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(B) Given $f(x) = (a^2 - 3a + 2) \cos \frac{x}{2} + (a - 1)x + \sin 1$.Taking the derivative with respect to $x$:$f'(x) = -(a^2 - 3a + 2) \sin \frac{x

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$(0, 1) \cup (1, 4)$

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(B) Given $f(x) = (a^2 - 3a + 2) \cos \frac{x}{2} + (a - 1)x + \sin 1$.Taking the derivative with respect to $x$:$f'(x) = -(a^2 - 3a + 2) \sin \frac{x}{2} \cdot \frac{1}{2} + (a - 1)$$f'(x) = -(a - 1)(a - 2) \cdot \frac{1}{2} \sin \frac{x}{2} + (a - 1)$$f'(x) = (a - 1) \left[ 1 - \frac{a - 2}{2} \sin \frac{x}{2} ight]$.For $f(x)$ to have no critical points,$f'(x) 
eq 0$ for all $x \in \mathbb{R}$.This implies $a - 1 
eq 0$ (so $a 
eq 1$) and $1 - \frac{a - 2}{2} \sin \frac{x}{2} 
eq 0$ for all $x \in \mathbb{R}$.If $a = 2$,$f'(x) = 1 
eq 0$,which is valid.If $a 
eq 2$,then $\sin \frac{x}{2} = \frac{2}{a - 2}$ must have no solution in $\mathbb{R}$.This occurs when $\left| \frac{2}{a - 2} ight| > 1$,which means $|a - 2| $-2 Combining $a 
eq 1$ and $0 < a < 4$,we get $a \in (0, 1) \cup (1, 4)$.

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

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