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:a種類數量兩位數93,84,…,397三位數9cdH23=48cdH24=5⋯⋯3cdH29=102cd,需扣除(c,d)=(10,0),(0,10)H210−2=91cd,需扣除(c,d)=(11,0),(0,11),(10,1),(1,10)H211−4=8四位數9bcdH33=108bcdH34=15⋯⋯3bcdH39=552bcd,需扣除(10,0,0)的排列數H310−3=631bcd,需扣除(11,0,0),(10,1,0)的排列數9H311−9=69五位數9abcdH43=208abcdH44=35⋯⋯3abcdH49=2202abcd,需扣除(a,b,c,d)=(10,0,0,0)的排列數4H410−4=2821abcd,需扣除(a,b,c,d)=(11,0,0,0),(10,1,0,0)的排列數16H411−16=348合計 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)⇒平面E=△ABC:x+y=1假設垂心H(a,b,c)⇒{→AH⊥→BC→BH⊥→AC→CH⊥→AB⇒{(a,b−1,c−2)⋅(2,−2,0)=0(a+1,b−2,c−1)⋅(1,−1,−1)=0(a−1,b,c−1)⋅(−1,1,−1)=0⇒H(a,a+1,3)又H需在△ABC平面上⇒a+a+1=1⇒a=0⇒H(0,1,3)
解答:
∠A=∠C=90∘⇒ABCD共圓,假設此外接圓圓心O,半徑=a,如上圖因此∠ABO=45∘⇒∠OBC=60∘−45∘=15⇒{¯BC=2acos15∘¯CD=2asin15∘⇒¯BC+¯CD=2a(cos15∘+sin15∘)=2a⋅√62=√6a=k⇒a=k√6⇒ABCD周長=2√2a+k=2√2√6k+k=(3+2√33)k
由上圖可知:cosθ=13⇒sinθ=2√23⇒h=10+∞∑k=0(102k+102k+1)sinθ=10+10(2+1)⋅2√23=10+20√2
解答:第一個聯立方程式除了明顯解(0,0,0)外,還有一解(3,−1,2),代表有無限多組解第二個聯立方程式是第一個方程組三個平面的平移,顯然其解為過(6,2,0)方向向量為(3,−1,2)的直線即x−63=y−2−1=z2,也就是(3t+6,−t+2,2t),t∈R
解答:迴歸直線y=28+25x一定通過(μx,μy)⇒μy=28+25⋅60=52又{p=−x/2+8q=y/4−7⇒{μp=−60/2+8=−22σp=σx/2μq=52/4−7=6σq=σy/4⇒q,p迴歸直線斜率b=−25×σy/4σx/2=−15⇒q,p迴歸直線:q=−15(p+22)+6⇒−15p+85⇒(a,b)=(85,−15)
解答:f(x)=x3+9x2+8x+5⇒f″
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
解答:迴歸直線y=28+25x一定通過(μx,μy)⇒μy=28+25⋅60=52又{p=−x/2+8q=y/4−7⇒{μp=−60/2+8=−22σp=σx/2μq=52/4−7=6σq=σy/4⇒q,p迴歸直線斜率b=−25×σy/4σx/2=−15⇒q,p迴歸直線:q=−15(p+22)+6⇒−15p+85⇒(a,b)=(85,−15)
解答:f(x)=x3+9x2+8x+5⇒f″
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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