Question

Determine the range of the function $f(x) = \frac{45^{\tan x}}{2025^{\tan x}+1}$.

• A. (0, 1/2]
• B. [0, 1/2]
• C. (0, 1)
• D. [1/2, \infty)

Answer 1

Answer: (A)

(0, 1/2]

Answer 2

Let $y = \tan x$. The range of $\tan x$ is $(-\infty, \infty)$. The function becomes $f(x) = \frac{45^y}{2025^y+1}$. Since $2025 = 45^2$, we can rewrite the function as $f(x) = \frac{45^y}{(45^2)^y+1} = \frac{45^y}{45^{2y}+1}$. Let $u = 45^y$. As $y$ ranges over $(-\infty, \infty)$, $u = 45^y$ ranges over $(0, \infty)$ because the base $45 > 1$. The function in terms of $u$ is $g(u) = \frac{u}{u^2+1}$, where $u \in (0, \infty)$. To find the range of $g(u)$, we can use the AM-GM inequality for $u > 0$. For the terms $u^2$ and $1$, the arithmetic mean is $\frac{u^2+1}{2}$ and the geometric mean is $\sqrt{u^2 \cdot 1} = u$. By AM-GM, $\frac{u^2+1}{2} \ge u$, which implies $u^2+1 \ge 2u$. Since $u > 0$, $u^2+1 > 0$. Dividing by $u^2+1$, we get: $g(u) = \frac{u}{u^2+1} \le \frac{u}{2u} = \frac{1}{2}$. The maximum value is $1/2$. Equality holds when $u^2 = 1$, which for $u > 0$ means $u=1$. Now, we consider the limits as $u$ approaches the boundaries of its domain $(0, \infty)$: As $u \to 0^+$, $\lim_{u \to 0^+} g(u) = \lim_{u \to 0^+} \frac{u}{u^2+1} = \frac{0}{0+1} = 0$. As $u \to \infty$, $\lim_{u \to \infty} g(u) = \lim_{u \to \infty} \frac{u}{u^2+1} = \lim_{u \to \infty} \frac{1/u}{1+1/u^2} = \frac{0}{1+0} = 0$. Since $u > 0$, $g(u) = \frac{u}{u^2+1}$ is always positive. Thus, $g(u)$ approaches $0$ but never reaches it. Combining the maximum value and the limits, the range of $g(u)$ for $u \in (0, \infty)$ is $(0, 1/2]$. This is also the range of $f(x)$.