JEE Main 2023 — Mathematics Algebra

Given:

$$S={x:{(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}+{(\sqrt{3}-\sqrt{2})}^{{x}^{2}-4}=10}$$S=x:(3​+2​)x2−4+(3​−2​)x2−4=10

Now,

$$\sqrt{3}-\sqrt{2}=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}}$$3​−2​=3​+2​(3​+2​)(3​−2​)​=3​+2​1​

So,

$${(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}+{(\sqrt{3}-\sqrt{2})}^{{x}^{2}-4}=10$$(3​+2​)x2−4+(3​−2​)x2−4=10

$$\Rightarrow {(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}+{(\frac{1}{\sqrt{3}+\sqrt{2}})}^{{x}^{2}-4}=10$$⇒(3​+2​)x2−4+(3​+2​1​)x2−4=10

Put ${(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}=u$(3​+2​)x2−4=u, then

$$u+\frac{1}{u}=10$$u+u1​=10

$$\Rightarrow {u}^{2}-10u+1=0$$⇒u2−10u+1=0

$$\Rightarrow u=\frac{10\pm \sqrt{100-4}}{2}$$⇒u=210±100−4​​

$$\Rightarrow u=5\pm \sqrt{24}$$⇒u=5±24​

$$\Rightarrow u=5\pm 2\sqrt{3}\sqrt{2}$$⇒u=5±23​2​

$$\Rightarrow u={(\sqrt{3}\pm \sqrt{2})}^{2}$$⇒u=(3​±2​)2

$$\Rightarrow {(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}={(\sqrt{3}\pm \sqrt{2})}^{2}$$⇒(3​+2​)x2−4=(3​±2​)2

So, ${(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}={(\sqrt{3}+\sqrt{2})}^{2}$(3​+2​)x2−4=(3​+2​)2 and ${(\sqrt{3}+\sqrt{2})}^{{x}^{2}-4}={(\frac{1}{\sqrt{3}+\sqrt{2}})}^{2}$(3​+2​)x2−4=(3​+2​1​)2

Therefore,

{x}^{2}-4=2&{x}^{2}-4=-2

\Rightarrow x=\pm \sqrt{6}&x=\pm \sqrt{2}

So, $S={\sqrt{6},\sqrt{2},-\sqrt{6},-\sqrt{2}}$S=6​,2​,−6​,−2​

Hence, $n(S)=4$n(S)=4