CONCEPT:\(\smallint {\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{f}}\left( {\rm{x}} \right)\) Calculation:\({\rm{f'}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2}}}{2} - {\rm{kx}} + 1\)\(\smallint {\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = \smallint \left( {\frac{{{{\rm{x}}^2}}}{2} - {\rm{kx}} + 1} \right){\rm{dx}}\) (INTEGRATE)\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^3}}}{6} - \frac{{{\rm{k}}{{\rm{x}}^2}}}{2} + {\rm{x}} + {\rm{c}}\) Now, f (0) = 0∴ \(\frac{0}{6} - \frac{{{\rm{k}}\left( 0 \right)}}{2} + 0 + {\rm{c}} = 0\)\(\Rightarrow {\rm{c}} = 0\)f (3) = 15\(\THEREFORE \frac{{{3^3}}}{6} - \frac{{{\rm{k}}{{\left( 3 \right)}^2}}}{2} + 3 = 15\)\(\Rightarrow \frac{{27}}{6} - \frac{{9{\rm{k}}}}{2} + 3 = 15\)\(\Rightarrow \frac{{9{\rm{k}}}}{2} = \frac{9}{2} - 12\)\(= - \frac{{15}}{2}\)\(\Rightarrow {\rm{k}} = - \frac{{15}}{9}\)\(= - \frac{5}{3}\)Hence, option (3) is correct