JEE 2023 — Mathematics PYQ

To find $n(S)$, we need to determine the number of real values of $x$ that satisfy the equation.

Step 1: Observe the relationship between the bases

Notice that the two bases are conjugates of each other. If we multiply them:

This means that:

Step 2: Substitution

Let $a=\sqrt{3}+\sqrt{2}$ and $y=x^2-4$.

The equation becomes:

Let $t=a^y$. Then the equation is:

Step 3: Solve for $t$

Using the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

Now, notice that $5\pm 2\sqrt{6}$ can be written as squares:

So, we have two cases for $t=a^y$:

1. $a^y=(\sqrt{3}+\sqrt{2}{)}^2⟹(\sqrt{3}+\sqrt{2}{)}^y=(\sqrt{3}+\sqrt{2}{)}^2⟹y=2$

2. $a^y=(\sqrt{3}-\sqrt{2}{)}^2⟹(\sqrt{3}+\sqrt{2}{)}^y=(\sqrt{3}+\sqrt{2}{)}^{-2}⟹y=-2$

Step 4: Solve for $x$

Recall that $y=x^2-4$.

Case 1: $y=2$

(2 real solutions)

Case 2: $y=-2$

(2 real solutions)

Conclusion

The set $S=\{\sqrt{6},-\sqrt{6},\sqrt{2},-\sqrt{2}\}$.

The number of elements in $S$, denoted by $n(S)$, is 4.

Correct Option: (1)