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Find Min. value of f(x) = sin6 x + cos6x.
Dear Man111,Solution:- f(x) = (sin2 x)3 + (cos2 x)3 = (sin2 x+ cos2x) [ (sin2 x)2 + sin2 x cos2x +(cos2x)2 ]
= [ (sin2 x + cos2x)2 - 2sin2 x cos2x + sin2 x cos2x ]
= [1 - sin2 x cos2x]
f(x) is minimum when (sin2 x cos2x) is maximum and maximum value of (sin2 x cos2x) is 1/2 when x =450
So, min. value of f(x) is 1/2 [ANS]. Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Man111!!!Regards,Askiitians ExpertsPriyansh Bajaj
f(x)=sin6x + cos6x
= (sin2x)3 + (cos2x)
f(x)= (sin2x+cos2x)(sin4x+cos4x-sinxcosx) a3 + b3 =(a+b)(a2 +b2 -ab)
f(x)={[(1-cos2x)/2 ]2 +[(1+cos2x)/2]2 -sin2x/2} sin2x=(1-cos2x)/2 ,cos2x=(1+cos2x)/2
f(x)=[1-(sin22x+sin2x)]/2 .......................1
for f(x) to be minimum ,the subtracting term (sin22x + sin2x) should be maximum ,
and its maximum value is 2 at x=pi/4,
so minimum value of f(x)=1-2/2=-1/2
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