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JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 9)

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 :
4
0
2
1

Explanation

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).

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