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2022年6月15日 星期三

108年新竹市國中教甄聯招-數學詳解

新竹市108學年度國民中學數校聯合教師甄選

解答:$$a \star b=a^2+b^3+1 \Rightarrow \cases{2\star 1=2^2+1^3+1 =6\\ 3\star (-2)=3^2+(-2)^3+1 =2} \\\Rightarrow (2\star 1)\star (3\star (-2)) =6\star 2 = 6^2+2^3+1 =45,故選\bbox[red, 2pt]{(D)}$$
解答:$${\sin 10^\circ + \sin 40^\circ \over \cos 10^\circ +\cos 40^\circ} ={2\sin 25^\circ \cos 15^\circ \over 2\cos 25^\circ \cos 15^\circ} ={\sin 25^\circ \over \cos 25^\circ} =\tan 25^\circ,故選\bbox[red, 2pt]{(B)}$$
解答:$$4^n = (3+1)^n = 1+ 3C^n_1 + 3^2C^n_2  + 3^3C^n_3 +\cdots + 3^nC^n_n\\ \Rightarrow {4^n\over 3}={1\over 3}+ C^n_1 +3C^n_2 +3^2C^n_3 +\cdots +3^{n-1}C^n_n \\ \Rightarrow C^n_1 +3C^n_2 +3^2C^n_3 +\cdots +3^{n-1}C^n_n={4^n\over 3}-{1\over 3} ={4^n-1\over 3},故選\bbox[red, 2pt]{(A)}$$
解答:$$a_1=2 \Rightarrow a_2= {2-1\over 2+1}= {1\over 3} \Rightarrow a_3= -{1\over 2} \Rightarrow a_4= -3 \Rightarrow a_5=2 \Rightarrow 循環數=4\\ 2019= 4\times 504+3 \Rightarrow a_{2019}=a_3= -{1\over 2},故選\bbox[red, 2pt]{(B)}$$
解答:$$y=8^n x^2-2^n(2^n+1)x+1 = (2^{2n}x-1)(2^nx-1) \Rightarrow L_n =\overline{A_nB_n}={1\over 2^n}-{1\over 2^{2n}} \\ \Rightarrow \sum_{n=1}^\infty L_n=\sum_{n=1}^\infty \left( {1\over 2^n}-{1\over 2^{2n}} \right) ={1/2\over 1-1/2}-{1/4\over 1-1/4} =1-{1\over 3}={2\over 3},故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{ 3的倍數有\lfloor{100\over 3}\rfloor= 33個\\5的倍數有\lfloor{100\over 5}\rfloor= 20個   \\ 15的倍數有\lfloor{100\over 15}\rfloor= 6個 } \Rightarrow 與15不互質的數有33+20-6=47 \\\Rightarrow 與15互質的數有100-47=53個,因此機率為{53\over 100},故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{\angle B=90^\circ\\ \overline{AB}=5\\ \overline{BC}=12} \Rightarrow \overline{AC}=\sqrt{5^2+12^2} =13 ,又\cases{\overline{AD}=5 \\\overline{CE}=12}\\\Rightarrow \overline{DE}= \overline{AD}-\overline{AE} =5-(13-12)=4 ,故選\bbox[red, 2pt]{(A)}$$
解答:$$(\sqrt 2+\sqrt 3+\sqrt 5)(\sqrt 2+\sqrt 3-\sqrt 5)(\sqrt 2-\sqrt 3+\sqrt 5)(-\sqrt 2+\sqrt 3+\sqrt 5)\\ =((\sqrt 2+\sqrt 3)^2-(\sqrt 5)^2)((\sqrt 5)^2-(\sqrt 2-\sqrt 3)^2) =2\sqrt 6\cdot 2\sqrt 6=24,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{z_1= \cos x+ i\sin x =e^{ix} \\ z_2=\cos y +i\sin y= e^{iy}} \Rightarrow z_1+z_2= {\sqrt 6\over 2}+i{\sqrt 2\over 2} \Rightarrow (z_1+ z_2)^2 =|z_1|^2 +|z_2|^2 +2z_1z_2 \\= 2+2z_1z_2 =2+2e^{i(x+y)}= 2+2\cos(x+y)+2\sin(x+y)i;\\ 同時 (z_1+ z_2)^2=({\sqrt 6\over 2}+i{\sqrt 2\over 2})^2 = 1 + \sqrt 3i\\ 兩者虛部相同,即2\sin(x+y)=\sqrt 3 \Rightarrow \sin(x+y)={\sqrt 3\over 2},故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{n+292=a^2\\ n+200=b^2} \Rightarrow a^2-b^2 = 92 = 2\times 46 = 4\times 23\\ 由於a,b\in \mathbb{Z},因此a^2-b^2=2\times 46 \Rightarrow \cases{a=(46+2)\div 2= 24\\ b=(46-2)\div 2=22} \Rightarrow n=22^2-200= 284\\,故選\bbox[red, 2pt]{(A)}$$
解答:$$f(x)={1\over 1-x} =1+x+x^2 +\cdots+ x^n+ \cdots \Rightarrow f'(x)={1\over (1-x)^2}=1+ 2x+ 3x^2 +\cdots +nx^{n-1}+ \cdots\\ \Rightarrow f'({1\over 2019})={1\over (2018/2019)^2}= 1+2\times({1\over 2019})+3 \times({1\over 2019})^2+\cdots \\ \Rightarrow 1+2\times({1\over 2019})+3 \times({1\over 2019})^2+\cdots = \left({2019\over 2018}\right)^2,故選\bbox[red, 2pt]{(B)}$$
解答:$$\tan 1^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 89^\circ \\=\tan 1^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 44^\circ \tan 45^\circ \tan (90^\circ-44^\circ) \tan (90^\circ-43^\circ)\cdots  \tan (90^\circ-1^\circ) \\  =(\tan 1^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 44^\circ) \tan 45^\circ (\cot  44^\circ \cot  43^\circ \cdots  \cot 1^\circ) \\ =\tan 45^\circ =1\\ 因此\log_{10}(\tan 1^\circ) +\log_{10}(\tan 2^\circ) + \cdots +\log_{10}(\tan 89^\circ) \\ =\log_{10}(\tan 1^\circ\tan 2^\circ \tan 3^\circ \cdots \tan 89^\circ) =\log_{10} 1= 0,故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{正三角形每一內角為60^\circ \\ 正五邊形每一內角為(5-2)\cdot 180\div 5=108^\circ\\ 正八邊形每一內角為(8-2)\cdot 180\div 5=135^\circ} \Rightarrow \cases{\angle ALK=60^\circ \\ \angle HLK = 108^\circ \\ \angle BLH= 135^\circ } \\ \Rightarrow \angle ALB= 360^\circ -60^\circ -108^\circ -135^\circ = 57^\circ,故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{3a^2+ 2019a+7=0 \\ 7b^2+2019b + 3=0} \Rightarrow \cases{a=\cfrac{-2019\pm \sqrt{k}}{6} \\b=\cfrac{-2019\pm \sqrt{k}}{14}},\text{where }k=2019^2-4\cdot 3\cdot 7 \\ \Rightarrow (a,b)=(\cfrac{-2019+ \sqrt{k}}{6},\cfrac{-2019+ \sqrt{k}}{14})  或(\cfrac{-2019- \sqrt{k}}{6},\cfrac{-2019- \sqrt{k}}{14}) \Rightarrow {b\over a} ={6\over 14}={3\over 7} \\,故選\bbox[red, 2pt]{(A)}$$
解答:$$5\lt x+y\le 10 \Rightarrow \cases{x+y=6 \Rightarrow 有H^2_4=C^5_4=5組正整數解\\ x+y=7 \Rightarrow 有H^2_5=C^6_5=6 組正整數解\\ x+y=8 \Rightarrow 有H^2_6=C^7_6=7組正整數解\\ x+y=9 \Rightarrow 有H^2_7=C^8_7=8組正整數解\\ x+y=10 \Rightarrow 有H^2_8=C^9_8=9組正整數解} \\ \Rightarrow 共有5+6+7 +8 +9 = 35組解,故選\bbox[red, 2pt]{(D)}$$
解答:$$甲+乙: 乙+丙: 甲+丙=2:3: 4\Rightarrow \cases{甲+乙= 2k\\ 乙+丙=3k\\ 甲+丙= 4k},k\in \mathbb{R}\\ 三式相加\Rightarrow 2(甲+乙+ 丙)=9k \Rightarrow 2\times 36=9k \Rightarrow k=8 \Rightarrow \cases{甲+乙= 16\\ 乙+丙= 24\\ 甲+丙= 32}\\ \Rightarrow \cases{甲= 36-24=12\\ 乙=36-32=4\\ 丙=36-16=20},故選\bbox[red, 2pt]{(C)}$$
解答:$$\log(3^2\times 4^{16} \times 5^{25})= 2\log 3+ 32\log 2+ 25(1-\log 2) =2\log 3+ 7\log 2+25\\ =2\times 0.4771+ 7\times 0.301 + 25= 28.0612 \Rightarrow 3^2\times 4^{16} \times 5^{25}是28+1=29位數,故選\bbox[red, 2pt]{(C)}$$
解答:$$7,17,37,47,67,97,共6個質數,故選\bbox[red, 2pt]{(B)}$$
解答:$${m+k\over nk} ={m\over n} \Rightarrow m+k=mk \Rightarrow k={m\over m-1}=1+{1\over m-1} \Rightarrow k=m=2\\ 又{1\over 3}\lt {m\over n}\lt 1 \Rightarrow {1\over 3}\lt {2\over n}\lt 1  \Rightarrow n=3,5(4不符,違反m,n互質) \Rightarrow {m\over n} ={2\over 3}或{2\over 5},共2 個\\,故選\bbox[red, 2pt]{(B)}$$
20. 試問方程式\((x^2-x-1)^{x+2}=1\)共有多少個整數解\(x\)? (A) 2  (B) 3  (C) 4   (D) 5
解答:$$(x^2-x-1)^{x+2}=1 \Rightarrow \cases{x+2=0 \Rightarrow x=-2\\ x^2-x-1=1 \Rightarrow x=2,-1\\ x^2-x-1=-1且x+2為偶數\Rightarrow x=0(1不合)} \\ \Rightarrow x=-2,-1,0,2,共四組解,故選\bbox[red, 2pt]{(C)}$$
21. 已知下面兩個四邊形,\(\overline{AB}=\overline{EF}、\overline{BC} = \overline{FG}、\overline{CD} =\overline{GH}、\overline{DA}=\overline{HE}\),請問增加哪一個選項的條件仍然無法讓兩個四邊形全等?(A) \(\angle A=\angle E\) (B) \(\overline{BD}=\overline{FH}\)  (C) 分別在\(\overline{AB}、\overline{AD}、\overline{EF}、\overline{EH}\)上取\(P、Q、R、S\)四個點,使得\(\overline{AP}=\overline{ER}\),\(\overline{AQ}=\overline{ES}\),\(\overline{PQ} = \overline{RS}\)。  (D) 以上都可以讓兩個四邊形全等。
解答:$$四邊形四邊長均相等不代表全等,如菱形與正方形可以有相同邊長但不全等\\需再加上任一頂角或相關限制皆可,故選\bbox[red,2pt]{(D)}$$
解答:$$(A)\times: ABCD為鳶形\Rightarrow \triangle ABD \lt \triangle BCD \Rightarrow E不是重心\\(B)\times: 理由同(A)\\ (C)\times: 鳶形是左右對稱、非上下對稱,H\ne {1\over 2}\overline{AC} \\(D)\bigcirc: 與距離成反比,故選\bbox[red, 2pt]{(D)}$$
解答:$$(A)與(B)在平面才成立,(D)空間的歪斜線不等距,故選\bbox[red, 2pt]{(C)}$$
解答:$$假設\cases{甲罐重量為a,且水酒比為3:1\\ 乙罐重量為b,且水酒比為1:1},混合後水酒比為({3\over 4}a+{1\over 2}b):({1\over 4}a+{1\over 2}b) =k:1 \\ 因此\cases{b=1 \Rightarrow k=3\\ a=0 \Rightarrow k=1}\Rightarrow 1\lt k\lt 3,故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{f(0)=3\Rightarrow c=3\\ f(3)=3 \Rightarrow 9a+ 3b=0\\ f(4)=8 \Rightarrow 16a+4b=5} \Rightarrow \cases{a=5/4\\ b=-15/4\\ c=3} \Rightarrow f(x)= {5\over 4}x^2-{15\over 4}x+3  ={5\over 4}(x-{3\over 2})^2+{3\over 16}\\ \Rightarrow x={3\over 2}時有最小值(不是最大值),故選\bbox[red, 2pt]{(C)}$$
解答:$$假設回程花了a分鐘,則早上去程花了a+20分鐘;\\回程三分之一時間騎車、三分之二時間走路 \Rightarrow 小明家距廟{1\over 3}a\times 240+ {2\over 3}a\times 80 ={400\over 3}a公尺\\去程{400\over 3}a公尺,一半距離走路一半騎車,共花了\cfrac{200a/3}{240}+ \cfrac{200a/3}{80} =a+20 \Rightarrow a=180\\ (A)\times: 小明家與廟的距離:{400\over 3}a={400\over 3}\times 180 = 24000公尺=24公里\ne 36公里\\(B) \times:早上花了a+20=200分鐘,其中一半時間,即100分鐘走路\ne 160\\ (C) \times:下午花了a=180分鐘,其中{1\over 3}a=60分鐘騎車\ne 50\\ (D)\bigcirc:下午走路花了{2\over 3}a = 120分鐘=2小時\\,故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)=\sqrt x-\sqrt{x-2} \Rightarrow f'(x)={\sqrt{x-2}-\sqrt x\over 2\sqrt{x(x-2)}} \lt 0 \Rightarrow f(x)為嚴格遞減\\ \Rightarrow f(13)\gt f(15) \gt f(17)\Rightarrow a\gt b \gt \sqrt{17}-\sqrt{15}\gt \sqrt{17}-4=c \Rightarrow a\gt b\gt c ,故選\bbox[red, 2pt]{(C)}$$
解答:$$保持平衡代表力矩相等,並非重量相等,故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{\angle 1+\angle 2=180^\circ\\ \angle 2+\angle 3=90^\circ \\ \angle 3=25^\circ } \Rightarrow \angle 2=65^\circ \Rightarrow \angle 1=115^\circ,故選\bbox[red, 2pt]{(B)}$$
解答:$$500-2\times 60=380 \Rightarrow 每趟可搬9箱,需搬12趟;前11趟搭電梯11\times 2=22次\\,最後一趟搬到十樓就結束了,共23次,故選\bbox[red, 2pt]{(D)}$$
解答:$$H^3_{10}= C^{12}_{10} =66,故選\bbox[red, 2pt]{(B)}$$
解答:$$(A)\cases{a\circledcirc b=a \\ b\circledcirc a= b} \Rightarrow a\circledcirc b \ne b\circledcirc a \Rightarrow 違反交換律\\(B)\cases{(a\circledcirc b) \circledcirc c=2(a+b) \circledcirc c= 2(2(a+b)+c) = 4a+4b+2c \\ a\circledcirc (b \circledcirc c)= a \circledcirc 2(b+c) = 2(a+2(b+c))= 2a+ 4b+4c} \Rightarrow 違反結合律\\ (C)\cases{a\circledcirc b=2a+b \\ b\circledcirc a= 2b+a} \Rightarrow a\circledcirc b \ne b\circledcirc a \Rightarrow 違反交換律 \\(D) \cases{a\circledcirc b= b\circledcirc a =a+b+1\\ (a\circledcirc b)\circledcirc c= a\circledcirc (b\circledcirc c)= a+b+c+2}\\,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{C1:(x+3)^2 +y^2 = 5^2 \Rightarrow \cases{圓心O_1(-3,0)\\ 半徑r_1=5}\\ C2:(x-3)^2 +(y+8)^2 =5^2 \Rightarrow \cases{圓心O_2(3,-8)\\ 半徑r_2=5}} \Rightarrow \overline{O_1O_2}=10 = r_1+r_2 \Rightarrow 相切,故選\bbox[red, 2pt]{(D)}$$
解答:$$(C)90^\circ \lt \theta\lt 180^\circ \Rightarrow \cos \theta \lt 0,而\sqrt{1-\sin^2\theta} \gt 0,故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)= 4x^3-6x^2+3x-2 \Rightarrow f'(x)=12x^2-12x+3 =3(2x-1)^2 \ge 0 \Rightarrow f(x)遞增\\ 又\cases{f(0)=-2 \lt 0\\ f(1)=-1\lt 0\\ f(2)=12 \gt 0} \Rightarrow  f(x)=0,\exists x\in (1,2),故選\bbox[red, 2pt]{(C)}$$
解答:$$原式=\sum_{m=2}^{100} \cfrac{1}{\sum_{k=1}^m k} =\sum_{m=2}^{100}\cfrac{2}{m(m+1)} =\sum_{m=2}^{100} 2({1\over m}-{1\over m+1}) =2({1\over 2}-{1\over 101})= {99\over 101},故選\bbox[red, 2pt]{(B)}$$
解答:$$(A) {101\over 100}\gt 1 \\(B)\left|-{2\over 3}\right| \lt 1 \\(C)\left| -2\right| \gt 1 \\(D)\left| 3-{1\over 2}\right|\gt 1\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$(A)若a\lt 1,則f(x)為遞減非遞減\\(C)\cases{f(x_1\cdot x_2)= a^{x_1\cdot x_2} \\ f(x_1)\cdot f(x_2)= a^{x_1}\cdot a^{x_2}= a^{x_1+ x_2}} \Rightarrow f(x_1\cdot x_2)\ne f(x_1)\cdot f(x_2) \\(D)f(x)與y軸恰交於一點(0,1)\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$S=2^{50}-1 \Rightarrow \log S\approx \log 2^{50}= 50\log 2 =50\times 0.301 = 15.05 \Rightarrow S為15+1=16位數,故選\bbox[red, 2pt]{(B)}$$
解答:$$x=3+i \Rightarrow (x-3)^2=-1 \Rightarrow x^2-6x=-10 \Rightarrow f(3+i)= (-10)^2-2(-10)-3=117\\,故選\bbox[red, 2pt]{(D)}$$
解答:$$f(x)=ax^5+bx^3+ cx-5 \Rightarrow g(x)=f(x)+5=ax^5+bx^3+ cx \Rightarrow g(3)=f(3)+5=8+5=13\\ g(-3)=-g(3)=f(-3)+5=-13 \Rightarrow f(-3)=-18,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{\sin 70^\circ, \sin 146^\circ, \sin 365^\circ 皆為正值\\ \sin 219^\circ ,\sin 292^\circ 皆為負值} \Rightarrow 中位數為正值中最小的數,即\sin 365^\circ  =\sin 5^\circ,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cos^2 A+ \cos^2 B\gt 1+\cos^2 C \Rightarrow 1-\sin^2 A+ 1-\sin^2 B\gt 1+ 1-\sin^2 C\\ \Rightarrow \sin^2 C\gt \sin^2 A+\sin ^2 B \Rightarrow ({c\over 2R})^2 \gt  ({a\over 2R})^2 + ({b\over 2R})^2 (\because {a\over \sin A}={b\over \sin B}= {c\over \sin C}=2R)\\ \Rightarrow c^2 \gt a^2 +b^2 \Rightarrow \cos C={a^2+b^2-c^2\over 2ab}\lt 0 \Rightarrow C為鈍角,故選\bbox[red, 2pt]{(C)}$$
解答

$$範圍小,直接手算: 4+4+ 4+ 4+ 3+ 3+ 2 + 1+1=26,故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{E(X)=40\\ \sigma(X)=8} \Rightarrow \cases{E(2X+5) =2E(X)+5=85\\ \sigma(2X+5) = 2\sigma(X)=16},故選\bbox[red, 2pt]{(C)}$$
解答:$$2\log_2 x= x \Rightarrow x=2,4 \Rightarrow 小明說得對,故選\bbox[red, 2pt]{(B)}$$
解答:$$百位數有5種選擇(不能選0)、十位數有5種選擇、個位數有4種選擇,共有5\times 5\times 4=100\\,故選\bbox[red, 2pt]{(B)}$$
解答:$${1\over 4} ={12\over 48} ={13\over 52} ={14\over 56} ={15\over 60} ={17\over 68} ={18\over 72} ={19\over 76}  ={21\over 84}  ={24\over 96},共九種,故選\bbox[red, 2pt]{(C)}$$
解答:$$原式=\sum_{k=4}^{15} k^3 =\sum_{k=1}^{15} k^3 -\sum_{k=1}^{3} k^3 = \left({15\cdot 16\over 2} \right)^2 -\left({3\cdot 4 \over 2} \right)^2 = 120^2-6^2= 14400-36=14364\\,故選\bbox[red, 2pt]{(C)}$$
解答:$$y=f(x)=2x^3 \Rightarrow f(x-1)=2(x-1)^3 = 2x^3-6x^2+6x-2 \\\Rightarrow f(x-1)+1=2x^3-6x^2+6x-1 \Rightarrow 向右1單位,再向上1單位,故選\bbox[red, 2pt]{(A)}$$

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