If `x = ( 4 lambda)/( 1+ lambda^(2)) ` and `y = ( 2 - 2 lambda^(2))/( 1+ lambda^(2))`, where `lambda` is a real parameter then `Z = x^(2) - xy + y^(2)` lies between

Options

A. [2,6]
B. [2,4]
C. [4,6]
D. none

Answer

To solve the problem, we need to find the range of the expression \( Z = x^2 - xy + y^2 \) given the definitions of \( x \) and \( y \) in terms of the parameter \( \lambda \).

### Step 1: Substitute \( \lambda \) with \( \tan(\alpha) \)

Given:
\[ x = \frac{4\lambda}{1 + \lambda^2}, \quad y = \frac{2 - 2\lambda^2}{1 + \lambda^2} \]
Let \( \lambda = \tan(\alpha) \). Then:
\[ x = \frac{4\tan(\alpha)}{1 + \tan^2(\alpha)} = 4\sin(\alpha)\cos(\alpha) = 2\sin(2\alpha) \]
\[ y = \frac{2 - 2\tan^2(\alpha)}{1 + \tan^2(\alpha)} = 2\frac{1 - \tan^2(\alpha)}{1 + \tan^2(\alpha)} = 2\cos(2\alpha) \]

### Step 2: Substitute \( x \) and \( y \) into \( Z \)

Now we substitute \( x \) and \( y \) into the expression for \( Z \):
\[ Z = (2\sin(2\alpha))^2 - (2\sin(2\alpha))(2\cos(2\alpha)) + (2\cos(2\alpha))^2 \]
\[ Z = 4\sin^2(2\alpha) - 4\sin(2\alpha)\cos(2\alpha) + 4\cos^2(2\alpha) \]

### Step 3: Factor out the common term

Factoring out the 4 from the expression:
\[ Z = 4(\sin^2(2\alpha) - \sin(2\alpha)\cos(2\alpha) + \cos^2(2\alpha)) \]

### Step 4: Simplify the expression inside the parentheses

Using the identity \( \sin^2(2\alpha) + \cos^2(2\alpha) = 1 \):
\[ Z = 4(1 - \sin(2\alpha)\cos(2\alpha)) \]

### Step 5: Use the double angle identity

Recall that \( \sin(2\alpha)\cos(2\alpha) = \frac{1}{2}\sin(4\alpha) \):
\[ Z = 4\left(1 - \frac{1}{2}\sin(4\alpha)\right) = 4 - 2\sin(4\alpha) \]

### Step 6: Determine the range of \( Z \)

The function \( \sin(4\alpha) \) varies between -1 and 1. Therefore:
\[ -2 \leq -2\sin(4\alpha) \leq 2 \]
Adding 4 to all parts of the inequality:
\[ 4 - 2 \leq Z \leq 4 + 2 \]
\[ 2 \leq Z \leq 6 \]

### Conclusion

Thus, the value of \( Z \) lies in the interval \([2, 6]\).