115年高雄中學教甄-數學詳解

高雄中學 115 年正式教師甄選數學科試題

※本份試卷共有 15 大題，每題皆為計算或證明題。

解答:$$若\Delta\ne 0, 依克拉瑪公式，該方程組有唯一解(x,y,z)=({\Delta_x\over \Delta}, {\Delta_y\over \Delta}, {\Delta_z\over \Delta} ),\\ 但該三平面並未交於一點，因此 \Delta=0\\ 假設\Delta_x=\Delta_y= \Delta_z=0,加上\Delta=0 \Rightarrow 三平面交於一直線,即無限多解 \Rightarrow 與題意違背\\ 因此\Delta_x, \Delta_y, \Delta_z不能同時為0，也就是至少有一個不為0 \quad \bbox[red,2pt]{故得證}$$

解答:$$\textbf{法一:}f(x)為一次函數(線性) \Rightarrow f(1),f(2),f(3)成等差\Rightarrow 2f(2)=f(1)+f(3) \\\qquad \Rightarrow f(3)=2f(2)-f(1) ; 又\cases{1\le f(1) \le 4\\ -2 \le f(2)\le 7} \Rightarrow -8\le f(3) \le 13 \\\textbf{法二:} 假設f(x)=ax+b \Rightarrow \cases{f(1)=a+b\\ f(2)=2a+b\\ f(3)=3a+b} \Rightarrow f(3)= \alpha f(1)+\beta f(2) \Rightarrow \cases{\alpha+ 2\beta=3\\ \alpha +\beta=1} \\\qquad \Rightarrow \cases{\alpha= -1 \\ \beta=2} \Rightarrow f(3)=2f(2)-f(1) 又\cases{1\le f(1) \le 4\\ -2 \le f(2)\le 7} \Rightarrow -8\le f(3) \le 13 $$

解答:$$\bbox[cyan,2pt]{題目有誤 }f(x)= (3x^5+2x^4 -4x-2)^{115} 的最高次應該是5\times 115=575, 不是115$$

解答:$$a_n = \int_0^1 x^n(1-x)^2 \,dx = \int_0^1 (x^{n+2}-2x^{n+1}+x^n) \,dx =  \left. \left[ {1\over n+3}x^{n+3} -{2\over n+2}x^{n+2} +{1\over n+1}x^{n+1}\right] \right|_0^1  \\\qquad ={1\over n+3}-{2\over n+2} +{1\over n+1} ={2\over (n+1)(n+2)(n+3)} \\ \textbf{(1) } \sum_{k=1}^na_k = \sum_{k=1}^n  \left( {1\over k+3}-{2\over k+2} +{1\over k+1}  \right)= \sum_{k=1}^n  \left[ \left( {1\over k+1}-{1\over k+2} \right)- \left( {1\over k+2}-{1\over k+3} \right) \right] \\\qquad =\left[ \left( {1\over 2}-{1\over 3} \right)- \left( {1\over 3}-{1\over 4} \right) \right] + \left[ \left( {1\over 3}-{1\over 4} \right)- \left( {1\over 4}-{1\over 5} \right) \right] + \cdots \\\qquad + \left[ \left( {1\over n+1}-{1\over n+2} \right)- \left( {1\over n+2}-{1\over n+3} \right) \right] =\left( {1\over 2}-{1\over 3} \right)- \left( {1\over n+2}-{1\over n+3} \right)  \\\qquad  = \bbox[red, 2pt]{{1\over 6}-{1\over (n+2)(n+3)}} \\\textbf{(2) }a_n= {2\over (n+1)(n+2)(n+3)} \Rightarrow \lim_{n\to \infty} n^3\cdot a_n = \lim_{n\to \infty} {2n^3\over (n+1)(n+2)(n+3)} = \bbox[red, 2pt] 2$$

解答:$$正弦定理: {a\over \sin A} =2R \Rightarrow R={a\over 2\sin A} = {a\sqrt{bc} \over b+c} \\ \Rightarrow \sin A={b+c\over 2\sqrt {bc}} \ge {1\over \sqrt{bc}}\cdot \sqrt{bc} =1 \Rightarrow \sin A=1 \Rightarrow \angle A=90^\circ \\ 算機不等式等號成立條件: b=c \Rightarrow \triangle ABC為等腰直角三角形 \\ \Rightarrow \cos A+\cos B+\cos C =0+ {\sqrt 2\over 2}+{\sqrt 2\over 2} =\bbox[red, 2pt] {\sqrt 2}$$

解答:

$$假設\cases{\overline{BD}=a\\ \overline{AC}=b\\ \angle ACD= \theta} \Rightarrow \cases{ \overline{BC}=2a\\ \angle ABC=2\theta} \Rightarrow \cases{\tan \theta=1/b\\ \tan 2\theta=b/(a+1)} \Rightarrow {2/b\over 1-1/b^2} ={b\over a+1} \\ \Rightarrow b^2-1=2a+2 \Rightarrow b^2=2a+3 \cdots(1) \\ 直角\triangle ABC \Rightarrow (2a)^2=(a+1)^2+b^2 \Rightarrow b^2=3a^2-2a-1 \cdots(2) \\ \Rightarrow 2a+3=3a^2-2a-1 \Rightarrow 3a^2-4a-4=0 \Rightarrow (3a+2)(a-2)=0 \Rightarrow a=2 \Rightarrow b=\sqrt 7 \\ \Rightarrow \triangle BCD={1\over 2} \cdot \overline{BD} \cdot \overline{AC} ={1\over 2}\cdot 2\cdot \sqrt 7=\bbox[red ,2pt]{\sqrt 7}$$

解答:$$取g(x)-4f(x)=r(x)=mx+n \Rightarrow g(x)=4f(x)+mx+n \\ \Rightarrow (g(x))^2=16(f(x))^2+8f(x)r(x)+(r(x))^2  \Rightarrow (r(x))^2= m^2f(x)+28x-40 \cdots(1) \\ f(x)={1\over 4}(g(x)-r(x)) \Rightarrow (f(x))^2 = {1\over 16}(g(x))^2 -{1\over 8}g(x)r(x)+{1\over 16}(r(x))^2 \\\Rightarrow (r(x))^2={m^2\over 4}g(x)+44x-60 \cdots(2) \\ 由(1)及(2) \Rightarrow  m^2f(x)+28x-40={m^2\over 4}g(x)+44x-60  ={m^2\over 4}(4f(x)+ r(x))+44x-60 \\ \Rightarrow r(x)=-{64\over m^2}x+{80\over m^2} =mx+n \Rightarrow m^3=-64 \Rightarrow m=-4 \Rightarrow n=5 \\ \Rightarrow r(x)=-4x+5代入(1) \Rightarrow (-4x+5)^2 =16f(x)+28x-40 \Rightarrow \bbox[red, 2pt]{f(x)=x^2-{17\over 4}x+{65\over 16}}$$

解答:$$8個學生擲硬幣，有2^8=256種。為符合沒有相鄰二人皆為反面，反面數n最多為4\\ n=0: 所有人皆正面，只有1種情形\\n=1: 有C^8_1=8種情形\\ n=2:從8個位置選出2個不相鄰，任選有C^8_2=28種，扣除2個反面剛好相鄰有8種\\\qquad 所以共有28-8=20種\\ n=3: 任選有C^8_3=56種，\cases{三個反面全相連，有8種\\ 恰好兩個反面相連，第三個反面不與他們相連，有32種} \\\qquad 共有56-8-32=16種\\ n=4:正反正反...或反正反正...，只有2種情形\\ 符合條件共有1+8 +20+16+2= 47 \Rightarrow 機率為 \bbox[red, 2pt]{47\over 256}$$

解答:$$S: (x-1)^2+y^2+z^2=1 \Rightarrow Q\in S \Rightarrow (a-1)^2+b^2+ c^2=1 \\ 平面E的法向量\vec n= \overrightarrow{PQ} =(a-1,b,c) \Rightarrow E:(a-1)(x-a)+ b(y-b)+ c(z-c)=0 \\ \Rightarrow (a-1)x+ by+cz= a(a-1)+b^2+c^2 =(a-1)^2+b^2+c^2+a-1 =1+a-1=a \\ \Rightarrow E: (a-1)x+by+cz =a \Rightarrow \cases{A(a/(a-1),0,0) \\ B(0,a/b,0) \\C(0,0,a/c)} \Rightarrow \cases{\overrightarrow{AB} =(-{a\over a-1},{a\over b},0) \\ \overrightarrow{AC} =(-{a\over a-1},0,{a\over c})}  \\ \Rightarrow \triangle ABC面積= {1\over 2} |\overrightarrow{AB}\times \overrightarrow{AC}  | = {a^2\over 2bc(a-1)} \\ 由於(a-1)^2+b^2+c^2=1 \Rightarrow b^2+c^2=1-(a-1)^2 =2a-a^2 \\ 又{b^2+c^2} \ge 2bc \Rightarrow 2a-a^2 \ge 2bc \Rightarrow \triangle ABC面積\ge {a^2\over (2a-a^2)(a-1)} ={a\over -a^2+3a-2}\\ 欲求{a\over -a^2+3a-2} 的最小值，即{-a^2+3a-2\over a}=-a+3-{2\over a}的最大值,即3-2\sqrt 2 \\ \Rightarrow 最小面積={1\over 3-2\sqrt 2} = \bbox[red, 2pt]{3+2\sqrt 2}$$

解答:$$f(x)=x^3+ax^2 -a^2x \Rightarrow f'(x)=3x^2+2ax-a^2 =(3x-a)(x+a) =0 \Rightarrow x=a/3,-a\\ \Rightarrow f''(x)=6x+2a \Rightarrow \cases{f''(-a)=-4a \lt 0 \Rightarrow f(-a) \text{ is local maximum} \\ f''(a/3)=4a \gt 0 \Rightarrow f(a/3) \text{ is local minimum}} \Rightarrow b={a\over 3} \\ f(b)\text{ is local minimum } \Rightarrow f'(b)=0 \Rightarrow \text{ tangent line: }y=f(a/3)=-{5a^3\over 27} \\ \Rightarrow f(x)=f(b) \Rightarrow x^3+ax^2-a^2 x=-{5a^3\over 27} \Rightarrow x^3+ax^2-a^2x+{5a^3\over 27}=0 \\ \Rightarrow x^3+ax^2-a^2x+{5a^3\over 27} =(x-{a\over 3})^2 (x+k) \Rightarrow k={5a\over 3} \Rightarrow  \\ \Rightarrow \text{Area }A= \int_{-5a/3}^{a/3} [f(x)-f(b)]\,dx = \int_{-5a/3}^{a/3}  (x-{a\over 3})^2\left( x+{5a\over 3} \right)\,dx  = \bbox[red, 2pt]{{4\over 3}a^4}$$

解答:$$26+\sqrt{2026-n}=k^2 \Rightarrow \sqrt{2026-n} =k^2-26 \Rightarrow 0\le k^2-26 \le \sqrt{2026-1}=45 \\ \Rightarrow 26\le k^2 \le 71 \Rightarrow \cases{k^2=36 \Rightarrow m=10 \Rightarrow n=2026-100=1926\\ k^2=49 \Rightarrow m=23 \Rightarrow n=2026-529=1497\\ k^2=64 \Rightarrow m=38 \Rightarrow n=2026-1444=582}   \\ \Rightarrow n= \bbox[red, 2pt]{582,1497,1926}$$

解答:$$四面體P的體積V={1\over 6} \begin{vmatrix} 10& 20& 30\\ 20& 20& 30\\ 30& 10& 10 \end{vmatrix} ={500\over 3} \\ 將頂點座標縮小10倍，即\cases{O(0,0,0) \to O'(0,0,0) \\ A(10,20,30) \to A'(1,2,3) \\ B(20,20,30) \to B'(2,2,3) \\ C(30,10,10) \to C'(3,1,1)} \\\Rightarrow 縮小後的四面體P'的體積V'={1\over 6} \begin{vmatrix} 1& 2& 3\\ 2& 2& 3\\3& 1& 1 \end{vmatrix} ={1\over 6} \\ 由於P'只有四個頂點是格子點，內部均無格子點，因此可套用\text{ Ehrhart Polynomials} \\ \Rightarrow L(P,t=10) =C^{t+3}_3 =C^{13}_3= \bbox[red, 2pt]{286}$$

解答:$$\overleftrightarrow{AC}:{x-7\over 2} ={y+1\over -1} ={z-5\over 2} \Rightarrow \overleftrightarrow{AC}的方向向量\vec u_a=(2,-1,2) \\\overleftrightarrow{BD}:{x-1\over 1} ={y+7\over 4} ={z-1\over 1} \Rightarrow \overleftrightarrow{BD}的方向向量\vec u_b=(1,4,1) \\ 又\cases{ \overline{AC} \bot L\\ \overline{BD}\bot L} \Rightarrow L的方向向量\vec u_L平行 (\vec u_a \times \vec u_b) =(-9,0,9) ,取\vec u_L=(1,0,-1) \\ \cases{P(7,-1,5) \in \overleftrightarrow{AC} \\Q(1,-7,1) \in \overleftrightarrow{BD}} \Rightarrow \overrightarrow{PQ} =(-6,-6,-4) \Rightarrow \overline{CD} =\text{proj}_{\vec u_L}  \overline{PQ} = {|-2|\over \sqrt 2} =\sqrt 2 \\ \overrightarrow{AB} =\overrightarrow{AC}+ \overrightarrow{CD}-\overrightarrow{BD} \Rightarrow \overline{AB}^2 = (\overrightarrow{AC}+ \overrightarrow{CD}-\overrightarrow{BD}) \cdot )\overrightarrow{AC}+ \overrightarrow{CD}-\overrightarrow{BD} =4+36+2=42 \\ \Rightarrow \overline{AB}= \bbox[red, 2pt]{\sqrt{42}}$$

解答:$$假設\cases{A(4\cos \alpha, \sqrt 3\sin \alpha) \\C(4\cos \beta, \sqrt 3 \sin \beta)} \Rightarrow \overrightarrow{OB} =\overrightarrow{OA} +\overrightarrow{OC} \Rightarrow B(4(\cos \alpha+ \cos \beta), \sqrt3(\sin \alpha+\sin \beta)) 在橢圓上\\ \Rightarrow {[4(\cos \alpha+ \cos \beta)]^2\over 16}+{ [\sqrt3(\sin \alpha+\sin \beta)]^2 \over 3}=1 \Rightarrow (\cos \alpha+ \cos \beta)^2 +(\sin \alpha+\sin \beta)^2=1 \\ \Rightarrow 2+2\cos(\alpha-\beta)=1 \Rightarrow \cos(\alpha-\beta)=-{1\over 2} \Rightarrow \sin^2(\alpha-\beta) ={3\over 4}\\ 矩形面積= |\overrightarrow{OA}\times \overrightarrow{OB}| = 4\sqrt 3|\cos \alpha \sin \beta- \sin \alpha \cos \beta|= 4\sqrt 3|\sin (\alpha-\beta)| =4\sqrt 3\cdot {\sqrt 3\over 2} = \bbox[red, 2pt]6$$

解答:$$ 柯西不等式:  \left( (\sqrt{a+b})^2+ (\sqrt{c+d})^2 \right) (1^2+1^2) \ge  (\sqrt{a+b} +\sqrt{c+d})^2\\ \Rightarrow 2(a+b+c+d) \ge  (\sqrt{a+b} +\sqrt{c+d})^2 \Rightarrow \sqrt{2(a+b +c +d)} \ge \sqrt{a+b}+\sqrt{c+d} \cdots(1)\\ 又(\sqrt{a^2+b^2} +\sqrt{c^2+d^2})^2 =a^2+b^2 +c^2+d^2 + 2\sqrt{(a^2+b^2)(c^2+d^2)} \\ \ge a^2+b^2+c^2+d^2 +  2 \sqrt{(2ab)(2cd)} =a^2+b^2+c^2+d^2 +4=(a^2+1)+(b^2+1)+(c^2+1)+(d^2+1) \\ \ge 2a+ 2b+2c+2d =2(a+b +c+d) \Rightarrow (\sqrt{a^2+b^2} +\sqrt{c^2+d^2})^2 \ge 2(a+b+c +d) \\ \Rightarrow \sqrt{a^2+b^2} +\sqrt{c^2+d^2} \ge \sqrt{2(a+b+ c+ d)} \cdots(2) \\ 由\cases{\sqrt{2(a+b +c +d)} \ge \sqrt{a+b} +\sqrt{c+d} \cdots(1) \\\sqrt{a^2+b^2} +\sqrt{c^2+d^2} \ge \sqrt{2(a+b+ c+ d)} \cdots(2)} \Rightarrow \sqrt{a^2+b^2} +\sqrt{c^2+d^2} \ge\sqrt{a+b}+\sqrt{c+d}\\ \bbox[red, 2pt]{故得證}$$

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