JEE Main 2023 (Online) 1st February Morning ShiftQuestion 1
Let $S=\left\{x:x\in\mathbb R \text{ and } \left(\sqrt3+\sqrt2\right)^{x^2-4}+\left(\sqrt3-\sqrt2\right)^{x^2-4}=10\right\}$. Then $n(S)$ is equal to
Quadratic Equation · Equations Reducible to Quadratic
Solution
Concept. $(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=1$, so the two powers are reciprocals. Formulas used. (1) $(\sqrt3+\sqrt2)^{\pm2}=5\pm2\sqrt6$Step 1 - Substitute. With $t=(\sqrt3+\sqrt2)^{x^2-4}$: $t+\frac1t=10$, so $t=5\pm2\sqrt6$. Step 2 - Exponent. By (1), $x^2-4=\pm2$, so $x^2=6$ or $x^2=2$. Step 3 - Count. $x=\pm\sqrt6,\ \pm\sqrt2$: four solutions. Answer: (B)