Grade 11Trigonometry
Given A=sin^2x+cos^4x,then for all real x, min≤A≤max
Given A=sin^2x+cos^4x,then for all real x, min≤A≤max
2 Answers
Ritesh Khatri
We Know that
Sin^2x
Also Cos^2x >= Cos^4x
Add sin^2x both sides
Cos^2x + sin^2x >= Cos^4x + sin^2x
1 >= Cos^4x + sin^2x
Therefore max of A is 1
Also both terms
Cos^4x + sin^2x
= (1 – sin^2x)^2 + sin^2x
= 1 + sin^4x – 2sin^2x + sin^2x
=Sin^4x – sinn^2x + 1
= (sin^2x – 1/2)^2 + 3/4
now Min =3/4 and Max = 1
Rishi Sharma
Dear Student,
Please find the solution to your problem in the attached image.
Thanks and Regards
Please find the solution to your problem in the attached image.
Thanks and Regards
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