115學年度中區縣市政府教師甄選策略聯盟
選擇題【共50題，每題2分，共100分】
解答:$$x-y=1 \Rightarrow y=x-1代入x+y+z=3 \Rightarrow z=2x-4 \\\Rightarrow x^2+y^2+z^2=x^2+(x-1)^2+(2x-4)^2 =6(x-{3\over 2})^2+{7\over 2} \Rightarrow 最小值{7\over 2}，故選\bbox[red, 2pt]{(B)}$$
解答:$$u=x^2+2x+6 \Rightarrow u=5\sqrt {u+6} \Rightarrow u^2-25u-150=0有二實根u_1,u_1 \\ \Rightarrow \cases{x^2+2x+6=u_1\\ x^2+2x+6=u_2} \Rightarrow x^2+2x+k=0 \Rightarrow 兩根之和=-2 (與k無關)，故選\bbox[red, 2pt]{(D)}$$
解答:$$(A)\times: 沒請假不代表下午有研討會\\(B)\times: 有開會不能請假，沒說不開會可不可以請假\\ (C)\bigcirc: 有人請假代表沒有開會\\ (D)\times:有開會不能請假，沒說不開會可不可以請假 \\，故選\bbox[red, 2pt]{(C)}$$
解答:$$甲說實話\Rightarrow \cases{乙說謊話 \\丙說謊話:乙在說謊} \Rightarrow 乙丙矛盾\\ 乙說實話 \Rightarrow \cases{甲說謊話:是乙打破\Rightarrow 不是乙\\ 乙說實話:是丙打破\\丙說謊話:乙在說謊\Rightarrow 乙說實話} \Rightarrow 丙打破窗戶\\ 丙說實話\Rightarrow \cases{甲說謊話:是乙打破的 \Rightarrow 不是乙\\ 乙說謊話:是丙打破的\Rightarrow 不是丙 \\丙說實話:乙在說謊 \Rightarrow 乙說謊}\Rightarrow 甲打破窗戶\\ \\，故選\bbox[red, 2pt]{(D)}，但公布的答案是\bbox[cyan,2pt]{(A)}$$
解答:$$\cases{x\ge 3 \Rightarrow x+5+x-3\le 12 \Rightarrow 2x\le 10\Rightarrow x\le 5 \Rightarrow x=3,4,5\\ -5\le x\le 3 \Rightarrow x+5+3-x\le 12 \Rightarrow 4\ge 0 \Rightarrow x=-5,-4,\dots,3\\ x\le -5\Rightarrow -x-5-x+3\le 12 \Rightarrow x\ge -7 \Rightarrow x=-7,-6,-5} \\ \Rightarrow x=-7,-6,\dots,5，共13個整數，故選\bbox[red, 2pt]{(C)}$$
解答:$$算幾不等式: {x+x+ y+y+ y\over 5}\ge \sqrt[5]{x^2y^3} \Rightarrow {12\over 5}\ge \sqrt[5]{x^2y^3} \Rightarrow x^2y^3 \le \left( {12\over 5} \right)^5，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{\log_2 (4-x^2) \Rightarrow 4-x^2\gt 0 \Rightarrow -2\lt x\lt 2\\ \log_2 x \Rightarrow x\gt 0} \Rightarrow 0\lt x\lt 2 \\ \log_2(4-x^2)-\log_2 x=\log_2 {4-x^2\over x}=3 \Rightarrow {4-x^2\over x}=8 \Rightarrow x^2+8x-4=0 \Rightarrow x={-8\pm \sqrt{80}\over 2} \\ \Rightarrow \cases{x=-4+2\sqrt 5 \approx 0.5\\ x=-4-2\sqrt 5\lt 0 不合}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{(1+x^2)^2 =1+2x^2+x^4 \\ (1+x)^{10} = \sum_{k=0}^{10} {10\choose k}x^k} \Rightarrow (1+x^2)^2 (1+x)^{10} 展開式中x^4係數= {10\choose 0}+ 2{10\choose 2}+ {10\choose 4} \\=1+ 90+210 =301，故選\bbox[red, 2pt]{(C)}$$
解答:$$每本書有三種分法，共有3^5=243種，需扣除有人沒拿到書的情形\\1人沒拿到書:將5本書分給二個人,有2^5=32，共有C^3_1\times 32=96 種分法\\ 2人沒拿到書:將5本書分給一個人,只有一種分法，共有C^3_2\times 1=3種分法\\ 3人沒拿到書:不可能發生\\ 符合要求的分法:243-96+3=150種，故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x) =\int_{-x}^x \sin s\,ds = \left. \left[ -\cos s \right] \right|_{-x}^x=-\cos x-(-\cos(-x)) =0 \Rightarrow f(x)=0 \Rightarrow f'(1)=0\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{P\in L_1 \\Q\in L_2} \Rightarrow \cases{P=(s+3,2s+4,as-2) \\Q=(-t,t+1,2t+2)} \Rightarrow P=Q \Rightarrow \cases{s+3=-t\\ 2s+4=t+1\\ as-2=2t+2} \Rightarrow \cases{s=-2\\ t=-1} \\ \Rightarrow -2a-2=-2+2 \Rightarrow a=-1，故選\bbox[red, 2pt]{(A)}$$
解答:$$最後一球是白球的機率:{白球數\over 全部}={4\over 7}，故選\bbox[red, 2pt]{(D)}$$
解答:$$假設\cases{a=\overline{BC} \\b= \overline{CA}\\ c=\overline{AB}} \Rightarrow 正弦定理:a:b= \sin A:\sin B=3:5 \Rightarrow \cases{a=3k\\ b=5k} \Rightarrow c=30-a-b=30-8k \\ s={1\over 2}(a+b+c)=15 \Rightarrow \triangle ABC面積=\triangle = rs={abc\over 4R} \Rightarrow {R\over r}={15abc\over 4\triangle^2} ={14\over 3} \\ \Rightarrow \triangle ^2={45\over 56}abc \Rightarrow s(s-a)(s-b)(s-c)=15(15-a)(15-b)(15-c)={45\over 56}abc \\ \Rightarrow 15(15-3k)(15-5k)(15-(30-8k)) ={45\over 56}(3k)(5k)(30-8k) \Rightarrow k=2 \\ \Rightarrow \triangle =\sqrt{15(15-6)(15-10)(15-14)} =15\sqrt 3，故選\bbox[red, 2pt]{(A)}$$
解答:$$\sin^2 x-4\sin x+5=(\sin x-2)^2+1 \Rightarrow 當\sin x=1時，最小值=2，故選\bbox[red, 2pt]{(B)}$$
解答:$$3x^2+2y^2+xy=8 \Rightarrow 6x+4yy'+y+xy'=0 \Rightarrow y'=-{6x+y\over 4y+x} \\ \Rightarrow y'(0,2)=-{2\over 8} =-{1\over 4}，故選\bbox[red, 2pt]{(A)}$$
解答:$$群的運算不一定滿足交換率，故選\bbox[red, 2pt]{(B)}$$
解答:$$(1+i)^2=2i \Rightarrow \sum_{k=0}^\infty(1+i)^{2k} z^k= \sum_{k=0}^\infty (2iz)^k \Rightarrow |2iz|\lt 1 \Rightarrow |z|\lt {1\over |2i|} ={1\over 2} \Rightarrow R={1\over 2}，故選\bbox[red, 2pt]{(B)}$$
解答:$$y'={dy\over dx}=2xy^2 \Rightarrow {1\over y^2}dy=2x \,dx \Rightarrow \int {1\over y^2}\,dy= \int {2x\,dx \Rightarrow -{1\over y}}=x^2+C \\ \Rightarrow y=-{1\over x^2+C} \Rightarrow y(1)=-{1\over 1+C}=1 \Rightarrow C=-2 \Rightarrow y=-{1\over x^2-2} ={1\over 2-x^2}，故選\bbox[red, 2pt]{(A)}$$
解答:$$|z_1+z_2|^2 = |z_1|^2+ |z_2|^2 +2 |z_1||z_2| \cos \theta =2^2+3^2 +2 \cdot 2\cdot 3 \cos 120^\circ= 7 \Rightarrow |z_1+z_2|= \sqrt 7\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$a_{n+1}={1\over 2}a_n+3 \Rightarrow r={1\over 2}r+3 \Rightarrow r=6 \Rightarrow a_{n+1}-6= {1\over 2}a_n+3-6 ={1\over 2}(a_n-6)\\ 取b_n= a_n-6, 則b_{n+1} ={1\over 2}b_n \Rightarrow \langle b_n \rangle為等比數列,公比r={1\over 2} \Rightarrow b_1=a_1-6=-5 \\ \Rightarrow |b_n|= |a_n-6|= \left|-5\times ({1\over 2})^{n-1} \right| =5\times {1\over 2^{n-1}} \lt 10^{-3} \Rightarrow 10\times {1\over 2^n} \lt 10^{-3} \\ \Rightarrow 1-n\log 2 \lt -3 \Rightarrow n\gt {4\over \log 2}={4\over 0.301} \approx 13.3 \Rightarrow n=14，故選\bbox[red, 2pt]{(B)}$$
解答:$$a_n= \sin{n\pi\over 3} \cos {n\pi\over 4} \Rightarrow a_{n+12}=\sin{(n+12)\pi\over 3} \cos{(n+12)\pi\over 4} =\sin \left( {n \pi\over 3}+4\pi \right) \cos \left( {n \pi\over 4}+3\pi \right) \\= -\sin {n \pi\over 3}\cos {n \pi\over 4} =-a_n \Rightarrow a_{n+12}=-a_n \Rightarrow \sum_{n=1}^{48}a_{n} =0 \Rightarrow \sum_{n=1}^{50}a_{n} =a_{49}+a_{50} \\=a_1+a_2= \sin{\pi\over 3}\cos {\pi\over 4}+\sin{2\pi\over 3}\cos {\pi\over 2}= {\sqrt 6\over 4}+0={\sqrt 6\over 4}，故選\bbox[red, 2pt]{(B)}$$
解答:$$3+2+a+1+8=14+a為9的倍數\Rightarrow a=4，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{2^1= 2\\ 2^2=4\\ 2^3 \equiv 3 \text{ (mod 5)} \\2^4 \equiv 1 \text{ (mod 5)} \\} \Rightarrow 循環數4 \\ 3 \equiv -1 \text{ (mod 4)} \Rightarrow 3^{100} \equiv (-1)^{100} \equiv 1\text{ (mod 4)} \Rightarrow 2^{3^{100}} =(2^4)^k\cdot 2\equiv 2 \text{ (mod ) 5}，故選\bbox[red, 2pt]{(B)}$$
解答:$$已知\cases{ n \equiv 2\text{ (mod 5)} \cdots(1) \\ n \equiv 3 \text{ (mod 7)} \cdots(2)}, 式(2) \Rightarrow n=7k+3 代入(1) \Rightarrow 7k+3 \equiv 2 \text{ (mod 5)} \\ \Rightarrow 2k+3 \equiv 2 \text{ (mod 5)} \Rightarrow 2k \equiv -1 \text{ (mod 5)}\Rightarrow 2k \equiv 4 \text{ (mod 5)} \Rightarrow k=5m+2 \\ \Rightarrow n=7(5m+2)+3=35m+17，故選\bbox[red, 2pt]{(C)}$$
解答:$$假設k={3n+2\over 2n-1} \in \mathbb Z \Rightarrow 2k=3+{{7\over 2n-1}} \Rightarrow {7\over 2n-1} \in \mathbb Z \\ \Rightarrow \cases{2n-1=1 \Rightarrow n=1\\ 2n-1=7 \Rightarrow n=4\\ 2n-1=-1 \Rightarrow n=0\\ 2n-1=-7 \Rightarrow n=-3} \Rightarrow 1+4+0-3=2，故選\bbox[red, 2pt]{(D)}$$
解答:$$2026=2\times 1013 \Rightarrow 尤拉函數 \phi(2026)=20206\times \left( 1-{1\over 2} \right) \times \left( 1-{1\over 1013} \right) =1012，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{a=bq_1+r_1 \Rightarrow gcd(a,b)= gcd(b,r_1) \\ b=r_1q_2+r_2 \Rightarrow gcd(b,r_1) =gcd(r_1,r_2) \\ r_1=r_2q_3 \Rightarrow gcd(r_2,r_1) =r_2} \Rightarrow gcd(a,b)=gcd(b,r_1) = gcd(r_1,r_2)=r_2，故選\bbox[red, 2pt]{(B)}$$
解答:$$(A)\times: \left\lfloor{100\over 5} \right \rfloor+ \left\lfloor {100\over 25}\right \rfloor=20+4=24 \Rightarrow 24個0,不是25個0\\ (B)\times: \left\lfloor{100\over 3} \right \rfloor +\left\lfloor {100\over 9}\right \rfloor+ \left\lfloor {100\over 27}\right \rfloor+\left\lfloor {100\over 81}\right \rfloor=48 \ne 33 \\(C)\bigcirc: 對於任何小於或等於 100 的質數 p,p一定是100!的因數 \Rightarrow 100! \equiv 0 \text{ (mod p)}\\\qquad \Rightarrow 100! +1\equiv 1 \text{ (mod p)} \Rightarrow (100! + 1) 無法被任何小於或等於 100 的質數整除\\ \qquad \Rightarrow 如果它有質因數，那個質因數必定大於 100 \\(D)\times: 97是質數, 1至100的整數中，97的倍數只有一個，因此100!不是完全平方數\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$A= \begin{bmatrix}2&1\\4& 3 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2-5\lambda+2 \Rightarrow A^2-5A+2I=0 \Rightarrow A^4-5A^3+2A^2=0 \\ \Rightarrow A^4-5A^4+3A^2+7A =A^2+7A= \begin{bmatrix}8&5\\20&13 \end{bmatrix}+ \begin{bmatrix}14& 7\\28& 21 \end{bmatrix}= \begin{bmatrix}22& 12\\48& 34 \end{bmatrix}，故選\bbox[red, 2pt]{(D)}$$
解答:$$A\in W \Rightarrow A= [a_{ij}]_{n\times n} 滿足a_{ij}=a_{ji} \Rightarrow A 由1+2+3+ \cdots+n={n(n+1)\over 2}個元素組成 \\，故選\bbox[red, 2pt]{(C)}$$
解答:$$(A)\bigcirc: tr(BA)=tr(AB)=1+1+0=2\\ (B)\times: \det(AB)=0 \ne 1 \\(C)\times :A不是方陣 \\(D)\times: tr(BA)=2 \ne 0 \Rightarrow BA不是零矩陣\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$u=\sin x\Rightarrow du=\cos x\,dx \Rightarrow \int_0^{\pi/2} \sin^3x \cos x\,dx =\int_0^1 u^3\,du = \left. \left[ {1\over 4}u^4 \right] \right|_0^1={1\over 4}，故選\bbox[red, 2pt]{(C)}$$
解答:$$x=e^t \Rightarrow t=\ln x \Rightarrow y=t^2=(\ln x)^2 \Rightarrow {dy\over dx}=y'={2 \ln x\over x} \Rightarrow {d^2y\over dx^2}=y''={2\over x^2}-{2\ln x\over x^2} \\={2\over e^{2t}}-{2t\over e^{2t}} \Rightarrow y''(t=2)={2\over e^4}-{4\over e^4}=-2e^{-4}，故選\bbox[red, 2pt]{(C)}$$
解答:$$V=\int_0^4 (\sqrt x)^2 \pi\,dx =\pi \int_0^4 x\,dx =\pi \left. \left[ {1\over 2}x^2 \right] \right|_0^4=8\pi，故選\bbox[red, 2pt]{(D)}$$
解答:$$\lim_{n\to \infty} \sum_{k=1}^n {2(n+2k)^3\over n^4} =\lim_{n\to \infty} \sum_{k=1}^n {2(1+2(k/n))^3\over n} = \int_0^1 2(1+2x)^3\,dx = \left. \left[{1\over 4} (1+2x)^4 \right] \right|_0^1 \\={1\over 4}(81-1) =20，故選\bbox[red, 2pt]{(B)}$$
解答:$$f(x) =\int_0^{\sin x} e^{-t^2+1}\,dt \Rightarrow f'(x)=e^{-\sin^2x+1}\cdot\cos x \Rightarrow f'(0)=e，故選\bbox[red, 2pt]{(A)}$$
解答:$$A= \begin{bmatrix}2& 1\\0& 2 \end{bmatrix} \Rightarrow A^2 = \begin{bmatrix}4& 4\\0& 4 \end{bmatrix} \Rightarrow A^4= \begin{bmatrix}16&32\\0&16 \end{bmatrix} \Rightarrow A^{8} = \begin{bmatrix}256&1024\\0& 256 \end{bmatrix} \\ \Rightarrow A^{10}=A^8\cdot A^2 =\begin{bmatrix}256&1024\\0& 256 \end{bmatrix} \begin{bmatrix}4& 4\\0& 4 \end{bmatrix} = \begin{bmatrix} 1024 &5120 \\0& 1024\end{bmatrix} = \begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22} \end{bmatrix} \Rightarrow a_{12}=5120\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{x+y+z=1\\ 2x+2y+3z=2 \\2x+3y+kz=r} \Rightarrow A= \left[ \begin{array}{rrr|r}1& 1& 1& 1\\1& 2& 3& 2\\2& 3& k& r \end{array} \right] \Rightarrow rref(A)= \left[ \begin{array}{rrr|r}1& 1& 1& 1\\0& 1& 2& 1\\0& 0& k-4& r-3 \end{array} \right] \\ (A) \times:\cases{k=4\\r=3} \Rightarrow \left[ \begin{array}{rrr|r}1& 1& 1& 1\\0& 1& 2& 1\\0& 0& 0& 0 \end{array} \right] \Rightarrow 無限多解 \\(B) \times: \cases{k=4\\r=5} \Rightarrow \left[ \begin{array}{rrr|r}1& 1& 1& 1\\0& 1& 2& 1\\0& 0& 0& 2 \end{array} \right] \Rightarrow 無解 \\(C)\bigcirc: \cases{k=4\\r=3} \Rightarrow無限多解代表交於一直 \\(D)\times: \cases{k=4\\r=5} \Rightarrow 無平行平面\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$(A)\times: A \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} y\\x\end{pmatrix} \Rightarrow 對稱直線:y=x \\(B)\bigcirc: A= \begin{pmatrix} 3/5& 4/5\\-4/5& 3/5\end{pmatrix} =\begin{pmatrix} \cos \theta& -\sin \theta\\ \sin\theta& \cos \theta\end{pmatrix} 為一旋轉矩陣\\ (C)\times: A \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix}x\\0\end{pmatrix} \Rightarrow 投影到x軸,不是y軸\\(D) \times: A \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} x\\ -y\end{pmatrix} \Rightarrow 對稱x軸,不是y軸\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$取u=x-3 \Rightarrow (x-5)^2+(x-1)^4 =(u-2)^4+(u+2)^4 =2u^4+48u^2+32=196\\ \Rightarrow u^4+24u^2-82=0 \Rightarrow u^2= -12+\sqrt{226} \Rightarrow \cases{u_1= \sqrt{\sqrt{226}-12} \\u_2=-\sqrt{\sqrt{226}-12}}\\\Rightarrow u_1+u_2 =0 \Rightarrow x的兩根和=u_1+3+u_2+3=6，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{x(t)=t^2\\ y(t)= 2t^3/3} \Rightarrow \cases{x'(t)=2t\\y'(t)=2t^2} \Rightarrow 弧長L=\int_0^1 \sqrt{(2t)^2+(2t^2)^2}\,dt = \int_0^1 2t\sqrt{1+t^2}\,dt\\ 取u=1+t^2 \Rightarrow du=2tdt \Rightarrow L= \int_1^2\sqrt u\,du = \left. \left[{2\over 3} u^{3/2} \right] \right|_1^2= {2\over 3}(2\sqrt 2-1)，故選\bbox[red, 2pt]{(B)}$$
解答:$$\text{費馬小定理 (Fermat's Little Theorem)}:若 p 為質數，且整數 a 不是 p 的倍數，\\則 a^{p-1} \equiv 1 \pmod p \Rightarrow 2^{101-1} \equiv 1\pmod {101} \Rightarrow 2^{100} \equiv 1\pmod {101} \\ \Rightarrow 2^{103} \equiv 8\pmod {101}，故選\bbox[red, 2pt]{(C)}$$
解答:$$u=x^2 \Rightarrow du=2x\,dx \Rightarrow \int_0^\infty {x\over 1+x^4}\,dx =\int_0^\infty{1/2\over 1+u^2}\,du ={1\over 2} \left. \left[ \tan^{-1}u \right] \right|_0^\infty ={1\over 2} \cdot {\pi\over 2}={\pi\over 4}\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$f(x,y,z)=x^2y+yz \Rightarrow \nabla f=(f_x,f_y,f_z)=(2xy,x^2+z,y) \Rightarrow \nabla f(1,-1,2)=(-2,3,-1) \\ \Rightarrow \nabla f(1,-1,2)\cdot ({2\over 3}, {1\over 3}, {2\over 3}) ={1\over 3}(-4+3-2) =-1，故選\bbox[red, 2pt]{(B)}$$
解答:$$取y=kx^2 \Rightarrow \lim_{(x,y)\to (0,0)}f(x,y) =\lim_{x\to 0} {kx^4\over (k^2+1)x^4}={k\over k^2+1} 非定數 \Rightarrow f在(0,0)不連續 \Rightarrow (B),(C)皆錯誤\\ \cases{f_x(0,0)=\lim_{h\to 0} {f(h,0)-f(0,0)\over h} =\lim_{h\to 0} {0-0\over h}=0 \\ f_y(0,0) =\lim_{h\to 0}{f(0,h)-f(0,0)\over h}=\lim_{h\to 0}{0-0\over h}=0} \Rightarrow f_x(0,0),f_y(0,0)皆存在 \Rightarrow (D)錯誤\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$T^2=3T \Rightarrow T^2-3T=0 \Rightarrow 特徵值\lambda=0或3 \Rightarrow 假設\cases{\lambda=0的重數a\\ \lambda=3的重數b}\\\cases{dim(V)=5 \Rightarrow a+b=5\\ tr(T)=6\Rightarrow 0\cdot a+3\cdot b=6} \Rightarrow \cases{a=3\\ b=2} \Rightarrow \text{dim ker}(T)=a=3，故選\bbox[red, 2pt]{(C)}$$
解答:$$u=x^3+x^2+1 \Rightarrow du=(3x^2+2x)\,dx \Rightarrow \int_0^1 {3x^2+2x\over x^3+x^2+1}\,dx = \int_1^3 {1\over u}du = \left. \left[ \ln u \right] \right|_1^3 =\ln 3\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$改變積分順序\Rightarrow \int_0^1 \int_y^1 e^{x^2}\,dx\,dy = \int_0^1 \int_0^x e^{x^2}\,dy\,dx = \int_0^1 xe^{x^2}\,dx = \left. \left[ {1\over 2}e^{x^2} \right] \right|_0^1 \\={1\over 2}(e-1)，故選\bbox[red, 2pt]{(C)}$$
解題僅供參考，其他教甄試題及詳解
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