If `I=int(tan^(-1)(e^(x)))/(e^(x)+e^(-x))dx=([tan^(-1)(f(x))]^(2))/(2)+C` (where C is the constant of integration), then the range of `y=f(x) AA x in R` is

Options

A. `(-oo, oo)`
B. `[0, oo)`
C. `(o, oo)`
D. `(-oo, 0)`

Answer

To solve the given problem, we need to analyze the integral and the relationship provided in the question. Let's break it down step by step.

### Step 1: Understand the Integral
We have:
\[ I = \int \frac{\tan^{-1}(e^x)}{e^x + e^{-x}} \, dx \]
This integral is equal to:
\[ \frac{(\tan^{-1}(f(x)))^2}{2} + C \]
where \(C\) is the constant of integration.

### Step 2: Differentiate Both Sides
To find \(f(x)\), we differentiate both sides with respect to \(x\):
\[ \frac{dI}{dx} = \frac{\tan^{-1}(e^x)}{e^x + e^{-x}} \]
Using the Fundamental Theorem of Calculus, we differentiate the right-hand side:
\[ \frac{d}{dx}\left(\frac{(\tan^{-1}(f(x)))^2}{2}\right) = \tan^{-1}(f(x)) \cdot \frac{f'(x)}{1 + (f(x))^2} \]

### Step 3: Set the Derivatives Equal
Setting the derivatives equal gives us:
\[ \frac{\tan^{-1}(e^x)}{e^x + e^{-x}} = \tan^{-1}(f(x)) \cdot \frac{f'(x)}{1 + (f(x))^2} \]

### Step 4: Analyze the Function
From the expression for \(f(x)\), we can see that \(f(x)\) is related to \(e^x\). Since \(e^x\) ranges from \(0\) to \(\infty\) as \(x\) varies over the reals, we need to analyze how \(\tan^{-1}(e^x)\) behaves.

### Step 5: Determine the Range of \(\tan^{-1}(e^x)\)
The function \(\tan^{-1}(e^x)\):
- As \(x \to -\infty\), \(e^x \to 0\) and \(\tan^{-1}(0) = 0\).
- As \(x \to \infty\), \(e^x \to \infty\) and \(\tan^{-1}(\infty) = \frac{\pi}{2}\).

Thus, the range of \(\tan^{-1}(e^x)\) is \((0, \frac{\pi}{2})\).

### Step 6: Relate to \(f(x)\)
Since \(\tan^{-1}(f(x))\) also takes values in the range \((0, \frac{\pi}{2})\), we can conclude that \(f(x)\) must be a function that maps to values such that \(\tan^{-1}(f(x))\) stays within this range.

### Step 7: Find the Range of \(f(x)\)
To find the range of \(f(x)\):
- Since \(\tan^{-1}(f(x))\) is in \((0, \frac{\pi}{2})\), it implies that \(f(x)\) must be positive and can take any positive value.
- Therefore, the range of \(f(x)\) is \((0, \infty)\).

### Final Answer
The range of \(y = f(x)\) as \(x\) varies over \(\mathbb{R}\) is:
\[ (0, \infty) \]