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If m= sin6x + cos6x, then m belongs to the interval
(a) [7/8, 5/4] (b) [1/2 , 5/8] (c) [1/4, 1]
Explain pls...
m=sin6x+cos6x=(sin2x+cos2x)(sin4x+cos4x-sin2xcos2x)
as a3+b3=(a+b)(a2-ab+b2)
we know that,
sin2x+cos2x=1
m=((sin2x+cos2x)2-3sin2xcos2x)
m=1-3sin2xcos2x
m=1-3/4(sin2x)2
we know 0<=sin2A<=1,
so -1<=-sin22x<=0 multiply with -1,
-3/4<=-3/4(sin22x)<=0 multiply with 3/4,
1/4<=1-3/4(sin22x)<=1 add 1,
1/4<=m<=1
therefore "m" belongs to[1,1/4].
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