log 3 = 2n\times 0.4771 \\ \Rightarrow 0.0458n \gt 2 \Rightarrow n\gt 43.67 \Rightarrow n=44，故選\bbox[red, 2pt]{(C)}$$

解答： $$令\cases{a=\sqrt[3]{2+\sqrt 3} \\ b=\sqrt[3]{2-\sqrt 3}} \Rightarrow x=a-b \Rightarrow x^3=(a-b)^3 = (a^3-b^3)-3ab(a-b) =2\sqrt 3 -3\cdot 1\cdot x \\ \Rightarrow x^3+3x=2\sqrt 3 \Rightarrow x^3+3x+1 = 2\sqrt 3+1，故選\bbox[red, 2pt]{(A)}$$

解答： $$x={1-i\over \sqrt 3-i} ={(\sqrt 3+1) -(\sqrt 3-1)i\over 4} = {1\over \sqrt 2}\left({\sqrt 6+\sqrt 2\over 4}- {\sqrt 6 -\sqrt 2\over 4}i\right) ={1 \over \sqrt 2}(\cos (15^\circ)-i\sin(15^\circ)) \\ \Rightarrow x^{10} ={1\over 32}(\cos( 150^\circ )-i\sin(150^\circ)) = {1\over 32}(-{\sqrt 3\over 2}-{1\over 2}i) \Rightarrow 64x^{10} =-\sqrt 3-i，故選\bbox[red, 2pt]{(A)}$$

解答： $$(A)\bigcirc: 66\equiv 1(\mod 13) \Rightarrow 66^{111}\equiv 1(\mod 13) \Rightarrow 66^{111}-1是13的倍數\\(B) \times: 2^{999} \equiv 5^{111} \equiv 8^{37}  (\mod 13) ，而\cases{ 8 \equiv 8 (\mod 13)\\ 8^2 \equiv 12 (\mod 13)\\ 8^3\equiv 5(\mod 13)\\ 8^4 \equiv 1(\mod 13)\\ 8^5 \equiv 8(\mod 13)} \Rightarrow 循環數為4 \\\qquad \Rightarrow 8^{37} \equiv 8(\mod 13)\Rightarrow 2^{999}+5 是13的倍數\\(C) \times: 68^{333} \equiv 3 ^{333} \equiv 1^{111} (\mod 13) \Rightarrow 2^{999}+1 不是13的倍數\\(D) \bigcirc: 18^{50}=(18^2)^{25} \equiv (13-1)^{25} (\mod 13) \Rightarrow 18^{50}+1是13的倍數\\，故選\bbox[red, 2pt]{(C)}$$

解答： $$令p(x)=1+x+x^2 +x^3+x^4,則(1+x+x^2 +x^3+x^4)(x+x^2+\cdots +x^{10})^2 \\=p(x)(xp(x)+x^6p(x))^2 =p(x)(x^6+x)^2p(x)^2 =p^3(x)(x^2+2x^7+x^{12}) \\ 由於\cases{p^3(x)的x^3係數=10\\ p^3(x)的x^8係數=15} \Rightarrow 原式x^{10}係數=2\times 10+15=35，故選\bbox[red, 2pt]{(C)}\\ 註:p(x)的係數=1,1,1,1,1 \Rightarrow p^2(x)的係數1,2,3,4,5,4,3,2,1\\ \Rightarrow p^3(x)的係數1,3,6,10,15,18,19,18,15,10,6,3,1$$

解答： $$\log x+ 2\log y=1 \Rightarrow \log xy^2=1 \Rightarrow xy^2 =10 \Rightarrow {3x+2y^2\over 2} \ge \sqrt{6xy^2}= \sqrt{60}\\ \Rightarrow 3x+2y^2 \ge 2\sqrt{60}=4\sqrt{15} \Rightarrow 3x+2y^2的最小值為4\sqrt{15}，故選\bbox[red, 2pt]{(C)}$$

解答： $$兩根為3\pm \sqrt 5 \Rightarrow 兩根之和=6=9\cos \theta \Rightarrow \cos \theta ={2\over 3} \Rightarrow 1-2\sin^2{\theta \over 2}={2\over 3} \\ \Rightarrow \sin^2{\theta\over 2}={1\over 6} \Rightarrow \sin{\theta\over 2}={\sqrt 6\over 6}，故選\bbox[red, 2pt]{(A)}$$

解答： $$\cases{1+2x -x^2+ x^3 \\ 3+x+x^2 +2x^3\\ -1+x-2x^2+ 2x^3\\ 0-2x+x^2 +2x^3} \Rightarrow A=\left[\begin{matrix}1 & 2 & -1 & 1\\3 & 1 & 1 & 2\\-1 & 1 & -2 & 2\\0 & -2 & 1 & 2\ \end{matrix}\right] \Rightarrow \text{rref}(A)=\left[\begin{matrix}1 & 0 & 0 & 3\\0 & 1 & 0 & -3\\0 & 0 & 1 & -4\\0 & 0 & 0 & 0\end{matrix}\right] \\ \Rightarrow Rank(A)=3，故選\bbox[red, 2pt]{(C)}$$

解答： $$\cases{3x+3y-z=10 \\4x-y-3z=2m\\ nx-4y-2z=m-2} \Rightarrow \left[\begin{matrix}3 & 3 & -1\\4 & -1 & -3\\n & -4 & -2\end{matrix}\right] \begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 10\\ 2m\\ m-2\end{bmatrix}\equiv Ax=b\\ 無解 \Rightarrow \det\left( \left[\begin{matrix}3 & 4 & -1\\4 & -1 & -3\\n & -4 & -2\end{matrix}\right]\right)=0 \Rightarrow 10-10n=0 \Rightarrow n=1 \\\Rightarrow \cases{3x+3y-z=10 \cdots(1)\\4x-y-3z=2m \cdots(2)\\ x-4y-2z=m-2 \cdots(3)} \Rightarrow  \cases{3\times(3)-(1) \Rightarrow -15y-5z=3m-16\\ 4\times(3)-(2)\Rightarrow  -15y-5z=2m-8} \\ \Rightarrow 3m-16=2m-8 \Rightarrow m=8 \Rightarrow m-n=8-1=7，故選\bbox[red, 2pt]{(D)}$$

解答：

$$|(z-1)(z+{1\over 2})| = |z-1||z+{1\over 2}| = \overline{PA} \times \overline{PB},其中\cases{P在圓周上\\ A(1,0)\\ B(-1/2,0)}，見上圖；\\ \cases{\cos \angle POA = \cos \theta=\cfrac{\overline{OA}^2 +\overline{OP}^2 - \overline{PA}^2}{2\cdot \overline{OA}\cdot \overline{OP}} =\cfrac{2-\overline{PA}^2}{2} \Rightarrow \overline{PA}^2 = 2-2\cos\theta\\ \cos \angle POB= -\cos \theta =\cfrac{\overline{OB}^2 +\overline{OP}^2 - \overline{PB}^2}{2\cdot \overline{OB}\cdot \overline{OP}} ={5\over 4}-\overline{PB}^2 \Rightarrow \overline{PB}^2 = \cos\theta+{5\over 4} } \\ \Rightarrow \alpha^2 =f(\theta)= \overline{PA}^2\times \overline{PB}^2 = (2-2\cos\theta)(\cos\theta+{5\over 4})= -2\cos^2\theta-{1\over 2}\cos \theta+{5\over 2}\\\Rightarrow \alpha^2的最大值=f(-{1\over 8})=-{2\over 64}+{1\over 16} +{5\over 2}={81\over 32}，故選\bbox[red, 2pt]{(B)}$$

解答： $$\lim_{x\to 0}{\sin x-x\over x} =\lim_{x\to 0}{\cos x-1\over 1} ={1-1\over 1}=0，故選\bbox[red, 2pt]{(C)}$$

解答： $$f(x)=\cos(\sin (x)) \Rightarrow f'(x)=-\sin(\sin(x))\cos(x) \Rightarrow f'(0)= 0，故選\bbox[red, 2pt]{(B)}$$

解答： $$y=x^2 \Rightarrow y'=\left. 2x \right|_{(1,1)}=2 \Rightarrow L斜率為2 \Rightarrow L: y=2(x-1)+1 \Rightarrow y=2x-1 與x軸交於({1\over 2},0)\\ \Rightarrow x坐標={1\over 2}，故選\bbox[red, 2pt]{(D)}$$

解答： $$f(x,y)=x^2+y^2 \Rightarrow \cases{f_x=2x\\ f_y=2y} \Rightarrow \nabla f(x,y)=(2x,2y) \Rightarrow \nabla f(1,1)=(2,2)，故選\bbox[red, 2pt]{(C)}$$

解答： $$取u=x^2 \Rightarrow du=2xdx \Rightarrow \int_0^\sqrt \pi x\cos x^2\,dx = \int_0^\pi {1\over 2}\cos u\,du =\left. \left[ {1\over 2}\sin u \right] \right|_{0}^\pi =0，故選\bbox[red, 2pt]{(B)}$$

解答： $$\lim_{n\to \infty}{1\over n} \sum_{k=0}^n \left({k\over n}\right)^4 =\int_0^1 x^4\,dx ={1\over 5}，故選\bbox[red, 2pt]{(D)}$$

解答： $$本題與42題一樣，都是採用羅必達\text{(l'Hospital rule)}，再加萊布尼茲積分法:\\\lim_{s\to \pi}\cfrac{\int_\pi^s \cos^3 x\,dx}{s-\pi} =\lim_{s\to \pi}\cfrac{  \cos^3 s }{1} = -1，故選\bbox[red, 2pt]{(D)}$$

解答： $$x_k=1+{1\over k^2} \gt 1  \Rightarrow f(x_k)=1，故選\bbox[red, 2pt]{(D)}$$

解答： $${1\over 4}+\left({1\over 4}\right)^2 +\left({1\over 4}\right)^3+\cdots ={1/4\over 1-1/4} ={1\over 3}，故選\bbox[red, 2pt]{(C)}$$

解答： $$顯然是(D)，故選\bbox[red, 2pt]{(D)}$$

解答： $$\sin x=x-{x^3\over 3!}+{x^5\over 5!}-{x^7\over 7!}+\cdots 沒有偶次項，故選\bbox[red, 2pt]{(B)}$$

解答： $$y=\sum_{k=0}^\infty {x^k\over k!} \Rightarrow y'= \sum_{k=1}^\infty {x^{k-1}\over (k-1)!} =\sum_{k=0}^\infty {x^k\over k!}=y \Rightarrow y'=y，故選\bbox[red, 2pt]{(B)}$$

解答： $$y=\sum_{k=0}^\infty {kx^k\over k!} \Rightarrow y'= \sum_{k=1}^\infty {k^2x^{k-1}\over k!}   \Rightarrow y''=\sum_{k=2}^\infty {k^2(k-1)x^{k-2}\over k!} \Rightarrow y'''= \sum_{k=3}^\infty {k^2(k-1)(k-2)x^{k-3}\over k!} \\\Rightarrow y'''(0)={9\cdot 2\cdot 1\over 3!} =3，故選\bbox[red, 2pt]{(C)}$$

解答： $$(2\times 3)(m\times n)(5\times 8) \Rightarrow \cases{m=3\\ n=5}，故選\bbox[red, 2pt]{(C)}$$

解答： $$\begin{vmatrix} 3 & 4\\ 6 & 8\end{vmatrix} =3\cdot 8-6\cdot 4=0，其他選項行列式皆不為0，故選\bbox[red, 2pt]{(C)}$$

解答： $$AX=0 \Rightarrow \cases{x+y+z =0\\ x+2y +3z=0} ，兩平面相交為一直線，故選\bbox[red, 2pt]{(B)}$$

解答： $$A=\begin{bmatrix} 1 & 1\\ 1& -1\end{bmatrix} /\sqrt 2=\begin{bmatrix} 1/\sqrt 2 & 1/\sqrt 2\\ 1/\sqrt 2 & -1/\sqrt 2\end{bmatrix} =\begin{bmatrix} \cos 45^\circ & \sin 45^\circ \\ \sin 45^\circ & -\cos 45^\circ\end{bmatrix} 為直線鏡射矩陣，故選\bbox[red, 2pt]{(C)}$$

解答： $$A=\left[\begin{matrix}1 & 1 & 2\\1 & 3 & 4\\1 & 1 & 2\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 0\end{matrix}\right] \Rightarrow \text{Rank}(A)=2 \Rightarrow \text{Null}(A)=1，故選\bbox[red, 2pt]{(D)}$$

解答： $$2\times 3=6，故選\bbox[red, 2pt]{(D)}$$

解答： $$取\cases{u=x^2 \Rightarrow du=2xdx\\ dv = \cos x\,dx \Rightarrow v=\sin x} \Rightarrow \int x^2\cos x\,dx = x^2\sin x-2\int x\sin x\,dx\\ 再取\cases{u=x \Rightarrow du=dx\\ dv=\sin x\,dx \Rightarrow v=-\cos x}，則 \int x^2\cos x\,dx = x^2\sin x-2(-x\cos x+\sin x) \\ \Rightarrow \int_0^\pi x^2\cos x\,dx = \left. \left[ x^2\sin x+2x\cos x-2\sin x\right]\right|_0^\pi =-2\pi，故選\bbox[red, 2pt]{(D)}$$

解答： $$E(X+1)=E(X)+1=3 \Rightarrow E(X)=2;\\E[(X+1)^2] =E(X^2+2X+1) =E(X^2)+2E(X)+1 = E(X^2)+4+1=20\Rightarrow E(X^2)=15\\ Var(2X-1)= E[(2X-1)^2]-(E(2X-1))^2 =E(4X^2-4X+1)-(2E(X)-1)^2 \\ =4E(X^2)-4E(X)+1-(2\cdot 2-1)^2=60-8+1-9=44，故選\bbox[red, 2pt]{(C)}$$

解答： $$取\cases{u=f(x) \Rightarrow du=f'(x)dx\\ dv = \sin x dx \Rightarrow v= -\cos x} \Rightarrow \int f(x)\sin x\,dx = -\cos xf(x)+ \int f'(x)\cos x\,dx\\ 再取\cases{u= f'(x) \Rightarrow df=f''(x)dx\\ dv= \cos x\,dx \Rightarrow v=\sin x} \Rightarrow \int f'(x)\cos x\,dx=f'(x)\sin x-\int f''(x)\sin x\,dx\\ 因此\int f(x)\sin x\,dx = -\cos xf(x)+ f'(x)\sin x-\int f''(x)\sin x\,dx\\ 原式\int_0^\pi (f''(x)+f(x))\sin x\,dx = \left.\left[ -\cos xf(x)+ f'(x)\sin x\right] \right|_0^\pi =f(\pi)-{1\over 2}=0 \Rightarrow f(\pi)={1\over 2}\\，故選\bbox[red, 2pt]{(B)}$$

解答： $$點數和為5的情況:(1,4),(4,1),(2,3),(3,2)，共四種情況，機率p={4\over 36} ={1\over 9}\\ 因此P(X=1)=p, P(X=2)=(1-p)p,..,P(X=n)=(1-p)^{n-1}p,n\in \mathbb{N}\\\Rightarrow X\sim \text{Geometric}(p) \Rightarrow 期望值E(X)= {1\over p} \Rightarrow E(5X-1)= 5E(X)-1={5\over p}-1 =45-1=44\\，故選\bbox[red, 2pt]{(D)}$$

解答： $$f(x)=\cos^2 2x+ 4\cos^2x-4 \Rightarrow f'(x)= -4\cos 2x\sin 2x-8\cos x\sin x =-4\cos 2x\sin 2x-4\sin 2x\\ =-4\sin 2x(\cos 2x+1) =0 \Rightarrow \cases{\sin 2x=0 \Rightarrow x=k\pi/2, k\in \mathbb{Z}\\ \cos 2x=-1 \Rightarrow x=(2k-1)\pi/2, k\in \mathbb{Z}} \\ \Rightarrow \cos 2x=-1 且\cos^2x =0 ,則f(x)有最小值\Rightarrow f(\pi/2)=1+0-4=-3，故選\bbox[red, 2pt]{(B)}$$

解答： $$\lim_{x\to 0} x^{-3}\int_0^{2x} {t^2\over 1+t^3}\,dt =\lim_{x\to 0} \cfrac{\int_0^{2x} {t^2\over 1+t^3}\,dt }{x^3}  =\lim_{x\to 0} \cfrac{  {4x^2\over 1+8x^3}\cdot 2 }{3x^2}= \lim_{x\to 0} \cfrac  {8x^2}{3x^2+24x^5} \\= \lim_{x\to 0} \cfrac  {16x}{6x+ 120x^4} = \lim_{x\to 0} \cfrac  {16}{6+ 480x^3} ={16\over 6}={8\over 3}，故選\bbox[red, 2pt]{(D)}$$

解答： $$x+{1\over x}=2\cos \theta \Rightarrow (x+{1\over x})^2 =x^2+{1\over x^2}+2 =4\cos^2\theta \Rightarrow x^2+{1\over x^2}=4\cos^2\theta -2\\ \Rightarrow x^3+{1\over x^3} =(x+{1\over x})(x^2-1+{1\over x^2}) =2\cos \theta(4\cos^2\theta -3)= 2(4\cos^3\theta-3\cos\theta) =2\cos 3\theta\\，故選\bbox[red, 2pt]{(A)}$$

解答： $${a+b\over 4}=\cfrac{{a\over 3} +{a\over 3} +{a\over 3} +b}{4} \ge \sqrt[4]{a^3b\over 27} \Rightarrow 1\ge \sqrt[4] {a^3b\over 27} \Rightarrow a^3b\le 27 \Rightarrow \log a^3b \le \log 27\\ \Rightarrow 3\log a+\log b\le 3\log 3，故選\bbox[red, 2pt]{(D)}$$

解答： $$\cfrac{3^8 +3^6+3^4 +3^2+1}{3^4 +3^3+3^2 +3 +1} =\cfrac{1-3^{10}\over 1-3^2}{1-3^5\over 1-3} ={3^{10}-1\over 8}\cdot {2\over 3^{5}-1} ={3^5+1\over 4} =61，故選\bbox[red, 2pt]{(B)}$$

解答： $$假設x=p為其共同根 \Rightarrow \cases{x^2+ kx+1 = (x-p)(x-q)\\ x^2+x+k = (x-p)(x-r)},兩式相除\Rightarrow {x^2+ kx+1 \over x^2+x+k} ={x-q\over x-r}\\ \Rightarrow x^3+(k-r)x^2+(1-kr)x-r = x^3+(1-q)x^2+(k-q)x-kq\\ \Rightarrow \cases{k-r= 1-q \cdots(1)\\ 1-kr= k-q\cdots(2)\\ r=kq \cdots(3)},將(3)代入(1)及(2) \Rightarrow \cases{k-kq=1-q \Rightarrow q=1 \cdots(4)\\ 1-k^2q= k-q \cdots(5)}\\將(4)代入(5) \Rightarrow 1-k^2 =k-1 \Rightarrow k^2+k-2=0 \Rightarrow (k+2)(k-1)=0 \\ \Rightarrow k=1,-2 \Rightarrow 1-2=-1，故選\bbox[red, 2pt]{(A)}$$

解答： $$\begin{array} {cc|cc|c} |ab| & |a+b| & a & b & 小計\\\hline 0 & 5 & 0 & 5 \\ & & 0 & -5\\ & & 5 & 0\\ & & -5 & 0 & 4\\\hdashline 1 & 4 & & & 0\\\hdashline 2 & 3 & 1& 2& \\ & & 2 & 1\\ & & -1 & -2\\ & & -2 & -1 & 4\\\hdashline 3 & 2 & 3& -1& \\ & & -1 & 3 \\ & & -3 & 1\\ & & 1& -3& 4\\\hdashline 4 & 1 & & &0 \\ 5 & 0 & & & 0\\\hline\end{array}\\ \Rightarrow 共12組答案，故選\bbox[red, 2pt]{(D)}$$

解答： $$d=32 \Rightarrow 最小的\sum a_i = 32\times 50=1600 \gt 1504 \Rightarrow d\ne 32\\ d=16 \Rightarrow 最小的\sum a_i = 16\times 50=800 \lt 1504，又(1504-800)\div 16=44 \\ \Rightarrow a_1=a_2=\cdots = a_6=16,a_7=a_8=\cdots =a_{50}=32 滿足 16\mid a_i,i=1-50\\ 因此最大的d=16，故選\bbox[red, 2pt]{(C)}$$

解答：

$$\cases{y^2-4x^2\le 0 \Rightarrow (y+2x)(y-2x) \le 0\\|x|\le 2 \Rightarrow -2\le x\le 2}為兩條斜直線與兩條垂直線所圍區域，見上圖;\\因此兩面積之和=8\times 2=16，故選\bbox[red, 2pt]{(B)}$$

解答：

$$斜線區域面積=正方形扣除右上角的\triangle BPC=1-{1\over 2}(2-a)^2 = -{a^2\over 2}+2a-1，故選\bbox[red, 2pt]{(C)}$$

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