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2023年6月23日 星期五

112年屏東高中教甄-數學詳解

國立屏東高級中學112學年度正式教師甄試

一、填充題(共10題,每題6,共60分)

解答a93,84,,3979cdH23=48cdH24=53cdH29=102cd,(c,d)=(10,0),(0,10)H2102=91cd,(c,d)=(11,0),(0,11),(10,1),(1,10)H2114=89bcdH33=108bcdH34=153bcdH39=552bcd,(10,0,0)H3103=631bcd,(11,0,0),(10,1,0)9H3119=699abcdH43=208abcdH44=353abcdH49=2202abcd,(a,b,c,d)=(10,0,0,0)4H4104=2821abcd,(a,b,c,d)=(11,0,0,0),(10,1,0,0)16H41116=348 7+(4+5+6+7+8+9+10+9+8)+(10+15+21+28+36+45+55+63+69)+(20+35+56+84+120+165+220+282+348)=7+66+342+1330=1745
解答{AB=(1,1,1)AC=(1,1,1)n=AB×AC=(2,2,0)E=ABC:x+y=1H(a,b,c){AHBCBHACCHAB{(a,b1,c2)(2,2,0)=0(a+1,b2,c1)(1,1,1)=0(a1,b,c1)(1,1,1)=0H(a,a+1,3)HABCa+a+1=1a=0H(0,1,3)
解答
A=C=90ABCD,O,=aABO=45OBC=6045=15{¯BC=2acos15¯CD=2asin15¯BC+¯CD=2a(cos15+sin15)=2a62=6a=ka=k6ABCD=22a+k=226k+k=(3+233)k
解答a,b滿a2+b2=43
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解答
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:cosθ=13sinθ=223h=10+k=0(102k+102k+1)sinθ=10+10(2+1)223=10+202
解答(0,0,0)(3,1,2)(6,2,0)(3,1,2)x63=y21=z2(3t+6,t+2,2t),tR
解答y=28+25x(μx,μy)μy=28+2560=52{p=x/2+8q=y/47{μp=60/2+8=22σp=σx/2μq=52/47=6σq=σy/4q,pb=25×σy/4σx/2=15q,p:q=15(p+22)+615p+85(a,b)=(85,15)
解答f(x)=x3+9x2+8x+5f
解答令f(x)={x^2-x+1\over x^2+x+1} \Rightarrow f'(x)={x^2-1\over (x^2+x+1)^2} =0 \Rightarrow x=\pm 1 \Rightarrow \cases{f(1)=1/3\\ f(-1)=3} \\ \Rightarrow \cases{M=\log_2 3 \\ m=-\log 3} \Rightarrow M-m=2\log_2 3 =\log_2 9 \Rightarrow 8^{M-m} =8^{\log_2 9} =2^{\log_2 9^3} \\=9^3 = \bbox[red, 2pt]{729}
解答令a_3=a_4=a \Rightarrow a_5=2a \Rightarrow a_3^2+a_4^2+a_5^2=6a^2\lt 2344 \Rightarrow a  \le 19\\ 由於a_5=2a 是完全平方數,取a=18\Rightarrow a_3=a_4=18, a_5=36 \\ \Rightarrow a_1^2+a_2^2 =2344-18^2-18^2-36^2=400 \cdots(1)\\ 又\sum_{i=1}^5 a_i =a_1+a_2+18+18+36=a_1+a_2+72=k^2 \cdots(2)\\ 由(1)及(2) \Rightarrow \cases{a_2=16\\ a_1=12} \Rightarrow a_1+a_5=12+36=\bbox[red, 2pt]{48}
解答x+{2\over x-1}=x-1+{2\over x-1}+1=(x-1)+{x+1\over x-1} \ge 2\sqrt{(x-1)\cdot {x+1\over x-1}} =2\sqrt{x+1}\\  同理可得y+{2\over y-1}=(y+1)+{y+1\over y-1} \ge 2\sqrt{y+1} \\ 因此我們有x+{2\over x-1}+y+{2\over y-1}\ge 2(\sqrt{x+1}+\sqrt{y+1}),依題意,等號成立\\ \Rightarrow \cases{x-1={x+1\over x-1}\\ y-1={y+1\over y-1}} \Rightarrow \cases{x^2-3x=0\\ y^2-3y=0} \Rightarrow \bbox[red, 2pt]{\cases{x=3\\y=3}}
解答令a=x+y,其中\cases{x\in \mathbb z\\ 0\le y\lt 1} \Rightarrow \left[ a+{19\over 100}\right] + \left[ a+{20\over 100}\right] +\cdots + \left[ a+{85\over 100}\right] \\= 67x +\left[ y+{19\over 100} \right] + \left[ y+{20\over 100}\right] +\cdots + \left[ y+{85\over 100}\right] =500\\ \Rightarrow x=7 且\left[ y+{19\over 100} \right] + \left[ y+{20\over 100}\right] +\cdots + \left[ y+{85\over 100}\right]=500-67\times 7=31 \\ \Rightarrow \cases{\left[ y+{19\over 100} \right] = \left[ y+{20\over 100}\right] =\cdots = \left[ y+{54\over 100}\right]=0 \\\left[ y+{55\over 100} \right] = \left[ y+{56\over 100}\right] =\cdots + \left[ y+{85\over 100}\right]=1} \Rightarrow \cases{y+0.54\lt 1\\ y+0.55\ge 1} \\ \Rightarrow 0.45\le y\lt 0.46 \Rightarrow [100a] =[100x+100y] =\bbox[red, 2pt]{745}
解答\cases{O(0,0)\\A(-4,2)} \Rightarrow \cases{\overline{OA}=2\sqrt 5\\ \overline{OA}的中垂線L:2x-y+5=0},由於C\in L \Rightarrow C(a,2a+5)\\ 正五邊形每一內角為{(5-2)\times 180\over 5}=108^\circ =60^\circ+45^\circ \Rightarrow \cos \angle B={(2\sqrt 5)^2+(2\sqrt 5)^2-\overline{AC}^2\over 2\cdot 2\sqrt 5\cdot 2\sqrt 5} \\ \Rightarrow {\sqrt 2-\sqrt 6\over 4} ={40-\overline{AC}^2\over 80} \Rightarrow \overline{AC}^2=40-20(\sqrt 2-\sqrt 6) =\overline{OC}^2=a^2+(2a+5)^2 \\ \Rightarrow a^2+4a-3+4(\sqrt 2-\sqrt 6)=0 \Rightarrow a=
解答f(x)=(3x+2)^3 (7-2x)^2 \\\Rightarrow f'(x)=9(3x+2)^2(7-2x)^2 -4(3x+2)^3(7-2x) = 5(3x+2)^2(7-2x)(-6x+11) \\\Rightarrow f''(x)=30(3x+2)(7-2x)(-6x+11) -10(3x+2)^2(-6x+11)-30(3x+2)^2 (7-2x)\\ 若f'(x)=0 \Rightarrow \cases{x=-2/3\\ x=7/2\\ x=11/6} \Rightarrow \cases{f''(-2/3)=0\\ f''(7/2)\gt 0\\ f''(11/6) \lt 0}\\ \Rightarrow f(11/6)= \left( {11\over 2}+2\right)^3\left( 7-{11\over 3} \right)^2 =\bbox[red, 2pt]{9375\over 2}為範圍內的最大值

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