金門縣 115 學年度國民中學正式教師聯合甄選筆試
解答:$$\lim_{x\to 0}{\sin x-x\cos x\over x^3} = \lim_{x\to 0}{{d\over dx}(\sin x-x\cos x)\over {d\over dx}x^3} = \lim_{x\to 0}{ x\sin x\over 3x^2} = \lim_{x\to 0}{ \sin x\over 3x} =\lim_{x\to 0}{ {d\over dx} \sin x\over {d\over dx}3x} \\ =\lim_{x\to 0}{ \cos x\over 3} ={1\over 3}，故選\bbox[red, 2pt]{(B)}$$
解答:$$f(x)=e^x \cos x \Rightarrow f'(x)=e^x(\cos x-\sin x) \Rightarrow f''(x)=e^x(-2\sin x)\\ \Rightarrow f'''(x)=e^x(-2\sin x-2\cos x) \Rightarrow f^{[4]}(x)=e^x(-4\cos x) \Rightarrow f^{[4]}(0)=-4\\ \Rightarrow x^4係數={-4\over 4!}=-{1\over 6}，故選\bbox[red, 2pt]{(A)}$$
解答:$$F(x)= \int_x^{x^2}e^{t^2}\,dt \Rightarrow F'(x)=2xe^{x^4}-xe^{x^2} \Rightarrow F'(1)=2e-e=e，故選\bbox[red, 2pt]{(B)}$$
解答:$$依交錯級數審歛法，該級數收斂；但依極限比較審斂法，\lim_{n\to \infty} {1/n\over n/n^2+1} =\lim_{n\to \infty} {n^2+1\over n^2} =1 \\ \Rightarrow \sum {n\over n^2+1} 與\sum {1\over n}同為發散級數 \Rightarrow 級數收斂，但不是絕對收斂，故選\bbox[red, 2pt]{(B)}$$
解答:$$\iint_D (x+y)\,dA =\int_0^1 \int_0^x (x+y)\,dydx = \int_0^1 {3\over 2}x^2\,dx = \left. \left[ {1\over 2}x^3 \right] \right|_0^1 ={1\over 2}，故選\bbox[red, 2pt]{(B)}，\\但公布的答案是\bbox[cyan,2pt]{(C)}$$
解答:$$\cases{x=t^2+1\\ y=t^3-t} \Rightarrow \cases{dx/dt=2t\\ dy/dt =3t^2-1} \Rightarrow {dy\over dx}={dy\over dt}\cdot {dt\over dx}={3t^2-1\over 2t} \\ \Rightarrow t=1 \Rightarrow \cases{(x,y)=(2,0) \\ y'=1} \Rightarrow 切線方程式: y=x-2，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{x=3 \Rightarrow a_n={(x-1)^n\over n2^n} ={1\over n} \Rightarrow 調合級數, 發散\\ x=-1 \Rightarrow a_n= {(x-1)^n\over n2^n} = {(-2)^n\over n2^n} ={(-1)^n\over n} \Rightarrow 交錯級數審斂法,收斂} \Rightarrow 收斂區間為[-1,3)\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{u=xy\\ v=y/x} \Rightarrow {\partial(u,v)\over \partial(x,y)} = \begin{vmatrix} y& x\\-y/x^2& 1/x \end{vmatrix}={2y\over x} =2v\\ \Rightarrow \iint_R {x^2+y^2\over xy}\,dA =\iint_R \left( {x\over y}+{y\over x} \right)\,dA = \int_1^2 \int_1^4 \left( {1\over v}+v \right) \cdot {1\over 2v}\,du \,dv = \int_1^2 \int_1^4 \left( {1\over 2v^2}+{1\over 2} \right) \,du \,dv \\={3\over 2} \int_1^2 \left( {1\over v^2}+1 \right)\,dv= {3\over 2} \left. \left[ -{1\over v}+v \right] \right|_1^2= {9\over 4}，故選\bbox[red, 2pt]{(C)}$$
解答:$$\begin{vmatrix} 1& 1& 0\\1& a& 1\\0& 1& 1 \end{vmatrix} =0 \Rightarrow a-2=0 \Rightarrow a=2，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{A= \begin{bmatrix}1& 1\\1& 0\\0& 1 \end{bmatrix} \\[1ex] b=\begin{bmatrix} 1\\2\\2 \end{bmatrix}} \Rightarrow \cases{A^TA= \begin{bmatrix}2& 1\\1& 2 \end{bmatrix} \\ A^Tb= \begin{bmatrix}3\\3 \end{bmatrix}} \Rightarrow A(A^TA)^{-1}A^Tb = \begin{bmatrix}1& 1\\1& 0\\0& 1 \end{bmatrix} \begin{bmatrix}2/3 & -1/3\\ -1/3& 2/3 \end{bmatrix} \begin{bmatrix} 3\\3 \end{bmatrix} \\= \begin{bmatrix}1& 1\\1& 0\\0& 1 \end{bmatrix} \begin{bmatrix}1\\1 \end{bmatrix} = \begin{bmatrix}2\\1\\1 \end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{B=\{1,x,x^2\} \\ T(p)=p'+p(0)}\Rightarrow \cases{T(1) =1 = \begin{bmatrix}1\\0\\0 \end{bmatrix}\\ T(x)=1 = \begin{bmatrix}1\\0\\0 \end{bmatrix}\\ T(x^2)=2x = \begin{bmatrix}0\\2\\0 \end{bmatrix}} \Rightarrow M = \begin{bmatrix}1& 1& 0\\0& 0& 2\\0& 0& 0 \end{bmatrix}，故選\bbox[red, 2pt]{(A)}$$
解答:$$v_1=u_1=(1,1,0) \\ v_2= u_2-{u_2\cdot v_1\over v_1\cdot v_1}v_1= (1,0,1)-{1\over 2}(1,1,0) =({1\over 2},-{1\over 2},1) \Rightarrow 2v_2=(1,-1,2)，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設A的特徵值為\lambda_1, \lambda_2 \Rightarrow \cases{\lambda_1+\lambda_2= tr(A)=5\\ \lambda_1\lambda_2= \det(A)=6} \Rightarrow A^{-1}的特徵值為{1\over \lambda_1}, {1\over \lambda_2} \\ \Rightarrow tr(A^{-1}) ={1\over \lambda_1}+ {1\over \lambda_2} ={\lambda_1+\lambda_2\over \lambda_1 \lambda_2} ={5\over 6}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\lim_{n\to \infty} n(\sqrt{n^2+1}-n) =\lim_{n\to \infty}{ n(\sqrt{n^2+1}-n) (\sqrt{n^2+1}+n) \over \sqrt{n^2+1}+n}= \lim_{n\to \infty}{n\over \sqrt{n^2+1}+n} \\= \lim_{n\to \infty}{1 \over \sqrt{1 +1/n}+1} ={1\over 2}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\text{compact 必須是 bounded 且 closed} \\(A)\times: (0,1)為\text{bounded 但不是 closed} \\(B)\times: [0,\infty) 不是 \text{bounded} \\(C) \times: 0 \not \in S=\{ {1\over n}: n\in \mathbb N, n\ge 1\}不是 \text{closed}\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$在該區間內\cases{上界\text{Supremum }M_i=1\\ 下界\text{Infimum }m_i=0} \Rightarrow \cases{上積分=1\\ 下積分=0} \Rightarrow 1\ne 0\Rightarrow f\; 不是黎曼可積\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$S為\mathbb R的子集，故選\bbox[red, 2pt]{(C)}$$
解答:$$y'+2y=e^{-x} \Rightarrow y'e^{2x} +2ye^{2x}=e^x \Rightarrow \left( ye^{2x} \right)'=e^x \Rightarrow ye^{2x}=\int e^x\,dx=e^x+C \\ \Rightarrow y=e^{-x}+Ce^{-2x} \Rightarrow y(0)=1+C=0\Rightarrow C=-1\Rightarrow y=e^{-x} -e^{-2x}，故選\bbox[red, 2pt]{(C)}$$
解答:$$y'={dy\over dx}=y(1-y) \Rightarrow \int {1\over y(1-y)}dy = \int 1\,dx \Rightarrow \ln \left|{y\over 1-y} \right| =x+C_1 \\ \Rightarrow {y\over 1-y}=-1+{1\over 1-y}= C_2e^x \Rightarrow {1\over 1-y}=C_2e^x+1 \Rightarrow 1-y={1\over C_2e^x+1} \\ \Rightarrow y=1-{1\over C_2e^x+1} \Rightarrow y(0)=1-{1\over C_2+1}={1\over 2} \Rightarrow C_2=1 \Rightarrow y=1-{1\over e^x+1} \\ \Rightarrow y={e^x \over e^x+1} ={1\over 1+e^{-x}}，故選\bbox[red, 2pt]{(B)}$$
解答:$$公式: order(a)={n\over gcd(a,n)}={12\over gcd(8,12)}={12\over 4}=3，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{3^1 \equiv 3\text{ mod 7} \\ 3^2 \equiv 2\text{ mod 7} \\ 3^3 \equiv 6\text{ mod 7} \\ 3^4 \equiv 4\text{ mod 7} \\ 3^5 \equiv 5\text{ mod 7} \\3^6 \equiv 1\text{ mod 7} \\ 3^7 \equiv 3\text{ mod 7} } \Rightarrow 循環數6 \Rightarrow 100=6\times 16+4 \Rightarrow 3^{100} \equiv 3^4 \text{ mod 7} =4，故選\bbox[red, 2pt]{(D)}$$
解答:$$(A)\times :x^2-2=0 \Rightarrow x=\pm \sqrt 2 \not \in \mathbb Q \\(B)\times: x^2+1=0 \Rightarrow x=\pm i \not \in \mathbb Q \\(C)\times :x^2+x+1=0 \Rightarrow x={-1 \pm \sqrt 3i\over 2} \not \in \mathbb Q \\ (D)\bigcirc: x^2-1=0 \Rightarrow x=\pm 1 \in \mathbb Q\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{L_1:(x,y,z)=s(1,0,1) \Rightarrow \cases{L_1方向向量\vec u=(1,0,1) \\ L_1通過O(0,0,0)} \\L_2:(x,y,z) =(0,1,0)+t(0,1,1) \Rightarrow \cases{L_2方向向量\vec v=(0,1,1) \\ L_2通過P(0,1,0)}} \Rightarrow \cases{\vec n= \vec u\times \vec v=(-1,-1,1) \\ \overrightarrow{OP}=(0,1,0)} \\ \Rightarrow d(L_1, L_2)=\left| {\overrightarrow{OP} \cdot \vec n\over |\vec n|}\right| = {1\over \sqrt 3}，故選\bbox[red, 2pt]{(C)}$$
解答:$$|\vec a\times \vec b|=|(4,-2,-3)| =\sqrt{16+4+9}= \sqrt{29}，故選\bbox[red, 2pt]{(C)}$$
解答:$$a_n=3a_{n-1}-2a_{n-2} \Rightarrow a_n-a_{n-1}=2(a_{n-1}-a_{n-2}) \\取b_n=a_n-a_{n-1} \Rightarrow b_n=2b_{n-1} \Rightarrow \langle b_n \rangle 為等比數列,公比r=2, 且b_1=a_1-a_0=1 \\ \Rightarrow \cases{a_n-a_{n-1}=2^{n-1} \\ a_{n-1}-a_{n-2}= 2^{n-2} \\ \cdots\\a_2-a_1=2} \Rightarrow 全部相加可得a_n-a_1=2^{n-1}+2^{n-2} +\cdots+2 \\ \Rightarrow a_n=2^0+2^1+\cdots+2^{n-2}+2^{n-1} =2^n-1，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設取出的 1 元、5 元、10 元硬幣數量分別為 x、y、z 枚 \Rightarrow x+y+z=8\\ 不考慮限制條件，有H^3_8=C^{10}_8= 45種，需扣除y\gt 6的情形\\\cases{ y=7 \Rightarrow x+z=1 \Rightarrow (x,z)=(1,0),(0,1) \Rightarrow 2組解\\ y=8\Rightarrow x+z=0\Rightarrow (x,z)=(0,0) \Rightarrow 1組解} \Rightarrow 45-2-1=42，故選\bbox[red, 2pt]{(A)}$$
解答:$$共有10+6+9=25 枚硬幣，總面額10\times 1+6\times 5+9\times 10=130 \\ \Rightarrow 取出一枚硬幣的期望值:E(X) ={130\over 25}=5.2 \\\Rightarrow 取出二枚硬幣的期望值:E(X_1+X_2)=E(X_1)+E(X_2)=5.2+5.2=10.4，故選\bbox[red, 2pt]{(C)}$$
解答:$$假設家豪位於原點(0,0) \Rightarrow \cases{淑芬在A(10,0) \\ 怡君在B(-{10\sqrt 2\over \sqrt 2},-{10\sqrt 2\over \sqrt 2}) =(-10,-10)} \Rightarrow \cases{\overrightarrow{BO}=(10,10) \\ \overrightarrow{BA}=(20,10)} \\ \Rightarrow \cos \alpha ={\overrightarrow{BO} \cdot \overrightarrow{BA} \over |\overrightarrow{BO}| |\overrightarrow{BA}|} ={300\over 10\sqrt 2\cdot 10\sqrt 5}={3\over \sqrt{10}} \Rightarrow \sin \alpha ={1\over \sqrt{10}}，故選\bbox[red, 2pt]{(D)}$$
解答:$$A= \begin{bmatrix}2&-3\\-2& 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2-3\lambda-4= (\lambda-4)(\lambda+1)=0 \Rightarrow \cases{\alpha=4\\ \beta=-1} \\ \Rightarrow \alpha^2+\beta^2=16+1=17，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{A(10,0) \\B(-10,-10)} \Rightarrow \overline{AB} =\sqrt{400+100} =10\sqrt 5，故選\bbox[red, 2pt]{(B)}$$
解答:$$a_1=1 \Rightarrow a_2=2+1=3 \Rightarrow a_3=2\cdot 3+2=8 \Rightarrow a_4=2\cdot 8+3=19 \Rightarrow a_5=2\cdot 19+4=42\\ \Rightarrow a_6=2\cdot 42+5=89 \Rightarrow a_7=2\cdot 89+6=184 \Rightarrow a_8=2\cdot 184+7=375 \Rightarrow a_9=2\cdot 375+8=758\\ \Rightarrow a_{10}=2\cdot 758+9=1525，故選\bbox[red, 2pt]{(A)}\\ 當然也可以用:a_{n+1}= 2a_n+n \Rightarrow a_{n+1} +(n+1)+1= 2(a_{n}+n+1)\Rightarrow b_{n+1}=2b_n 等比級數去計算$$
解答:$$f(x) = 4x^3 - 30x^2 + 72x + 7 \Rightarrow f'(x)=12x^2-60x+72= 12(x-2)(x-3) \\ \Rightarrow \begin{cases} f'(x)\gt 0& x\gt 3或x\lt 2\\ f'(x)\lt 7& 2\lt x\lt 3\end{cases} \Rightarrow 在區間2\lt x\lt 3遞減，故選\bbox[red, 2pt]{(C)}$$
解答:$$3x+5y=150 \Rightarrow y=30-{3\over 5}x \gt 0 \Rightarrow x\lt 50且x是5的倍數 \\ \Rightarrow x=5,10,\dots, 45\Rightarrow 共9個，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{u=\ln x\\ dv=x\,dx} \Rightarrow \cases{du=dx/x\\ v={1\over 2}x^2} \Rightarrow \int_1^2 x\ln x\,dx = \left. \left[ {1\over 2}x^2\ln x \right] \right|_1^2 -{1\over 2}\int_1^2 x\,dx = \left. \left[ {1\over 2}x^2\ln x -{1\over 4}x^2\right] \right|_1^2 \\= \left( 2\ln 2-1 \right)+{1\over 4}=2\ln 2-{3\over 4}，故選\bbox[red, 2pt]{(B)}$$
解答:$$(A)\bigcirc: T(v)=\lambda v \Rightarrow T^2(v)=\lambda T(v) \Rightarrow T(v)=\lambda T(v) \Rightarrow \lambda=1,0 \\(B)\times: 滿足T^2=T的矩陣可對角化 \Rightarrow trace(T)=rank(T) \\ (C)\bigcirc: rank(T)+null(T)=3 \Rightarrow trace(T)+null(T)=3\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$假設定價增加了 x 個 10 元 \Rightarrow 新定價=200+10x \Rightarrow 新的銷售量為(5000-200x) 包 \\ \Rightarrow 營業額f(x)= (200+10x)(5000-200x) =-2000x^2+10000x+ 1000000 \\ \Rightarrow 當x={10000\over 4000} =2.5時, f(x)有最大值，此時定價為200+10\cdot 2.5=225，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)= |x-1|+ |x-2|+ |x-3| \Rightarrow \cases{x\ge 3 \Rightarrow f(x)=3x-6=2.5 \Rightarrow x=17/6 \not \ge 3\\ 2\le x\lt 3 \Rightarrow f(x)=x=2.5\\ 1\le x\lt 2 \Rightarrow f(x)=-x+4=2.5 \Rightarrow x=1.5\\ x\lt 1 \Rightarrow f(x)=-3x+6=2.5 \Rightarrow x=7/6 \not \lt 1} \\ \Rightarrow x=2.5,1.5 \Rightarrow 2.5+1.5=4，故選\bbox[red, 2pt]{(A)}$$
解答:$$7^4=2401 \Rightarrow 7^{9999}=7^3\times (7^4)^{2499} =343\times (2400+1)^{2499} \\= 343 \times \left[ 1+ {2499\choose 1}2400+ {2499\choose 2}2400^2+ \cdots \right] \equiv 343 \times \left[ 1+2499\times 2400 \right] \text{ mod 1000} \\ \equiv 343\times 601 \text{ mod 1000} \equiv 206143 \text{ mod 1000}=143，故選\bbox[red, 2pt]{(B)}$$
解答:$$邊長為 a 的正 n 邊形面積公式: S_n={na^2\over 4}\cot {\pi\over n}\\ a=1代入上式\Rightarrow S_n={n\over 4}\cot {\pi\over n} \Rightarrow \cases{S_3=\sqrt 3/4 \approx 0.433\\ S_4=1\\ S_6= 3\sqrt 3/2 \approx 2.6\\ S_8=2+2\sqrt 2 \approx 4.8} \Rightarrow \cases{S_4S_6=2.6\\ S_3S_8= 2.08} \Rightarrow S_4S_6\gt S_3S_8\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$正方形對角線\overline{AC} =\sqrt{4+36}=2\sqrt{10} \Rightarrow 正方形邊長={2\sqrt{10} \over \sqrt 2}=2\sqrt 5 \\ \Rightarrow \cases{\overline{AB}^2=(x-1)^2+(y-1)^2=(2\sqrt 5)^2 \\ \overline{BC}^2 = (x-3)^2+(y-7)^2=(2\sqrt 5)^2} \Rightarrow x+3y=14 \Rightarrow x=14-3y \\\Rightarrow (13-3y)^2+(y-1)^2=20 \Rightarrow y^2-8y+15 =(y-5)(y-3)=0 \\\Rightarrow \cases{y=5\Rightarrow B=(-1,5) 不在第一象限\\ y=3 \Rightarrow B=(5,3)} \Rightarrow B=(5,3) \Rightarrow x-2y=5-6=-1，故選\bbox[red, 2pt]{(B)}$$
解答:$$P\in L:6-x={y+18\over 5}=z \Rightarrow P(-t+6,5t-18,t) 代入S \Rightarrow (-t+6)^2+(5t-18)^2=2t^2 \\ \Rightarrow t^2-8t+15=0 \Rightarrow (t-5)(t-3)=0\Rightarrow \cases{t=5 \Rightarrow A(1,7,5)\\ t=3 \Rightarrow B(3,-3,3)} \\ \Rightarrow \overline{AB}= \sqrt{4+100+4} =\sqrt{108} =6\sqrt 3，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{F_1 = 1 \equiv 1 \pmod 7\\ F_2 = 1 \equiv 1 \pmod 7\\F_3 = 2 \equiv 2 \pmod 7 \\F_4 = 3 \equiv 3 \pmod 7 \\F_5 = 5 \equiv 5 \pmod 7 \\F_6 = 8 \equiv 1 \pmod 7\\F_7 = 13 \equiv 6 \pmod 7\\F_8 = 21 \equiv 0 \pmod 7} \Rightarrow F_8是7的倍數\\ 2880=360\times 8 \Rightarrow 2880是8的倍數 \Rightarrow F_{2880} 是7的倍數，故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)={1\over x^3+x} \Rightarrow f(-x)=-{1\over x^3+x} =-f(x) \ne f(x) \Rightarrow 與y軸不對稱，故選\bbox[red, 2pt]{(C)}$$
解答:$$x^2-3xy+2y^2=3 \Rightarrow 2x-3y-3xy'+4yy'=0 \Rightarrow y'={3y-2x\over 4y-3x} \Rightarrow y'(1,2)={6-2\over 8-3}={4\over 5}\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$將 4 個島嶼排列，共有 4! = 24 種順序。\\這 4 個島嶼之間有 3 個空隙。我們只需在這 3 個空隙中任選 1 個地方切一刀，刀的左邊是第一天的行程，\\刀的右邊是第二天的行程。共有 3 種切法。因此有3\times 24=72種安排方式，故選\bbox[red, 2pt]{(C)}$$
解答:$$\det(A) =-(1+\sqrt 3i)^2-(1-\sqrt 3i)^2=4 \Rightarrow \det(A^5) =4^5=1024，故選\bbox[red, 2pt]{(C)}$$
解答:$$(x-5)^2+5=6\sqrt{(x-5)^2} \Rightarrow \cases{(x-5)^2+5=6(x-5) \Rightarrow (x-5)^2-6(x-5)+5=0\\ (x-5)^2+5=-6(x-5) \Rightarrow (x-5)^2+6(x-5)+5=0} \\ \Rightarrow \cases{(x-5-5)(x-5-1)=(x-10)(x-6) =0\\ (x-5+5)(x-5+1)=x(x-4) =0} \Rightarrow x=0,4, 6,10 \Rightarrow 0+4+6+10=20\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$99=3^2\times 11 \Rightarrow 與99互質必須不是3的倍數,也不是11的倍數 \\ \Rightarrow \cases{ \lfloor {1000\over 3} \rfloor =333 \\ \lfloor {1000\over 11} \rfloor =90 \\ \lfloor {1000\over 33} \rfloor =30 } \Rightarrow (3 的倍數) + (11 的倍數) - (33 的倍數) =333+90-30=393 \\ \Rightarrow 1 到 1000 之間，有 393 個數字和 99 不互質\Rightarrow 與 99 互質的數字個數:1000-393=607，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)={1\over 1-x}=1+x+x^2+\cdots= \sum_{k=0}^\infty x^k \Rightarrow f'(x)= {1\over (1-x)^2} = \sum_{k=0}^\infty kx^{k-1} \\ \Rightarrow g(x)=xf'(x)={x\over (1-x)^2} = \sum_{k=0}^\infty kx^k \Rightarrow {1\over 115}+ 2\left( {1\over 115} \right)^2+ 3 \left( {1\over 115} \right)^3+\cdots = g({1\over 115}) \\={1/115\over (114/115)^2} = {115\over 114^2}，故選\bbox[red, 2pt]{(A)}$$
解答:$$\tan \theta={1\over 2} \Rightarrow 三邊的斜率乘積=\tan \theta\cdot \tan(\theta+60^\circ)\cdot \tan(\theta- 60^\circ) = \frac{\tan^3\theta - 3\tan\theta}{1 - 3\tan^2\theta} \\={1/8-3/2\over 1-3\cdot 1/4}= -{11\over 2}，故\bbox[red, 2pt]{(無解)}, 公布的答案是\bbox[cyan,2pt]{(B)}$$
解題僅供參考，其他教甄試題及詳解
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