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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...

4 years ago

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Answers : (1)

										

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].

4 years ago

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