QJEEM24Ma1322: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1322 Solution: Let . Then the number of elements in is :

Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because :

• $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$

Using the Reciprocal Property :

• This means $(\sqrt{3} - \sqrt{2})^x = \frac{1}{(\sqrt{3} + \sqrt{2})^x}$

Let $a = (\sqrt{3} + \sqrt{2})^x$ . The equation given in the problem becomes :

• $a + \frac{1}{a} = 10$

Simplifying

Multiplying both sides by *a*, we get a quadratic equation :

• $a^2 + 1 = 10a$


• $a^2 - 10a + 1 = 0$

Solving the Quadratic

Using the quadratic formula, we find :

• $a = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$

Possible values of x

Since $a = (\sqrt{3} + \sqrt{2})^x$ , we have two cases :

1. $(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$ = $(\sqrt{3} + \sqrt{2})^2$ . There is one real solution for x in this case which is x = 2.


2. $(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$ = $(\sqrt{3} - \sqrt{2})^2= \frac{1}{(\sqrt{3} + \sqrt{2})^2}$ = $(\sqrt{3} + \sqrt{2})^{-2}$ . There is one real solution for x in this case which is x = - 2.

Conclusion

There are two real solutions for *x*. Therefore, the number of elements in the set S is 2. This corresponds with option (C).