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If-k-is-an-integer-which-satisfies-2sin-2-10cos-2-2-7-2k-then-k-A-0-1-2-3-B-1-0-1-2-3-C-2-1-0-1-2-3-D-3-2-1-0-

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Question Number 117791 by ZiYangLee last updated on 13/Oct/20
If k is an integer which satisfies 2sin^2 θ+10cos^2 (θ/2)=7−2k, then k∈? A.{0,1,2,3} B.{−1,0,1,2,3} C.{(−2,−1,0,1,2,3} D.{−3,−2,−1,0}
Ifkisanintegerwhichsatisfies2sin2θ+10cos2θ2=7−2k,thenk∈?A.{0,1,2,3}B.{−1,0,1,2,3}C.{(−2,−1,0,1,2,3}D.{−3,−2,−1,0}
Answered by floor(10²Eta[1]) last updated on 13/Oct/20
2(1−cos^2 θ)+10(((1+cosθ)/2))=7−2k 2−2cos^2 θ+5+5cosθ=7−2k 2cos^2 θ−5cosθ−2k=0 cosθ=((5±(√(25+16k)))/4) −1≤((5±(√(25+16k)))/4)≤1 −9≤±(√(25+16k))≤−1 ⇒−9≤−(√(25+16k))≤−1 9≥(√(25+16k))≥1 81≥25+16k≥1 3.5≥k≥−1.5 Ans: B
2(1−cos2θ)+10(1+cosθ2)=7−2k2−2cos2θ+5+5cosθ=7−2k2cos2θ−5cosθ−2k=0cosθ=5±25+16k4−1⩽5±25+16k4⩽1−9⩽±25+16k⩽−1⇒−9⩽−25+16k⩽−19⩾25+16k⩾181⩾25+16k⩾13.5⩾k⩾−1.5Ans:B
Answered by TANMAY PANACEA last updated on 13/Oct/20
cos^2 (θ/2)=a 2(2sin(θ/2)cos(θ/2))^2 +10cos^2 (θ/2) 8(1−cos^2 (θ/2))cos^2 (θ/2)+10cos^2 (θ/2) 8(1−a)a+10a 8a−8a^2 +10a 18a−8a^2 −8a^2 +18a −8(a^2 −((9a)/4)) −8(a^2 −2×a×(9/8)+((81)/(64))−((81)/(64))) 8×((81)/(64))−8(a−(9/8))^2 ((81)/8)−8(a−(9/8))^2 1≥cos(θ/2)≥−1 but 1≥cos^2 (θ/2)≥0 1≥a≥0 ((81)/8)−8(0−(9/8))^2 =0 ((81)/8)−8(1−(9/8))^2 =((81)/8)−8×(1/(64))=((81−1)/8)=10 now 10≥7−2k≥0 3≥−2k≥−7 −3≤2k≤7 ((−3)/2)≤k≤(7/2) −1.5≤k≤3.5
cos2θ2=a2(2sinθ2cosθ2)2+10cos2θ28(1−cos2θ2)cos2θ2+10cos2θ28(1−a)a+10a8a−8a2+10a18a−8a2−8a2+18a−8(a2−9a4)−8(a2−2×a×98+8164−8164)8×8164−8(a−98)2818−8(a−98)21⩾cosθ2⩾−1but1⩾cos2θ2⩾01⩾a⩾0818−8(0−98)2=0818−8(1−98)2=818−8×164=81−18=10now10⩾7−2k⩾03⩾−2k⩾−7−3⩽2k⩽7−32⩽k⩽72−1.5⩽k⩽3.5
Answered by 1549442205PVT last updated on 14/Oct/20
2sin^2 θ+10cos^2 (θ/2)=7−2k (1) ⇔2(sin(θ/2)cos(θ/2))^2 +10cos^2 (θ/2)=7−2k ⇔2sin^2 (θ/2)cos^2 (θ/2)+10cos^2 (θ/2)=7−k •If cos^2 x=0 then sin^2 x=1⇒(1)⇔k=2.5∉Z •If cos^2 x≠0 then divide both sides of equation(1)by cos^2 x we get (1)⇔2tan^2 (θ/2)+10=(7−2k)(1+tan^2 (θ/2)) ⇔(5−2k)tan^2 (θ/2)=3+2k⇒k≠2.5 ⇔tan^2 (θ/2)=((3+2k)/(5−2k))≥0⇔−1.5≤k<2.5 k∈Z⇒k∈{−1,0,1,2}⇒Choose B
2sin2θ+10cos2θ2=7−2k(1)⇔2(sinθ2cosθ2)2+10cos2θ2=7−2k⇔2sin2θ2cos2θ2+10cos2θ2=7−k∙Ifcos2x=0thensin2x=1⇒(1)⇔k=2.5∉Z∙Ifcos2x≠0thendividebothsidesofequation(1)bycos2xweget(1)⇔2tan2θ2+10=(7−2k)(1+tan2θ2)⇔(5−2k)tan2θ2=3+2k⇒k≠2.5⇔tan2θ2=3+2k5−2k⩾0⇔−1.5⩽k<2.5k∈Z⇒k∈{−1,0,1,2}⇒ChooseB

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