109年高雄市高中職教甄聯招-數學科詳解

高雄市109學年度市立高級中等學校
聯合教師甄選數學科試題

(一)

解：

$$f(x)=x^5 +ax^4+bx^3 +cx^2+dx-217且\cases{f(1)=1 \\ f(2)=3 \\ f(3)=5 \\ f(4)=7} \\\Rightarrow \cases{1+a+b+c+d-217 =1 \\ 32+16a +8b+4c+2d-217 =3 \\ 243+81a+ 27b+9c+3d-217=5 \\ 1024+256a+ 64b+16c+4d-217=7} \Rightarrow \cases{a+b+c+d=217  \\ 8a +4b +2c+ d  = 94 \\ 27a+ 9b+3c+d =-7 \\ 64a+ 16b+ 4c+ d=-200} \\ \Rightarrow \cases{a=-19 \\ b=125 \\ c= -365 \\d=476}  \Rightarrow  f(x)= x^5 -19x^4+125x^3 -365x^2+476x-217 \\ \Rightarrow f(5)=3125 -11875 +15625-9125+ 2380-217 = \bbox[red, 2pt]{-87}$$ 另解(網友提供)： $$由題意可假設f(x)=(x-1)(x-2)(x-3)(x-4)(x-a)+2x-1\\  常數項=-217 =f(0)=24(-a)-1 \Rightarrow a=9 \Rightarrow f(5)=4\cdot 3\cdot 2\cdot 1\cdot (-4)+10-1 = \bbox[red,2pt]{-87}$$

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解：

$$f(x)=x^3+ax^2+bx+c \Rightarrow f'(x)=3x^2+2ax+b;\\由題意知:f'(x)的最小值出現在x=-2，即{-2a\over 6}=-2 \Rightarrow a=6\\ \Rightarrow f'(-2)=12-4a+b=b-12 \Rightarrow \int_{-a}^ag(x)\;dx= \int_{-6}^6(b-12)(x+2)+10\;dx \\ =\int_{-6}^6 (b-12)x+(2b-14)\;dx =\left. \left[ {b-12\over 2} x^2+(2b-14)x \right] \right|_{-6}^6 =12(2b-14) =0\\ \Rightarrow b=7;又 f'(-2)=b-12 =-5 \Rightarrow g(x)=-5(x+2)+10 = -5x \Rightarrow g(-2)=10 \\ \Rightarrow y=f(x)圖形經過(-2,10) \Rightarrow f(-2)=10 \Rightarrow -8+4\times 6-2\times 7+c=10 \\ \Rightarrow c=8 \Rightarrow \bbox[red, 2pt]{f(x) =x^3+6x^2+7x+8}$$

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解：

$$\cases{p=\sqrt 3+i \\ q=1-i} \Rightarrow \cases{p^3=8i \\ q^2=-2i}\\ 令f(x)=(x+\sqrt 3)^{21}+(1-x)^{32} \Rightarrow f(i)=(\sqrt 3+i)^{21}+(1-i)^{32} = (8i)^7 +(-2i)^{16} =-8^7i +2^{16}\\ f(i)=a_0 + a_1i+a_2i^2 +a_3i^3+ \dots + a_{32}i^{32} \Rightarrow f(i)的實部=\cases{2^{16}\\ a_0-a_2+a_4-a_6 + \cdots+a_{32}} \\ \Rightarrow a_0-a_2+a_4-a_6  + \cdots+a_{32} =2^{16} \Rightarrow k=\bbox[red, 2pt]{16}$$

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解：
$$\begin{array}{} x_1 & x_2 & x_3 & 數量\\\hline 1 & 5 & 10-20 & 11\\ & 6 & 11-20 & 10\\ & \cdots & \cdots & \cdots \\ & 15 & 20 & 1\\\hdashline 2 & 6 & 11-20 & 10 \\ & 7 & 12-20 & 9 \\ & \cdots & \cdots & \cdots \\ & 15 & 20 & 1\\\hdashline \cdots & \cdots & \cdots & \cdots \\\hdashline 11 & 15 & 20 & 1\\\hline\end{array} \\\Rightarrow 總數量= \sum_{k=1}^{11} {k(k+1)\over 2} ={1\over 2}(506+66) =286 \Rightarrow 機率={總數量\over C^{20}_3} ={286 \over 1140} = \bbox[red, 2pt]{143 \over 570}$$

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解：
$$\left[{1\over 3} \right] +\left[{2\over 3} \right]+\left[{2^2\over 3} \right] + \cdots ++\left[{2^{100}\over 3} \right] = 0+ 0 + 1 + 2 + 5+ 10 + 21 + 42 + \cdots \\= a_1+a_2 +\cdots + a_{101} \Rightarrow a_n=\begin{cases} 2a_{n-1} & n 是偶數\\ 2a_{n-1}+1 & n是奇數\\ a_1=0\end{cases} \\ \Rightarrow \begin{array}{}n & a_n \\\hline 1 & 0 \\ 2 & 2a_1=0 \\ 3 & 2\cdot 0+1=1 \\ 4 & 2 \cdot 1=2 \\ 5 & 2\cdot 2 +1 =2^2+1 \\ 6 & 2(2^2+1) = 2^3+2 \\ 7 & 2(2^3+2)+1 =2^4+2^2+1 \\ 8& 2(2^4+2^2+1) =2^5+2^3+2 \\ \cdots & \cdots \\ 100 & 2^{97}+2^{95}+\cdots + 2\\ 101 & 2^{98}+2^{96}+\cdots + 2^2+1\\\hline\end{array} \\\Rightarrow S= \sum_{k=0}^{98} 2^k +\sum_{k=0}^{96} 2^k + \cdots +\sum_{k=0}^{2} 2^k +\sum_{k=0}^{0} 2^k \\ =(2^{99}-1)+(2^{97}-1) +\cdots +(2^3-1)+(2^1-1)\\ =(2^{99}+2^{97}+\cdots + 2^1)-50 ={2^{101}-2\over 3}-50 =\bbox[red, 2pt]{2^{101}-152\over 3}$$

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解： $$\cases{{a\over 5}+ {b\over 8}+ {c\over 11} =2\cdots(1) \\ {a\over 6}+ {b\over 9}+ {c\over 12} =2 \cdots(2)\\ {a\over 7}+ {b\over 10}+ {c\over 13} =2 \cdots(3)} \underrightarrow{}\Rightarrow \cases{(1)\times 11-(2)\times 12 \\ (2)\times 12-(3)\times 13} \Rightarrow \cases{{1\over 5}a+{1\over 24}b=-2 \cdots(4)\\ {1\over 7}a+{1\over 30}b=-2\cdots(5)} \\  \Rightarrow \cases{a=70/3\\ b=-160}代入(1)\Rightarrow c=572/3 \Rightarrow a+b+c=70/3-160+572/3=\bbox[red,2pt]{54}$$

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解：

$$直線y+2x=0上的點可表示成(a,-2a)，該點至直線y-2x=0的距離=12 \Rightarrow {|-4a|\over \sqrt 5}=12\\ \Rightarrow a=\pm 3\sqrt 5 , 令\cases{A(-3\sqrt 5,6\sqrt 5)\\ C(3\sqrt 5,-6\sqrt 5)}，見上圖;\\ 直線y-2x=0上的點可表示成(a,2a)，該點至直線y+2x=0的距離=12 \Rightarrow {|4a|\over \sqrt 5}=12\\ \Rightarrow a=\pm 3\sqrt 5 , 令\cases{B(3\sqrt 5,6\sqrt 5)\\ D(-3\sqrt 5,-6\sqrt 5)}，見上圖;\\因此矩形ABCD面積= \overline{AB}\times \overline{CD}= 6\sqrt 5\times 12\sqrt 5 = \bbox[red, 2pt]{360}$$

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解：
$$令\cases{y_1=x^3-3x+1 \\ y_2=x^3-3x+33} \Rightarrow \cases{y_1'=3x^2-3 \\ y_2'=3x^2-3}，並假設公切線在兩曲線的切點為\cases{A(a,a^3-3a+1)\\ B(b,b^3-3b+33)} \\ \Rightarrow \overline{AB}的斜率m={b^3-3b-a^3+3a+32\over b-a} = y_1'(a)=y_2'(b) \Rightarrow \cases{{b^3-3b-a^3+3a+32\over b-a}=3a^2-3 \\ {b^3-3b-a^3+3a+32\over b-a}=3b^2-3} \\ \Rightarrow \cases{a=-2 \\b=2} \Rightarrow y'_1(-2)=y_2'(2)=9 \Rightarrow A(-2,-1) \Rightarrow 公切線斜率為9且過A \Rightarrow  \bbox[red, 2pt]{y=9x+17}$$

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解：
$$\sum_{k=1}^{12} C^{12}_k\left( {1\over 4}\right)^k\left( {3\over 4}\right)^{12-k}k(k+2)= EX^2 +2EX = (n^2p^2+npq)+2np \\ =12^2\times {1\over 4^2} +12\times {1\over 4}\times{3\over 4}+2\times 12\times{1\over 4} = 9+{9\over 4}+6 = \bbox[red,2pt]{69\over 4}$$

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解：

$$假設球半徑為a\Rightarrow 半球體積(V_1)為{2\over 3}\pi a^3\\ 剩下體積(V_2)=上圖著色面積繞X軸旋轉體積\\=\pi\int_{a\sin\theta}^a (\sqrt{a^2-x^2})^2 \;dx = a^2\pi\int_{a\sin\theta}^a 1\;dx-\pi\int_{a\sin\theta}^a x^2\;dx\\ =a^2\pi(a-a\sin\theta)-{1\over 3}\pi(a^3-a^3\sin^3\theta) = \left({2\over 3}a^3-a^3\sin \theta+{1\over 3}a^3\sin^3 \theta \right)\pi\\ {V_2\over V_1} = {4\over 27} \Rightarrow {{2\over 3}-\sin \theta+{1\over 3}\sin^3 \theta \over {2\over 3}}={4\over 27} \Rightarrow {2\over 3}-\sin \theta+{1\over 3}\sin^3 \theta ={8\over 81}\\\Rightarrow 27\sin^3\theta -81\sin \theta+46=0\\ \Rightarrow (3\sin\theta -2)(9\sin^2\theta +6\sin \theta-23)=0 \Rightarrow \sin \theta=\bbox[red, 2pt]{2\over 3}$$

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解：
$$令\cases{a=2^x\\ b=3^y \\ c=5^z} \Rightarrow \cases{a+b+c=7 \cdots(1)\\ {1\over 2}a+b+5c=11\cdots(2) } ;由(2)\Rightarrow b+5c=11-{1\over 2}a 代入 t \\ \Rightarrow t=2a+(b+5c)=2a+(11-{1\over 2}a) =11+{3\over 2}a >11 (\because a>0) \Rightarrow t> 11\\ 由(1)及(2)可得a={48-8b\over 9} 代入 t=11+{3\over 2}a =11 +{3\over 2}\times {48-8b\over 9} =19-{4\over 3}b < 19 (\because b>0)\\ 因此\bbox[red, 2pt]{11 < t < 19}$$

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解：

$$分成兩堆，只有一種分法: \begin{array}{c|cc} 鋼珠& 第1堆數量&第2堆數量\\\hline 3g & 1 & 2\\ 2g & 3 & 1 \\ 1g & 1 & 2\\\hline & & &\end{array}\\ 由於鋼珠每顆顏色不同，因此有3\times 4\times 3=\bbox[red, 2pt]{36}種分法$$

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(二)證明題

解：

$$(ab-ac)^2+(bc-ab)^2+(ac-bc)^2 \ge 0 \\ \Rightarrow 2a^2b^2+ 2b^2c^2 +2a^2c^2-2a^2bc-2ab^2c-2abc^2 \ge 0\\ \Rightarrow a^2b^2+ b^2c^2 +a^2c^2-a^2bc-ab^2c-abc^2 \ge 0 \\ \Rightarrow a^2b^2+ b^2c^2 +a^2c^2+2(a^2bc+ab^2c+ abc^2) -3(a^2bc +ab^2c+abc^2) \ge 0\\ \Rightarrow (ab+bc+ca)^2 -3abc(a+b+c) \ge 0 \\ \Rightarrow (ab+bc+ca)^2 \ge 3abc(a+b+c)$$

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解：

$$f(x)=x^8-x^5+x^2+x+1=(x^2+x+1)(x^6-x^5+1)\\由於x^2+x+1=(x+{1\over 2})^2+{3\over 4} > 0 \Rightarrow x^2+x+1=0無實根\\ 令g(x)= x^6-x^5+1 \Rightarrow g'(x)=6x^5-5x^4= x^4(6x-5) \Rightarrow g''(x)=30x^4-20x^3\\因此g'(x)=0 \Rightarrow x=0或5/6 \Rightarrow \cases{g''(0)=0 \Rightarrow (0,g(0))為反曲點\\ g''(5/6)={625\over 216} > 0 \Rightarrow g(5/6)\approx 0.93 > 0}\\ \Rightarrow g(x)=0無實根 \Rightarrow f(x)=0無實根$$

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