Let $\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is :

<p>Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because :</p> <ul> <li>$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$</li> </ul> <br/><strong>Using the Reciprocal Property :</strong> <br/><ul> <br/><li>This means $(\sqrt{3} - \sqrt{2})^x = \frac{1}{(\sqrt{3} + \sqrt{2})^x} $</li> </ul> <br/><strong>Let</strong> $a = (\sqrt{3} + \sqrt{2})^x $. The equation given in the problem becomes : <ul> <br/><li> $a + \frac{1}{a} = 10$</li> </ul> <br/><strong>Simplifying</strong> <p>Multiplying both sides by *a*, we get a quadratic equation :</p> <ul> <li> $a^2 + 1 = 10a$</li><br> <li> $a^2 - 10a + 1 = 0$</li> </ul> <br/><strong>Solving the Quadratic</strong> <p>Using the quadratic formula, we find :</p> <ul> <li> $a = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$</li> </ul> <br/><strong>Possible values of x</strong> <p>Since $a = (\sqrt{3} + \sqrt{2})^x $, we have two cases :</p> <ol> <li>$(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$ = $(\sqrt{3} + \sqrt{2})^2$. There is one real solution for x in this case which is x = 2.</li><br> <li>$(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$ = $(\sqrt{3} - \sqrt{2})^2= \frac{1}{(\sqrt{3} + \sqrt{2})^2} $ = $(\sqrt{3} + \sqrt{2})^{-2}$. There is one real solution for x in this case which is x = - 2.</li> </ol> <br/><strong>Conclusion</strong> <p>There are <strong>two</strong> real solutions for *x*. Therefore, the number of elements in the set S is <strong>2</strong>. This corresponds with option <strong>(C)</strong>.</p> <p></p>