Let `S={x:x in IR` and `(sqrt(3)+sqrt(2))^(x^(2)-4)+(sqrt(3)-sqrt(2))^(x^(2)-4)=10}`.Then `n(S)` is equal to

Options

A. `2`
B. `4`
C. `0`
D. `6`

Answer

To solve the equation given in the problem, we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ (\sqrt{3} + \sqrt{2})^{x^2 - 4} + (\sqrt{3} - \sqrt{2})^{x^2 - 4} = 10 \]

### Step 2: Let \( y = x^2 - 4 \)
We can simplify our equation by letting:
\[ y = x^2 - 4 \]
Thus, the equation becomes:
\[ (\sqrt{3} + \sqrt{2})^y + (\sqrt{3} - \sqrt{2})^y = 10 \]

### Step 3: Analyze the Terms
Notice that \( \sqrt{3} + \sqrt{2} > 1 \) and \( \sqrt{3} - \sqrt{2} < 1 \). As \( y \) increases, \( (\sqrt{3} + \sqrt{2})^y \) increases, while \( (\sqrt{3} - \sqrt{2})^y \) decreases.

### Step 4: Find the Values of \( y \)
We can analyze the function:
\[ f(y) = (\sqrt{3} + \sqrt{2})^y + (\sqrt{3} - \sqrt{2})^y \]
We need to find \( y \) such that \( f(y) = 10 \).

### Step 5: Calculate \( f(0) \) and \( f(2) \)
1. For \( y = 0 \):
\[ f(0) = (\sqrt{3} + \sqrt{2})^0 + (\sqrt{3} - \sqrt{2})^0 = 1 + 1 = 2 \]

2. For \( y = 2 \):
\[ f(2) = (\sqrt{3} + \sqrt{2})^2 + (\sqrt{3} - \sqrt{2})^2 \]
Calculate \( (\sqrt{3} + \sqrt{2})^2 \) and \( (\sqrt{3} - \sqrt{2})^2 \):
\[ (\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6} \]
\[ (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6} \]
Therefore,
\[ f(2) = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10 \]

### Step 6: Check for Other Possible Values
Since \( f(y) \) is continuous and strictly increasing, there will be two values of \( y \) that satisfy \( f(y) = 10 \):
1. \( y = 2 \)
2. \( y = -2 \) (since \( f(-2) = 10 \) as well)

### Step 7: Solve for \( x \)
Recall that \( y = x^2 - 4 \):
1. For \( y = 2 \):
\[ x^2 - 4 = 2 \implies x^2 = 6 \implies x = \pm \sqrt{6} \]

2. For \( y = -2 \):
\[ x^2 - 4 = -2 \implies x^2 = 2 \implies x = \pm \sqrt{2} \]

### Step 8: Count the Solutions
The solutions for \( x \) are:
- \( x = \sqrt{6} \)
- \( x = -\sqrt{6} \)
- \( x = \sqrt{2} \)
- \( x = -\sqrt{2} \)

Thus, the total number of solutions \( n(S) = 4 \).

### Final Answer
\[ n(S) = 4 \]