If f.(0,π)→R is given by f(x)=∑k=1n [1+sinkx],[x] denotes the greatest integer function, then the range of f(x) is
{n-1, n+1}
{n}
{n, n+1}
{n-1, n}
f(x)=∑k=1n 1+[sinkx]=n+[sinx]+[sin2x]+⋯+sin nx. If kx≠π2 for any k=1, 2 .........n then 0[sinkx]=0,k=1,2…n. i.e. f(x)=n. If kx=π2for some k then x=π2k, hence sinx,sin2x,…sin(k−1)x will lie between 0 and 1 so sinjx=01≤j≤k−1; sinkx=1 so f(x) can be n+1 or n.