QJEEM23Maab5b: JEE Main 2023 Mathematics - Class 11 JEEM23Maab5b Solution: Let S = { x : x ∈ R and ( 3 + 2 ) x 2 − 4 + ( 3 − 2 ) x 2 − 4 = …

Question

Verified Solution • 2025–26 Session moderate

Let $S = {x : x \in R and {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10}$ . Then $n (S)$ is equal to

Accepted Answer

Let

(\sqrt{3}+\sqrt{2})^{x^{2}-4}=t

\begin{aligned} t + \frac{1}{t} = 10 \\ \Rightarrow & t^{2} - 10 t + 1 = 0 \\ \Rightarrow & t = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm 2 \sqrt{6} \end{aligned}

Case-I

\begin{aligned} t = 5 + 2 \sqrt{6} = (\sqrt{3} + \sqrt{2})^{2} \\ \Rightarrow & (\sqrt{3} + \sqrt{2})^{x^{2} - 4} = (\sqrt{3} + \sqrt{2})^{2} \\ \Rightarrow & x^{2} - 4 = 2 \Rightarrow x^{2} = 6 \Rightarrow x = \pm \sqrt{6} \end{aligned}

Case-II :

t=5-2 \sqrt{6}

=

(\sqrt{3}-\sqrt{2})^{2}

(\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}

\Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}

\Rightarrow 4-x^{2}=2

\Rightarrow x^{2}=2

\Rightarrow x=\pm \sqrt{2}

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