If xi>0, 1≤i≤n, and x1+x2+x3…+xn=π, then the greatest value of sinx1+sinx2+sinx3+...+sinxn=

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nsinπn

answer is C.

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Detailed Solution

We have sinx1+sinx2+sinx3+….+sinxnn ≤sinx1+x2+x3+….+xnn since sin is concave in 0,π ≤sinπn ∵x1+x2+x3+….+xn=π⇒sinx1+sinx2+sinx3+….+sinxn≤nsinπn

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