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2019年9月1日 星期日

108年身心障礙學生大學甄試-數學甲-詳解


108學年度身心障礙學生升學大專校院甄試試題
甄試類(群)組別:大學組
考試科目(編號):數學甲
單選題,共 20 題,每題 5 分

{a>0ax3+b>x2,xa<0ax3+b>x2,xa=0,x=0x2>ax3+b0>b(a=0,b<0)(D)



¯AO=1{A(1){B(3){C(6)C(0)B(1){C(2)C(4)A(1){B(1){C(4)C(2)B(3){C(0)C(6)(A,B,C)={(1,3,6)(1,3,0)(1,1,2)(1,1,4)(1,1,4)(1,1,2)(1,3,0)(1,3,6)C=0,2,4,6,2,4,67(C)


{f(x)=p(x)(x1)(x2)+(2x2)f(x)=q(x)(x2)(x3)+(ax+2){f(2)=0+(42)=2f(2)=0+(2a+2)2a+2=2a=0f(x)=r(x)(x1)(x3)+(bx+c){f(1)=b+c=0+(22)=0f(3)=0+(3b+c)=0+(3a+2)=2{b+c=03b+c=2{b=1c=1:bx+c=x1(A)


解:

{A=(1,1)B=(1/2,1)C=(2,2)D=(1,2)ABDC=(¯AB+¯CD)×dist(L,M)2=(1/2+1)×12=34(A)


{2x+3y=112x5y=3{x=4y=1O=(4,1)r=O3x+4y=6r=|12+4632+42|=105=2(x4)2+(y1)2=22(D)


{xy+z=0x+2y+z=3x+y=4[111012131104]r1+r2,r1+r3[111003030214]r2/3[111001010214]r2+r1,2r2+r3[101101010012]r3+r1[100301010012]r3[100301010012](C)




θ=DAB2θ=Acos2θ=cosA=¯AC2+¯AB2¯BC22¯AC¯AB=102+521222105=19100cos2θ=19100=2cos2θ1cos2θ=81200cosθ=9102=9220(B)


16(200+500+1000+0+200+500)=16×2400=400(B)


M=[abcd]{M[12]=[21]M[21]=[42]{[abcd][12]=[21][abcd][21]=[42]{a+2b=2c+2d=12a+b=42c+d=2{a=2b=0c=1d=0M=[2010][1111]M[11]=[1111][2010][11]=[1010][11]=[11](C)


L:y=mx+b{L(1,3)d(O,L)=3{3=m+b|bm2+1|=3|3mm2+1|=3(m3)2=9(m2+1)8m2+6m=0m(4m+3)=0m=34,0(,


\vec{u}\cdot \vec{v}=5 \Rightarrow (1,-2,2)\cdot(a,b,0)=a-2b=5 \Rightarrow a^2+b^2 =(2b+5)^2 +b^2= 5b^2+20b +25\\ = 5(b^2+4b+4)+5 =5(b+2)^2+5 \Rightarrow b=-2時, a^2+b^2有最小值5\\ \Rightarrow |\vec{v}|=\sqrt{a^2+b^2}的最小值為=\sqrt{5},故選\bbox[red,2pt]{(A)}


{甲不拿A且乙拿B \over 甲不拿A} = {C^4_2C^3_2 \over C^5_2C^4_2C^2_2} = {6\times 3 \over 10\times 6} ={18\over 60} ={3\over 10},故選\bbox[red,2pt]{(D)}




餘弦定理\Rightarrow \cos{\angle COA}={\overline{AO}^2 +\overline{CO}^2-\overline{AC}^2 \over 2\overline{AO}\times \overline{CO}} = {10+10-16 \over 20} ={1\over 5}\\ \Rightarrow \sin{\angle BOC} =\cos{\angle COA} ={1\over 5},故選\bbox[red,2pt]{(A)}


|\vec{a}||\vec{b}|\sin{\theta} =15^2\times {7\over 25}=63,故選\bbox[red,2pt]{(B)}


L:\begin{cases} x=1+t\\ y=-1-2t\\ z=3+2t \end{cases} \Rightarrow L的方向向量\vec{u}=(1,-2,2) \\ 假設平面的法向量為\vec{n},若平面與L不相交,則\vec{u}\cdot\vec{n}=0,且L不在平面上\\ (A) \vec{n}=(2,2,1) \Rightarrow \vec{u}\cdot \vec{n}=0, 但2(1+t)+2(-1-2t)+(3+2t)=3\Rightarrow L在平面上\\ (B)\vec{n}=(2,-1,-2) \Rightarrow \vec{u}\cdot \vec{n}=0, 且2(1+t)-(-1-2t)-2(3+2t)=-3\ne 3\Rightarrow L不在平面上\\ (C)\vec{n}=(2,2,-1) \Rightarrow \vec{u}\cdot \vec{n}= 2-4-2=-4\ne 0\\ (D)\vec{n}=(2,1,-1) \Rightarrow \vec{u}\cdot \vec{n}=2-2-2=-2\ne 0,\\故選\bbox[red,2pt]{(B)}


\text{實際支持甲的比率為 }p \Rightarrow \text{實際不支持甲的比率為 }1-p \Rightarrow p(1-0.1)+(1-p)\times 0.3=0.54\\ \Rightarrow 0.3+0.6p=0.54 \Rightarrow p=0.4,故選\bbox[red,2pt]{(C)}


\log{E(r)}=5.24+1.44r \Rightarrow \log{E(6)}=5.24+1.44\times 6= 13.88 \Rightarrow E(6)=10^{13.88}\\ \Rightarrow 100\times 10^{13.88} = 10^{15.88} \Rightarrow \log{E(a)} = 15.88 =5.24+1.44\times a \Rightarrow a={15.88-5.24 \over 1.44} \approx 7.39\\,故選\bbox[red,2pt]{(B)}



f(x)=-\sqrt{3}\cos{x} +\sqrt{6}\sin{x}-2 = -3\left({\sqrt{3} \over 3}\cos{x} -{\sqrt{6}\over 3} \sin{x}\right) -2  \\= -3\left(\sin{y}\cos{x} -\cos{y} \sin{x}\right) -2 = -3\sin{(y-x)}-2 \Rightarrow -3-2\le f(x)\le 3-2 \\\Rightarrow -5 \le f(x)\le 1 \Rightarrow f(x)最大值為1,故選\bbox[red,2pt]{(A)}


\begin{cases} f(x)=a\cdot 2^{bx}為凹口向下 \Rightarrow a<0;\\ f(-1)>f(0) \Rightarrow a\cdot 2^{-b}> a \Rightarrow 2^{-b}<1 \Rightarrow b>0 \end{cases} \Rightarrow \begin{cases} a<0\\ b>0 \end{cases},故選\bbox[red,2pt]{(C)}


z=\sqrt{2}\left(\cos{\pi \over 12}+i\sin{\pi \over 12} \right) \Rightarrow z^6= (\sqrt{2})^6\left(\cos{\pi \over 12}\times 6+i\sin{\pi \over 12}\times 6 \right) =8\left(\cos{\pi \over 2}+i\sin{\pi \over 2} \right)\\ =8(0+i)= 8i,故選\bbox[red,2pt]{(D)}


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