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Question Number 48484 by Meritguide1234 last updated on 24/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18

trying to solve  let in place of (√3)  putting  a     a=(√3)   finding the values of x  so that  sinax=sinx  ax=kπ+(−1)^k x  ax=π+(−1)^1 x  for k=1   x=(π/(a+1))  ax=2π+(−1)^2 x         for k=2   x=((2π)/(a−1))  ax=3π+(−1)^3 x   for k=3  x=((3π)/(a+1))  ax=4π+(−1)^4 x  for k=4   x=((4π)/(a−1))  ax=5π+(−1)^5 x   for k=5    x=((5π)/(a+1))  ax=(n−1)π+(−1)^(n−1) x   k=n−1  x=(((n−1)π)/(a−(−1)^(n−1) ))  ax=nπ+(−1)^n x  k=n   x=((nπ)/(a−(−1)^n ))  now  in between two consequtive interval the value of  max{sinx,sinax} is=1  ∫_0 ^(π/(a+1)) 1.dx+∫_(π/(a+1)) ^((2π)/(a−1))  dx+∫_((2π)/(a−1 )) ^((3π)/(a+1))  dx+∫_((3π)/(a+1)) ^((4π)/(a−1))  dx+...+    ∫_(((n−1)π)/(a−(−1)^(n−1) )) ^((nπ)/(a−(−1)^n ))  dx  ={(π/(a+1))−0}+{((2π)/(a−1))−(π/(a+1))}+{((3π)/(a+1))−((2π)/(a−1))}+...+  {((nπ)/(a−(−1)^n ))−(((n−1)π)/(a−(−1)^(n−1) ))}   all terms cacelled each other   the value of intregal is  =((nπ)/(a−(−1)^n ))−(((n−1)π)/(a−(−1)^(n−1) ))  now   lim_(n→∞)  (1/n)[((nπ)/(a−(−1)^n ))−(((n−1)π)/(a−(−1)^(n−1) ))]  =lim_(n→∞)  (1/n)[n{(π/(a−(−1)^n ))−(π/(a−(−1)^(n−1) ))}+(π/(a−(−1)^(n−1) ))]  so ans is =(π/(a+1)) or (π/(a−1))  =(π/((√3) +1))  or (π/((√3) −1))  i have tried to solve...              ...  ...

$${trying}\:{to}\:{solve} \\ $$$${let}\:{in}\:{place}\:{of}\:\sqrt{\mathrm{3}}\:\:{putting}\:\:{a}\:\:\:\:\:{a}=\sqrt{\mathrm{3}}\: \\ $$$${finding}\:{the}\:{values}\:{of}\:{x}\:\:{so}\:{that} \\ $$$${sinax}={sinx} \\ $$$${ax}={k}\pi+\left(−\mathrm{1}\right)^{{k}} {x} \\ $$$${ax}=\pi+\left(−\mathrm{1}\right)^{\mathrm{1}} {x}\:\:{for}\:{k}=\mathrm{1}\:\:\:{x}=\frac{\pi}{{a}+\mathrm{1}} \\ $$$${ax}=\mathrm{2}\pi+\left(−\mathrm{1}\right)^{\mathrm{2}} {x}\:\:\:\:\:\:\:\:\:{for}\:{k}=\mathrm{2}\:\:\:{x}=\frac{\mathrm{2}\pi}{{a}−\mathrm{1}} \\ $$$${ax}=\mathrm{3}\pi+\left(−\mathrm{1}\right)^{\mathrm{3}} {x}\:\:\:{for}\:{k}=\mathrm{3}\:\:{x}=\frac{\mathrm{3}\pi}{{a}+\mathrm{1}} \\ $$$${ax}=\mathrm{4}\pi+\left(−\mathrm{1}\right)^{\mathrm{4}} {x}\:\:{for}\:{k}=\mathrm{4}\:\:\:{x}=\frac{\mathrm{4}\pi}{{a}−\mathrm{1}} \\ $$$${ax}=\mathrm{5}\pi+\left(−\mathrm{1}\right)^{\mathrm{5}} {x}\:\:\:{for}\:{k}=\mathrm{5}\:\:\:\:{x}=\frac{\mathrm{5}\pi}{{a}+\mathrm{1}} \\ $$$${ax}=\left({n}−\mathrm{1}\right)\pi+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}\:\:\:{k}={n}−\mathrm{1}\:\:{x}=\frac{\left({n}−\mathrm{1}\right)\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} } \\ $$$${ax}={n}\pi+\left(−\mathrm{1}\right)^{{n}} {x}\:\:{k}={n}\:\:\:{x}=\frac{{n}\pi}{{a}−\left(−\mathrm{1}\right)^{{n}} } \\ $$$${now}\:\:{in}\:{between}\:{two}\:{consequtive}\:{interval}\:{the}\:{value}\:{of} \\ $$$${max}\left\{{sinx},{sinax}\right\}\:{is}=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{{a}+\mathrm{1}}} \mathrm{1}.{dx}+\int_{\frac{\pi}{{a}+\mathrm{1}}} ^{\frac{\mathrm{2}\pi}{{a}−\mathrm{1}}} \:{dx}+\int_{\frac{\mathrm{2}\pi}{{a}−\mathrm{1}\:}} ^{\frac{\mathrm{3}\pi}{{a}+\mathrm{1}}} \:{dx}+\int_{\frac{\mathrm{3}\pi}{{a}+\mathrm{1}}} ^{\frac{\mathrm{4}\pi}{{a}−\mathrm{1}}} \:{dx}+...+ \\ $$$$\:\:\int_{\frac{\left({n}−\mathrm{1}\right)\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }} ^{\frac{{n}\pi}{{a}−\left(−\mathrm{1}\right)^{{n}} }} \:{dx} \\ $$$$=\left\{\frac{\pi}{{a}+\mathrm{1}}−\mathrm{0}\right\}+\left\{\frac{\mathrm{2}\pi}{{a}−\mathrm{1}}−\frac{\pi}{{a}+\mathrm{1}}\right\}+\left\{\frac{\mathrm{3}\pi}{{a}+\mathrm{1}}−\frac{\mathrm{2}\pi}{{a}−\mathrm{1}}\right\}+...+ \\ $$$$\left\{\frac{{n}\pi}{{a}−\left(−\mathrm{1}\right)^{{n}} }−\frac{\left({n}−\mathrm{1}\right)\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }\right\} \\ $$$$\:{all}\:{terms}\:{cacelled}\:{each}\:{other}\: \\ $$$${the}\:{value}\:{of}\:{intregal}\:{is} \\ $$$$=\frac{{n}\pi}{{a}−\left(−\mathrm{1}\right)^{{n}} }−\frac{\left({n}−\mathrm{1}\right)\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} } \\ $$$${now}\: \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\:\frac{\mathrm{1}}{{n}}\left[\frac{{n}\pi}{{a}−\left(−\mathrm{1}\right)^{{n}} }−\frac{\left({n}−\mathrm{1}\right)\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }\right] \\ $$$$={li}\underset{{n}\rightarrow\infty} {{m}}\:\frac{\mathrm{1}}{{n}}\left[{n}\left\{\frac{\pi}{{a}−\left(−\mathrm{1}\right)^{{n}} }−\frac{\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }\right\}+\frac{\pi}{{a}−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }\right] \\ $$$${so}\:{ans}\:{is}\:=\frac{\pi}{{a}+\mathrm{1}}\:{or}\:\frac{\pi}{{a}−\mathrm{1}} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{3}}\:+\mathrm{1}}\:\:{or}\:\frac{\pi}{\sqrt{\mathrm{3}}\:−\mathrm{1}} \\ $$$${i}\:{have}\:{tried}\:{to}\:{solve}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$... \\ $$$$... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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