臺北市立大直高級中學 113 學年度第一次專任教師甄選
一、填充題(48 分)
解答:2310=2×3×5×7×11,在5個因數中,Case I 挑3個連乘積當作a,剩下2個作為b及c,有C53=10種可能Case II 挑2個連乘積當作a,剩下3個再挑2個乘積作為b,有C52C32=30種可能共有10+30=40種可能解答:an=Sn−Sn−1=(√Sn+√Sn−1)(√Sn−√Sn−1)=√Sn+√Sn−1⇒(√Sn+√Sn−1)(√Sn−√Sn−1−1)=0⇒√Sn=√Sn−1+1取bn=√Sn,則bn=bn−1+1=bn−2+2=⋯=b1+(n−1)=n⇒Sn=n2⇒S20−S19+S18=202−192+182=39+182=363
解答:an=|nn+10n+2n+1n+2n+20n+2|=|nn+1020n+2n+20n+2|=(n+1)(n+2)2−2(n+1)(n+2)=n(n+1)(n+1)⇒1ak=1k(k+1)(k+2)=12(1k(k+1)−1(k+1)(k+2))⇒limn→∞n∑k=11ak=limn→∞12n∑k=1(1k(k+1)−1(k+1)(k+2))=limn→∞12(12−16+16−112+⋯−1(n+1)(n+2))=limn→∞12(12−1(n+1)(n+2))=14
解答:取{O(0,0)A(0,4)B(4,0)⇒C(0,2)⇒D=(4C+3B)/7=(127,87)⇒{L1=↔AD:y=−53x+4L2=↔AB:x+y=4L3=↔OD:y=23x⇒{E=L1∩x軸=(125,0)F=L2∩L3=(125,2415)⇒S△CEFS△OAB=‖02112/50112/524/151‖‖041001401‖=962516=625
解答:
解答:
取{A(0,20)B(0,0)C(5,0)P(5,a)⇒L1=↔AP:y=a−205x+20⇒Q(10020−a,0)⇒S△ADP+S△PQC=12(5(20−a)+(10020−a−5)a)取f(a)=5(20−a)+(10020−a−5)a⇒f′(a)=0⇒−10+10020−a+100a(20−a)2=0⇒a2−40a+200=0⇒a=20−10√2
解答:x29+y225=1⇒{a=5b=3⇒c=4⇒繞長軸(y軸)旋轉體積=2π∫40(√9(1−y225))2dy=2π[9y−3y425]|40=141625π
解答:{A(−1,3,2)B(3,3,4)⇒{→AB=(4,0,2)P=¯AB中點=(1,3,3)⇒過P且法向量為→AB的平面Ep:2x+z=5⇒球心O在直線L=Ep∩E上⇒O(t,10−52t,5−2t),t∈R⇒球半徑R=¯OA=√(t+1)2+(7−52t)2+(3−2t)2令f(t)=(t+1)2+(7−52t)2+(3−2t)2⇒f′(t)=0⇒t=2⇒{R=√14O(2,5,1)⇒球方程式:(x−2)2+(y−5)2+(z−1)2=14
解答:
解答:x29+y225=1⇒{a=5b=3⇒c=4⇒繞長軸(y軸)旋轉體積=2π∫40(√9(1−y225))2dy=2π[9y−3y425]|40=141625π
解答:{A(−1,3,2)B(3,3,4)⇒{→AB=(4,0,2)P=¯AB中點=(1,3,3)⇒過P且法向量為→AB的平面Ep:2x+z=5⇒球心O在直線L=Ep∩E上⇒O(t,10−52t,5−2t),t∈R⇒球半徑R=¯OA=√(t+1)2+(7−52t)2+(3−2t)2令f(t)=(t+1)2+(7−52t)2+(3−2t)2⇒f′(t)=0⇒t=2⇒{R=√14O(2,5,1)⇒球方程式:(x−2)2+(y−5)2+(z−1)2=14
二、非選題(52 分)
解答:柯西不等式:xy=((x−z)+z)(z+(y−z))≥(√z(x−z)+√z(y−z))2⇒√xy≥√z(x−z)+√z(y−z),故得證解答:
直線L:{√3x−z−6=0y=0⇒{A=L∩xy平面=(2√3,0,0)B=L∩z軸=(0,0,−6)假設{球心P(0,0,−r),r為球半徑Q為球與L的切點⇒△ABO∼△PBQ(AAA)⇒¯OA¯AB=¯PQ¯PB⇒2√34√3=r6−r⇒r=2⇒P(0,0,−2)
解答:n=7,參考資料:中學生通訊解題第五十九期
解答:略
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解答:n=7,參考資料:中學生通訊解題第五十九期
解答:略
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