Question

Let \(S=x: x∈R\) and $\left(\sqrt{3}+\sqrt{2})^{x^2-4}+(\sqrt{3}-\sqrt{2})^{x^2-4}=10\right\}$Then $n(S)$ is equal to

• A. 2
• B. 4
• C. 0
• D. 6

Answer 1

The Correct Option is B

To solve this problem, we need to determine \( n(S) \), where the set \( S = \{ x : x \in \mathbb{R} \} \) makes the given equation true:

\[ (\sqrt{3} + \sqrt{2})^{x^2 - 4} + (\sqrt{3} - \sqrt{2})^{x^2 - 4} = 10 \]

Let us denote \( a = (\sqrt{3} + \sqrt{2})^{x^2 - 4} \) and \( b = (\sqrt{3} - \sqrt{2})^{x^2 - 4} \). Therefore, we have:

\[ a + b = 10 \]

Also, note that:

\[ ab = (\sqrt{3}+\sqrt{2})^{x^2-4} \cdot (\sqrt{3}-\sqrt{2})^{x^2-4} = ((\sqrt{3})^2 - (\sqrt{2})^2)^{x^2-4} = (1)^{x^2-4} = 1 \]

We now have a system of equations:

• \( a + b = 10 \)
• \( ab = 1 \)

These are similar to the sum and product of roots. We can use these to find potential solutions for \( a \) and \( b \), which are the roots of the quadratic equation:

\[ t^2 - 10t + 1 = 0 \]

The solution to this quadratic equation is given by the quadratic formula:

\[ t = \frac{10 \pm \sqrt{10^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} \]

Hence, the roots are:

\[ t = 5 + 2\sqrt{6} \quad \text{and} \quad t = 5 - 2\sqrt{6} \]

These represent the possible values for \( a \) and \( b \) since they satisfy both conditions.

This implies that:

\[ (\sqrt{3} + \sqrt{2})^{x^2 - 4} = 5 + 2\sqrt{6} \quad \text{or} \quad (\sqrt{3} - \sqrt{2})^{x^2 - 4} = 5 + 2\sqrt{6} \]

Since both equations are of the form above, both must work, producing the same possible values \\( x^2 - 4 \\). Solving for \( x^2 - 4 \), we substitute into the possible conditions, remembering that since the terms involve power functions, we consider real \( x \) values.

Conclusion: Each base has two valid solutions for the quadratic formula corresponding \\( x^2 - 4 = 2 \\) and \\( x^2 - 4 = -2 \\) but letting \( x \) range confirms that indeed the appropriate \( x^2 \) makes all solutions valid:

\[ x^2 = 6 \quad \Rightarrow x = \pm \sqrt{6} \]

\[ x^2 = 2 \quad \Rightarrow x = \pm \sqrt{2} \]

Thus, we have four possible values of \( x \). Therefore, \( n(S) = 4 \).