lf `x >=0` and `theta = sin^(-1)x + cos^(-1)x-tan^(-1) x`, then

Options

A. `(pi)/2le theta le (3pi)/4`
B. `0le theta le (pi)/4`
C. `0le theta lt (pi)/2`
D. `(pi)/4 le theta le (pi)/2`

Explanation

To solve the problem, we need to analyze the expression for \( \theta \) given by:

\[ \theta = \sin^{-1}(x) + \cos^{-1}(x) - \tan^{-1}(x) \]

where \( x \geq 0 \).

### Step 1: Determine the Range of \( x \)

Since \( x \) is given to be greater than or equal to 0, we also need to consider the domains of the inverse trigonometric functions involved:

- The domain of \( \sin^{-1}(x) \) is \( x \in [0, 1] \).
- The domain of \( \cos^{-1}(x) \) is \( x \in [0, 1] \).
- The domain of \( \tan^{-1}(x) \) is \( x \in [0, \infty) \).

Thus, combining these, we conclude that \( x \) must lie in the interval:

\[ x \in [0, 1] \]

### Step 2: Simplify \( \theta \)

Using the identity:

\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]

we can rewrite \( \theta \):

\[ \theta = \frac{\pi}{2} - \tan^{-1}(x) \]

### Step 3: Analyze the Behavior of \( \tan^{-1}(x) \)

Now, we need to find the maximum and minimum values of \( \theta \) as \( x \) varies from 0 to 1.

- When \( x = 0 \):
\[ \tan^{-1}(0) = 0 \quad \Rightarrow \quad \theta = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

- When \( x = 1 \):
\[ \tan^{-1}(1) = \frac{\pi}{4} \quad \Rightarrow \quad \theta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Step 4: Determine the Range of \( \theta \)

From the above calculations, we find that:

- The maximum value of \( \theta \) is \( \frac{\pi}{2} \) (when \( x = 0 \)).
- The minimum value of \( \theta \) is \( \frac{\pi}{4} \) (when \( x = 1 \)).

Thus, as \( x \) varies from 0 to 1, \( \theta \) varies from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \).

### Conclusion

Therefore, the range of \( \theta \) is:

\[ \theta \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \]

This means that the correct option is:

**Option 3: \( \theta \) lies between \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \)**.