Find the extreme values of 3sin^2x+5cos^2x
Answers
Given : 3sin²x+5cos²x
To Find : extreme values
Solution:
3sin²x+5cos²x
= 3sin²x + 5(1 - sin²x)
= 3sin²x + 5 - 5 sin²x
= 5 - 2 (sin²x)
sin²x minimum value = 0
Hence
5 - 0 = 5 Max Value
sin²x maximum value = 1
Hence 5 - 2 = 3 is Minimum Value
extreme values are 3 & 5
A = 3sin²x+5cos²x
dA/dx = 6Sinxcosx + 10 Cos(-sinx)
=> dA/dx = - 4sinxcosx
=> dA/dx = - 2Sin2x
dA/dx = 0
=> - 2Sin2x = 0
=> x = 0 , x = π/2
d²A/dx² = -4Cos2x
d²A/dx² = -4 < 0 at x = 0 hence maximum value
= 3sin²0+5cos²0
= 0 + 5
= 5
d²A/dx² = 4 > 0 at x = π/2 hence minimum value
= 3sin²90+5cos²90
= 3 + 0
= 3
Extreme values are 5 & 3
Learn More:
Find the maximum and minimum values of sin 2x - cos 2x - Brainly.in
https://brainly.in/question/7029769
Answer:
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