Statement-1: `tan22 (1/2)^(@)` is a root of the equation `(1+x^(2))/(1-x^(2)) = sqrt(2)`.
Statement-2: `sin alpha = x+ k/x` is possible for real value x if `k le 1/4`.
Statement -3: `|sin nx| le n|sin x|` is valid for all natural numbers n.

Options

A. TTT
B. TFT
C. TFF
D. FFF

Answer

To solve the problem, we need to analyze each of the three statements provided and determine their validity step by step.

### Statement 1:
**tan(22.5°) is a root of the equation (1 + x²)/(1 - x²) = √2.**

1. **Understanding the Equation**: The equation can be rewritten using the identity for tangent. We know that:
\[ \tan(45°) = 1 \]
Therefore, we can express the left side of the equation in terms of tangent:
\[ \frac{1 + \tan^2(22.5°)}{1 - \tan^2(22.5°)} = \frac{1 + x^2}{1 - x^2} \]

2. **Substituting x**: Let \( x = \tan(22.5°) \). We can use the double angle formula for cosine:
\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Here, \( \theta = 22.5° \), so \( 2\theta = 45° \).

3. **Evaluating**: We know that:
\[ \cos(45°) = \frac{1}{\sqrt{2}} \]
Thus, substituting \( x = \tan(22.5°) \):
\[ \cos(45°) = \frac{1 - \tan^2(22.5°)}{1 + \tan^2(22.5°)} = \frac{1 - x^2}{1 + x^2} \]
Rearranging gives:
\[ \sqrt{2} = \frac{1 + x^2}{1 - x^2} \]
This confirms that \( \tan(22.5°) \) is indeed a root of the equation.

### Conclusion for Statement 1:
**True**

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### Statement 2:
**sin(α) = x + k/x is possible for real value x if k ≤ 1/4.**

1. **Rearranging the Equation**: We can rewrite the equation:
\[ \sin(α) = x + \frac{k}{x} \]
Multiplying both sides by \( x \) (assuming \( x \neq 0 \)):
\[ x^2 - \sin(α)x + k = 0 \]

2. **Applying the Quadratic Formula**: For \( x \) to be real, the discriminant must be non-negative:
\[ D = b^2 - 4ac = \sin^2(α) - 4k \geq 0 \]

3. **Finding the Condition**: This leads to:
\[ \sin^2(α) \geq 4k \]
The maximum value of \( \sin^2(α) \) is 1, hence:
\[ 1 \geq 4k \implies k \leq \frac{1}{4} \]

### Conclusion for Statement 2:
**True**

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### Statement 3:
**|sin(nx)| ≤ n|sin(x)| is valid for all natural numbers n.**

1. **Understanding the Sine Function**: The sine function has a maximum value of 1. Therefore:
\[ |sin(nx)| \leq 1 \]
and
\[ |sin(x)| \leq 1 \]

2. **Applying the Inequality**: The statement claims that:
\[ |sin(nx)| \leq n|sin(x)| \]
Since \( |sin(nx)| \) can take values up to 1, we can see that:
\[ |sin(nx)| \leq n \cdot 1 = n \]
This holds true since \( n \) is a natural number.

3. **Using the Property of Sine**: The inequality can be derived from the properties of sine and its periodicity, confirming that:
\[ |sin(nx)| \leq n|sin(x)| \]

### Conclusion for Statement 3:
**True**

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### Final Conclusion:
All three statements are true.