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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
	$trying to solve let in place of (√3) putting a a=(√3) finding the values of x so that sinax=sinx ax=kπ+(−1)^k x ax=π+(−1)^1 x for k=1 x=(π/(a+1)) ax=2π+(−1)^2 x for k=2 x=((2π)/(a−1)) ax=3π+(−1)^3 x for k=3 x=((3π)/(a+1)) ax=4π+(−1)^4 x for k=4 x=((4π)/(a−1)) ax=5π+(−1)^5 x for k=5 x=((5π)/(a+1)) ax=(n−1)π+(−1)^(n−1) x k=n−1 x=(((n−1)π)/(a−(−1)^(n−1) )) ax=nπ+(−1)^n x k=n x=((nπ)/(a−(−1)^n )) now in between two consequtive interval the value of max{sinx,sinax} is=1 ∫_0 ^(π/(a+1)) 1.dx+∫_(π/(a+1)) ^((2π)/(a−1)) dx+∫_((2π)/(a−1 )) ^((3π)/(a+1)) dx+∫_((3π)/(a+1)) ^((4π)/(a−1)) dx+...+ ∫_(((n−1)π)/(a−(−1)^(n−1) )) ^((nπ)/(a−(−1)^n )) dx ={(π/(a+1))−0}+{((2π)/(a−1))−(π/(a+1))}+{((3π)/(a+1))−((2π)/(a−1))}+...+ {((nπ)/(a−(−1)^n ))−(((n−1)π)/(a−(−1)^(n−1) ))} all terms cacelled each other the value of intregal is =((nπ)/(a−(−1)^n ))−(((n−1)π)/(a−(−1)^(n−1) )) now lim_(n→∞) (1/n)[((nπ)/(a−(−1)^n ))−(((n−1)π)/(a−(−1)^(n−1) ))] =lim_(n→∞) (1/n)[n{(π/(a−(−1)^n ))−(π/(a−(−1)^(n−1) ))}+(π/(a−(−1)^(n−1) ))] so ans is =(π/(a+1)) or (π/(a−1)) =(π/((√3) +1)) or (π/((√3) −1)) i have tried to solve... ... ...$
	$t r y i n g t o s o l v e$ $l e t i n p l a c e o f \sqrt{3} p u t t i n g a a = \sqrt{3}$ $f i n d i n g t h e v a l u e s o f x s o t h a t$ $s i n a x = s i n x$ $a x = k π + {(- 1)}^{k} x$ $a x = π + {(- 1)}^{1} x f o r k = 1 x = \frac{π}{a + 1}$ $a x = 2 π + {(- 1)}^{2} x f o r k = 2 x = \frac{2 π}{a - 1}$ $a x = 3 π + {(- 1)}^{3} x f o r k = 3 x = \frac{3 π}{a + 1}$ $a x = 4 π + {(- 1)}^{4} x f o r k = 4 x = \frac{4 π}{a - 1}$ $a x = 5 π + {(- 1)}^{5} x f o r k = 5 x = \frac{5 π}{a + 1}$ $a x = (n - 1) π + {(- 1)}^{n - 1} x k = n - 1 x = \frac{(n - 1) π}{a - {(- 1)}^{n - 1}}$ $a x = n π + {(- 1)}^{n} x k = n x = \frac{n π}{a - {(- 1)}^{n}}$ $n o w i n b e t w e e n t w o c o n s e q u t i v e i n t e r v a l t h e v a l u e o f$ $m a x {s i n x, s i n a x} i s = 1$ $\int_{0}^{\frac{π}{a + 1}} 1 . d x + \int_{\frac{π}{a + 1}}^{\frac{2 π}{a - 1}} d x + \int_{\frac{2 π}{a - 1}}^{\frac{3 π}{a + 1}} d x + \int_{\frac{3 π}{a + 1}}^{\frac{4 π}{a - 1}} d x + . . . +$ $\int_{\frac{(n - 1) π}{a - {(- 1)}^{n - 1}}}^{\frac{n π}{a - {(- 1)}^{n}}} d x$ $= {\frac{π}{a + 1} - 0} + {\frac{2 π}{a - 1} - \frac{π}{a + 1}} + {\frac{3 π}{a + 1} - \frac{2 π}{a - 1}} + . . . +$ ${\frac{n π}{a - {(- 1)}^{n}} - \frac{(n - 1) π}{a - {(- 1)}^{n - 1}}}$ $a l l t e r m s c a c e l l e d e a c h o t h e r$ $t h e v a l u e o f i n t r e g a l i s$ $= \frac{n π}{a - {(- 1)}^{n}} - \frac{(n - 1) π}{a - {(- 1)}^{n - 1}}$ $n o w$ $l i \underset{n \to \infty}{m} \frac{1}{n} [\frac{n π}{a - {(- 1)}^{n}} - \frac{(n - 1) π}{a - {(- 1)}^{n - 1}}]$ $= l i \underset{n \to \infty}{m} \frac{1}{n} [n {\frac{π}{a - {(- 1)}^{n}} - \frac{π}{a - {(- 1)}^{n - 1}}} + \frac{π}{a - {(- 1)}^{n - 1}}]$ $s o a n s i s = \frac{π}{a + 1} o r \frac{π}{a - 1}$ $= \frac{π}{\sqrt{3} + 1} o r \frac{π}{\sqrt{3} - 1}$ $i h a v e t r i e d t o s o l v e . . .$ $. . .$ $. . .$
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