106年臺北市華江高中教甄-數學詳解

 臺北市立華江高中106學年度正式教師甄選數學科試題

解：

全部-三數字相同-三數字全不同\(=900-9-9\cdot 9\cdot 8=900-9-648=\bbox[red,2pt]{243}\)

解：$$過A(3,5)直線方程式:y=a(x-3)+5, a為斜率且a< 0 \Rightarrow \cases{P({3a-5\over a},0)\\ Q(0,5-3a)} \\ \Rightarrow \overline{OP} +\overline{OQ} = {3a-5\over a}+5-3a = 8-3a-{5\over a} \equiv f(a) \\ \Rightarrow f'(a)=0 \Rightarrow -3+{5\over a^2}=0 \Rightarrow a^2={5\over 3} \Rightarrow a=-\sqrt{5\over 3} \\\Rightarrow f(-\sqrt{5\over 3})= 8+ 3\sqrt{5\over 3} +5\sqrt{ 3\over 5} =8+\sqrt{15}+ \sqrt{15} =\bbox[red,2pt]{8+2\sqrt{15}}$$

解：

$$\cases{A(4,0)\\ B(10,15) \\C(16,12)} \Rightarrow \cases{L_1=\overleftrightarrow{AB}: 5x-2y=20\\ L_2= \overleftrightarrow{AC}: -x+y=-4\\ L_3=\overleftrightarrow{AB}:x+2y=40\\ L_3':2x+y=44} \Rightarrow \cases{L_3'與L_2交於C(16,12)\\ L_3'與L_1交於B'(12,20)}\\ 又\cases{f(C)=96\\ f(A)=12} \Rightarrow \cases{16p+12q=96 \\ 4p=12} \Rightarrow \cases{p=3\\ q=4} \Rightarrow f(x,y)=3x+4y; \\ \Rightarrow \cases{f(A)=12\\ f(B')=116\\ f(C)=96} \Rightarrow 最大值為\bbox[red,2pt]{116}$$

解：

$$令\cases{\angle A=2\theta \\ \overline{CD}=1\\ \overline{BD}=k}，又\cos \angle A={4\sqrt 3\over 7} \Rightarrow \sin \angle A={1\over 7}\\在\triangle DAB中，\frac{k}{\sin \theta}= \frac{\overline{AD}}{ \sin 30^\circ} \Rightarrow \overline{AD}= \frac{k}{2\sin \theta}\cdots(1);\\ \triangle ACD: \frac{\overline{CD}}{\sin \theta} =\frac{\overline{AD}}{\sin \angle C} \Rightarrow {1\over \sin \theta}= {\overline{AD} \over \sin (150^\circ- \angle A)} \Rightarrow \overline{AD} = {\sin (150^\circ- \angle A)\over \sin \theta} \cdots(2)\\ (1)=(2) \Rightarrow k=2\sin (150^\circ- \angle A) =2(\sin 150^\circ \cos \angle A-\sin \angle A\cos 150^\circ) \\=2({1\over 2}\cdot {4\sqrt 3\over 7}-{1\over 7}\cdot (-{\sqrt 3\over 2})) =  \bbox[red,2pt]{5\sqrt 3\over 7}$$

解：

$$(x^2+y^2-4x)(y^2-x-7)=0 \Rightarrow \cases{x^2+y^2-4x=0 \\y^2-x-7=0} \Rightarrow \cases{(x-2)^2+y^2=2^2 \\ y^2=x+7}，\\其中(x-2)^2+y^2=2^2為一圓，圓心O(2,0),半徑r=2; x=2代入y^2=x+7=9 \Rightarrow y=\pm 3 \\ \Rightarrow 該圓完全在拋物線內，如上圖；\\又直線L:mx-y+4-2m=0 \Rightarrow y=m(x-2)+4 \Rightarrow L過P(2,4);\\ 因此L與拋物線與圓有四個交點，代表L介於P於圓的兩切線之間；\\ L為切線\Rightarrow \text{dist}(O,L)=r \Rightarrow \frac{4}{\sqrt{m^2+1}} =2 \Rightarrow m=\pm \sqrt 3 \Rightarrow \bbox[red,2pt]{m\gt \sqrt 3或m\lt -\sqrt 3}有4個相異交點$$

解：$$\cos C={a^2+b^2-c^2\over 2ab} ={a^2+b^2-{1\over 3}(a^2+b^2) \over 2ab} ={a^2+b^2\over 3ab } \ge {2ab\over 3ab}={2\over 3}\\ \Rightarrow \cos C \ge {2\over 3} \Rightarrow \sin C \le \sqrt{1-({2\over 3})^2}= \bbox[red,2pt]{\sqrt 5\over 3}$$

解：$$假設\cases{a:選到水餃的機率\\ b:選到便當的機率\\ c:選到湯麵的機率} \\\Rightarrow \cases{a_n=0\cdot a_{n-1}+{1\over 4}b_{n-1}+ {1\over 3}c_{n-1} \\ b_n={2\over 5}a_{n-1}+ 0\cdot b_{n-1} +{2\over 3}c_{n-1} \\ c_n ={3\over 5}a_{n-1}+{3\over 4}b_{n-1}+0\cdot c_{n-1}} \Rightarrow \begin{bmatrix} 0 & 1/4 & 1/3\\ 2/5& 0 & 2/3 \\ 3/5 & 3/4 & 0 \end{bmatrix}\begin{bmatrix} a\\b \\c\end{bmatrix}= \begin{bmatrix} a\\b\\ c\end{bmatrix} \\ \Rightarrow \begin{bmatrix} 0 & 1/4 & 1/3\\ 2/5& 0 & 2/3 \\ 3/5 & 3/4 & 0 \end{bmatrix}\begin{bmatrix} a\\b \\1-a-b\end{bmatrix}= \begin{bmatrix} a\\b\\ 1-a-b\end{bmatrix}\Rightarrow \cases{a=5/22 \\ b=4/11} \\ \Rightarrow 選到便當的機率為4/11 \Rightarrow 選到排骨便當的機率為\bbox[red,2pt]{2\over 11}$$

解：

$$\frac{x^2}{16} -\frac{y^2}{12}=1 \Rightarrow \cases{a=4\\b=2\sqrt 3} \Rightarrow c=\sqrt{16+12} =2\sqrt 7 \Rightarrow \cases{F_1(-2\sqrt 7,0)\\ F_2(2\sqrt 7,0)} \\ 令\cases{\overline{PF_2}=m \\ \overline{QF_2}=n} \Rightarrow \cases{\overline{PF_1}=m+2a= m+8 \\ \overline{QF_1}=n+2a = n+8} \\ \triangle PF_1F_2: \cos \angle F_1PF_2 = \cos 60^\circ =\frac{1}{2} =\frac{(m+8)^2+m^2 -(4\sqrt 7)^2}{2m(m+8)} \Rightarrow m=4\\ \triangle PQF_1: \cos \angle F_1PF_2 =\frac{1}{2} =\frac{(m+8)^2+(m+n)^2 -(n+8)^2}{2(m+n)(m+8)} =\frac{144+(n+4)^2-(n+8)^2}{24(n+4)}\\ \Rightarrow n={12\over 5}  \Rightarrow \triangle PQF_1周長= (m+8)+m+n+(n+8)= 2(m+n)+16 = {64\over 5}+16\\ =\bbox[red,2pt]{144\over 5}$$

解：$$\cases{A=7 = \int_3^0 f'(x)\;dx = f(0)-f(3)=10-f(3) \Rightarrow f(3)=3 \\ B=6=\int_6^3 f'(x)\;dx=f(3)-f(6) = 3-f(6) \Rightarrow f(6)=-3\\ C=4 = \int_6^7 f'(x)\;dx = f(7)-f(6)=f(7)+3 \Rightarrow f(7)=1}\\ 切點(7,g(7))的切線方程式:\;y=g'(7)(x-7)+g(7)= {d\over dx}[f(7)]^2(x-7) +[f(7)]^2 \\=2f(7)f'(7)(x-7)+f(7) = 2\cdot 1\cdot 2(x-7)+1=4x-27\\ \Rightarrow 切線方程式:\;\bbox[red,2pt]{y=4x-27}$$

解：$$正弦定理:\frac{\overline{AC}}{\sin \angle B} = \frac{\overline{BC}}{\sin \angle A} \Rightarrow \frac{3}{\sin \angle B} = \frac{\overline{BC}}{\sin 2\angle B} \Rightarrow \overline{BC} = \frac{3\sin 2\angle B}{\sin \angle B} =6\cos \angle B\\ 餘弦定理: \cos \angle B = \frac{\overline{AB}^2+ \overline{BC}^2-\overline{AC}^2}{2\cdot \overline{AB}\cdot \overline{BC}} =\frac{25 +36\cos^2\angle B-9}{60 \cos \angle B} \Rightarrow \cos \angle B=\frac{\sqrt 6}{3}\\ \Rightarrow \overline{BC}= 6\cos \angle B=2\sqrt 6 \Rightarrow S=(\overline{AB} +\overline{BC}+\overline{CA})\div 2= 4+\sqrt 6 \\ \triangle ABC 面積= \sqrt{S(S-\overline{AB})(S-\overline{BC})(S-\overline{CA})} =\sqrt{(4+\sqrt 6)(4-\sqrt 6)(\sqrt 6+1)(\sqrt 6-1)} \\ =\sqrt{50}=5\sqrt 2 ={1\over 2}r(\overline{AB} +\overline{BC}+\overline{CA}) =r(4+\sqrt 6) \Rightarrow r=\frac{5\sqrt 2}{4+\sqrt 6} =\bbox[red,2pt]{2\sqrt 2-\sqrt 3}$$

解：

此題相當求\(a+b+c+d+e \le 10\)的非負整數解的個數，又相當於\(a+b+ c+d+e+f=10\)的非負整數解的個數;也就是\(H^6_{10}= C^{15}_{10}=3003\)；需扣除(a,b,c,d,e)=(0,0,0,0,0),(10,0,0,0,0),(0,10,0,0,0),(0,0,10,0,0),(0,0,0,10,0),(0,0,0,0,10)六組解，再加上六位數100000這一組解，即\(n=3003-6+1= \bbox[red,2pt]{2998}\)

解：$$2節體育課可選(1,3),(1,4),(1,5),(2,4),(2,5),(3,5)，有6種排法；\\
若音樂課與美術課都與體育課同一天:2種排法；\\
若音樂課與美術課都不與體育課同一天:3\times 3=9種排法；\\
若音樂課與體育課同一天，但美術課不與體育課同一天:3\times 2=6種排法\\
若美術課與體育課同一天，但音樂課不與體育課同一天:3\times 2=6種排法\\
共有6(2+9+6+6)= \bbox[red,2pt]{138}種排法$$

解：$$\cases{\stackrel{\rightharpoonup}{a}\times \stackrel{\rightharpoonup}{b} =(-2,2,1)\\ \stackrel{\rightharpoonup}{a}\times \stackrel{\rightharpoonup}{c} =(2,1,2)} \Rightarrow (-2,2,1) \times (2,1,2) = (3,6,-6) \Rightarrow \stackrel{\rightharpoonup}{a}=(k,2k,-2k)\\ 又|\stackrel{\rightharpoonup}{a}|=6 \Rightarrow \sqrt{k^2+4k^2+4k^2} =6 \Rightarrow k^2= 4\\ 平行四邊形面積= \sqrt{|\stackrel{\rightharpoonup}{a}|^2|\stackrel{\rightharpoonup}{u}|^2-(\stackrel{\rightharpoonup}{a} \cdot \stackrel{\rightharpoonup}{u})^2} = \sqrt{36\cdot 14-81k^2} =\sqrt{504-324} =\bbox[red,2pt]{6\sqrt 5}$$

解：$$\cases{y=\log_2(kx^2)+{3x\over 4} \\ y=2^{|x|}+{3x\over 4}}，交集\Rightarrow \log_2(kx^2)+{3x\over 4}=2^{|x|}+{3x\over 4} \\ \Rightarrow \log_2(kx^2)=2^{|x|} ，相當於求兩圖形\cases{y=\log_2(kx^2) \\y=2^{|x|}}的交點；\\由於兩圖形皆對稱Y軸，因此可令兩交點為\cases{A(x,\log_2(kx^2)+{3x\over 4})\\ B(-x,\log_2(kx^2)-{3x\over 4})}\\ \overline{AB}=10 \Rightarrow \sqrt{(2x)^2 + ({3x\over 2})^2} =10 \Rightarrow x^2=16 \Rightarrow x=\pm 4 \Rightarrow \log_2(16k)=2^4 \\ \Rightarrow 4+\log_2 k=16 \Rightarrow k=2^{12}= \bbox[red,2pt]{4096}$$

解：$$\sqrt{x^4-15x^2-4x+68} -\sqrt{x^4-3x^2-2x+5} \\=\sqrt{(x^4-16x^2+64)+(x^2-4x+4)} -\sqrt{(x^4-4x^2+4)+(x^2-2x+1)} \\= \sqrt{(x^2-8)^2 +(x-2)^2} -\sqrt{(x^2-2)^2 +(x-1)^2} =\overline{PA}-\overline{PB},其中\cases{P(x,x^2)\\ A(2,8)\\ B(1,2)}\\ P(x,x^2)代表拋物線\;\Gamma:y=x^2上的點，因此最大值發生在P為\overleftrightarrow{AB}與\Gamma的交點；\\ 直線L:\overleftrightarrow{AB}方程式\;y=6(x-1)+2 \Rightarrow y=6x-4 \Rightarrow \cases{y=x^2\\ y=6x-4}的交點(3\pm \sqrt 5,14\pm 6\sqrt 5)\\ 若P(3+\sqrt 5,14+6\sqrt 5)，則\overline{PA}-\overline{PB} < 0;故取P(3-\sqrt 5,14-6\sqrt 5)，因此x=\bbox[red,2pt]{3-\sqrt 5}$$