Question

If $f(x) = \log_e (\frac{x^2+e}{x^2+1})$, then the number of integers in the range of $f(x)$ is

• A. 0
• B. 1
• C. 2
• D. Infinite

Answer 1

Answer: (B)

1

Answer 2

The problem asks us to find the number of integers in the range of the function $f(x) = \log_e \left(\frac{x^2+e}{x^2+1}\right)$.

1. Analyze the argument of the logarithm: Let $g(x) = \frac{x^2+e}{x^2+1}$. To find the range of $g(x)$, let $t = x^2$. Since $x$ is a real number, $x^2 \ge 0$, so $t \ge 0$. Now, $g(t) = \frac{t+e}{t+1}$. We can rewrite $g(t)$ by performing polynomial division or by algebraic manipulation: $g(t) = \frac{t+1 + e-1}{t+1} = \frac{t+1}{t+1} + \frac{e-1}{t+1} = 1 + \frac{e-1}{t+1}$.

2. Determine the range of $g(t)$ for $t \ge 0$: Since $t \ge 0$: $t+1 \ge 1$. Taking the reciprocal of $t+1$: $0 < \frac{1}{t+1} \le 1$. (As $t \to \infty$, $\frac{1}{t+1} \to 0$. When $t=0$, $\frac{1}{t+1}=1$).

The value of $e$ is approximately $2.718$. So, $e-1 \approx 1.718$, which is a positive constant. Multiply the inequality $0 < \frac{1}{t+1} \le 1$ by $(e-1)$: $0 \cdot (e-1) < \frac{e-1}{t+1} \le 1 \cdot (e-1)$ $0 < \frac{e-1}{t+1} \le e-1$.

Now, add 1 to all parts of the inequality: $1+0 < 1 + \frac{e-1}{t+1} \le 1 + (e-1)$ $1 < g(t) \le e$.

So, the range of $g(x) = \frac{x^2+e}{x^2+1}$ is $(1, e]$.

3. Find the range of $f(x) = \log_e(g(x))$: The natural logarithm function, $h(u) = \log_e u$, is a strictly increasing function for $u > 0$. Since the range of $g(x)$ is $(1, e]$, we can apply the logarithm to this interval: $\log_e(1) < \log_e(g(x)) \le \log_e(e)$. We know that $\log_e(1) = 0$ and $\log_e(e) = 1$. Therefore, the range of $f(x)$ is $(0, 1]$.

4. Count the number of integers in the range of $f(x)$: The range of $f(x)$ is $(0, 1]$. The integers that lie within this interval are values greater than 0 and less than or equal to 1. The only integer satisfying this condition is 1.

Thus, there is only one integer in the range of $f(x)$.