f(x)= 1+2sinx + 3(cosx)^2, 0<=x<=2Π/3 is

a) min at x= Π/2

b) min at x= Π/6

Dear Shefali Sharma,

Ans:- f(x)=1+2sinx +3cos²x

Differentiating WRT x we get = 2cosx-6sinx cosx and

the 2nd order differential is= -2sinx-6cos2x

putting dy/dx=0 we get sinx=1/3 && cosx=0

Hence applying the conditions for max and minima we get the values as

Max value:-13/3

Min value:-3

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