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2020年7月22日 星期三

109年臺北市聯合轉學考-普通型高中(升高二)數學詳解


臺北市高級中等學校109學年度
聯合轉學考招生考試高中數學科試題

選擇題: 共 14 題,總分 70 分,每題 5 分

解:
$$A=10^{\log 2} \Rightarrow \log A=\log 10^{\log 2}=\log 2 \Rightarrow A=2 \Rightarrow 1000A =2000,故選\bbox[red, 2pt]{(D)}$$


解:
$${2a+b \over 2} \ge \sqrt{2ab} \Rightarrow 5 \ge \sqrt{2ab} \Rightarrow {25\over 2} \ge ab \Rightarrow {75\over 2} \ge 3ab,故選\bbox[red, 2pt]{(E )}$$


解:
$$\cases{P(兩白球)=C^3_2/C^5_2 =3/10\\ P(兩黑球)=C^2_2/C^5_2 =1/10\\ P(一白一黑)=C^3_1C^2_1/C^5_2 =6/10} \Rightarrow 期望值=100(3/10+1/10)+ 50\times 6/10-10=60\\,故選\bbox[red, 2pt]{(B )}$$


解:
$$\sin(90^\circ -\alpha) + \cos(-\beta) = \cos \alpha+\cos \beta={4\over 5} +{12\over 13}= {112\over 65},故選\bbox[red, 2pt]{( E)}$$


解:
$$甲乙丙三人排列有2種排法:\cases{ 甲乙丙\\乙甲丙};再考慮戊庚 \Rightarrow \cases{戊甲乙丙,則庚有4種選擇\\甲戊乙丙,則庚有3種選擇\\甲乙戊丙,則庚有2種選擇\\甲乙丙戊,則庚有1種選擇 }\\\Rightarrow 有4+3+2+1=10種排法; 因此甲乙丙戊庚共有2\times 10=20種排法;\\剩下丁有6個位置可選,然後己有7個位置可選,總共有20\times 6\times 7=840種排法\\,故選\bbox[red, 2pt]{(B )}$$


解:
$$x^2-2x+y^2-4y=20 \Rightarrow (x-1)^2+(y-2)^2=5^2 \Rightarrow \cases{圓心(1,2)\\ 半徑r=5} \\ \Rightarrow 圓心至直線L距離\text{dist}(O,L)={ 12+10+4\over \sqrt{12^2 +5^2}}=2 \Rightarrow L與圓相交兩點\\\Rightarrow 與L平行且與L相距3的直線L'與L''與圓相交於三點A,B,C見上圖,故選\bbox[red, 2pt]{( C)}$$


解:
$$\begin{array}{} a_n & 數量k& 累積\sum k \\\hline 1 & 1 & 1\\ 1,3 & 2 & 3\\ 1,3,5 & 3 & 6\\ \cdots & & \cdots\\ 1,3, \dots, 19 & 10 & 55 \\ 1,3,5,7,9 & 5 & 60\\\hline \end{array} \Rightarrow a_{60}=9 \Rightarrow S_{60} =\sum_{n=1}^{10}\sum_{k=1}^n(2k-1) +(1+3+5+7+9) \\ \sum_{n=1}^{10}\sum_{k=1}^n(2k-1) = \sum_{n=1}^{10}(n(n+1)-n) = \sum_{n=1}^{10} n^2 ={1\over 6}\times 10\times 11\times 21=385 \\ \Rightarrow S_{60}=385 +(1+3+5+7+9)=410,故選\bbox[red, 2pt]{(D )}$$


解:
$$\cases{a=\sqrt{7+\sqrt{13}} \\ b=\sqrt{7-\sqrt{13}}} \Rightarrow \cases{a^2=7+\sqrt{13} \\ b^2=7-\sqrt{13} \\ ab=\sqrt{49-13}=6} \Rightarrow (a+b)^2 =a^2+b^2+2ab = 14+12=26 \\ \Rightarrow a+b=\sqrt{26} \approx \sqrt{25}=5,故選\bbox[red, 2pt]{(B )}$$


解:

$$\cases{4的倍數有500/4=125個\\ 5的倍數有500/5=100個\\ 20的倍數有500/20=25個\\  12的倍數有500/12=41個\\ 30的倍數有500/30=16個\\ 60的倍數有500/60=8個  } \Rightarrow 125+100-25-41-16+8=151,故選\bbox[red, 2pt]{(A )}$$


解:
$$假設大樓頂部為T,底部為B,則樓高\overline{TB}=h \Rightarrow \cases{\angle TPB=30^\circ \\ \angle TQB=45^\circ} \Rightarrow \cases{\overline{BP}= \sqrt 3h\\ \overline{BQ}=h};\\ 在\triangle PBQ中,\cos \angle PBQ = {\overline{BP}^2+ \overline{BQ}^2 -\overline{PQ}^2 \over 2\times \overline{BP}\times \overline{BQ}} \Rightarrow \cos 30^\circ={\sqrt 3\over 2} ={3h^2+h^2-100^2 \over 2\sqrt 3h^2} \\\Rightarrow h=100,故選\bbox[red, 2pt]{( C)}$$


解:
$$\cases{(1,135) \\ (4,40)} \Rightarrow \cases{135=k\cdot a\cdots(1)\\ 40= k\cdot a^4 \cdots(2)\\ s=k\cdot a^6\cdots (3)} \Rightarrow {(2)\over (1)}= {40\over 135}={8\over 27} =a^3 \Rightarrow a= {2\over 3}\\ {(3)\over (2)} ={s\over 40} =a^2 ={4\over 9} \Rightarrow s= {160\over 9} \approx 18,故選\bbox[red, 2pt]{(C )}$$


解:
$$\cases{a\ne c且b\ne d:有6\times 5\times 4\times 3=360種\\ a\ne c且b= d:有6\times 5\times 4\times 1=120種\\ a= c且b\ne d:有6\times 5\times 1\times 4=120種\\ a= c且b= d:有6\times 5\times 1\times 1=30種}\\ \Rightarrow 共有360+120+120+30=630 \Rightarrow 機率為{630\over 6^4} ={630\over 1296} = {35\over 72},故選\bbox[red, 2pt]{(D )}$$



解:
$$r_1=r_2=0,r_3 > r_4,r_4 \le 0.5,故選\bbox[red, 2pt]{(E)}$$


解:
$$x^3-3x^2+x+7=2x+k \Rightarrow x^3-3x^2-x+7-k=0 的三根為\alpha,\beta,\gamma \Rightarrow \cases{\alpha+\beta+\gamma =3 \cdots(1) \\ \alpha \beta+\beta\gamma + \gamma\alpha=-1 \cdots(2)\\ \alpha\beta\gamma=k-7 \cdots(3)};\\ 由於2\beta = \alpha+\gamma,代入(1)可得: 3\beta=3 \Rightarrow \cases{\beta=1 \\ \alpha+\gamma =2} 代入(2)及(3) \Rightarrow \cases{2+\gamma \alpha=-1 \cdots(4)\\ \alpha\gamma=k-7 \cdots(5)}\\ 由(4)\Rightarrow \gamma\alpha=-3 代入(5) \Rightarrow -3=k-7 \Rightarrow k=4,故選\bbox[red, 2pt]{( A)}$$


解:
$$(A)\times: L_1斜率大於L_2斜率\Rightarrow -a > -c \Rightarrow a< c \\(B)\times: L_1 的Y截距 小於L_2的Y截距\Rightarrow b< d \\(C)\bigcirc: \cases{L_1的Y截距小於0 \Rightarrow b< 0\\ L_1的斜率-a> 0 \Rightarrow a<0} \Rightarrow ab >0 \\(D)\times: \cases{L_2的Y截距大於0 \Rightarrow d> 0\\ L_2的斜率-c> 0 \Rightarrow c<0} \Rightarrow cd < 0\\(E) \bigcirc: \cases{a< 0 \\ c < 0 \\ b < 0 \\ d >0} \Rightarrow \cases{ac >0\\ bd < 0} \Rightarrow ac > bd\\,故選\bbox[red, 2pt]{(CE )}$$


解:
$$(A)\bigcirc: f(x)=(2x+3)Q(x)+r(x),r(x)為一常數 \Rightarrow f(-3/2)=0+r(-3/2)=r(x)\\ (B)\times: f(x)=(2x+3)Q(x)+r(x) =2(x+{3\over 2})Q(x)+r(x)\Rightarrow 商式為2Q(x)\\ (C)\times:f(x)=(x-{3\over 2})2Q(x)+r(x)\Rightarrow 餘式仍為r(x) \\ (D)\bigcirc: f(x)=(2x+3)Q(x)+r(x) \Rightarrow f(x/2)=(x+3)Q(x/2)+r(x) \Rightarrow 商式為Q(x/2) \\ (E)\bigcirc: f(x)= 2(x+{3\over 2})Q(x)+r(x) \Rightarrow 2f(x) =4(x+{3\over 2})Q(x)+ 2r(x) \Rightarrow 餘式為2r(x)\\,故選\bbox[red, 2pt]{( ADE)}$$


解:
$$\begin{array}{llll}& a_1& = & 2\\ & a_2& = & 2^2-1\\ & a_3 & = & 2^3-2-1 \\ & a_4 & = & 2^4-2^2-2-1\\ & \cdots &  & \cdots  \\ & a_{n-1} &=& 2^{n-1}-2^{n-3}-2^{n-4}-\cdots -1\\ & a_n & = & 2^n-2^{n-2}-2^{n-3}-\cdots -1\\ & & =&2^n-2^{n-1}+1 =2^{n-1}+1\\\hline\end{array}\\(A) \bigcirc: A_4=2^3+1=8+1=9\\ (B) \bigcirc: a_7=2^6+1=65 \\(C) \times:a_{10}=2^9+1=  512+1=513 \not \gt 1000 \\(D)\bigcirc: a_{11}+a_{12}+a_{13} =2^{10}+1+2^{11}+1+2^{12}+1 =2^2(2^8+2^9+2^{10})+3 \\(E) \times:\cases{a_2=3\\a_1=2} \Rightarrow a_2-a_1=1不是偶數\\,故選\bbox[red, 2pt]{(ABD )}$$


解:
$$(A)\bigcirc: X越大則Y越小,為負相關\\ (B)\bigcirc: \cases{EX=(1+2+3+4+5)\div 5=3\\ EX^2=(1^2+2^2+3^2 +4^2+5^2) \div 5=11} \Rightarrow \sigma_X=\sqrt{EX^2-(EX)^2} =\sqrt{11-9}=\sqrt 2 \\ (C)\bigcirc: \cases{EY=(45+55+35+15+25)\div 5=35\\ EY^2=(45^2+55^2+35^2 +15^2+25^2) \div 5=1425} \\\qquad \Rightarrow \sigma_Y=\sqrt{EY^2-(EY)^2} =\sqrt{1425-1225}=10\sqrt 2 \\(D)\times: EXY=(45+110+105+60+125)\div 5= 89 \Rightarrow EXY -EXEY=89-3\times 35=-16 \\ \qquad \Rightarrow \rho = {-16\over 2} =-8 \Rightarrow 迴歸直線 y-EY=-8(x-EX) \Rightarrow y=-8x+59\\ (E)\bigcirc: y=-8\times 7+59=3\\,故選\bbox[red, 2pt]{(ABCE )}$$


解:
$$(A)\times: \triangle 面積={1\over 2}\times \overline{AB}\times \overline{AC} \times\sin \angle A ={1\over 2}\times 8\times 6\times \sin 60^\circ =24\times {\sqrt 3\over 2} =12\sqrt 3\\(B)\bigcirc: \overline{AD}為\angle A的角平分線(見上圖),則\overline{DE}=\overline{DF}=h \Rightarrow \triangle ABC面積={1\over 2}h(\overline{AB}+\overline{AC}) \\\qquad \Rightarrow 7h = 12\sqrt 3  \Rightarrow h={12\over 7}\sqrt 3 \Rightarrow \overline{AD}=2h={24\over 7}\sqrt 3\\(C)\times: \cos \angle A ={\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\times \overline{AB}\times \overline{AC}} \Rightarrow \cos 60^\circ ={64+36-\overline{BC}^2 \over 2\times 8\times 6} \Rightarrow {1\over 2}={100-\overline{BC}^2 \over 96} \\\qquad \Rightarrow \overline{BC}^2 =52 \Rightarrow \overline{BC}=2\sqrt{13}\\ (D)\times: {\overline{BC}\over \sin \angle A}=2R \Rightarrow {2\sqrt{13} \over \sqrt 3/2} \Rightarrow R=2\sqrt{13\over 3} ={2\over 3}\sqrt{39} \\(E)\bigcirc: 若A(0,0) \Rightarrow \cases{B(8,0)\\ C(6\cos 60^\circ,6\sin 60^\circ)=(3,3\sqrt 3)} \Rightarrow B,C中點P(11/2,3\sqrt 3/2) \\\qquad \Rightarrow 中線長\overline{OP} =\sqrt{{121\over 4}+{27\over 4}} =\sqrt{148\over 4}=\sqrt{37}\\,故選\bbox[red, 2pt]{(BE )}$$


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