112年台師大附中教甄-數學詳解(待續)

國立台灣師大附中 111 學年度 數學科 教師甄選

一．填充題(1-6每格7分，7-12每格8分共90分)

解答: $$\sin(41^\circ)= \sin(60^\circ-19^\circ)={\sqrt 3\over 2} \cos 19^\circ-{1\over 2}\sin 19^\circ \\ \Rightarrow \cases{\sin^2(41^\circ) = {3\over 4}\cos^2 19^\circ-{\sqrt 3\over 2}\sin 19^\circ \cos 19^\circ +{1\over 4} \sin^2 19^\circ \\ \sin 41^\circ \sin 19^\circ = {\sqrt 3\over 2}\sin 19^\circ \cos 19^\circ-{1\over 2}\sin^2 19^\circ} \\ \Rightarrow \sin^2 41^\circ +\sin^2 19^\circ+ \sin 41^\circ \sin 19^\circ ={3\over 4}\cos^2 19^\circ+ {3\over 4} \sin^2 19^\circ= \bbox[red, 2pt]{3\over 4}$$

解答: $$\cot A,\cot B,\cot C成等差\Rightarrow \cos A+\cot C= 2\cot C\\ 左式:{\cos A\over \sin A}+ {\cos C\over \sin A}= {\sin (A+C)\over \cos A\cos C} ={\sin B\over \cos A\cos C};\quad 右式:2\cot B ={2\cos B\over \sin B}\\ 左式=右式\Rightarrow \cos B={\sin^2 B\over 2\sin A\sin C} \Rightarrow {a^2+c^2 -b^2 \over 2ac} ={b^2 \over 2ac} (左式用餘弦定理,右式用正弦定理) \\ \Rightarrow a^2+c^2 =2b^2,將\cases{a=\overline{BC}=9\\ c=\overline{AB}=5}代入左式可得: 2b^2=9^2+5^2 \Rightarrow b^2=53 \Rightarrow \cos B={53\over 59} \\\Rightarrow \tan^2{B\over 2} ={1-\cos B\over 1+\cos B} ={90-53\over 90+53} =\bbox[red, 2pt]{37\over 143}$$

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$$\angle C=\theta \Rightarrow \angle B=2\theta \Rightarrow \angle A=\pi-\theta-2\theta=\pi -3\theta\\ 正弦定理: {\overline{AB}\over \sin \angle A} ={\overline{AC} \over \sin \angle B} \Rightarrow {25\over \sin \theta} ={30 \over \sin 2\theta} \Rightarrow \cos \theta =   {3 \over 5}\\ 又對同弧的圓周角相等\Rightarrow \cases{\angle D=\angle A\\ \angle B=\angle DCB} \Rightarrow \triangle ABC \cong \triangle DCB(ASA) \\ \Rightarrow \triangle BCD面積=\triangle ABC 面積= {1\over 2}\overline{AB}\cdot \overline{AC}\sin \angle A={1\over 2}\cdot 25\cdot 30 \cdot \sin(\pi-3\theta) \\ =375\sin 3\theta =375(3\sin \theta-4\sin^3\theta) =375({12\over 5}-4\cdot {64\over 125})=\bbox[red, 2pt]{132}$$

解答: $$取\cases{A(0,0,0) \\B(6,0,0)\\ C(6,2\sqrt 3,0)\\ D(0,2\sqrt 3,0)},又D'垂足在\overline{AB}上 \Rightarrow D'(a,0,b) \\ \cases{\overline{D'A} =\overline{DA}=2\sqrt 3 \\ \overline{D'C} =\overline{DC}=6} \Rightarrow \cases{a^2+b^2 =12\\ (a-6)^2+12+b^2=36} \Rightarrow a=2 \Rightarrow b=2\sqrt 2 \Rightarrow D'(2,0,2\sqrt 2) \\ \Rightarrow \cases{\overrightarrow{D'A} =(-2,0,-2 \sqrt 2)\\ \overrightarrow{D'C} =(-4,-2\sqrt 3, 2\sqrt 2)} \Rightarrow \vec n_1=\overrightarrow{D'A} \times \overrightarrow{D'C} =(-4\sqrt 6,12\sqrt 2, 4\sqrt 3)\\ \triangle ABC平面法向量\vec n_2=(0,0,1) \Rightarrow \cos \theta={\vec n_1\cdot \vec n_2\over |\vec n_1||\vec n_2}={1\over 3} \Rightarrow \tan \theta =\bbox[red, 2pt]{2\sqrt 2}$$

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$$\cases{x^2+y^2 =18 \\y^2+\sqrt 3yz+z^2 =13 \\ x^2+xz+ z^2=19}\Rightarrow \cases{x^2+y^2-2xy\cos 90^\circ=(3\sqrt 2)^2\\ y^2+z^2-2yz\cos 150^\circ=(\sqrt{13})^2 \\ x^2+z^2-2xz\cos 120^\circ= (\sqrt{19})^2}\\ 因此我們可以建構\triangle ABC滿足\cases{\overline{AB}=\sqrt{18}\\ \overline{BC}= \sqrt{13} \\ \overline{AC}=\sqrt{19}}且\cases{\angle AOB=90^\circ\\ \angle BOC=150^\circ \\ \angle AOC=120^\circ} 及\cases{\overline{OA}=x\\ \overline{OB}=y \\ \overline{OC}=z},如上圖\\ 令s=(\sqrt{18}+\sqrt{13}+ \sqrt{19})\div 2 \Rightarrow \triangle ABC面積=\sqrt{s(s-\sqrt{18})( s-\sqrt{13})(s-\sqrt{19})} =3\sqrt{11\over 2} \\ \triangle ABC面積也等於={1\over 2}(xy\sin 90^\circ+yz\sin 150^\circ +zx\sin 120^\circ) ={1\over 2}( xy+ {1\over 2}yz+ {\sqrt 3\over 2}xz)\\ 因此{1\over 2}( xy+ {1\over 2}yz+ {\sqrt 3\over 2}xz) =3\sqrt{11\over 2} \Rightarrow 2xy+yz+ \sqrt 3xz = 4\cdot {3\sqrt{11}\over \sqrt 2}= \bbox[red, 2pt]{6 \sqrt{22}}$$

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解答: $$假設三次多項式的根為{z_1\over z_2},{z_2\over z_3},{z_3\over z_1} \Rightarrow \cases{三根之和={z_1\over z_2}+{z_2\over z_3}+{z_3\over z_1}=1 \\ 三根之積={z_1\over z_2}\cdot {z_2\over z_3} \cdot {z_3\over z_1} =1}\\ 又三根兩兩積之和={z_1\over z_2}\cdot {z_2\over z_3}+ {z_2\over z_3}\cdot{z_3\over z_1}+ {z_3\over z_1}\cdot {z_1\over z_2}={z_1\over z_3}+ {z_2\over z_1}+ {z_3\over z_2} \\=\overline{({z_3\over z_1})} + \overline{({z_1\over z_2})} +\overline{({z_2\over z_3})}   =\overline{\left( {z_1\over z_2}+{z_2\over z_3}+{z_3\over z_1}\right)} =\bar 1=1 \\因此該多項式為f(x)=x^3-x^2+x-1 =(x^2+1)(x-1)=0 \Rightarrow x=1,\pm i\\ 為求|z_1+ 2z_2+ 3z_3|之最大化,取\cases{z_2/z_3=1\\ z_3/z_1=i\\ z_1/z_2=-i} \Rightarrow \cases{z_2=z_3=1\\ z_1=-i} \\\Rightarrow |z_1+ 2z_2+ 3z_3|=|-i+5|= \bbox[red, 2pt]{\sqrt{26}}$$

解答: $$柴比雪夫不等式:P(\mu-k\sigma\lt X\lt \mu+k\sigma)\gt 1-{1\over k^2}\\ P(50\lt X\lt 70)=P(60-2\cdot 5\lt X\lt 60+2\cdot 5)\gt 1-{1\over 2^2}={3\over 4}\\ \Rightarrow 人數\gt 36\times {3\over 4}= 27 \Rightarrow 至少\bbox[red, 2pt]{28}人$$

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解答: $$X\sim B(n=50,p={4\over 10}) 從試題的附表可得\cases{P(X\le 12)=0.0133\\ P(X\le 13)=0.028\\ P(X\le 14) =0.054} \\ 也就是只要\bbox[red,2pt]{0\le X\le13},其機率就會\le5\%,就能拒絕H_0$$

解答: $$\mathbf{(1)}\; f(x)=x^2-2 \Rightarrow \cases{f'(x)=2x \\ f(a_1=2)=2}\Rightarrow f'(a_1=2)=4 \\ \Rightarrow 經過(a_1,f(a_1))的切線L_1:y=4(x-2)+2 \Rightarrow L_1與x軸交於({3\over 2},0) \Rightarrow a_2={3\over 2}\\ 同理\cases{f(a_2)={1\over 4} \\ f'(a_2)=3} \Rightarrow 經過(a_2,f(a_2))的切線L_2: y=3(x-{3\over 2})+{1\over 3} \\ \Rightarrow L_2與x軸交於({17\over 12},0) \Rightarrow a_3={17\over 12},因此\bbox[red,2pt]{\cases{a_2=3/2\\ a_3=17/12}}\\ \mathbf{(2)} f(x)=4x^3-12x^2+12x-3 \Rightarrow f'(x)=12x^2-24x+12 \\ 若f'(x)=0 \Rightarrow f'(x)=12(x-1)^2=0 \Rightarrow x=1\\若切線為水平線就無法找到a_3,也就是f'(a_2)=0 \Rightarrow a_2=1\\ 假設過(a_1,f(a_1))的切線L:y=mx+b,由於L通過(1,0) \Rightarrow b=-m\\ \Rightarrow L:y=mx-m \Rightarrow y=f'(x)x-f'(x) ,將(a_1,f(a_1))代入L \Rightarrow f(a_1)=f'(a_1)a_1-f'(a_1) \\ \Rightarrow 4a_1^3-12a_1^2+12a_1-3 = (12a_1^2-24a_1+12)(a_1-1) \\ \Rightarrow (2a_1-3)(4a_1^2-6a_1+3) =0 \Rightarrow a_1=\bbox[red, 2pt]{3\over 2}$$

 

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解題僅供參考,其他試題及詳解