臺北市立中崙高級中學 113 學年度第 1 次數學科正式教師甄選
壹、填充題(每題 5 分)
解答:¯BC=2⇒¯AD=√3⇒¯AE=√33cos∠DAB=cos30∘=√32=¯AE2+¯AB2−¯BE22⋅¯AE⋅¯AB=1/3+4−¯BE24√3/3⇒¯BE2=73⇒cos∠DEB=43+73−143√7=2√7⇒sin∠DEB=3√7⇒¯BF=¯BDcos∠DBE=¯BDsin∠EDB=1⋅3√7=√217
解答:56×5×4×(24+24)+56×5×(24+48)=73920
解答:可能n值之和=138+140+141+142+143+145=849
解答:x2+y2=1⇒x2=1−y2代回曲線⇒(1−y2)+2xy+2y2−2x−2y=0⇒y2−2y+1+2xy−2x=0⇒(y−1)2+2x(y−1)=0⇒(y−1)(y−1+2x)=0⇒{y=1⇒x=0y=1−2x⇒x(5x−4)=0⇒交點(x,y)=(0,1),(45,−35)
解答:z+w=2√33i⇒取{z=a+(√33+b)iw=−a+(√33−b)i⇒z2+w2=2a2−23−2b2+4abi=2√33i⇒{a2−b2=13ab=√36⇒a2−(√36a)2=13⇒12a4−4a2−1=0⇒(6a2+1)(2a2−1)=0⇒a2=12⇒a=√22
解答:(3x)2−2(m+1)⋅3x−(m−1)=0⇒3x=(m+1)±√m2+3m由於{兩相異實根⇒m2+3m>0⇒m>0或m<−33x>0⇒m+1>√m2+3m⇒m<1⇒0<m<1
解答:
解答:56×5×4×(24+24)+56×5×(24+48)=73920
解答:可能n值之和=138+140+141+142+143+145=849
解答:x2+y2=1⇒x2=1−y2代回曲線⇒(1−y2)+2xy+2y2−2x−2y=0⇒y2−2y+1+2xy−2x=0⇒(y−1)2+2x(y−1)=0⇒(y−1)(y−1+2x)=0⇒{y=1⇒x=0y=1−2x⇒x(5x−4)=0⇒交點(x,y)=(0,1),(45,−35)
解答:z+w=2√33i⇒取{z=a+(√33+b)iw=−a+(√33−b)i⇒z2+w2=2a2−23−2b2+4abi=2√33i⇒{a2−b2=13ab=√36⇒a2−(√36a)2=13⇒12a4−4a2−1=0⇒(6a2+1)(2a2−1)=0⇒a2=12⇒a=√22
解答:(3x)2−2(m+1)⋅3x−(m−1)=0⇒3x=(m+1)±√m2+3m由於{兩相異實根⇒m2+3m>0⇒m>0或m<−33x>0⇒m+1>√m2+3m⇒m<1⇒0<m<1
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圓C:(x−2)2+(y−2)2=8−k⇒{圓心M(2,2)圓半徑r=√8−kA(−3,3)對稱於x軸的對稱點A′(−3,−3)⇒L=↔A′M:x=y⇒L為∠PA′Q的角平分線⇒L∩x軸=O(0,0)⇒{¯PO=a¯QO=7/4−a,其中{P(−a,0),Q(−a+74,0)⇒¯A′P¯A′Q=¯PO¯QO⇒√(−3+a)2+9√(a−19/4)2+9=a7/4−a⇒4a2−31a+21=0⇒a=3/4(a=7不合)⇒P(−34,0)⇒L2=↔A′P:3y=4x+3⇒r=d(M,L2)=1=√8−k⇒k=7
解答:{A在L1:x+2y=0上⇒A(−2a,a)B在L2:x+y=0上⇒B(b,−b)⇒交點為重心G(0,0)⇒C(2a−b,−a+b)⇒{→CA=(−4a+b,2a−b)→CB=(−2a+2b,a−2b)⇒→CA⋅→CB=0⇒10a2−15ab+4b2=0⋯(1)又¯AB=36⇒(−2a−b)2+(a+b)2=362⇒5a2+6ab+2b2=362⋯(2)由(1)及(2)可得ab=2⋅36227⇒△ABC面積=12‖−2aa1b−b12a−b−a+b1‖=32ab=32⋅2⋅36227=144
解答:cosθ=(1+cos2θ2)1/2⇒a0=cosπ3=12⇒a1=cosπ3⋅2=(1+(1/2)2)1/2=√32⇒a2=cosπ3⋅22⇒⋯⇒an=cosπ3⋅2n=1−(π3⋅2n)22!+(π3⋅2n)44!−⋯ (Taylor series)⇒lim
解答:3\sin 2x+4\cos 2x=1 \Rightarrow (4\cos 2x)^2= (1-3\sin 2x)^2 \Rightarrow 16\cos^2 2x=1-6\sin 2x+9\sin^2 2x \\ \Rightarrow 16(1-\sin^2 2x)=1-6\sin 2x+9\sin^2 2x \Rightarrow 25\sin^2 2x-6\sin 2x-15=0 \\\Rightarrow \sin 2\alpha \sin 2\beta=-{3\over 5} \cdots(1)\\ 同理,(3\sin 2x)^2=(1-4\cos 2x)^2 \Rightarrow 9\sin^2 2x= 9(1-\cos^2 2x)=1-8\cos 2x+ 16\cos^2 2x \\ \Rightarrow 25\cos^2 2x-8\cos 2x-8=0 \Rightarrow \cos 2\alpha \cos 2\beta=-{8\over 25} \cdots(2)\\ 由(1)及(2)可得 \cos 2(\alpha+\beta)= \cos 2\alpha \cos 2\beta-\sin 2\alpha \sin 2\beta={7\over 25} \\ \Rightarrow \sin 2(\alpha+\beta)= \sqrt{1-\cos^2 2(\alpha+\beta)} =\sqrt{1-{49\over 625}} = \bbox[red, 2pt]{24\over 25}
解答:X=7 \Rightarrow 111111排列數=1\\ X=6 \Rightarrow 211111排列數={6!\over 5!}=6\\ X=5\Rightarrow \cases{11113排列數=5!/4!=5\\ 11122排列數= 5!/(3!2!)= 10} \Rightarrow 合計15\\ X=4 \Rightarrow \cases{1114排列數=4!/3!=4\\ 1123排列數=4!/2!=12\\ 1222排列數=4} \Rightarrow 合計20 \\ X=3 \Rightarrow \cases{115排列數=3!/2!=3\\ 124排列數=6\\ 133排列數=3\\ 223排列數=3} \Rightarrow 合計15 \\X=2 \Rightarrow \cases{16 排列數=2\\ 25排列數=2\\ 34排列數=2} \Rightarrow 合計6\\ X=1 \Rightarrow 7排列數=1\\ \Rightarrow E(X)={1\cdot 1+2\cdot 6+ 3 \cdot 15+ 4\cdot 20+ 5\cdot 15+6\cdot 6+ 7\cdot 1\over 1+6+15+20+15+6+1} ={256\over 64} =\bbox[red, 2pt]4
解答:f(x)-2f({1\over x})-{3\over x}=0 \Rightarrow f({1\over x})={1\over 2}\left( f(x)-{3\over x}\right) \cdots(1)\\又f({1\over x})-2f(x)-3x=0 \Rightarrow f({1\over x})=2f(x)+3x \cdots(2)\\ 由(1)及(2)可得f({1\over x})={1\over 2}\left( f(x)-{3\over x}\right) =2f(x)+3x \Rightarrow f(x)=-2x-{1\over x} \\ 令g(x)=(f(x))^2 =4x^2+{1\over x^2}+4 \Rightarrow g'(x)=0 \Rightarrow x^2={1\over 2} \Rightarrow g(x)=4\cdot {1\over 2}+2+4= \bbox[red, 2pt]8
解答:a_1=2^{20}-24(偶數) \Rightarrow a_4=2^{17}-3(奇數) \Rightarrow a_5=2^{17}-4 \Rightarrow a_7=2^{15}-1 \\ \Rightarrow a_8=2^{15}-2 \Rightarrow a_9=2^{14}-1 \Rightarrow a_{11}=2^{13}-1 \Rightarrow a_{13}= 2^{12}-1 \Rightarrow a_{33}=2^2-1\\ \Rightarrow a_{35}=2-1=1 \Rightarrow M=\bbox[red, 2pt]{35}
解答:
解答:{A在L1:x+2y=0上⇒A(−2a,a)B在L2:x+y=0上⇒B(b,−b)⇒交點為重心G(0,0)⇒C(2a−b,−a+b)⇒{→CA=(−4a+b,2a−b)→CB=(−2a+2b,a−2b)⇒→CA⋅→CB=0⇒10a2−15ab+4b2=0⋯(1)又¯AB=36⇒(−2a−b)2+(a+b)2=362⇒5a2+6ab+2b2=362⋯(2)由(1)及(2)可得ab=2⋅36227⇒△ABC面積=12‖−2aa1b−b12a−b−a+b1‖=32ab=32⋅2⋅36227=144
解答:cosθ=(1+cos2θ2)1/2⇒a0=cosπ3=12⇒a1=cosπ3⋅2=(1+(1/2)2)1/2=√32⇒a2=cosπ3⋅22⇒⋯⇒an=cosπ3⋅2n=1−(π3⋅2n)22!+(π3⋅2n)44!−⋯ (Taylor series)⇒lim
解答:3\sin 2x+4\cos 2x=1 \Rightarrow (4\cos 2x)^2= (1-3\sin 2x)^2 \Rightarrow 16\cos^2 2x=1-6\sin 2x+9\sin^2 2x \\ \Rightarrow 16(1-\sin^2 2x)=1-6\sin 2x+9\sin^2 2x \Rightarrow 25\sin^2 2x-6\sin 2x-15=0 \\\Rightarrow \sin 2\alpha \sin 2\beta=-{3\over 5} \cdots(1)\\ 同理,(3\sin 2x)^2=(1-4\cos 2x)^2 \Rightarrow 9\sin^2 2x= 9(1-\cos^2 2x)=1-8\cos 2x+ 16\cos^2 2x \\ \Rightarrow 25\cos^2 2x-8\cos 2x-8=0 \Rightarrow \cos 2\alpha \cos 2\beta=-{8\over 25} \cdots(2)\\ 由(1)及(2)可得 \cos 2(\alpha+\beta)= \cos 2\alpha \cos 2\beta-\sin 2\alpha \sin 2\beta={7\over 25} \\ \Rightarrow \sin 2(\alpha+\beta)= \sqrt{1-\cos^2 2(\alpha+\beta)} =\sqrt{1-{49\over 625}} = \bbox[red, 2pt]{24\over 25}
解答:X=7 \Rightarrow 111111排列數=1\\ X=6 \Rightarrow 211111排列數={6!\over 5!}=6\\ X=5\Rightarrow \cases{11113排列數=5!/4!=5\\ 11122排列數= 5!/(3!2!)= 10} \Rightarrow 合計15\\ X=4 \Rightarrow \cases{1114排列數=4!/3!=4\\ 1123排列數=4!/2!=12\\ 1222排列數=4} \Rightarrow 合計20 \\ X=3 \Rightarrow \cases{115排列數=3!/2!=3\\ 124排列數=6\\ 133排列數=3\\ 223排列數=3} \Rightarrow 合計15 \\X=2 \Rightarrow \cases{16 排列數=2\\ 25排列數=2\\ 34排列數=2} \Rightarrow 合計6\\ X=1 \Rightarrow 7排列數=1\\ \Rightarrow E(X)={1\cdot 1+2\cdot 6+ 3 \cdot 15+ 4\cdot 20+ 5\cdot 15+6\cdot 6+ 7\cdot 1\over 1+6+15+20+15+6+1} ={256\over 64} =\bbox[red, 2pt]4
解答:f(x)-2f({1\over x})-{3\over x}=0 \Rightarrow f({1\over x})={1\over 2}\left( f(x)-{3\over x}\right) \cdots(1)\\又f({1\over x})-2f(x)-3x=0 \Rightarrow f({1\over x})=2f(x)+3x \cdots(2)\\ 由(1)及(2)可得f({1\over x})={1\over 2}\left( f(x)-{3\over x}\right) =2f(x)+3x \Rightarrow f(x)=-2x-{1\over x} \\ 令g(x)=(f(x))^2 =4x^2+{1\over x^2}+4 \Rightarrow g'(x)=0 \Rightarrow x^2={1\over 2} \Rightarrow g(x)=4\cdot {1\over 2}+2+4= \bbox[red, 2pt]8
貳、計算證明題(每小題 10 分)
解答:x^2+x+1=0的兩根為\omega,\omega^2 且\omega^3=1\\ 由於\cases{(\omega+2)^2=\omega^2+4\omega+4=3\omega+3=3(\omega+1)\\ (\omega+1)^2=\omega^2+2\omega +1=\omega}, 因此f(\omega)=(\omega+2)^{24}-(\omega+1)^{24} \\=(3(\omega+1))^{12} -(\omega+1)^{24} =3^{12}(\omega+1)^{12}-(\omega+1)^{24}=3^{12}\omega^6-\omega^{12}= \bbox[red, 2pt]{ 3^{12}-1}解答:a_1=2^{20}-24(偶數) \Rightarrow a_4=2^{17}-3(奇數) \Rightarrow a_5=2^{17}-4 \Rightarrow a_7=2^{15}-1 \\ \Rightarrow a_8=2^{15}-2 \Rightarrow a_9=2^{14}-1 \Rightarrow a_{11}=2^{13}-1 \Rightarrow a_{13}= 2^{12}-1 \Rightarrow a_{33}=2^2-1\\ \Rightarrow a_{35}=2-1=1 \Rightarrow M=\bbox[red, 2pt]{35}
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