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Question-54856




Question Number 54856 by Meritguide1234 last updated on 13/Feb/19
Commented by Meritguide1234 last updated on 14/Feb/19
Commented by Meritguide1234 last updated on 14/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19
((cosx)/(1+2cosx))  (((1/2)cosx)/((1/2)+cosx))→((cos(π/3)cosx)/(cos(π/3)+cosx))→((ab)/(a+b))→((ab)/(((√a) −(√b) )^2 +2(√a) (√b) ))  max value of ((ab)/(a+b)) when D_r =min → (√a) −(√b) =0  cosx=cos(π/3)→x=(π/3)  cos^(−1) (((cosx)/(1+2cosx)))→cos^(−1) (((1/2)/(1+2×(1/2))))→cos^(−1) ((1/4))  cos^(−1) (((cosx)/(1+2cosx)))→cos^(−1) ((1/(1+2×1))) atx=0  cos^(−1) (((cos(π/2))/(1+2cos(π/(2 )))))→cos^(−1) (0) at x=(π/2)    cos^− (0)∫_0 ^(π/2) dx> cos^(−1) ((1/4))∫_0 ^(π/2) dx> cos^(−1) ((1/3))∫_0 ^(π/2) dx  so (π/2)×(π/2)>I>cos^(−1) ((1/3))×(π/2)  pls check...
$$\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cosx}} \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}}{cosx}}{\frac{\mathrm{1}}{\mathrm{2}}+{cosx}}\rightarrow\frac{{cos}\frac{\pi}{\mathrm{3}}{cosx}}{{cos}\frac{\pi}{\mathrm{3}}+{cosx}}\rightarrow\frac{{ab}}{{a}+{b}}\rightarrow\frac{{ab}}{\left(\sqrt{{a}}\:−\sqrt{{b}}\:\right)^{\mathrm{2}} +\mathrm{2}\sqrt{{a}}\:\sqrt{{b}}\:} \\ $$$${max}\:{value}\:{of}\:\frac{{ab}}{{a}+{b}}\:{when}\:{D}_{{r}} ={min}\:\rightarrow\:\sqrt{{a}}\:−\sqrt{{b}}\:=\mathrm{0} \\ $$$${cosx}={cos}\frac{\pi}{\mathrm{3}}\rightarrow{x}=\frac{\pi}{\mathrm{3}} \\ $$$${cos}^{−\mathrm{1}} \left(\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cosx}}\right)\rightarrow{cos}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}\right)\rightarrow{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${cos}^{−\mathrm{1}} \left(\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cosx}}\right)\rightarrow{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}×\mathrm{1}}\right)\:{atx}=\mathrm{0} \\ $$$${cos}^{−\mathrm{1}} \left(\frac{{cos}\frac{\pi}{\mathrm{2}}}{\mathrm{1}+\mathrm{2}{cos}\frac{\pi}{\mathrm{2}\:}}\right)\rightarrow{cos}^{−\mathrm{1}} \left(\mathrm{0}\right)\:{at}\:{x}=\frac{\pi}{\mathrm{2}}\: \\ $$$$\:{cos}^{−} \left(\mathrm{0}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}>\:{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}>\:{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$${so}\:\frac{\pi}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}>{I}>{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)×\frac{\pi}{\mathrm{2}} \\ $$$${pls}\:{check}… \\ $$
Commented by Meritguide1234 last updated on 13/Feb/19
no ..ans is 5π^2 /24
$$\mathrm{no}\:..\mathrm{ans}\:\mathrm{is}\:\mathrm{5}\pi^{\mathrm{2}} /\mathrm{24} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19
upload the answer pls with working...  i tried to reach upto the....  (π^2 /4)>I>(π/2)cos^(−1) ((1/3))  ((6π^2 )/(24))>I>(π/2)cos^(−1) ((1/3))
$${upload}\:{the}\:{answer}\:{pls}\:{with}\:{working}… \\ $$$${i}\:{tried}\:{to}\:{reach}\:{upto}\:{the}…. \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}>{I}>\frac{\pi}{\mathrm{2}}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\frac{\mathrm{6}\pi^{\mathrm{2}} }{\mathrm{24}}>{I}>\frac{\pi}{\mathrm{2}}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19
thsnk you sir...let me unlock the secret...
$${thsnk}\:{you}\:{sir}…{let}\:{me}\:{unlock}\:{the}\:{secret}… \\ $$

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