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(B) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.Thus,we require $\left|\frac{x^{2}-4x+2}{x^{2}+3}\right| \leq 1$.This implies $-1 \leq \frac{x^{2}

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$[-\frac{1}{4}, \infty)$

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(B) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.Thus,we require $\left|\frac{x^{2}-4x+2}{x^{2}+3}\right| \leq 1$.This implies $-1 \leq \frac{x^{2}-4x+2}{x^{2}+3} \leq 1$.Since $x^{2}+3 > 0$ for all $x \in \mathbb{R}$,we can multiply by $(x^{2}+3)$:$-(x^{2}+3) \leq x^{2}-4x+2 \leq x^{2}+3$.First inequality: $-x^{2}-3 \leq x^{2}-4x+2 \implies 2x^{2}-4x+5 \geq 0$. The discriminant $D = (-4)^{2} - 4(2)(5) = 16 - 40 = -24  0$ for all $x \in \mathbb{R}$.Second inequality: $x^{2}-4x+2 \leq x^{2}+3 \implies -4x \leq 1 \implies x \geq -\frac{1}{4}$.Therefore,the domain is $[-\frac{1}{4}, \infty)$.

Considering only the principal values of the inverse trigonometric functions,the domain of the function $f(x) = \cos^{-1}\left(\frac{x^{2}-4x+2}{x^{2}+3}\right)$ is:

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