Question

Let $f(x) = \sqrt{\frac{\log(x^2+kx+k+1)}{x^2+k}}$. If $f(x)$ is continuous for all $x \in \mathbb{R}$, then the range of $k$ is

• A. $(-\infty, 0) \cup [4, \infty)$
• B. [0,4]
• C. [0,4]
• D. [0, 4]

Answer 1

Answer: (B)

[0,4]

Answer 2

For the function $f(x) = \sqrt{\frac{\log(x^2+kx+k+1)}{x^2+k}}$ to be continuous for all $x \in \mathbb{R}$, its domain must be $\mathbb{R}$. This requires two main conditions:

1. The argument of the logarithm must be positive for all $x \in \mathbb{R}$: $x^2+kx+k+1 > 0$. This is a quadratic in $x$ with a positive leading coefficient. For it to be always positive, its discriminant must be negative. $\Delta_1 = k^2 - 4(1)(k+1) < 0$ $k^2 - 4k - 4 < 0$. The roots of $k^2 - 4k - 4 = 0$ are $k = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$. Thus, for the inequality to hold, $2 - 2\sqrt{2} < k < 2 + 2\sqrt{2}$.

2. The expression under the square root must be non-negative for all $x \in \mathbb{R}$: $\frac{\log(x^2+kx+k+1)}{x^2+k} \ge 0$.

   We analyze this based on the denominator $x^2+k$:

   • Case 1: $k > 0$. In this case, $x^2+k > 0$ for all $x \in \mathbb{R}$ (since $x^2 \ge 0$). The condition simplifies to $\log(x^2+kx+k+1) \ge 0$. Since the base of the logarithm is $e$ (or any base $>1$), this implies $x^2+kx+k+1 \ge e^0 = 1$. $x^2+kx+k \ge 0$. For this quadratic to be always non-negative, its discriminant must be less than or equal to zero. $\Delta_2 = k^2 - 4(1)(k) \le 0$ $k(k-4) \le 0$. This inequality holds for $0 \le k \le 4$. Combining with the condition $k > 0$, we get $0 < k \le 4$.

   • Case 2: $k = 0$. The function becomes $f(x) = \sqrt{\frac{\log(x^2+1)}{x^2}}$. The denominator $x^2$ is zero at $x=0$. For continuity, we check the limit as $x \to 0$: $\lim_{x \to 0} \frac{\log(x^2+1)}{x^2}$. Using L'Hôpital's rule or the standard limit $\lim_{y \to 0} \frac{\log(1+y)}{y} = 1$, we get the limit as 1. So, $\lim_{x \to 0} f(x) = \sqrt{1} = 1$. The function can be defined as $f(0)=1$ to be continuous. For $x \neq 0$, $x^2 > 0$ and $x^2+1 > 1$, so $\log(x^2+1) > 0$. Thus, $\frac{\log(x^2+1)}{x^2} > 0$. The condition $\frac{\log(x^2+kx+k+1)}{x^2+k} \ge 0$ is satisfied for $k=0$.

   • Case 3: $k < 0$. If $k < 0$, then for $x=0$, the denominator $x^2+k = k$, which is negative. This means the denominator can be negative, leading to potential issues with the expression under the square root. For the function to be defined for all $x$, the denominator $x^2+k$ must not be zero for any $x$. This implies $x^2 = -k$ must have no real solutions, which means $-k < 0$, or $k > 0$. Thus, $k < 0$ is not allowed for the denominator to be consistently positive or non-negative in a way that allows continuity for all $x$.

   Combining the valid cases for the second condition, we have $k=0$ or $0 < k \le 4$, which gives $0 \le k \le 4$.

   Finally, we intersect the ranges from both conditions: $(2 - 2\sqrt{2}, 2 + 2\sqrt{2}) \cap [0, 4]$. Since $2 - 2\sqrt{2} \approx -0.828$ and $2 + 2\sqrt{2} \approx 4.828$, the interval $[0, 4]$ is fully contained within $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$. Therefore, the intersection is $[0, 4]$.