The set of all values of x for which the function ;$\displaystyle f\left ( x \right )= \left ( k^{2}-3k+2 \right )\left ( \cos ^{2}\frac{x}{4}-\sin ^{2}\frac{x}{4} \right )+\left ( k-1 \right )x+\sin 1$& ;does not posses critical points is 

$\displaystyle f\left ( x \right )= \left ( k^{2}-3x+2 \right )\cos \frac{x}{2}+\left ( k-1 \right )x+\sin 1$

$\displaystyle {f}'\left ( x \right )= \left ( k-1 \right )\left ( k-2 \right )\left ( -\frac{1}{2}\sin \frac{x}{2} \right )+\left ( k-1 \right )$

$=\displaystyle \left ( k-1 \right )\left [ 1-\frac{k-2}{2}\sin \frac{x}{2} \right ]$

Since f(x) does not possess critical points therefore f'(x) is not equal to zero.

i.e.,

$\displaystyle k\neq 1$ or $\displaystyle 1-\frac{k-2}{2}\sin \frac{x}{2}= 0$ does not posses a solution or $\displaystyle \sin \frac{x}{2}= \frac{2}{k-2}$ does not have a solution.

Hence we must

have $\displaystyle \left | \frac{2}{k-2} \right |> 1$ as

$\displaystyle \left | \sin \frac{x}{2} \right |< 1.$

Above implies that $\displaystyle \left | k-2 \right |^{2}\leq 4$

or $\displaystyle -2< \left ( k-2 \right )< 2$

$\displaystyle \because x^{2}< a^{2}\Rightarrow \left ( x^{2}-a^{2} \right )= -ive$ or $\displaystyle -a< x< a$

$\displaystyle \therefore 0 < k< 4$. Also $\displaystyle k\neq 1.$

$\displaystyle \therefore k \epsilon \left ( 0,1 \right )\cup \left ( 1,4 \right )$