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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}

$$\mathrm{If}\:{k}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{which}\:\mathrm{satisfies}\: \\ $$$$\mathrm{2sin}^{\mathrm{2}} \theta+\mathrm{10cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\mathrm{7}−\mathrm{2}{k},\:\mathrm{then}\:{k}\in? \\ $$$${A}.\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}\:\:\:{B}.\left\{−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$${C}.\left\{\left(−\mathrm{2},−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}\:\:{D}.\left\{−\mathrm{3},−\mathrm{2},−\mathrm{1},\mathrm{0}\right\}\right. \\ $$

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

$$\mathrm{2}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta\right)+\mathrm{10}\left(\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{2}}\right)=\mathrm{7}−\mathrm{2k} \\ $$$$\mathrm{2}−\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{5}+\mathrm{5cos}\theta=\mathrm{7}−\mathrm{2k} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{5cos}\theta−\mathrm{2k}=\mathrm{0} \\ $$$$\mathrm{cos}\theta=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}+\mathrm{16k}}}{\mathrm{4}} \\ $$$$−\mathrm{1}\leqslant\frac{\mathrm{5}\pm\sqrt{\mathrm{25}+\mathrm{16k}}}{\mathrm{4}}\leqslant\mathrm{1} \\ $$$$−\mathrm{9}\leqslant\pm\sqrt{\mathrm{25}+\mathrm{16k}}\leqslant−\mathrm{1} \\ $$$$\Rightarrow−\mathrm{9}\leqslant−\sqrt{\mathrm{25}+\mathrm{16k}}\leqslant−\mathrm{1} \\ $$$$\mathrm{9}\geqslant\sqrt{\mathrm{25}+\mathrm{16k}}\geqslant\mathrm{1} \\ $$$$\mathrm{81}\geqslant\mathrm{25}+\mathrm{16k}\geqslant\mathrm{1} \\ $$$$\mathrm{3}.\mathrm{5}\geqslant\mathrm{k}\geqslant−\mathrm{1}.\mathrm{5} \\ $$$$\mathrm{Ans}:\:\mathrm{B} \\ $$

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

$${cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}={a} \\ $$$$\mathrm{2}\left(\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{10}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{8}\left(\mathrm{1}−{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right){cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{10}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{8}\left(\mathrm{1}−{a}\right){a}+\mathrm{10}{a} \\ $$$$\mathrm{8}{a}−\mathrm{8}{a}^{\mathrm{2}} +\mathrm{10}{a} \\ $$$$\mathrm{18}{a}−\mathrm{8}{a}^{\mathrm{2}} \\ $$$$−\mathrm{8}{a}^{\mathrm{2}} +\mathrm{18}{a} \\ $$$$−\mathrm{8}\left({a}^{\mathrm{2}} −\frac{\mathrm{9}{a}}{\mathrm{4}}\right) \\ $$$$−\mathrm{8}\left({a}^{\mathrm{2}} −\mathrm{2}×{a}×\frac{\mathrm{9}}{\mathrm{8}}+\frac{\mathrm{81}}{\mathrm{64}}−\frac{\mathrm{81}}{\mathrm{64}}\right) \\ $$$$\mathrm{8}×\frac{\mathrm{81}}{\mathrm{64}}−\mathrm{8}\left({a}−\frac{\mathrm{9}}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{81}}{\mathrm{8}}−\mathrm{8}\left({a}−\frac{\mathrm{9}}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$\:\:\mathrm{1}\geqslant{cos}\frac{\theta}{\mathrm{2}}\geqslant−\mathrm{1}\:{but}\:\mathrm{1}\geqslant{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\:\mathrm{1}\geqslant{a}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{81}}{\mathrm{8}}−\mathrm{8}\left(\mathrm{0}−\frac{\mathrm{9}}{\mathrm{8}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{81}}{\mathrm{8}}−\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{81}}{\mathrm{8}}−\mathrm{8}×\frac{\mathrm{1}}{\mathrm{64}}=\frac{\mathrm{81}−\mathrm{1}}{\mathrm{8}}=\mathrm{10} \\ $$$${now}\:\:\mathrm{10}\geqslant\mathrm{7}−\mathrm{2}{k}\geqslant\mathrm{0} \\ $$$$\mathrm{3}\geqslant−\mathrm{2}{k}\geqslant−\mathrm{7} \\ $$$$−\mathrm{3}\leqslant\mathrm{2}{k}\leqslant\mathrm{7} \\ $$$$\frac{−\mathrm{3}}{\mathrm{2}}\leqslant{k}\leqslant\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$−\mathrm{1}.\mathrm{5}\leqslant{k}\leqslant\mathrm{3}.\mathrm{5} \\ $$$$ \\ $$

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

$$\mathrm{2sin}^{\mathrm{2}} \theta+\mathrm{10cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\mathrm{7}−\mathrm{2}{k}\:\left(\mathrm{1}\right) \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{sin}\frac{\theta}{\mathrm{2}}\mathrm{cos}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{10cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\mathrm{7}−\mathrm{2k} \\ $$$$\Leftrightarrow\mathrm{2sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{10cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\mathrm{7}−\mathrm{k} \\ $$$$\bullet\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}=\mathrm{0}\:\mathrm{then}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\mathrm{1}\Rightarrow\left(\mathrm{1}\right)\Leftrightarrow\mathrm{k}=\mathrm{2}.\mathrm{5}\notin\mathrm{Z} \\ $$$$\bullet\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\neq\mathrm{0}\:\mathrm{then}\:\mathrm{divide}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{of} \\ $$$$\mathrm{equation}\left(\mathrm{1}\right)\mathrm{by}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\mathrm{2tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{10}=\left(\mathrm{7}−\mathrm{2k}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\left(\mathrm{5}−\mathrm{2k}\right)\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\mathrm{3}+\mathrm{2k}\Rightarrow\mathrm{k}\neq\mathrm{2}.\mathrm{5} \\ $$$$\Leftrightarrow\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}+\mathrm{2k}}{\mathrm{5}−\mathrm{2k}}\geqslant\mathrm{0}\Leftrightarrow−\mathrm{1}.\mathrm{5}\leqslant\mathrm{k}<\mathrm{2}.\mathrm{5} \\ $$$$\mathrm{k}\in\mathrm{Z}\Rightarrow\mathrm{k}\in\left\{−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2}\right\}\Rightarrow\mathrm{Choose}\:\mathrm{B} \\ $$

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