JEE Main 2024 (Online) 1st February Morning ShiftQuestion 1
Let $S=\left\{x\in\mathbf{R}:(\sqrt{3}+\sqrt{2})^{x}+(\sqrt{3}-\sqrt{2})^{x}=10\right\}$. Then the number of elements in $S$ is :
Quadratic Equation · Equations Reducible to Quadratic
Solution
Concept. $(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=1$, so the two powers are reciprocals. Formulas used. (1) $(\sqrt3\pm\sqrt2)^2=5\pm2\sqrt6$Step 1 - Substitute. With $y=(\sqrt3+\sqrt2)^x>0$: $y+\frac1y=10$, i.e. $y^2-10y+1=0$, $y=5\pm2\sqrt6$. Step 2 - Back to $x$. By (1), $5+2\sqrt6=(\sqrt3+\sqrt2)^2$ and $5-2\sqrt6=(\sqrt3+\sqrt2)^{-2}$. So $x=2$ or $x=-2$. Step 3 - Count. $S$ has $2$ elements. Answer: (C)