Correct option: B.$\left\{ {\frac{{{n^2}\, + \,n\, - \,2}}{2}\,,\,\frac{{{n^2}\, + \,3n}}{2}\,,\,\frac{{{n^{2\,}} + \,n}}{2}} \right\}$
b
$f\left( x \right) = \sum\limits_{r = 1}^n {r + } \sum\limits_{r = 1}^n {\left[ {\cos \frac{x}{r}} \right]} $
$f\left( x \right) = \frac{{n\left( {n + 1} \right)}}{2} + \sum\limits_{r = 1}^n {\left[ {\cos \frac{x}{r}} \right]} $
$ = \frac{{n\left( {n + 1} \right)}}{2} + \left[ {\cos x} \right] + \left[ {\cos \frac{x}{2}} \right] + \left[ {cod\frac{x}{3}} \right] + .... + \left[ {\cos \frac{x}{n}} \right]$
$x=0,$
$f\left( 0 \right) = \frac{{n\left( {n + 1} \right)}}{2} + 1 + 1 + ....n\,time = \frac{{{n^2} + 3n}}{2}$
$x = \frac{\pi }{2},$ $f\left( {\frac{\pi }{2}} \right) = \frac{{n\left( {n + 1} \right)}}{2}$
$x = \pi ,\,\,f\left( \pi \right) = \frac{{n\left( {n + 1} \right)}}{2} - 1 = \frac{{{n^2} + n - 2}}{2}$