JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 6)
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
6
4
0
2
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 :
$t=5-2 \sqrt{6}$ = $(\sqrt{3}-\sqrt{2})^{2}$
$(\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$
$\Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$
$\Rightarrow 4-x^{2}=2$
$\Rightarrow x^{2}=2$
$\Rightarrow x=\pm \sqrt{2}$
$$ \begin{aligned} & t+\frac{1}{t}=10 \\\\ \Rightarrow & t^{2}-10 t+1=0 \\\\ \Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6} \end{aligned} $$
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 :
$t=5-2 \sqrt{6}$ = $(\sqrt{3}-\sqrt{2})^{2}$
$(\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$
$\Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$
$\Rightarrow 4-x^{2}=2$
$\Rightarrow x^{2}=2$
$\Rightarrow x=\pm \sqrt{2}$
Comments (0)
