臺北市立麗山高級中學108學年度第1次教師甄選
數學科試題卷
解:五座島最多可以蓋C52=10條橋,只蓋4條橋有C104=210種蓋法;其中不連通的方式:(1)4+1:4島相連,另一島不相連;4島互連可蓋C42=6座橋,任取4座,共有C64=15種情形;而五島選四島有C54=5種選法,因此蓋4橋有5×15=75種蓋法;(2)2+3:2島互連,另3島互連,共有C52=10種情形,每一種情形剛好蓋4座橋;因此蓋4橋可相連的方式有210−75−10=125種。
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解:令三切點的x坐標分別為α,β,γ⇒x6−10x5+29x4−4x3+ax2−bx−c=(x−α)2(x−β)2(x−γ)2⇒α+β+γ=10÷2=5 ===========================================================
解:令log2x=t⇒log√2x=2t⇒6log√2x−81+4(log2x)2=12t−81+4t2≡g(t)⇒g′(t)=121+4t2−8a(12t−8)(1+4t2)2=−4(6t+1)(2t−3)(1+4t2)2=0⇒{t=−1/6t=3/2⇒{x=1/6√2x=2√2⇒{f(1/6√2)=(−2−8)/(1+4/36)f(2√2)=(24√2−8)/(1+32)⇒f(2√2)>f(1/6√2)⇒a=2√2有最大值 ===========================================================
解:logab=c⇒b=ac,若{a=2xb=2y,其x,y∈{1,2,...,25}且x≠y⇒2y=(2x)c=2xc⇒cx=y⇒x∣y(x是y的因數)⇒xy數量12−252424,5,..,241136,9,..,24748,12,..,245510,15,20,254612,18,243714,212816,2429181102011122112241⇒24+11+⋯+1=62⇒機率為6225×24=31300
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解:取3,32,…,32n的中間兩數3n及3n+1之間任一數,即可得最小之f(x),x∈[3n,3n+1]f(3n)=(3n−3)+(3n−32)+⋯+(3n−3n)+(3n+1−3n)+(3n+2−3n)+⋯(32n−3n)=n⋅3n−(3+32+⋯+3n)+(3n+1+3n+2+⋯+32n)−n⋅3n=3n(3+32+⋯+3n)−(3+32+⋯+3n)=(3n−1)(3+32+⋯+3n)=(3n−1)3n+1−33−1=(3n−1)3(3n−1)2=32(3n−1)2
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解:
假設圓半徑為1⇒¯OA=¯OB=¯OC=¯OD=1△BCD:∠DBC=180∘−45∘−75∘=60∘⇒∠DOC=2∠DBC=120∘⇒∠ODC=∠OCD=(180∘−120∘)÷2=30∘△OCD:¯OCsin30∘=¯CDsin120∘⇒11/2=¯CD√3/2⇒¯CD=√3△BCD:¯CDsin60∘=¯BCsin45∘⇒√3√3/2=¯BC1/√2⇒¯BC=√2△BCD:cos∠DBC=cos60∘=¯BD2+2−32√2ׯBD⇒¯BD=√2+√62又∠DAC=∠DBC=60∘⇒△OAD為一正△⇒¯AD=1直角△ABC:{¯AC=2¯BC=√2⇒¯AB=√2令{B為原點→BC為x軸→BA為y軸⇒{B(0,0)C(√2,0)A(0,√2)O(√2/2,√2/2),又↔BD斜率為tan60∘=√3⇒D(t,t√3)由¯AD=1⇒t2+(t√3−√2)2=1⇒t=√2+√64⇒D(√2+√64,3√2+√64)→BD=a→OA+b→OB⇒(√2+√64,3√2+√64)=a(−√22,√22)+b(−√22,−√22)⇒{a+b=−√2+√62√2a−b=√6+3√22√2⇒(a,b)=(12,−2−√32)... =============================================================
解:P為橢圓上的切點,其切線斜率與L斜率相同;9x2+4y2=36⇒ddx(9x2+4y2)=ddx36⇒18x+8yy′=0⇒y′=18x−8y=−34(L的斜率)⇒y=3x代回橢圓方程式⇒9x2+36x2=36⇒x2=45⇒x=−2√5(x=2√5為最近的切點)⇒y=3x=−6√5⇒P坐標為(−2√5,−6√5) ====================================================================
解:六面分別為1,1,1,2,2,3,其中有4個奇數,2個偶數⇒P1=26=13又偶數=偶數+偶數或奇數+奇數⇒Pn=Pn−1×13+(1−Pn−1)×23=23−13Pn−1⇒Pn=2(13−132+133−⋯+13n(−1)n−1)⋯(1)⇒13Pn=2(132−133+134−⋯+13n+1(−1)n−1)⋯(2)式(1)+式(2)⇒43Pn=2(13−13n+1(−1)n−1)⇒Pn=32(13−13n+1(−1)n−1)⇒limn→∞Pn=32(13−0)=12
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解:n+2n!+(n+1)!+(n+2)!=n+2n!(1+(n+1)+(n+2)(n+1))=n+2n!(n+2)2=1n!(n+2)=n+1(n+2)(n+1)n!=n+1(n+2)!=(n+2)−1(n+2)!=1(n+1)!−1(n+2)!⇒2017∑n=1n+2n!+(n+1)!+(n+2)!=2017∑n=1(1(n+1)!−1(n+2)!)=(12!−13!)+(13!−14!)+⋯+(12018!−12019!)=12!−12019! ===============================================================
解:x2+y2=x+y=n⇒(x+y)2=x2+y2+2xy⇒n2=n+2xy⇒xy=(n2−n)/2⇒(x−y)2=(x+y)2−4xy=n2−2(n2−n)=2n−n2⇒x−y=√2n−n2(負值不合,因為x>y)⇒x4−y4=(x2+y2)(x2−y2)=(x2+y2)(x+y)(x−y)=n2⋅√2n−n2=√2n5−n6令f(n)=2n5−n6⇒f′(n)=0⇒10n4−6n5=0⇒2n4(5−3n)=0⇒n=0,5/3⇒f″(n)=40n3−30n4⇒{f″(0)=0f″(5/3)<0⇒n=5/3時,f(n)有極大值⇒M=5232√103−5232=25√527⇒(n,M)=(53,25√527) =======================================================================
解:limn→∞1n√1−k24n2=2limn→∞12n√1−(k2n)2=2∫1/20√1−x2dx=2∫π/60√1−sin2ucosudu(其中sinu=x)=[12sin2u+u]|π/60=√34+π6 ==========================================================
解:\cases{P(x,y,z) \\A(1,-1,2) \\ B(1,1,0)\\ C(1,0,4)} \Rightarrow \overline{AP}^2+ \overline{BP}^2+\overline{CP}^2\\ = (x-1)^2 +(y+1)^2+(z-2)^2 +(x-1)^2+(y-1)^2+z^2 +(x-1)^2+y^2 +(z-4)^2\\ = 3((x-1)^2+y^2 +(z-2)^2) +10\\ 柯西不等式:((x-1)^2+y^2 +(z-2)^2) (1^2+1^2+1^2) \ge (x+y+z-3)^2 = (0-3)^2=9 \\ \Rightarrow (x-1)^2+y^2 +(z-2)^2\ge {9\over 3}=3\Rightarrow 3((x-1)^2+y^2 +(z-2)^2) +10 \ge 3\times 3+10= \bbox[red,2pt]{19} ============================================================
解:n^3+108 = (n+11)(n^2-11n+121)-1223 \Rightarrow {n^3+108\over n+11}=(n^2-11n+121)-{1223\over n+11} \\ \Rightarrow n=1223-11= \bbox[red,2pt]{1212}為最大的整數使用得{n^3+108\over n+11}也是整數
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解:\\ 令\sin u={x-3\over 4} \Rightarrow \cos u\;du = {1\over 4} dx,則\int \sqrt{-x^2+6x+7}\;dx = \int \sqrt{-(x-3)^2+4^2}\;dx \\=\int 4\sqrt{1-({x-3\over 4})^2} \;dx =\int 4\sqrt{1-\sin^2 u} \cdot 4 \cos u \;du =16\int\sqrt{\cos ^2u}\cdot \cos u\;du = 16\int \cos^2 u\;du \\=8\int \cos 2u+1 \;du= 4\sin 2u +8u =8\sin u\cos u+8u \\ = 8\cdot {x-3\over 4}\cdot {\sqrt{-x^2+6x+7}\over 4} +8\sin^{-1}{x-3\over 4} ={x-3\over 2}\cdot \sqrt{-x^2+6x+7} +8\sin^{-1}{x-3\over 4} \\ 因此\int_1^7 (-2+\sqrt{-x^2+6x+7})\;dx = \left. \left[ -2x+ {x-3\over 2}\cdot \sqrt{-x^2+6x+7} +8\sin^{-1}{x-3\over 4}\right] \right|_1^7\\ =(-14+0+8\sin^{-1}1) -(-2+(-1)\cdot 2\sqrt 3 +8\sin^{-1}{-1\over 2})\\ =-12+2\sqrt 3+ 8\cdot {\pi \over 2}-8\cdot {-\pi \over 6} = \bbox[red,2pt]{-12+2\sqrt 3+ {16\over 3}\pi} ============================================================
正八邊形每一內角為\;(8-2)\times 180\div 8 = 135^\circ \Rightarrow \angle EFB= 180^\circ-135^\circ =45^\circ \\ \Rightarrow \angle BEF = \angle ABC-\angle AFE=60^\circ- 45^\circ=15^\circ \Rightarrow \angle CED = 180^\circ -135^\circ-15^\circ = 30^\circ \\ \Rightarrow \angle CDE=30^\circ \Rightarrow \overline{CD}=\overline{CE}= 2;\\ \triangle CDE: \cos \angle CDE = \cos 120^\circ ={2^2+2^2-\overline{DE}^2 \over 2\times 2\times 2} \Rightarrow -{1\over 2}={8-\overline{DE}^2 \over 8} \Rightarrow \overline{DE}=2\sqrt 3;\\ \triangle BEF: {\overline{EF}\over \sin \angle EBF} = {\overline{EB}\over \sin \angle EFB} = {\overline{BF}\over \sin \angle BEF} \Rightarrow {2\sqrt 3\over \sin 120^\circ} = {\overline{EB}\over \sin 45^\circ} = {\overline{BF}\over \sin 15^\circ} \\\Rightarrow {2\sqrt 3\over \sqrt 3/2} = {\overline{EB}\over 1/\sqrt 2} = {\overline{BF}\over (\sqrt 6-\sqrt 2)/4} \Rightarrow \cases{\overline{EB}=2\sqrt 2\\ \overline{BF}=\sqrt 6-\sqrt 2} \\ \Rightarrow \overline{AF}= \overline{AB}+ \overline{BF} = \overline{CE}+ \overline{EB}+\overline{BF} =2+2\sqrt 2+\sqrt 6-\sqrt 2= \bbox[red,2pt]{2+\sqrt 6+\sqrt 2}===============================================================
解:(1) f(x+2y)=f(x)+g(y) \Rightarrow f'(x+2y)=f'(x) \Rightarrow f'(s)=f'(t)\;\forall s,t \Rightarrow f'(x)=c,c 為常數\\ (2)f'(x)=c\Rightarrow f(x)= cx+d, 其中c,d 為常數;又\cases{f(0)=1 \Rightarrow d=1\\ f'(0)=2 \Rightarrow c=2} \\ \Rightarrow f(x)=2x+1 \Rightarrow f(x+2y)=2x+4y+1 =f(x)+g(y)=2x+1+g(y) \Rightarrow g(y)=4y\\ \Rightarrow g(5)=\bbox[red,2pt]{20}
解:
假設\cases{O:外接圓圓心\\ P:內切圓圓心\\ R:外接圓半徑\\ r:內切圓半徑} \Rightarrow \overline{AG}=\overline{AO}\cos {\phi\over 2} =R\cos {\phi\over 2} \Rightarrow \overline{AB}=2\overline{AG} =2R\cos{\phi\over 2} \\ \Rightarrow \overline{AF}=\overline{AB} \cos{\phi \over 2} =2R\cos^2{\phi \over 2} \Rightarrow \overline{OP} =\overline{AF}-R-r =2R\cos^2{\phi \over 2}-R-r\\ \overline{GO}\parallel \overline{EP} \Rightarrow {\overline{GO} \over \overline{EP}} ={\overline{AO} \over \overline{AP}(=\overline{AO}+\overline{OP})} \Rightarrow {R\sin{\phi\over 2} \over r} ={R\over R+(2R\cos^2{\phi \over 2}-R-r)} ={R\over 2R(1-\sin^2{\phi \over 2})-r} \\ 令\cases{x=\sin{\phi\over 2} \\ r=k \\ R=(1+\sqrt 2) k},則上式為 {x\over 1}={1\over 2(1+\sqrt 2)(1-x^2)-1} \Rightarrow (2+2\sqrt 2)x^3-(1+2\sqrt 2)x+1=0 \\ \Rightarrow (x+1) ((2+2\sqrt 2)x^2-(2+2\sqrt 2)x+1)=0 \Rightarrow \cases{x=-1(不合,\because \phi/2 \ne 270^\circ)\\ x={(2+\sqrt 2)\pm 2 \over 2(2+2\sqrt 2)}= {\sqrt 2\over 2}或{2-\sqrt 2\over 2}} \\ \Rightarrow \sin{\phi\over 2}=x =\bbox[red,2pt]{{\sqrt 2\over 2},{2-\sqrt 2\over 2}} =============================================================
解:
假設立方體邊長為a,A為原點及P(x,y,z),則\cases{A(0,0,0)\\ B(a,0,0)\\ C(a,a,0)\\ D(0,a,0)\\ E(0,0,a)\\ F(a,0,a)\\ G(a,a,a)\\ H(0,a,a)},如圖;\\ \cases{\overline{PA}=\sqrt 2\\ \overline{PB}= \overline{PD}=\sqrt 3\\ \overline{PE}=1} \Rightarrow \cases{x^2+y^2+z^2=2\cdots(1) \\ (x-a)^2+y^2+z^2 = 3 \cdots(2)\\ x^2+(y-a)^2+z^2 = 3 \cdots(3)\\ x^2+y^2+(z-a)^2=1 \cdots(4)},\\由(1)得\cases{y^2+z^2=2-x^2 代入(2) \Rightarrow (x-a)^2+2-x^2=3 \Rightarrow x=(a^2-1)/ 2a\\ x^2+z^2=2-y^2代入(3) \Rightarrow 2-y^2+(y-a)^2=3 \Rightarrow y=(a^2-1)/2a \\ x^2+y^2= 2-z^2代入(4) \Rightarrow 2-z^2+(z-a)^2=1 \Rightarrow z=(a^2+1)/2a}\\ 將P({a^2-1\over 2a},{a^2-1\over 2a},{a^2+1\over 2a})代回(1) \Rightarrow {(a^2-1)^2\over 4a^2} +{(a^2-1)^2\over 4a^2}+{(a^2+1)^2\over 4a^2}= 2 \\ \Rightarrow 3a^4-2a^2+3=8a^2 \Rightarrow 3a^4-10a^2+3=0 \Rightarrow (3a^2-1)(a^2-3)=0 \\ \Rightarrow a=\sqrt 3(a=1/\sqrt 3\Rightarrow x<0,y<0 \Rightarrow P在立方體外) \Rightarrow 體積=a^3 =(\sqrt 3)^3 = \bbox[red,2pt]{3\sqrt 3}
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解:(1) f(x+2y)=f(x)+g(y) \Rightarrow f'(x+2y)=f'(x) \Rightarrow f'(s)=f'(t)\;\forall s,t \Rightarrow f'(x)=c,c 為常數\\ (2)f'(x)=c\Rightarrow f(x)= cx+d, 其中c,d 為常數;又\cases{f(0)=1 \Rightarrow d=1\\ f'(0)=2 \Rightarrow c=2} \\ \Rightarrow f(x)=2x+1 \Rightarrow f(x+2y)=2x+4y+1 =f(x)+g(y)=2x+1+g(y) \Rightarrow g(y)=4y\\ \Rightarrow g(5)=\bbox[red,2pt]{20}
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註:解題僅供參考
第5題,3的k次冪,應是3*k,解法就合理。若是3的k次冪,好像x=3的2n次冪,才會使和為最小。
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