Let S = \ x: x ∈ ℝand ( 3 + 2 ) ^ x ^ 2 - 4 + ( 3 - 2 ) ^ x ^ 2 - 4...

Given:

$S = \left\{x: {\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} + {\left(\sqrt{3} - \sqrt{2}\right)}^{{x}^{2} - 4} = 10\right\}$

Now,

$\sqrt{3} - \sqrt{2} = \dfrac{\left(\sqrt{3} + \sqrt{2}\right)\left(\sqrt{3} - \sqrt{2}\right)}{\sqrt{3} + \sqrt{2}} = \dfrac{1}{\sqrt{3} + \sqrt{2}}$

So,

${\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} + {\left(\sqrt{3} - \sqrt{2}\right)}^{{x}^{2} - 4} = 10$

$⇒ {\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} + {\left(\dfrac{1}{\sqrt{3} + \sqrt{2}}\right)}^{{x}^{2} - 4} = 10$

Put ${\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} = u$, then

$u + \dfrac{1}{u} = 10$

$⇒ {u}^{2} - 10u + 1 = 0$

$⇒ u = \dfrac{10 \pm \sqrt{100 - 4}}{2}$

$⇒ u = 5 \pm \sqrt{24}$

$⇒ u = 5 \pm 2\sqrt{3}\sqrt{2}$

$⇒ u = {\left(\sqrt{3} \pm \sqrt{2}\right)}^{2}$

$⇒ {\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} = {\left(\sqrt{3} \pm \sqrt{2}\right)}^{2}$

So, ${\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} = {\left(\sqrt{3} + \sqrt{2}\right)}^{2} and${\left(\sqrt{3} + \sqrt{2}\right)}^{{x}^{2} - 4} = {\left(\dfrac{1}{\sqrt{3} + \sqrt{2}}\right)}^{2}$</p><p>Therefore,</p><p>${x}^{2} - 4 = 2 \text{ and } {x}^{2} - 4 = - 2$</p><p>$⇒ x = \pm \sqrt{6} \text{ and } x = \pm \sqrt{2}$</p><p>So,$S = \left\{\sqrt{6}, \sqrt{2}, - \sqrt{6}, - \sqrt{2}\right\}$</p><p>Hence,$n\left(S\right) = 4$