Statement I Number of roots of the equation `cot^(-1)x cos^(-1) 2x + pi = 0` is zero.
Statement II Range of `cot^(-1) x " and " cos^(-1) x " is " (0, pi) " and " [0, pi]`, respectively.

Options

A. Statement I is True, Statement II is True, Statement II is a correct explanation for statement I
B. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I
C. Statement I is True, Statement II is False
D. Statement I is False, Statement II is True.

Answer

To solve the problem, we need to analyze the statements provided and verify their validity step by step.

### Step 1: Analyze Statement II
**Statement II** claims that the range of \( \cot^{-1} x \) is \( (0, \pi) \) and the range of \( \cos^{-1} x \) is \( [0, \pi] \).

- The function \( \cot^{-1} x \) is defined for all real numbers \( x \) and its range is indeed \( (0, \pi) \).
- The function \( \cos^{-1} x \) is defined for \( x \in [-1, 1] \) and its range is \( [0, \pi] \).

Since both ranges are correctly stated, **Statement II is true**.

### Step 2: Analyze Statement I
**Statement I** states that the number of roots of the equation \( \cot^{-1} x + \cos^{-1} (2x) + \pi = 0 \) is zero.

We can rearrange this equation as follows:
\[ \cot^{-1} x + \cos^{-1} (2x) = -\pi \]
However, since \( \cot^{-1} x \) is always positive (in the range \( (0, \pi) \)) and \( \cos^{-1} (2x) \) is non-negative (in the range \( [0, \pi] \)), the left-hand side of the equation \( \cot^{-1} x + \cos^{-1} (2x) \) will always be greater than or equal to \( 0 \).

Thus, it is impossible for the sum \( \cot^{-1} x + \cos^{-1} (2x) \) to equal \( -\pi \) (which is negative). Therefore, there are **no roots** for this equation.

Thus, **Statement I is true**.

### Conclusion
Both statements are true:
- **Statement I**: The number of roots of the equation is zero.
- **Statement II**: The ranges of \( \cot^{-1} x \) and \( \cos^{-1} x \) are correctly stated.

### Final Answer
- Statement I is true.
- Statement II is true.