Official solution
\[f\left( x \right) = x\sqrt {kx - {x^2}} \]
\[ \Rightarrow \] \[f'\left( x \right) = \sqrt {kx - {x^2}} + {{(k - 2x)x} \over {2\sqrt {kx - {x^2}} }}\]
\[ \Rightarrow {{2\left( {kx - {x^2}} \right) + kx - 2{x^2}} \over {2\sqrt {kx - {x^2}} }} = {{3kx - 4{x^2}} \over {2\sqrt {kx - {x^2}} }}\]
\[ \Rightarrow {{x(3k - 4x)} \over {2\sqrt {kx - {x^2}} }}\]
Now for increasing function
for f'(x) \[ \ge \] 0, \[\forall x \in [0,3]\]
\[ \Rightarrow kx - {x^2} \ge 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(3k - 4x) \ge 0,\,\forall x \in \left[ {0,3} \right]\,\]
\[ \Rightarrow x(x - k) \le 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(4x - 3k) \le 0,\,\forall x \in \left[ {0,3} \right]\,\]
k \[ \ge \] 3 and k \[ \ge \] 4 \[ \Rightarrow \] k \[ \ge \] 4
\[ \Rightarrow \] m = 4
So f(x) is maximum when k = 4 then \[3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3 = M\]
\[ \therefore \] (m, M) = (4, \[3\sqrt 3 \])