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f(x)=1+2sin x+3cos ²x[0,2p÷3) find max value of function

f(x)=1+2sin x+3cos ²x[0,2p÷3) find max value of function

Grade:upto college level

2 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
f(x)=1+2sin x+3cos ²x
f(x)=1+2sin x+ 3 (1 - sin2x)
f(x) = -3sin2x + 2 sin x + 4
df(x)/dx = -6sinxcosx + 2cosx = 0
hence, sin x = 1/3 or cosx = 0
d/dx(df(x)/dx) = -3cos2x - 2sinx
It is negative for sin x = 1/3 and positive for cos x = 0
Hence, we get a maxima at x = sin-1 (1/3)
Therefore, max value of f(x) is 13/3.
Thanks & Regards
Bharat Bajaj
askIITians Faculty
IIT Delhi
Parvez ali
askIITians Faculty 47 Points
7 years ago
f(x)= -3sin2x + 2sinx + 4
= -3(sin2x-2/3 sinx)+4
=-3(sinx-1/3)^2+3/9+4
=-3(sinx - 1/3)+13/3
now 0=<sin x<=1 when 0=<x=<2p/3
-1/3=<(sinx-1/3)=<2/3
-2=<-3(sinx-1/3)=<1
thus maximum value of f(x)=1+13/3=16/3
Parvez Ali
Askiitians Faculty
Btech ISM Dhanbad

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