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Question

If(x)=|cosxsinx|, then find f(π3)

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Solution

Given f(x)=|cosxsinx|
For 0<x<π4,f(x)=cosxsinx
For π4<x<π2,f(x)=sinxcosx
On differentiating
For 0<x<π4,f(x)=sinxcosx
For π4<x<π2,f(x)=cosx+sinx
f(π3)=cosπ3+sinπ3=12+32=3+12

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