國立北港高級中學115學年度第一次專任教師甄選
第一大題: 填充題(每題4分)
解答:
$$f(x)=x^2-4x-1=(x-2)^2-5 \Rightarrow 圖形凹向上，頂點A(2,-5) \\ g(x)= |f(x)|圖形A(2,-5) \to A'(2,5) \Rightarrow y=a = \bbox[red, 2pt]5與y=|f(x)|相交於三相異點$$
解答:$$g(x)=f(x)-x \Rightarrow \cases{g(1) =g(2)=g(5)=0 \\ g(x)為三次多項式} \Rightarrow g(x)=f(x)-x=(x-1)(x-2)(x-5) \\ \Rightarrow g(6)=f(6)-6=5\cdot 4\cdot 1=20 \Rightarrow f(6)= \bbox[red, 2pt]{26}$$
解答:$$x^2-2x+a= (x-1)^2+a-1 \gt 0 \Rightarrow a\gt 1, 又a^2-1\ne 1 \Rightarrow a^2\ne 2 \Rightarrow a\ne \pm \sqrt 2\\ 兩條件\cases{a\gt 1\\ a\ne \pm \sqrt 2}取交集 \Rightarrow \bbox[red, 2pt]{a\gt 1且a\ne \sqrt 2}$$
解答:$$f(x)= \sin^2x +2\sin x\cos x-\cos^2 x= -\cos 2x+\sin 2x =\sqrt 2\sin(2x-\pi/4)\\ \Rightarrow f(0) = \sqrt 2\sin(-\pi/4)=\sqrt 2\cdot {1\over -\sqrt 2} =\bbox[red, 2pt]{-1}$$
解答:
$$取\cases{A(3,\sqrt 3) \\B(-1, \sqrt 3) \\C(2, -2\sqrt 3)} \Rightarrow \cases{\overrightarrow{AB}= (-4,0) \\ \overrightarrow{AC}=(-1,-3\sqrt 3 )} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} =\bbox[red, 2pt]4$$
解答:$$假設\cases{A(a,0,0) \\ B(0,b,0)\\ C(0,0,c)},其中a,b,c \gt 0 \Rightarrow E:{x-a\over a}+{y\over b}+{z\over c}=0 通過P(2,3,1) \\ \Rightarrow {2-a\over a}+{3\over b}+{1\over c}=0 \Rightarrow {2\over a}+{3\over b}+{1\over c}=1\\ 柯西不等式: \left( {2\over a}+{3\over b}+{1\over c} \right) (2a+3b+4c) \ge \left( \sqrt{\frac{2}{a} \cdot 2a} + \sqrt{\frac{3}{b} \cdot 3b} + \sqrt{\frac{1}{c} \cdot 4c} \right)^2=(2+3+2)^2 \\ \Rightarrow 1\cdot (2a+3b+4c)\ge 49 \Rightarrow 2a+3b+4c\ge 49\\ 當最小值發時, \frac{\sqrt{2a}}{\sqrt{2/a}} = \frac{\sqrt{3b}}{\sqrt{3/b}} = \frac{\sqrt{4c}}{\sqrt{1/c}} \Rightarrow a= b=2c \Rightarrow {2\over 2c}+{3\over 2c}+{1\over c}=1 \Rightarrow c={7\over 2} \\ \Rightarrow a=b=7 \Rightarrow E:{x-7\over 7}+{y\over 7}+{z\over 7/2}=0 \Rightarrow \bbox[red, 2pt]{x+y+2z=7}$$
解答:$$假設\cases{長方體底面正方形的邊長為 x 公尺\\長方體的高為 y 公尺} \Rightarrow 8x+4y=20 \Rightarrow 2x+y=5\\ 長方體體積V=x^2 y, 利用算機不等式: {x+x+y\over 3} \ge \sqrt[3]{x\cdot x\cdot y} \Rightarrow {5\over 3}\ge \sqrt[3]{x^2y} \\ \Rightarrow V=x^2y \le \bbox[red, 2pt]{125\over 27}$$
解答:$$依題意\cases{電動車:T1, B2（共 2 輛）\\燃油車:T2, P1, P2（共 3 輛）\\油電混合車:B1（共 1 輛）}\Rightarrow 6車安排至6攤位, 有6!=720種\\ 事件A:A 攤位展示燃油車\\ \qquad 從 3 輛燃油車中選 1 輛放在 A 攤位，其餘 5 輛車在剩下的攤位任意排列，共有3\times 5!=360種\\ 事件B: F 攤位展示電動車\\ \qquad 從 2 輛電動車中選 1 輛放在 F 攤位，其餘 5 輛車任意排列,共有2\times 5!=240種\\ A\cap B:A 攤位展示燃油車，且 F 攤位展示電動車\\ \qquad 從3輛燃油車選 1 輛放 A 攤位，從 2 輛電動車選 1 輛放 F 攤位，剩下的 4 輛車在中間四個攤位任意排列 \\ \qquad 共有3\times 2\times 4!=144種\\ 依排容原理: 至少違反一個條件的排列數=360+240-144=456 \\ \Rightarrow 符合題意的排列數=720-456= \bbox[red, 2pt]{264}$$
解答:$$投保金額為 1000 萬元\Rightarrow \cases{全損的理賠全= 1000 萬\times 80\%=800萬\\ 半損的理賠全= 1000 萬\times 40\%= 400萬\\ 四分之一損的理賠全= 1000 萬\times 20\%=200萬 } \\ \Rightarrow \text{期望理賠金額} = \left(800萬 \times \frac{1}{1000}\right) + \left(400萬 \times \frac{5}{1000}\right) + \left(200萬 \times \frac{10}{1000}\right) =48,000元 \\ \Rightarrow \text{期望獲利} = \text{保費} - \text{期望理賠金額} \Rightarrow 20,000=保費-48,000 \Rightarrow 保費= \bbox[red, 2pt]{68,000}元$$
解答:$$\bar x={1+2+2+3\over 4}=2 \Rightarrow 迴歸直線經過(\bar x,\bar y) =(2, \bar y) \Rightarrow \bar y=2+3=5 \\ \Rightarrow {3+3+m+n\over 4}=5 \Rightarrow m+n=14 \\ \cases{S_{xx}= \sum(x_i-\bar x)^2 =2\\ S_{xy}= \sum(x_i-\bar x)(y_i-\bar y) = n-3} \Rightarrow 迴歸直線斜率b={S_{xy} \over S_{xx}} ={n-3\over 2}=1\Rightarrow n=5\\ \Rightarrow m+5=14 \Rightarrow m=9 \Rightarrow (m,n)= \bbox[red, 2pt]{(9,5)}$$
$$f(x)=g(x) \Rightarrow x={1\over 4}x^3 \Rightarrow x^3-4x=0 \Rightarrow x(x-2)(x+2)=0 \Rightarrow x=-2,0,2 \\ 取\cases{A(2,f(2)) =(2,2) \\O(0,f(0)) =(0,0) \\B(-2,f(-2)) =(-2,-2)} \Rightarrow 面積A= \int |f-g|\,dx= 2\int_0^2 \left( x-{1\over 4}x^3 \right)\,dx \\=2 \left. \left[ {1\over 2}x^2 -{1\over 16}x^4\right] \right|_0^2 =2\cdot 1=\bbox[red, 2pt]2$$
解答:$$S= \left( \sin{A\over 2}+ i\cos {A\over 2} \right) \left( \sin{B\over 2}+ i\cos {B\over 2} \right) \left( \cos{C\over 2}- i\sin {A\over 2} \right) \\ \Rightarrow (-i)\cdot (-i)\cdot S= \left( -i\sin{A\over 2}+ \cos {A\over 2} \right) \left( -i\sin{B\over 2}+ \cos {B\over 2} \right) \left( \cos{C\over 2}- i\sin {A\over 2} \right) \\ \Rightarrow -S=e^{-i{A\over 2}} e^{-i{B\over 2}} e^{-i{C\over 2}} =e^{-i(A+B+C)/2} =e^{-i\cdot {\pi\over 2}} =-i \Rightarrow S= \bbox[red, 2pt]i$$
解答:$$\cases{\lim_{x\to 2^+} f(x)=2^2+2=6\\ \lim_{x\to 2^-} f(x)=2a+3b} \Rightarrow 2a+3b=6\\ \cases{\lim_{x\to 2^+} f'(x) =\lim_{x\to 2^+} (2x+1)=5 \\ \lim_{x\to 2^-} f'(x)= \lim_{x\to 2^-} a=a} \Rightarrow a=5\Rightarrow 2\cdot 5+3b=6 \Rightarrow b=-{4\over 3} \\\Rightarrow (a,b) = \bbox[red, 2pt]{ \left(5, -{4\over 3} \right)}$$
解答:
$$假設P為原點\Rightarrow \cases{\overleftrightarrow{PA}: y=12x/5 \Rightarrow 12x-5y=0\\ \overleftrightarrow{PC}: y=3x/4 \Rightarrow 3x-4y=0 } \\ \Rightarrow 角平分線: \frac{12x - 5y}{\sqrt{12^2 + (-5)^2}} = \frac{3x - 4y}{\sqrt{3^2 + (-4)^2}} \Rightarrow 7x+9y=0 \Rightarrow 斜率為-{7\over 9} \\ \Rightarrow 另一條角平分線斜率為{9\over 7} \Rightarrow 9x-7y=0\\ 由圖形相對位置可知角平線為 9x-7y=0 \Rightarrow 斜率為\bbox[red, 2pt]{9\over 7}$$
解答:
第二大題:計算證明題(需寫出詳細過程，每題8分)
$$\textbf{(1) }假設\overline{MA}=\overline{MB}=a \Rightarrow \cos \angle AMB={a^2+a^2-\overline{AB}^2 \over 2\cdot a\cdot a} \Rightarrow -{1\over 8}={2a^2-900\over 2a^2} \Rightarrow a= \bbox[red, 2pt]{20} \\\textbf{(2) }假設\overline{NA}=\overline{NC}=b \Rightarrow \cos \angle ANC={b^2+b^2-\overline{AC}^2 \over 2\cdot b\cdot b} \Rightarrow -{1\over 8}={2b^2-324\over 2b^2} \Rightarrow b=12 \\ \quad \angle A= \angle BAC \Rightarrow \cases{\sin A=24/30 =4/5\\ \cos A= 18/30 =3/5} \\又 \angle MAN = \angle MAB + \angle BAC + \angle NAC = (90^\circ - \theta) + A + (90^\circ - \theta) = 180^\circ - (2\theta - A) \\ \Rightarrow \sin \angle MAN= \sin(180^\circ - (2\theta - A)) =\sin (2\theta - A) =\sin 2\theta \cos A-\cos 2\theta \sin A \\ ={3\sqrt 7\over 8}\cdot {3\over 5}- \left( -{1\over 8} \right)\cdot {4\over 5}={9\sqrt 7+4\over 40} \Rightarrow \triangle AMN= \frac{1}{2} \cdot \overline{AM} \cdot \overline{AN} \cdot \sin(\angle MAN) \\ =\frac{1}{2} \cdot 20 \cdot 12 \cdot \frac{9\sqrt{7} + 4}{40}= \bbox[red, 2pt]{27\sqrt 7+12}$$
解答:$$I_k = 10^{k+2} + 64 = 2^{k+2} \cdot 5^{k+2} + 2^6= 2^6 \left( 2^{k-4}\cdot 5^{k+2}+1 \right), k\ge 4 \\ 當k\gt 4時, 2^{k-4}\cdot 5^{k+2}為偶數 \Rightarrow 2^{k-4}\cdot 5^{k+2}+1為奇數 \Rightarrow N(k)=6 \\ 當k=4時, I_4=2^6 (1\cdot 5^6+1)=64\times 15626=64\times 2\times 7813 \Rightarrow N(4)=7 \\ \Rightarrow N(k)的最大值為\bbox[red, 2pt] 7$$
第三大題:教學情境應用題(每題8分)
請依各題情境，運用數學知識或教學專業完成作答，並適當說明理由或論證過程。
$$摺紙步驟: 先將紙張上下對摺，再展開，水平摺痕\overline{MN}即為紙張的平分線\\ \qquad 接著將A斜摺，使得A落在\overline{MN}上，即為A'，此時摺痕與紙張頂端交於D點\\ 證明\triangle ACD為30^\circ-60^\circ-90^\circ的三角形\\ 假設\cases{\overline{AC}=2a \\ A'在紙張底邊的垂足為H} \Rightarrow \overline{A'H}=a\\又 \overline{CD}為摺痕 \Rightarrow \triangle DAC \cong \triangle DA'C \Rightarrow \cases{\overline{CA'} =2a\\ \angle DA'C=\angle A=90^\circ} \Rightarrow \sin \angle A'CH ={a\over 2a}={1\over 2} \\ \Rightarrow \angle A'CH=30^\circ \Rightarrow \angle ACA'=90^\circ-30^\circ=60^\circ \Rightarrow \angle ACD=30^\circ (\overline{CD}為角平分線) \\ \Rightarrow \triangle ACD為30^\circ-60^\circ-90^\circ的三角形, \bbox[red, 2pt]{故得證}$$
解答:$$ \left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} C^n_k (1)^{n-k} \left(\frac{1}{n}\right)^k = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \frac{1}{n^k} \\ \Rightarrow 當 k=0: C^n_0 \left(\frac{1}{n}\right)^0 = 1 \\當 k=1: C^n_1 \left(\frac{1}{n}\right)^1 = n \cdot \frac{1}{n} = 1 = \frac{1}{1!} \\ 當 k=2:C^n_2 \left(\frac{1}{n}\right)^2 = \frac{n(n-1)}{2!} \cdot \frac{1}{n^2} = \frac{1}{2!} \left(\frac{n}{n}\right) \left(\frac{n-1}{n}\right) = \frac{1}{2!} \left(1 - \frac{1}{n}\right) \\當 k=3: C^n_3 \left(\frac{1}{n}\right)^3 = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{n^3} = \frac{1}{3!} \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\\ \Rightarrow \bbox[red, 2pt]{一般項: \frac{1}{k!} \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k-1}{n}\right), 其中1\le k\le n} \\ \bbox[red, 2pt]{最後一項(k=n): \frac{1}{n!} \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{n-1}{n}\right) }$$
解答:$$ \left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \cdots + \frac{1}{n!}\left(1 - \frac{1}{n}\right) \cdots \left(1 - \frac{n-1}{n}\right) \\ \lt 1+1+{1\over 2!}+{1\over 3!~}+ \cdots+{1\over n!} \lt 1+1+{1\over 2}+{1\over 2^2}+ \cdots+ {1\over 2^{n-1}} =1+ 2 \left( 1-{1\over 2^n} \right) \\ =1+2-{1\over 2^{n-1}} =3-{1\over 2^{n-1}} \lt 3\quad \bbox[red, 2pt]{故得證}$$
解題僅供參考，其他教甄試題及詳解
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