JEE Main 2024 — 1 February, Shift 1 — Question 9

Let the given equation be

$$(\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10$$

We observe that $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$. This implies that $\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}}$.

Let $y = (\sqrt{3} + \sqrt{2})^x$. Then the equation can be rewritten as:

$$y + \frac{1}{y} = 10$$

Multiply by $y$ to clear the denominator:

$$y^2 + 1 = 10y$$ $$y^2 - 10y + 1 = 0$$

This is a quadratic equation in $y$. We can solve for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)}$$ $$y = \frac{10 \pm \sqrt{100 - 4}}{2}$$ $$y = \frac{10 \pm \sqrt{96}}{2}$$

We can simplify $\sqrt{96}$: $\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$.

$$y = \frac{10 \pm 4\sqrt{6}}{2}$$ $$y = 5 \pm 2\sqrt{6}$$

So we have two possible values for $y$: $y_1 = 5 + 2\sqrt{6}$ $y_2 = 5 - 2\sqrt{6}$

Now, we need to solve for $x$ using $y = (\sqrt{3} + \sqrt{2})^x$.

Case 1: $y_1 = 5 + 2\sqrt{6}$ We notice that $( \sqrt{3} + \sqrt{2} )^2 = (\sqrt{3})^2 + (\sqrt{2})^2 + 2\sqrt{3}\sqrt{2} = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$. So, $y_1 = (\sqrt{3} + \sqrt{2})^2$. Setting this equal to $(\sqrt{3} + \sqrt{2})^x$: $(\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^2$ Therefore, $x = 2$.

Case 2: $y_2 = 5 - 2\sqrt{6}$ We know that $5 - 2\sqrt{6} = \frac{1}{5 + 2\sqrt{6}} = \frac{1}{(\sqrt{3} + \sqrt{2})^2} = (\sqrt{3} + \sqrt{2})^{-2}$. Setting this equal to $(\sqrt{3} + \sqrt{2})^x$: $(\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^{-2}$ Therefore, $x = -2$.

Both $x=2$ and $x=-2$ are real numbers. Thus, the set $S$ contains two elements: $S = \{-2, 2\}$. The number of elements in $S$ is 2.

The final answer is $\boxed{2}$.