Let `f(x)=tan^(-1)((sqrt(1+x^(2))-1)/(x))` then which of the following is correct :

Options

A. (a) f(x) has only one integer in its range
B. (b) Range of f(x) is `(-(pi)/(4), (pi)/(4))-{0}`
C. (c) Range of f(x) is `(-(pi)/(2), (pi)/(2))-{0}`
D. (d) Range of f(x) is `[ -(pi)/(4), (pi)/(4)]-{0}`

Answer

To solve the problem, we need to simplify the function \( f(x) = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \) and determine its range.

### Step 1: Substitute \( x \) with \( \tan \theta \)
Let \( x = \tan \theta \). Then, we can rewrite the function as:
\[ f(x) = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right) \]

### Step 2: Simplify the expression inside the arctangent
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we have:
\[ \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta \]
Thus, the expression becomes:
\[ f(x) = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) \]

### Step 3: Rewrite in terms of sine and cosine
We can express \( \sec \theta \) and \( \tan \theta \) in terms of sine and cosine:
\[ f(x) = \tan^{-1}\left(\frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}}\right) = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) \]

### Step 4: Simplify the fraction
This can be simplified as:
\[ f(x) = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) \]
Using the identity \( 1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right) \) and \( \sin \theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \), we can rewrite it as:
\[ f(x) = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) \]

### Step 5: Conclude the simplification
Thus, we have:
\[ f(x) = \frac{\theta}{2} \]
Since \( \theta = \tan^{-1}(x) \), we can express \( f(x) \) as:
\[ f(x) = \frac{1}{2} \tan^{-1}(x) \]

### Step 6: Determine the range of \( f(x) \)
The range of \( \tan^{-1}(x) \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Therefore, the range of \( \frac{1}{2} \tan^{-1}(x) \) will be:
\[ \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \]

### Step 7: Check for integers in the range
The range \( \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \) does contain the integer \( 0 \). However, we need to check if \( f(x) \) is defined at \( x = 0 \):
\[ f(0) = \tan^{-1}\left(\frac{\sqrt{1+0^2}-1}{0}\right) \text{ is of the form } \frac{0}{0}, \text{ which is undefined.} \]

### Conclusion
Since \( f(0) \) is undefined, the range \( \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \) does not contain any integers. Therefore, the correct option is:
**Option B: \( \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \)**.

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