Statement -1 If `x,y,z` be real variables satisfying `x+y+z=6` and `xy+yz+z=8`, the range of variables x,y and z are identical.
Statement -2 `x+y+z=6` and `xy+yz+zx=8` remains same if `x,y,z` interchange their positions.

Options

A. Statement -1 is true, Statement -2 is true, Statement -2 is a correct explanation for Statement-1
B. Statement -1 is true, Statement -2 is true, Statement -2 is not a correct explanation for Statement -1
C. Statement -1 is true, Statement -2 is false
D. Statement -1 is false, Statement -2 is true

Explanation

To solve the problem, we need to analyze the given statements and derive the necessary equations. Let's break it down step by step.

### Step 1: Understand the given equations
We have two equations:
1. \( x + y + z = 6 \) (Equation 1)
2. \( xy + yz + zx = 8 \) (Equation 2)

### Step 2: Express one variable in terms of the others
From Equation 1, we can express \( z \) in terms of \( x \) and \( y \):
\[ z = 6 - x - y \]

### Step 3: Substitute \( z \) into Equation 2
Now, substitute \( z \) into Equation 2:
\[ xy + y(6 - x - y) + x(6 - x - y) = 8 \]
Expanding this gives:
\[ xy + 6y - xy - y^2 + 6x - x^2 - xy = 8 \]
Combine like terms:
\[ 6x + 6y - x^2 - y^2 - xy = 8 \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ x^2 + y^2 + xy - 6x - 6y + 8 = 0 \]

### Step 5: Treat this as a quadratic in \( y \)
We can treat this as a quadratic equation in \( y \):
\[ y^2 + (x - 6)y + (x^2 - 6x + 8) = 0 \]
For \( y \) to have real solutions, the discriminant must be non-negative:
\[ D = (x - 6)^2 - 4(x^2 - 6x + 8) \geq 0 \]

### Step 6: Calculate the discriminant
Calculating the discriminant:
\[ D = (x - 6)^2 - 4(x^2 - 6x + 8) \]
Expanding this:
\[ D = x^2 - 12x + 36 - 4x^2 + 24x - 32 \]
Combine like terms:
\[ D = -3x^2 + 12x + 4 \geq 0 \]

### Step 7: Solve the inequality
To find the range of \( x \), we can solve the quadratic inequality:
\[ 3x^2 - 12x - 4 \leq 0 \]
Finding the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{144 + 48}}{6} = \frac{12 \pm \sqrt{192}}{6} = \frac{12 \pm 8\sqrt{3}}{6} = 2 \pm \frac{4\sqrt{3}}{3} \]

### Step 8: Determine the range for \( x, y, z \)
Thus, the range for \( x \) is:
\[ 2 - \frac{4\sqrt{3}}{3} \leq x \leq 2 + \frac{4\sqrt{3}}{3} \]
Since \( y \) and \( z \) are symmetric in the equations, they will have the same range.

### Conclusion
The ranges of \( x, y, \) and \( z \) are identical, confirming that **Statement 1 is true**.

**Statement 2** is also true since the equations remain the same when \( x, y, z \) are interchanged.