The set of all values of a for which the function f(x)=(a^ 2 -3...

(f(x)=\left(a^{2}=-3 a+2\right)\left(\cos ^{2}x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1) (\Rightarrow\mathrm{f}(\mathrm{x})=(\mathrm{a}-$1)($\mathrm{a}-$2) \cos$\mathrm{x}$/ 2+($\mathrm{a}-$1)$\mathrm{x}+\sin$1) (\Rightarrow \mathrmf^{\prime}($\mathrm{x})=-\frac{1}{2}(\mathrm{a}-$1)($\mathrm{a}-$2) \sin$\frac{\mathrm{x}}{2}+(\mathrm{a}-$1)) (\Rightarrow \mathrmf^{\prime}($\mathrm{x})=(\mathrm{a}-1)\left[1-\frac{(\mathrm{a}-2)}{2} \sin \frac{\mathrm{x}}{2}\right]) If $f(x)$ does not possess critical points, then (f^{\prime}(x)) = 0 for any (x \in R) (\Rightarrow(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]) = 0 for any (x \in R) (\Rightarrow $\mathrm{a}=$1) and (1-\left(\frac{\mathrm{a}-2}{2}\right) \sin \frac{\mathrm{x}}{2}=$0) must not have any solution in ($\mathrm{R}$). (\Rightarrow$\mathrm{a}=$1) and (\sin$\frac{\mathrm{x}}{2}=\frac{2}{\mathrm{a}-2}$) is not solvable in ($\mathrm{R}$). (\Rightarrow$\mathrm{a} =$1) and (\left|\frac{2}{\mathrm{a}-2}\right| \gt$1\left[\right.$) For ($\mathrm{a}=$2,$\mathrm{f}(\mathrm{x})=\mathrm{x}+\sin$1 \therefore \mathrmf^{\prime}($\mathrm{x})=$1) (\Rightarrow$\mathrm{a}=$1) and (0\lt$\mathrm{a}\lt$4 \Rightarrow$\mathrm{a} \in(0,1) \cup(1,4)).