115年松山家商教甄-數學詳解

 臺北市立松山高級商業家事職業學校115學年度第1次教師甄選

一、填充題（15 題，每題 5 分，合計 75 分）

解答:$$f(x)= x^{15}+ 3x^{10}-x^5-2 =(x^4-x^2)p(x)+ ax^3+bx^2+cx+d\\ \Rightarrow f'(x) =15x^{14}+30x^9-5x^4 =(4x^3-2x)p(x)+ (x^4-x^2)p'(x)+ 3ax^2+2bx+c \\    \Rightarrow \cases{f(0)=-2=d\\f(1)=1=a+b+c+d\\ f(-1)=1=-a+b-c+d\\ f'(0)=0=c} \Rightarrow \cases{a=0\\ b=3\\ c=0\\ d=-2} \Rightarrow 餘式: \bbox[red, 2pt]{3x^2-2}$$

解答:$${1\over n(n+1)(n+2)} ={1\over 2} \left( {1\over n(n+1)}-{1\over (n+1)(n+2)} \right) \\ \Rightarrow {1\over 1\times 2\times 3} +{1\over 2\times 3\times 4}+ {1\over 3\times 4\times 5} + \cdots +{1\over 13\times 14\times 15} \\={1\over 2} \left( {1\over 1\times 2}-{1\over 2\times 3}+{1\over 2\times 3}-{1\over 3\times 4} +{1\over 3\times 4}-{1\over 4\times 5} +\cdots +{1\over 13\times 14}-{1\over 14\times 15}\right) \\= {1\over 2} \left( {1\over 1\times 2}-{1\over 14\times 15} \right) ={1\over 2} \cdot {52\over 106}= \bbox[red, 2pt]{26\over 105}$$

解答:$$a_{n+1}=a_n+3 \Rightarrow {1\over a_n\times a_{n+1}} ={1\over 3} \left( {1\over a_n}-{1\over a_{n+1}} \right) \\ \Rightarrow \sum_{n=5}^\infty {1\over a_n\times a_{n+1}} = {1\over 3} \left({1\over a_5}-{1\over a_{6}} +{1\over a_6}-{1\over a_7} + \cdots\right) ={1\over 3a_5}={1\over 42} \Rightarrow a_5=14 =a_1+4\cdot 3 \\ \Rightarrow a_1= \bbox[red, 2pt]2$$

解答:$$1+2+3+ \cdots +15={16\cdot 15\over 2} =120 \\ \Rightarrow S=1+2+2+ 3+3+3+ \cdots +15 =1^2+2^2+\cdots+15^2 ={1\over 6}\cdot15\cdot 16\cdot 31 =\bbox[red, 2pt]{1240}$$

解答:$$假設\cases{f(x) =x^3+ax^2+2x-1 \\ g(x)=x^3+bx^2+1} , f(x)與g(x)的最高項係數皆為1 \Rightarrow 公因式D(x)=x^2+mx+n \\ \Rightarrow \cases{f(x)=D(x)\cdot (x-{1\over n})=x^3+(m-{1\over n})x^2+ (n-{m\over n})x-1 \\g(x)=D(x)\cdot (x+{1\over n}) =x^3+ (m+{1\over n})x^2+(n+{m\over n})x+1} \Rightarrow \cases{n-m/n=2\\ a=m-1/n\\m= -n^2 \\ b=m+1/n} \\ \Rightarrow \cases{m=-1\\ n=1\\a=-2\\ b=0} \Rightarrow (a,b)= \bbox[red, 2pt]{(-2,0)}$$

解答:

$$f(x)=x|x-2| =\begin{cases} f_1(x)=x^2-2x, & x\ge 2 \\f_2(x)=-x^2+2x, &x\le 2\end{cases} \Rightarrow \cases{f_1(x) 遞增, x\ge 2\\ f_2(x)=-(x-1)^2+1, 凹向下,頂點(1,1)} \\  \Rightarrow 兩圖形\cases{y=f(x) \\ y=k} 有\cases{1個交點, \bbox[red, 2pt]{k\gt 1或k\lt 0}\\3個交點, 0\lt k\lt 1}$$

解答:$$x=1+\sqrt[3]3+ \sqrt[3] 9 \Rightarrow (x-1)^3=(\sqrt[3]3+ \sqrt[3] 9 )^3 \\ \Rightarrow x^3-3x^2+3x-1=9x+3 \Rightarrow x^3-3x^2-6x=4 \Rightarrow x^3-3x^2-6x+5= \bbox[red, 2pt]9$$

解答:

$$f(x)=||x|-2| = \begin{cases} x-2,& x\ge 2\\ -x+2, &0\le x\le 2\\ x+2,&-2\le x\le 0\\ -x-2,& x\le -2\end{cases} \\ 直線y=mx+5 通過A(0,5)與y=f(x)相交於一點\Rightarrow 斜率介於兩直線\cases{y=x-2\\ y=-x-2}之間\\ 即 \bbox[red, 2pt]{m\ge 1或m\le -1}$$

解答:$$算幾不等式: 2^a+2^b \ge 2\sqrt{2^a\cdot 2^b} =2\sqrt{2^{a+b}} =2\sqrt 2 \Rightarrow 最小值y=2\sqrt 2\\ 端點值\Rightarrow \cases{a=0,b=1 \Rightarrow 2^a+2^b=1+2=3\\a=1,b=0 \Rightarrow 2^a+2^b=2+1=3} \Rightarrow 最大值x=3 \Rightarrow x+y =\bbox[red, 2pt]{3=2\sqrt 2}$$

解答:$$4^x+(m-5)\times 2^{x+1}+(3m-5)=0 \Rightarrow (2^x)^2+2(m-5)2^x+(3m-5)=0有相異實根 \\ \Rightarrow 實根2^x皆為正數 \Rightarrow \cases{判別式:4(m-5)^2-4(3m-5)\gt 0\\ 兩根之和:-2(m-5)\gt 0\\ 兩根之積: 3m-5\gt 0} \Rightarrow \cases{m\lt 3或m\gt 10\\ m\lt 5\\ m\gt {5\over 3}} \\ 三條件取交集 \Rightarrow \bbox[red, 2pt]{{5\over 3} \lt m\lt 3}$$

解答:$$\log x+\log y=1 \Rightarrow \log (xy)=1 \Rightarrow xy=10 \Rightarrow x+y\ge 2\sqrt{xy}=2\sqrt{10} \\ f(x,y)=x^2+y^2-4(x+y)+7 =(x+y)^2-2xy-4(x+y)+7=(x+y)^2-4(x+y)-13 \\ =(x+y-2)^2-17 \Rightarrow 當x+y=2時,f(x,y)有最小值-17,但x+y\ge 2\sqrt{10} \gt 2\\ 因此f(x,y)的最小值發生在端點,xy=10且x=y, 也就是 x=y=\sqrt{10} \\ \Rightarrow f(\sqrt{10},\sqrt{10}) =10+10-8\sqrt{10}+7= \bbox[red, 2pt]{27-8\sqrt{10}}, 公布的答案是\bbox[cyan,2pt]{47-12\sqrt{10}}$$

解答:$$假設\cases{\overline{BC}= a \\\overline{AC}=b\\ \overline{AB}=c \\ \triangle ABC面積=A} \Rightarrow A= {1\over 2} (3a)={1\over 2}(4b)={1\over 2}(6c) \Rightarrow \cases{a=2A/3\\ b= A/2\\ c=A/3} \\ \Rightarrow s={1\over 2}(a+ b+c )= {3A\over 4} \Rightarrow A=\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{{3A\over 4}\cdot {A\over 12}\cdot {A\over 4}\cdot {5A\over 12}} \\ \Rightarrow A={\sqrt{15}A^2\over 48} \Rightarrow A= \bbox[red, 2pt]{16\sqrt{15}\over 5}$$

解答:$$\Gamma:y^2=2x \Rightarrow B\in \Gamma \Rightarrow B(2t^2,2t) \Rightarrow \overline{AB} = \sqrt{(2t^2-1)^2+(2t-4)^2} =\sqrt{4t^4-16t+17} \\ 取f(t)=4t^4-16t+17 \Rightarrow f'(t)=16t^3-16=0 \Rightarrow t=1 \Rightarrow f(1)=5 \\ \Rightarrow \overline{AB}最小值= \bbox[red, 2pt]{\sqrt 5}$$

解答:$$假設\cases{梯腳到牆壁的距離為 x \\梯頂到地面的距離為y} \Rightarrow x^2+y^2=100 \Rightarrow 6^2+y^2=100 \Rightarrow y=8 \\ \Rightarrow 2x{dx\over dt}+2y{dy\over dt}=0 \Rightarrow {dy\over dt}=-{x\over y}{dx\over dt} \Rightarrow {dy\over dt}(x=6,y=8,{dx\over dt}=1) =-{6\over 8}\cdot 1= \bbox[red, 2pt]{-{3\over 4}}$$

解答:$$f(x)={x^2+3x+11\over x+2}= x+1+{9\over x+2} =x+2+{9\over x+2}-1 \ge 2\sqrt{(x+2) \cdot {9\over x+2}}-1=\bbox[red, 2pt]5$$

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