Integral Calculus> f(x)=1+2sin x+3cos ²x[0,2p÷3) find max va...

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

naveen , 12 Years ago

Grade upto college level

2 Answers

bharat bajaj

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

Last Activity: 12 Years ago

Parvez ali

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

Last Activity: 12 Years ago