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2021年1月25日 星期一

110年大學學測-數學詳解

110 學年度學科能力測驗試題

一、單選題

A=[1203]A2=[1809]A4=A2A2=[180081]=[abcd]a+b+c+d=1+80+0+81=162(2)
a1=log36=2log6{a2=2a1=4log6a2=12a1=log6<1()a2=4log6{a3=2a2=8log6a3=12a2=2log6{a4=2a3=16log6,4log6a4=12a3=4log6(log6<1,){a5=2a4=32log6,8log6a5=12a4=8log6,2log6a5=2log6,8log6,32log6(1)

OOCP=OBP=90BOC=180θA=12BOC=9012θcosA=cos(9012θ)=sin12θ(3)

{a=(m,n)b=(s,t){2a+b=(2m+s,2n+t)a+2b=(m+2s,n+2t)3a+b=(3m+s,3n+t)a+3b=(m+3s,n+3t)122m+s2n+tm+2sn+2t=6|mtns|=4123m+s3n+tm+3sn+3t=4|mtns|=4×4=16(5)
(x+1)f(x)=(x3+2)(ax+b)+x+2{(1+1)f(1)=(1+2)(a+b)1+2(0+1)f(0)=(0+2)(0+b)+0+2{ab=14=2b+2{a=2b=1(x+1)f(x)=(x3+2)(2x+1)+x+2(2+1)f(2)=105+4f(2)=18(4)

 
{A(3,0)D(3,0){B(32,332)C(32,332)E(32,332)F(32,332);x242+y2(7)2=1{P(4,0)Q(0,7)R(4,0)S(0,7){u=dist(O,¯BC)=332u2=274=6.75<(7)2¯OP>¯OA8(5)

二、多選題


(1)×:0.48(2):10.4=0.6(3):{6:10.4=0.68:10.4=0.69:10.5=0.59(4):0.40.4+0.3+0.2=49<0.5(5)×:0.50.2+0.1+0.5=58=0.625<23(=0.67)(2,3,4)


(1)×:A(3,4){¯OA=32+42=5Lm=43{Γ:x2+y2=25L:3y=4xLΓ(3,4)(2)×:OA=BCLBC¯BC=m=4/3(3):B(a,b)C(a+3,b+4)¯AC=a2+b2=5AOCAOC=60(4)×:¯OA=¯OB=¯AC=¯BC=5OACBCBA=12¯CA¯CBsinACB=1255sin120=2543(5):A60C120B{C=[cos60sin60sin60cos60][34]=[]B=[cos120sin120sin120cos120][34]=[]BCA60C120B(3,5)

10.4x0.6x20.55y0.45y(1)×:x+y0.4x+0.55y0.6x+0.45y(2):xy=k{:0.4x+0.55y=0.4ky+0.55y=(0.4k+0.55)y:0.6x+0.45y=0.6ky+0.45y=(0.6k+0.45)yk=0.5{:0.75y:0.75y{>k<1/2=k=1/2>k>1/2(3):x>yxy>1>12>(4):0.4x>0.55yxy>5540>12(5)×:0.45y>0.6xxy<4560=0.75{x/y<0.51/2<x/y<0.75(2,3,4)

¯BC=a>0(1):k=cosA=42+62a2246a2=5248ka=5248k,a(2):(1)(3)×:ABCABCC46¯BC¯BC¯BC(4)×:ABC=1246sinAsinAA(180A)2(5)×:asinA=2RsinA(4)A2(1,2)

¯CDE使¯AE=¯BCABCE(1):ADE¯AD>¯AEAED>D()AED+DAE>D+DAEBAE+DAE>AECA>B(2):B+D=AEC+D=180AED+D<180(AED>D)(3)×:B(4)×:k=2ADE2,3,523(5):cosC=cosAED=25+k2(k+1)210k=12k5kCBCD=15kcosC=15k×12k5k=363k363k<30k>2(k+1)+k>5k>2(125)

7顆球任排: {7!\over 2!2! 3!} =210\\A:\{黑球相鄰 \} \Rightarrow \#(A)={6!\over 2!3!}=60  \\ C:\{任2紅球不相鄰\} \Rightarrow \#(C)=全部-有2紅球相鄰+3紅球相鄰=210-{6!\over 2!2!} +{5!\over 2!2!} \\\qquad =210-180+30=60\\A\cap C:\{黑球相鄰且任2紅球不相鄰\} \Rightarrow \#(A\cap C)= {6!\over 2!3!}-{5!\over 2!}+{4!\over 2!}=60-60+12=12\\ B\cap C:\{黑球不相鄰且任2紅球不相鄰\}=\#(C)-\#(A\cap C)=60-12=48\\(1)\times: \cases{ P(A)=60/210\\ P(B)=1-P(A)=150/210 }\Rightarrow P(B)\gt P(A)\\(2)\bigcirc:P(C)=60/210=2/7\\(3)\times: 2P(C|A)+5P(C|B) = 2{P(A\cap C)\over P(A)} +5{P(B\cap C)\over P(B)} =2{12\over 60}+5{48\over 150}=2 \not \lt 2\\(4)\times: P(C|A)={P(A\cap C)\over P(A)}={12\over 60}={1\over 5}=0.2 \not \gt 0.2\\(5)\bigcirc: P(C|B)= {P(B\cap C)\over P(B)}={48\over 150}={32\over 100}=0.32 \gt 0.3\\ 故選\bbox[red,2pt]{(2,5)}
(1)\times: x^3+ax^2+bx+c = x^2+100 \Rightarrow x^3+(a-1)x^2+bx+c-100=0,三次式一定有實數解,即一定有交點\\(2)\bigcirc: f(0)f(1) \lt 0 \lt f(0)f(2) \Rightarrow f(0)f(1)f(0)f(2) < 0 \Rightarrow \cases{f(0)f(1) \lt 0 \\f^2(0)f(1)f(2) < 0 \Rightarrow f(1)f(2) < 0} \\ \qquad \Rightarrow \cases{有1根介於0與1之間\\ 有1根介於1與2之間},又f(x)為3次式,因此有3相異實根 \\(3)\bigcirc: x=1+3i \Rightarrow x^2-2x+10=0 \Rightarrow f(x)=(x^2-2x+10)(x-k),由於f(x)係數皆為有理數,因此k為有理數\\ (4)\times: \cases{f(1)=a+b+c+1\\ f(2)=4a+2b+c+8 \\f(3)=9a+3b+ c+27\\ f(4)=16a+4b+c+64} \Rightarrow \cases{f(2)-f(1)=3a+b+7 \cdots(1)\\ f(3)-f(2)=5a+b+19\cdots(2) \\ f(4)-f(3)=7a+b+37 \cdots(3)} \\ \qquad \Rightarrow \cases{(1)=(2) \Rightarrow 2a+12=0 \Rightarrow a=-6\\ (2)=(3) \Rightarrow 2a+18=0 \Rightarrow a=-9} 矛盾\\(5)\bigcirc: {f(2)\over f(1)} ={f(3)\over f(2)} ={f(4)\over f(3)}=r \Rightarrow \cases{f(2)=rf(1) \\ f(3)=rf(2) \\ f(4)=rf(3)} \Rightarrow \cases{ (4-r)a +(2-r)b+(1-r)c=r-8 \\ (9-4r)a+(3-2r)b+(1-r)c=8r-27 \\ (16-9r)a+(4-3r)b+(1-r)c =27r-64} \\ \qquad 可利用克拉瑪公式可求得a,b,c,由於各係數皆為有理數,其行列式亦為有理數\\\qquad 例:取r=2 \Rightarrow f(x)=x^3-3x^2+8x \Rightarrow \cases{f(1)=6\\ f(2)=12\\ f(3)=24\\ f(4)=48} \Rightarrow f(1),f(2),f(3),f(4)為等比數列\\故選\bbox[red,2pt]{(2,3,5)}

第貳部份、選填題

:每週期移動4\times 6=24單位,由於116=24\times 4+20;也就是歷經4週期(8\times 4=32秒)再加20\div 4=5秒,即32+5=\bbox[red,2pt]{37}秒。
\cases{L_1:(2s,-3s,-5s)\\ L_2:(1,1+2t,1+3t)} \Rightarrow \cases{L_1的方向向量\vec u=(2,-3,-5)\\ L_2的方向向量\vec v=(0,2,3)}  \Rightarrow E的法向量\vec n= \vec u\times \vec v=(1,-6,4)\\ E過L_1上的P(0,0,0) \Rightarrow  \bbox[red,2pt]{x-6y+4z=0}
\cases{4^2=16=1\times 2\times 8\\ 6^2=36=1\times 4\times 9 =2\times 3\times 6\\8^2= 64=2\times 4\times 8\\ 12^2=144=2\times 8\times 9 =3\times 6\times 8} ,共6組,機率為{6\over C^{9}_3} ={6\over 84} =\bbox[red,2pt]{1\over 14}
\cases{y=a\cdot 2^{x}經過(-1,3) \Rightarrow 3= a\cdot 2^{-1} \Rightarrow a=6\\ y=a\cdot 2^{x}經過(-1,-1) \Rightarrow -1= a\cdot 2^{-1} \Rightarrow a=-2} \Rightarrow \bbox[red,2pt]{-2 \le a\le 6}

\log (\sqrt[3]{49})^{100} =100\log \sqrt[3]{49} ={100\over 3}\log 49 ={200\over 3}\log 7 ={200\over 3}\times 0.8451 = 56.34 \Rightarrow n=56\\ 又 \log 2=0.301 \lt 0.34 \lt  0.4771=\log 3 \Rightarrow m=2\\ 因此(m,n)=\bbox[red,2pt]{(2,56)}

面積=1個正方形+4個等腰直角\triangle = (2+\sqrt 2)^2 + 4\times ({1\over 2}\times 1\times 1) =\bbox[red,2pt]{8+4\sqrt 2}


\cos \angle BAC ={\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\cdot \overline{AB}\cdot \overline{AC}} \Rightarrow {1\over 3} = {96+96  -\overline{BC}^2 \over 2\cdot 96}  \Rightarrow \overline{BC}=8\sqrt 2 \\ 令O為\overline{BC} 中點 \Rightarrow \overline{AC}^2=\overline{AO}^2+ \overline{CO}^2 \Rightarrow 96=\overline{AO}^2 +(4\sqrt 2)^2 \Rightarrow \overline{AO}=8\\ \triangle DBC為等腰\triangle \Rightarrow \overline{CD}^2 = \overline{DO}^2 +\overline{CO}^2 \Rightarrow 8^2=\overline{DO}^2+ (4\sqrt 2)^2 \Rightarrow \overline{DO}=4\sqrt 2\\ 令\cases{O(0,0,0)\\ B(-4\sqrt 2,0,0)\\ C(4\sqrt 2, 0,0)\\ A(0,8,0)\\ D(0,a,b)},由\cases{\overline{DO}=4\sqrt 2\\ \overline{DA}=4\sqrt 6} \Rightarrow \cases{a^2+b^2=32\\ (a-8)^2+b^2=96} \Rightarrow \cases{a=0\\ b=\pm 4\sqrt 2} \\ \Rightarrow D至\triangle ABC距離=\overline{OD}=|b|=\bbox[red,2pt]{4\sqrt 2}

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解題僅供參考,其他歷屆試題及詳解

7 則留言:

  1. 最後一句好像筆誤了 .... 應該是D到三角形ABC才對

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  2. 多選12題選項(5)詳解有誤,應該是P(C|B)

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  3. 回覆
    1. 最好用筆電/桌機看, 若用手機,某些版面會亂!!

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