國立新竹高商106學年度第一次數學科教師甄選
一、填充題
解:x4+ax2+bx+c=(x−1)3(x−c)=(x3−3x2+3x−1)(x−c)⇒x3係數=0=−c−3⇒c=−3⇒{a=3c+3=−6b=−3c−1=8⇒(a,b,c)=(−6,8,−3)解:
此題相當於求兩圖形{y=f(x)=x/3y=g(x)=cos(πx)有幾個交點{g(x)=1⇒x=2k,k∈Zg(x)=−1⇒x=2k+1,k∈Z及{f(x)≥1⇒x≥3f(x)≤−1⇒x≤−3⇒x在區間[−3,3)有交點且在x>0有3個交點,x<0有4個交點,共7個交點
解:{¯AB=10sinθ¯BO=10cosθ⇒¯BH=¯ABׯBO÷¯OA=100sinθcosθ÷10=10sinθcosθ⇒¯AB+¯BO+¯BH=10(sinθ+cosθ)+10sinθcosθ=10√2sin(θ+45∘)+5sin(2θ)顯然θ=45∘時,有最大值10√2+5
解:{n=1m=2⇒7n2−m2=7−4=3為最小值此題並無特定解法!
解:{21x+22y+27z=50⋯(1)22x+23y+28z=51⋯(2)23x+24y+25z=52⋯(3)(2)−(1),(3)−(2)→{x+y+z=1x+y−3z=1⇒{z=0x+y=1代回(1)⇒21x+22(1−x)=50⇒x=−28⇒y=1−x=29⇒(x,y,z)=(−28,29,0)
解:1x+1y=166⇒x+yxy=166⇒xy−66(x+y)=0⇒(x−66)(y−66)=612而612=212×312的因數有(1+12)(1+12)=169個,因此有169組解
解:取1,4,7,...,3k+1,...,2017,共有673個。
解:limn→∞1n2n∑k=1√16n2−(4k)2=limn→∞1nn∑k=1√16−16(kn)2=∫10√16−16x2dx=∫104√1−x2dx=∫π/204cos2θdθ(x=sinθ⇒dx=cosθdθ)=∫π/202cos(2θ)+2dθ=[sin(2θ)+2θ]|π/20=π
解:{a2+b2=4(c−5)2+(d−12)2=36⇒{(a2+b2)(122+(−5)2)≥(12a−5b)2(a2+(−b)2)((d−12)2+(c−5)2)≥(a(d−12)−b(c−5))2⇒{4×169≥(12a−5b)24×36≥(ad−bc−12a+5b)2⇒{26≥12a−5b≥−2612+12a−5b≥ac−bd≥−12+12a−b⇒12+26≥ac−bd≥−12−26⇒38≥ac−bd≥−38⇒ac−bd的最大值為38
解:{A+B+C+D=16A,B,C,D∈NA>B⇒{A+B+C+D=12A,B,C,D∈{0,1,2,⋯,12}A>B⇒ABC+D個數1011H211=C12112010H210=C1110⋯1200H20=C10219H29=C109⋯1110H20=C10⋯651H21=C21750H20=C10⇒總和=6∑n=12n∑k=1k=6∑n=1n(2n+1)=6∑n=12n2+6∑n=1n=282+21=303
解:1,sinθ,√3cosθ為x3+ax+b=0的三根⇒{1+sinθ+√3cosθ=0⋯(1)sinθ+√3sinθcosθ+√3cosθ=a⋯(2)√3sinθcosθ=−b⋯(3)(1)⇒sinθ+√3cosθ=−1⇒12sinθ+√32cosθ=−12⇒sin(θ+30∘)=−12⇒θ=150∘(∵
解:圓與坐標軸分別交於\cases{A(0,2)\\ B(-2,0)\\ C(0,-2) \\ D(2,0)},兩直線斜率均為1,因此兩直線即為\overleftrightarrow{AB}及\overleftrightarrow{CD}\\ \Rightarrow m^2+n^2=2^2+(-2)^2 =\bbox[red,2pt]{8}
解:
解:{¯AB=10sinθ¯BO=10cosθ⇒¯BH=¯ABׯBO÷¯OA=100sinθcosθ÷10=10sinθcosθ⇒¯AB+¯BO+¯BH=10(sinθ+cosθ)+10sinθcosθ=10√2sin(θ+45∘)+5sin(2θ)顯然θ=45∘時,有最大值10√2+5
解:{n=1m=2⇒7n2−m2=7−4=3為最小值此題並無特定解法!
解:{21x+22y+27z=50⋯(1)22x+23y+28z=51⋯(2)23x+24y+25z=52⋯(3)(2)−(1),(3)−(2)→{x+y+z=1x+y−3z=1⇒{z=0x+y=1代回(1)⇒21x+22(1−x)=50⇒x=−28⇒y=1−x=29⇒(x,y,z)=(−28,29,0)
解:1x+1y=166⇒x+yxy=166⇒xy−66(x+y)=0⇒(x−66)(y−66)=612而612=212×312的因數有(1+12)(1+12)=169個,因此有169組解
解:取1,4,7,...,3k+1,...,2017,共有673個。
解:limn→∞1n2n∑k=1√16n2−(4k)2=limn→∞1nn∑k=1√16−16(kn)2=∫10√16−16x2dx=∫104√1−x2dx=∫π/204cos2θdθ(x=sinθ⇒dx=cosθdθ)=∫π/202cos(2θ)+2dθ=[sin(2θ)+2θ]|π/20=π
解:{a2+b2=4(c−5)2+(d−12)2=36⇒{(a2+b2)(122+(−5)2)≥(12a−5b)2(a2+(−b)2)((d−12)2+(c−5)2)≥(a(d−12)−b(c−5))2⇒{4×169≥(12a−5b)24×36≥(ad−bc−12a+5b)2⇒{26≥12a−5b≥−2612+12a−5b≥ac−bd≥−12+12a−b⇒12+26≥ac−bd≥−12−26⇒38≥ac−bd≥−38⇒ac−bd的最大值為38
解:{A+B+C+D=16A,B,C,D∈NA>B⇒{A+B+C+D=12A,B,C,D∈{0,1,2,⋯,12}A>B⇒ABC+D個數1011H211=C12112010H210=C1110⋯1200H20=C10219H29=C109⋯1110H20=C10⋯651H21=C21750H20=C10⇒總和=6∑n=12n∑k=1k=6∑n=1n(2n+1)=6∑n=12n2+6∑n=1n=282+21=303
解:1,sinθ,√3cosθ為x3+ax+b=0的三根⇒{1+sinθ+√3cosθ=0⋯(1)sinθ+√3sinθcosθ+√3cosθ=a⋯(2)√3sinθcosθ=−b⋯(3)(1)⇒sinθ+√3cosθ=−1⇒12sinθ+√32cosθ=−12⇒sin(θ+30∘)=−12⇒θ=150∘(∵
解:圓與坐標軸分別交於\cases{A(0,2)\\ B(-2,0)\\ C(0,-2) \\ D(2,0)},兩直線斜率均為1,因此兩直線即為\overleftrightarrow{AB}及\overleftrightarrow{CD}\\ \Rightarrow m^2+n^2=2^2+(-2)^2 =\bbox[red,2pt]{8}
解:
C,F,D為切點 \Rightarrow \cases{\overline{AC}= \overline{AD}=a+2\\ \overline{BC}= \overline{BF}=a-2\\ \overline{ED}= \overline{EF}=1 } \Rightarrow \cases{\overline{AB}=4 \\ \overline{AE}=a+1 \\ \overline{BE}=a-1} \\ 直角\triangle ABE \Rightarrow \overline{AB}^2 = \overline{AE}^2 +\overline{BE}^2 \Rightarrow 4^2=(a+1)^2 +(a-1)^2 \Rightarrow a^2=7 \Rightarrow a= \bbox[red,2pt]{\sqrt{7}}
解:a+b+c \ge 3\sqrt[3]{abc} \Rightarrow {1\over 3} \ge \sqrt[3]{abc} \Rightarrow \sqrt[3]{abc}的最大值為{1\over 3}\\\sqrt{a^2+b^2} +\sqrt{b^2+c^2} +\sqrt{c^2+a^2} \ge 3\sqrt[3]{\sqrt{a^2+b^2}\sqrt{b^2+c^2}\sqrt{c^2+a^2}} \ge 3\sqrt[3]{\sqrt{2ab} \cdot \sqrt{2bc} \cdot \sqrt{2ca}}\\ = 3\sqrt 2\sqrt[3]{abc} \Rightarrow \sqrt{a^2+b^2} +\sqrt{b^2+c^2} +\sqrt{c^2+a^2} \ge 3\sqrt 2\sqrt[3]{abc} \ge \bbox[red,2pt]{\sqrt 2}
解:\cases{原數1000a+100b+10c+d \\ 新數1000d+100c+10b+a} \Rightarrow 1000d+100c+10b+a-(1000a+100b+10c+d)=4725\\ \Rightarrow 999d+90c-90b-999a=4725 \Rightarrow 111(d-a)-10(b-c)=525 \Rightarrow \cases{d-a=5 \\b-c=3} \\ \Rightarrow \cases{(d,a)=(6,1),(7,2),(8,4),(9,5)\\ (b,c)=(3,0),(4,1),(5,2), (6,3),(7,4),(8,5),(9,6)} \Rightarrow 共有4\times 7= \bbox[red,2pt]{28}個四位數
解:f(93)=90^2-f(90) = 90^2-(87^2-f(87))=90^2-87^2 +f(87)= 90^2-87^2 +84^2-f(81) \\ =90^2-87^2 +84^2-\cdots+30^2-f(30)=93 \Rightarrow f(30)=90^2-87^2 +84^2-\cdots+30^2-93 \\ =\sum_{k=0}^{10} (30+6k)^2-\sum_{k=0}^{10} (33+6k)^2+93^2-93=\sum_{k=0}^{10}(-36k-189)+93^2-93 \\= -36\times 55 -189\times 11+93^2-93= \bbox[red,2pt]{4497}
解:由長除法可知: f(x)=(x+1)(x^2+(a-1)x+(b-a+1))+c-b+a-1\\ 又\lim_{x\to -1}{f(x)\over x+1}=3 \Rightarrow \cases{f(x)=(x+1)g(x)\\ g(-1)=3} \Rightarrow g(x)= x^2+(a-1)x+(b-a+1) \\\Rightarrow g(-1)= 1+1-a+b-a+1=3 \Rightarrow b=2a\cdots(1)\\ y=f(x)無極值 \Rightarrow f'(x)=3x^2+2ax +b=0 無相異實根\Rightarrow 判別式\le 0 \Rightarrow 4a^2-12b \le 0 \\ \stackrel{式(1)}{\Rightarrow} 4a^2-12\cdot (2a) \le 0 \Rightarrow 4a(a-6) \le 0 \Rightarrow \bbox[red,2pt]{0\le a\le 6}
解:令\log_2{y\over x^2} =M為最大值 \Rightarrow y=2^Mx^2,依題意意相當於求\cases{y=2^Mx^2\\ 2x+y+2=0} 的交點;\\由於M為最大值,兩圖形僅交於一點,即相切,也就是2x+2^Mx^2+2=0僅有一解,\\判別式:2^2-8\cdot 2^M=0 \\ \Rightarrow 2^M={1\over 2} \Rightarrow M= \bbox[red,2pt]{-1}
解:內接拋物線\Gamma :y^2=4cx正\triangle 三頂點\cases{A({\alpha^2\over 4c},\alpha) \\ B({\beta^2 \over 4c},\beta) \\ C({\gamma^2\over 4c},\gamma) } \Rightarrow \cases{重心G({\alpha^2+\beta^2 +\gamma^2\over 12c},{\alpha +\beta +\gamma \over 3})\\ \overleftrightarrow{AB}斜率m_1={\beta-\alpha\over (\beta^2-\alpha^2)/4c}={4c\over \beta+\alpha} \\\overleftrightarrow{BC}斜率m_2= {\gamma-\beta\over (\gamma^2-\beta^2)/4c}={4c\over \gamma+\beta} \\ \overleftrightarrow{CA}斜率m_3={\alpha-\gamma\over (\alpha^2-\gamma^2)/4c}={4c\over \alpha+\gamma} }\\ \Rightarrow \tan 60^\circ = \sqrt 3= {m_2-m_1\over 1+m_1m_2 }= {m_3-m2\over 1+m_2m_3 }= {m_1-m_3\over 1+m_1m_3 } \\ \Rightarrow {{4c\over \gamma+\beta}-{4c\over \beta+\alpha} \over 1+{16c^2\over (\alpha+\beta)(\beta +\gamma)}} ={{4c\over \alpha+\gamma}-{4c\over \beta+\gamma} \over 1+{16c^2\over (\alpha+\gamma)(\beta +\gamma)}} = {{4c\over \alpha+\beta}-{4c\over \alpha+\gamma} \over 1+{16c^2\over (\alpha+\beta)(\alpha +\gamma)}} =\sqrt 3\\ \Rightarrow {4c(\alpha-\gamma) \over 16c^2+(\alpha+\beta)(\beta+\gamma)} ={4c(\beta-\alpha) \over 16c^2+(\beta+\gamma)(\gamma+\alpha)} = {4c(\gamma-\beta) \over 16c^2 +(\alpha+\beta)(\alpha+\gamma)} =\sqrt 3\\ \Rightarrow {4c(\alpha-\gamma)+4c(\beta-\alpha) + 4c(\gamma-\beta) \over 48c^2+(\alpha+\beta) (\beta+\gamma)+ (\beta+\gamma)(\gamma+\alpha) +(\gamma+\alpha) (\alpha+\beta)} =\sqrt 3\\ 上式分子為0,因此分母亦為0,即48c^2+(\alpha+\beta) (\beta+\gamma)+ (\beta+\gamma)(\gamma+\alpha) +(\gamma+\alpha) (\alpha+\beta)=0 \\ \Rightarrow 48c^2+ \alpha^2+\beta^2+\gamma^2 +3(\alpha\beta +\beta\gamma +\gamma\alpha)=0 \Rightarrow \alpha\beta +\beta\gamma +\gamma\alpha=-16c^2-(\alpha^2+\beta^2+\gamma^2)/3\\ 重心G({\alpha^2+\beta^2 +\gamma^2\over 12c}=x,{\alpha +\beta +\gamma \over 3}=y) \Rightarrow (\alpha+\beta +\gamma)^2 =9y^2 =\alpha^2+\beta^2+\gamma^2 +2(\alpha\beta +\beta\gamma +\gamma\alpha)\\ =12cx+2(\alpha\beta +\beta\gamma +\gamma\alpha) =12cx+2(-16c^2-(\alpha^2+\beta^2+\gamma^2)/3)=12cx+2(-16c^2-4cx) \\ =4cx-32c^2 \Rightarrow \bbox[red,2pt]{9y^2=4cx-32c^2}
解:a+b+c \ge 3\sqrt[3]{abc} \Rightarrow {1\over 3} \ge \sqrt[3]{abc} \Rightarrow \sqrt[3]{abc}的最大值為{1\over 3}\\\sqrt{a^2+b^2} +\sqrt{b^2+c^2} +\sqrt{c^2+a^2} \ge 3\sqrt[3]{\sqrt{a^2+b^2}\sqrt{b^2+c^2}\sqrt{c^2+a^2}} \ge 3\sqrt[3]{\sqrt{2ab} \cdot \sqrt{2bc} \cdot \sqrt{2ca}}\\ = 3\sqrt 2\sqrt[3]{abc} \Rightarrow \sqrt{a^2+b^2} +\sqrt{b^2+c^2} +\sqrt{c^2+a^2} \ge 3\sqrt 2\sqrt[3]{abc} \ge \bbox[red,2pt]{\sqrt 2}
解:\cases{原數1000a+100b+10c+d \\ 新數1000d+100c+10b+a} \Rightarrow 1000d+100c+10b+a-(1000a+100b+10c+d)=4725\\ \Rightarrow 999d+90c-90b-999a=4725 \Rightarrow 111(d-a)-10(b-c)=525 \Rightarrow \cases{d-a=5 \\b-c=3} \\ \Rightarrow \cases{(d,a)=(6,1),(7,2),(8,4),(9,5)\\ (b,c)=(3,0),(4,1),(5,2), (6,3),(7,4),(8,5),(9,6)} \Rightarrow 共有4\times 7= \bbox[red,2pt]{28}個四位數
解:f(93)=90^2-f(90) = 90^2-(87^2-f(87))=90^2-87^2 +f(87)= 90^2-87^2 +84^2-f(81) \\ =90^2-87^2 +84^2-\cdots+30^2-f(30)=93 \Rightarrow f(30)=90^2-87^2 +84^2-\cdots+30^2-93 \\ =\sum_{k=0}^{10} (30+6k)^2-\sum_{k=0}^{10} (33+6k)^2+93^2-93=\sum_{k=0}^{10}(-36k-189)+93^2-93 \\= -36\times 55 -189\times 11+93^2-93= \bbox[red,2pt]{4497}
解:由長除法可知: f(x)=(x+1)(x^2+(a-1)x+(b-a+1))+c-b+a-1\\ 又\lim_{x\to -1}{f(x)\over x+1}=3 \Rightarrow \cases{f(x)=(x+1)g(x)\\ g(-1)=3} \Rightarrow g(x)= x^2+(a-1)x+(b-a+1) \\\Rightarrow g(-1)= 1+1-a+b-a+1=3 \Rightarrow b=2a\cdots(1)\\ y=f(x)無極值 \Rightarrow f'(x)=3x^2+2ax +b=0 無相異實根\Rightarrow 判別式\le 0 \Rightarrow 4a^2-12b \le 0 \\ \stackrel{式(1)}{\Rightarrow} 4a^2-12\cdot (2a) \le 0 \Rightarrow 4a(a-6) \le 0 \Rightarrow \bbox[red,2pt]{0\le a\le 6}
解:令\log_2{y\over x^2} =M為最大值 \Rightarrow y=2^Mx^2,依題意意相當於求\cases{y=2^Mx^2\\ 2x+y+2=0} 的交點;\\由於M為最大值,兩圖形僅交於一點,即相切,也就是2x+2^Mx^2+2=0僅有一解,\\判別式:2^2-8\cdot 2^M=0 \\ \Rightarrow 2^M={1\over 2} \Rightarrow M= \bbox[red,2pt]{-1}
解:內接拋物線\Gamma :y^2=4cx正\triangle 三頂點\cases{A({\alpha^2\over 4c},\alpha) \\ B({\beta^2 \over 4c},\beta) \\ C({\gamma^2\over 4c},\gamma) } \Rightarrow \cases{重心G({\alpha^2+\beta^2 +\gamma^2\over 12c},{\alpha +\beta +\gamma \over 3})\\ \overleftrightarrow{AB}斜率m_1={\beta-\alpha\over (\beta^2-\alpha^2)/4c}={4c\over \beta+\alpha} \\\overleftrightarrow{BC}斜率m_2= {\gamma-\beta\over (\gamma^2-\beta^2)/4c}={4c\over \gamma+\beta} \\ \overleftrightarrow{CA}斜率m_3={\alpha-\gamma\over (\alpha^2-\gamma^2)/4c}={4c\over \alpha+\gamma} }\\ \Rightarrow \tan 60^\circ = \sqrt 3= {m_2-m_1\over 1+m_1m_2 }= {m_3-m2\over 1+m_2m_3 }= {m_1-m_3\over 1+m_1m_3 } \\ \Rightarrow {{4c\over \gamma+\beta}-{4c\over \beta+\alpha} \over 1+{16c^2\over (\alpha+\beta)(\beta +\gamma)}} ={{4c\over \alpha+\gamma}-{4c\over \beta+\gamma} \over 1+{16c^2\over (\alpha+\gamma)(\beta +\gamma)}} = {{4c\over \alpha+\beta}-{4c\over \alpha+\gamma} \over 1+{16c^2\over (\alpha+\beta)(\alpha +\gamma)}} =\sqrt 3\\ \Rightarrow {4c(\alpha-\gamma) \over 16c^2+(\alpha+\beta)(\beta+\gamma)} ={4c(\beta-\alpha) \over 16c^2+(\beta+\gamma)(\gamma+\alpha)} = {4c(\gamma-\beta) \over 16c^2 +(\alpha+\beta)(\alpha+\gamma)} =\sqrt 3\\ \Rightarrow {4c(\alpha-\gamma)+4c(\beta-\alpha) + 4c(\gamma-\beta) \over 48c^2+(\alpha+\beta) (\beta+\gamma)+ (\beta+\gamma)(\gamma+\alpha) +(\gamma+\alpha) (\alpha+\beta)} =\sqrt 3\\ 上式分子為0,因此分母亦為0,即48c^2+(\alpha+\beta) (\beta+\gamma)+ (\beta+\gamma)(\gamma+\alpha) +(\gamma+\alpha) (\alpha+\beta)=0 \\ \Rightarrow 48c^2+ \alpha^2+\beta^2+\gamma^2 +3(\alpha\beta +\beta\gamma +\gamma\alpha)=0 \Rightarrow \alpha\beta +\beta\gamma +\gamma\alpha=-16c^2-(\alpha^2+\beta^2+\gamma^2)/3\\ 重心G({\alpha^2+\beta^2 +\gamma^2\over 12c}=x,{\alpha +\beta +\gamma \over 3}=y) \Rightarrow (\alpha+\beta +\gamma)^2 =9y^2 =\alpha^2+\beta^2+\gamma^2 +2(\alpha\beta +\beta\gamma +\gamma\alpha)\\ =12cx+2(\alpha\beta +\beta\gamma +\gamma\alpha) =12cx+2(-16c^2-(\alpha^2+\beta^2+\gamma^2)/3)=12cx+2(-16c^2-4cx) \\ =4cx-32c^2 \Rightarrow \bbox[red,2pt]{9y^2=4cx-32c^2}
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