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

JEE · Math · previous-year question

  1. A.6
  2. B.4correct
  3. C.0
  4. D.2

Answer

B. 4

Explanation

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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