The function $f(x) = \dfrac{\sqrt{x^2 + kx + 1}}{x^2 - k}$ is continuous for all real $x$. Find the range of $k$.
Step-by-Step Solution
Key Concept: Need: (1) $x^2+kx+1\geq 0$ always (discriminant $\leq 0$: $k^2\leq 4$); (2) $x^2\neq k$ has no solution where denominator=0 coincides with numerator=0.
For $f$ to be continuous: $x^2+kx+1\geq 0$ for all $x$ requires $k^2-4\leq 0$, i.e. $k\in[-2,2]$. Also $x^2=k$ must either have no real solutions or numerator must also be 0. For $k>0$: $x=\pm\sqrt{k}$ gives $x^2+kx+1=k+k\sqrt{k}+1>0$, so numerator$\neq 0$ but denominator$=0$: discontinuous. For $k=0$: denominator $x^2=0$ at $x=0$, numerator=1, discontinuous. For $k<0$: $x^2=k<0$ has no real solutions, denominator never 0. So $k\in[-2,0)$.
Correct Answer: 2