114年高雄聯合轉學考-升高二-數學詳解

 高雄區公立高中 114 學年度聯合招考轉學生《升高二數學》

一、 單選題(60 分)：

解答:$$(A) y=x^2+2x+6 =(x+1)^2+5 \ge 5 \Rightarrow 最小值為5\\ (B)若x=0, y=1+9=10 \Rightarrow 最小值不是6\\ (C) x^2+1+{6\over x^2+1} \ge 2\sqrt{(x^2+1)\cdot {6\over x^2+1}} =2\sqrt 6 \lt 6  \Rightarrow 最小值不是6 \\ (D) 3^{x-1}+3^{3-x} \ge 2\sqrt{3^{x-1}\cdot 3^{3-x}} =6\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{y=0 \Rightarrow x\cos \theta=\sin \theta \Rightarrow \tan \theta =x \gt 0\\ x=0 \Rightarrow y=\sin \theta \lt 0} \Rightarrow \cases{\sin \theta \lt 0\\ \cos \theta\lt 0} \Rightarrow \theta為第三象限角，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{(a+a^{-1})^2 =a^2+a^{-2}+2=5+2=7 \Rightarrow a+a^{-1} =\sqrt 7\\ (a^2+a^{-2})^{2} =a^4+a^{-4}+2 \Rightarrow a^4+a^{-4}=5^2-2=23} \Rightarrow {a^4+a^{-4} \over a+a^{-1}} ={23\over \sqrt 7}，故選\bbox[red, 2pt]{(B)}$$

解答:$$假設一開始\cases{B細菌有1個\\A細菌有k個} \Rightarrow 210小時後\cases{B細菌有16^{210/3} =2^{280}個\\ A細菌有k\cdot 8^{210/7} =k\cdot 2^{90}} \\ \Rightarrow k\cdot 2^{90} =2^{280} \Rightarrow k=2^{190} \Rightarrow \log k=190\cdot \log 2=190\cdot 0.301=57.19 \Rightarrow k=10^{57.19}，故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\times: \log b=3.2 \Rightarrow b=10^{3.2} \gt 10^3=1000 \\(B) \bigcirc: \cases{a=10^{1.6} \\ b=10^{3.2}} \Rightarrow a^2b=10^{3.2} \cdot 10^{3.2} =10^{6.4} \Rightarrow \log a^2b=6.4=\log c \Rightarrow a^2b=c \\ (C)\times: \log abc=\log a+ \log b+\log c=1.6+3.2+6.4 =11.2 \ne 33\\ (D) \times: \cases{a=10^{1.6} \\b=10^{3.2} \\c=10^{6.4}} \Rightarrow \cases{b^2=10^{6.4} \\ac=10^{8}} \Rightarrow b^2\ne ac \Rightarrow a,b,c非等比數列\\，故選\bbox[red, 2pt]{(B)}$$

解答:

$$x^2+y^2+2x+6y-4=0 \Rightarrow (x+1)^2+ (y+3)^2=14 \Rightarrow \cases{圓心O(-1,-3)\\ 圓半徑r=\sqrt{14}} \\ 假設\overline{AB}中點為P, 則\overline{OP} = 直線與圓心的距離={|-a-3+4-a| \over \sqrt{a^2+1}}  ={|1-2a| \over \sqrt{a^2+1}} \\ \overline{OP}越大,則\overline{AB}越小, 欲求\overline{OP}之極大值 \\ 假設f(a)= \left( {|1-2a| \over \sqrt{a^2+1}}\right)^2=4-{4a+3\over a^2+1} \Rightarrow f'(a)={(2a-1)(a+2) \over (a^2+1)^2} =0 \Rightarrow a=1/2,-2 \\ \Rightarrow \overline{OP}之極大值=\sqrt{f(-2)} =\sqrt 5 \Rightarrow \overline{AP} =\sqrt{r^2-\overline{OP}^2} =3 \Rightarrow \overline{AB}=2\overline{AP}=6，故選\bbox[red, 2pt]{(A)}$$

解答:$$-3\lt x\lt 7 \Rightarrow (x+3)(x-7)\lt 0 \Rightarrow x^2-4x-21 \lt 0 \Rightarrow -x^2+4x+21\gt 0 \\ \Rightarrow \cases{a=-1\\ b=4\\ c=21} \Rightarrow -x^3+4x^2+252x\lt 0 \Rightarrow -x(x-18)(x+14)\lt 0 \\ \Rightarrow x(x-18)(x+14)\gt 0 \Rightarrow \cases{x\gt 18\\ -14\lt x\lt 0}，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)=(x+2)(x^5-x^3+1)+3x+4 =(x+2)(x^5-x^3+1)+3(x+2)-2 \\=(x+2)(x^5-x^3+4)-2 \Rightarrow \cases{商式:x^5-x^3+4\\ 餘式:-2}，故選\bbox[red, 2pt]{(C)}$$

解答:$$廣域特徵近似y=-2x^3 \Rightarrow a=-2 \Rightarrow f(x)=-2x^3+18x^2-50x+47 =-2(x+m)^3+p(x+m)+n \\=-2x^3-6mx^2+(p-6m^2)x-2m^3+pm+n \Rightarrow \cases{-6m=18\\ p-6m^2=-50\\ -2m^3+pm+n=47} \Rightarrow \cases{m=-3\\ p=4\\ n=5} \\ \Rightarrow m+n=-3+5=2，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)=-2x^3+18x^2-50x+47  \Rightarrow f'(x)=-6x^2+36x-50 \Rightarrow \cases{f(1)=13\\ f'(1)=-20} \\ \Rightarrow y=-20(x-1)+13 \Rightarrow -20x-y+33=0 \Rightarrow \cases{r=-20\\ t=33} \Rightarrow r-t=-53，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{\theta 為第三象限角\\ \tan \theta=12/5} \Rightarrow \cases{\sin \theta=-12/13\\ \cos \theta=-5/13} \Rightarrow \cos(\theta+90^\circ)-\cos(\theta+180^\circ) -\cos(\theta+270^\circ) \\= -\sin \theta+  \cos \theta-\sin \theta= {24\over 13}-{5\over 13} ={19\over 13}，故選\bbox[red, 2pt]{(B)}$$

解答:

$$\cases{\triangle NAC為等腰直角三角形\\ \triangle MAB為等腰直角三角形} \Rightarrow \cases{\overline{AN} =\overline{AC}/\sqrt 2=5\\ \overline{AM} =\overline{AB}/\sqrt 2=\sqrt{13}} \\ \Rightarrow \cos \angle MAN ={\overline{AM}^2+ \overline{AN}^2- \overline{MN}^2 \over 2\cdot \overline{AM} \cdot \overline{AN}} \Rightarrow \cos(45^\circ+ \angle BAC+ 45^\circ)=-\sin \angle BAC ={13+25-\overline{MN}^2 \over 10\sqrt{13}} \\ \Rightarrow -{2\sqrt 6\over 5\sqrt 2}={38-\overline{MN}^2\over 10\sqrt {13}} \Rightarrow \overline{MN}^2=38+4\sqrt{39}，故選\bbox[red, 2pt]{(A)}$$

解答:$$第一天5名學生取2名、第二天剩下3名學生取2名，共有C^5_2C^3_2\\ 欲求之機率C^6_1\cdot {C^5_2C^3_2\over C^6_3C^6_3} ={9\over 20}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{三顆球都放入相符的箱子:1種\\ 三顆球有二顆球放入相符的箱子:C^3_2=3種 } \Rightarrow 獲獎的情形有4種\Rightarrow 機率={4\over 4\times 3\times 2}={1\over 6}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{輸1次就成功, 機率為\displaystyle {1\over 5} \\ 輸2次才成功, 機率為\displaystyle {4\over 5}\times {1\over 4}={1\over 5} \\ 輸3次結束,前2次皆失敗,機率為\displaystyle {4\over 5}\cdot {3\over 4}={3\over 5}} \Rightarrow 期望值=1\cdot {1\over 5}+2\cdot {1\over 5}+3\cdot {3\over 5}={12\over 5}\\，故選\bbox[red, 2pt]{(C)}$$

二、 多選題(40 分)：

解答:$$(A)\times: \cases{a=3\\ b=2\\ c=3+\sqrt 3\\ d=2+\sqrt 3} \Rightarrow a-c=-\sqrt 3=b-d, 但c\ne d \\(C) \times:\cases{a=\sqrt 2\\ b=\sqrt 2} \Rightarrow ab=2為有理數\\，故選\bbox[red, 2pt]{(BDE)}$$

解答:$$假設\langle a_n\rangle 公差為d, \langle b_n \rangle 公比為r\\ (A)\bigcirc: (2a_n+3)-(2a_{n-1}+3) =2(a_n-a_{n-2})=2d \Rightarrow \langle 2a_n+3 \rangle 為公差為2d 的等差數列 \\(B)\times: 3^{a_n} -3^{a_{n-1}} =3^{a_1+(n-1)d} -3^{a_1+(n-2)d} =3^{a_1+nd} \left( 3^{-d} -3^{-2d}\right) 非定數 \\(C)\times: {b_{n+1}+2\over b_n+2} ={b_1r^n+2\over b_1r^{n-1}+2}非定數\\ (D) \bigcirc: {b_{n+1}^2 \over b_n^2} ={b_1^2r^{2n}\over b_1^2r^{2n-2}} =r^2 \Rightarrow \langle b_n^2\rangle 公比為r^2的等比數列\\ (E) {b_{n+3}-b_{n+1}\over b_{n+2}-b_n} ={b_1r^{n+2}-b_1r^n \over b_1r^{n+1}-b_1r^{n-1}} ={b_1r^n(r^2-1)\over b_1r^{n-1}(r^2-1)} =r \Rightarrow \langle b_{n+2}-b_n\rangle 公比為r的等比數列\\，故選\bbox[red, 2pt]{(ADE)}$$

解答:$$(A)\bigcirc: 三次多項式加二次多項式仍為三次多項式\\ (B)\times: 若g(x)=-f(x) \Rightarrow f(x)+g(x)=0\\ (C)\times: 餘式可能為0,不一定是一次多項式 \\(D)\times: f與g的三次項係數皆不為0, 兩者相乘亦不為0,因此fg為六次\\ (E)\bigcirc: f(x)=(6x-3)q(x) =(2x-1)3q(x)\\，故選\bbox[red, 2pt]{(AE)}$$

解答:$$(A)\bigcirc: g(x)=ax^2+bx+c \Rightarrow \cases{g(0)=c=1\\ g(-1) =a-b+c=3\\ g(-2)=4a-2b+c=7} \Rightarrow \cases{a=1\\ b=-1\\ c=1} \Rightarrow g(x)=x^2-x+1 \\\quad \Rightarrow g(3)=9-3+1=7 \\(B) \times: g(x)=x^2-x+1=(x-{1\over 2})^2+{3\over 4} \ge {3\over 4}為最小值 \\ (C)\times: 由(D)可知h(x)=x(x+1)(x+2)p(x)= f(x)-g(x) \Rightarrow f(x)=g(x)+x(x+1)(x+2)p(x)\\\quad 若g(x)是p(x)的因式, g(x)才是f(x)的因式, 即g(x)不一定是f(x)的因式\\(D)\bigcirc: 取h(x)=f(x)-g(x) 為n次多項式 \Rightarrow h(0)=h(-1)=h(-2) =0 \Rightarrow h(x)=x(x+1)(x+2)p(x) \\\quad \Rightarrow x(x+1)(x+2)是f(x)-g(x)的因式 \\(E)\bigcirc: f(x)=x(x+1)q(x)+ax+b \Rightarrow \cases{f(0)=b=1\\ f(-1)=-a+b=3} \Rightarrow \cases{a=-2\\ b=1} \Rightarrow 餘式:-2+1\\，故選\bbox[red, 2pt]{(ADE)}$$

解答:$$(A)\times: 7={1\over 2}\cdot 3\cdot 5 \sin \angle A \Rightarrow \sin A={14\over 15} \Rightarrow \angle A有兩種可能\\ (B)\bigcirc: \cos A={1\over 4}={3^2+5^2-\overline{BC}^2\over 2\cdot 3\cdot 5} \Rightarrow \overline{BC}= \sqrt{53\over 2} \Rightarrow 已知三邊長決定唯一三角形 \\(C)\bigcirc: \cos \angle A={1\over 2} ={\overline{AC}^2+\overline{AB}^2-\overline{BC}^2\over 2\cdot \overline{AB}\cdot \overline{AC}} \Rightarrow \overline{AB}={\sqrt{33\over 3}}\Rightarrow 已知三邊長決定唯一三角形 \\(D)\times: \cos \angle A={\sqrt 3\over 2}={\overline{AC}^2+\overline{AB}^2-\overline{BC}^2 \over 2\cdot \overline{AB}\cdot \overline{AC}} \Rightarrow \overline{AB}= 3\sqrt{11}\pm \sqrt{55} \Rightarrow 有兩種可能\\ (E)\bigcirc: \cos 40^\circ={\overline{AC}^2+\overline{AB}^2-\overline{BC}^2\over 2\cdot \overline{AB} \cdot \overline{AC}} \Rightarrow \overline{AB} ={6\cos 40^\circ+\sqrt{36\cos^2 40^\circ+64} \over 2}\Rightarrow 只有一種\\，故選\bbox[red, 2pt]{(BCE)}$$

解答:$$(A)\times: \cos B={a^2+c^2-b^2 \over 2ac} ={c\over a} \Rightarrow b^2+c^2=a^2 \Rightarrow \angle A=90^\circ \Rightarrow 直角三角形\\ (B)\bigcirc: \overline{BD}為\angle B的角平分線\Rightarrow {\overline{CD} \over \overline{AD} } ={\overline{BC} \over \overline{AB}} ={a\over c} ={1\over \cos B} =5 \Rightarrow \overline{CD}=5\overline{AB} \\(C)\times: \cos^2 {B\over 2} ={1\over 2}(\cos B+1)={3\over 5} \Rightarrow \cos {B\over 2}= \cos \angle ABD={\sqrt{15}\over 5}\\\quad  \Rightarrow \overline{AB}= \overline{BD}\cos {B\over 2} ={2\sqrt{15}\over 5} \ne {4\sqrt{15}\over 5} \\(D) \times: \cos B={1\over 5}={\overline{AB} \over \overline{BC}} \Rightarrow \overline{BC}=5\cdot {2\sqrt{15}\over 5} =2\sqrt{15} \ne 4\sqrt{15} \\(E)\bigcirc: \triangle ABC ={1\over 2}\overline{AB}\cdot \overline{BC}\sin B ={1\over 2}\cdot {2\sqrt{15}\over 5} \cdot 2\sqrt{15}\cdot {2\sqrt{6} \over 5} ={12\sqrt 6\over 5}\\，故選\bbox[red, 2pt]{(BE)}$$

解答:$$圓C: x^2+y^2+2mx+8y+12=0 \Rightarrow (x+m)^2+(y+4)^2=4+m^2 \Rightarrow \cases{圓心O(-m,-4)\\ 圓半徑r=\sqrt{m^2+4}} \\ 直線L:x+y-2=0與圓相切\Rightarrow d(O,L)=r \Rightarrow {|-m-6|\over \sqrt 2}= \sqrt{m^2+4} \\ \Rightarrow m^2-12m-28=0 \Rightarrow (m-14)(m+2)=0 \Rightarrow m=-2 (圓心需在第四象限, m=14不合)\\ (A) \times:圓心(2,-4)未通過2x+5y-8=0 \\(B)\bigcirc: (1-2)^2+(-5+4)^2=2\lt 8 \Rightarrow 在圓內\\ (C)\times: 圓心與原點距離=\sqrt{20} \Rightarrow 圓C上的點與原點距離為d \Rightarrow \sqrt{20}-\sqrt 8\le d\le \sqrt{20}+\sqrt 8 \\ \Rightarrow d= 2,3,4,5,6,7 \Rightarrow 共12個 \\ (D)\bigcirc:圓心(2,-4)與7x-y+2=0的距離={20\over \sqrt{50}} =2\sqrt 2=r \\(E) \times:圓心(2,-4)與(12,-2)距離\ne r\\，故選\bbox[red, 2pt]{(BD)}$$

解答:$$(A)\bigcirc: E(Y)=E(0.9X+15)=0.9E(X)+15=0.9\cdot 50+15=60\\ (B)\times: 若班級人數為偶數，調整後剛好一半的同學及格，沒有超過一半\\ (C)\bigcirc: \sigma(Y)= \sigma(0.9X+15) =0.9\sigma(X)=0.9\cdot 8=7.2\lt 8\\(D)\bigcirc: 標準差變小\Rightarrow 分數差異變小\\ (E) \times: 0.9=r\cdot {\sigma(Y)\over \sigma(X)} \Rightarrow 相關係數r= 0.9\cdot {8\over 7.2}=1\\，故選\bbox[red, 2pt]{(ACD)}$$

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