The set of values of $$k$$ for which $${x^2} - kx + {\sin ^{ - 1}}\left( {\sin 4} \right) > 0$$      for all real $$x$$ is

Solution :
$${\sin ^{ - 1}}\left( {\sin 4} \right) = {\sin ^{ - 1}}\left\{ {\sin \left( {\pi - 4} \right)} \right\} = \pi - 4\,\,\left( {\because \,\, - \frac{\pi }{2} < \pi - 4 < \frac{\pi }{2}} \right)$$
∴ we have, $${x^2} - kx + \pi - 4 > 0$$     for all $$x \in R.$$
$$\therefore \,\,D < 0,{\text{i}}{\text{.e}}{\text{., }}{k^2} - 4\left( {\pi - 4} \right) < 0$$
or, $${k^2} + 4\left( {4 - \pi } \right) < 0,$$    which is not true for any real $$k.$$