The function f(x)=2 \sin x+\sin 2 x, x \in[0,2 \pi] has absolute maximum and minimum at

NIMCET 2008 Question Paper with solutions

Question : 39 of 120

Marks: +1, -0

The function

f(x)=2 \sin x+\sin 2 x, \;\; x ∈[0,2 π]

has absolute maximum and minimum at

Solution:

Given,

f(x)=2 \sin x+\sin 2 x

⇒ \;\; f(x)=2 \cos x+2 \cos 2 x

For max or min of

f(x)

⇒ \;\; f(x)=0

⇒ \cos x+2 \cos ^{2} x-1=0

⇒\cos x=\;{-1 ± √{1+8}}/{4}=-1, \;{1}/{2}

⇒ Critical points are

π, \;{π}/{3}, \;{5 π}/{3}, 0,2 π

f(π)=0 ; f(\;{π}/{3})=\;{3 √{3}}/{2} ; f(\;{5 π}/{3})=\;{-3 √{3}}/{2}

f(0)=0 ; f(2 π)=0

\Rightarrow

Absolute maximum is at

\;{π}/{3}

and absolute minimum is at

\;{5 π}/{3}

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