1. Problem: Determine for which values of the parameter $k$ the function $f$ is defined for every real number $x \in \mathbb{R}$. We analyze each function separately, focusing on the domain restrictions caused by square roots and denominators. --- **a)** $f(x) = \sqrt{x^2 + (k + 2)x + 2k + 1}$ - The expression inside the square root must be non-negative for all $x$: $$x^2 + (k+2)x + 2k + 1 \geq 0 \quad \forall x \in \mathbb{R}$$ - This is a quadratic in $x$ with leading coefficient 1 (positive), so it opens upwards. - For it to be non-negative for all $x$, its discriminant must be less than or equal to zero: $$\Delta = (k+2)^2 - 4 \cdot 1 \cdot (2k+1) \leq 0$$ - Calculate: $$\Delta = k^2 + 4k + 4 - 8k - 4 = k^2 - 4k \leq 0$$ - Solve inequality: $$k(k - 4) \leq 0$$ - This holds when $k \in [0,4]$. --- Since the user asked for the first question only, we stop here. **Final answer:** The function $f$ is defined for all real $x$ if and only if $$k \in [0,4].$$