Let S = { x : x R , ( 3 + 2 ) x 2 - 4 + ( 3 - 2 ) x 2 - 4 = 10 } . Then n ( S ) is equal to [2023]

Question

Let $S = {x : x \in R, {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10} .$ Then $n (S)$ is equal to [2023]

Smart Achievers · Accepted Answer

(2)

$S = {x : x \in R and {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10}$

Let ${(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} = t and {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = \frac{1}{t}$

$∴ t + \frac{1}{t} = 10 \Rightarrow t^{2} - 10 t + 1 = 0 \Rightarrow t = \frac{10 \pm \sqrt{96}}{2}$

$\Rightarrow t = 5 \pm 2 \sqrt{6} = {(\sqrt{3} \pm \sqrt{2})}^{2}$

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

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

$\Rightarrow x^{2} - 4 = 2 and x^{2} - 4 = - 2 \Rightarrow x^{2} = 6 and x^{2} = 2$

$\Rightarrow x = \pm \sqrt{6}$ and $x = \pm \sqrt{2} \Rightarrow n (S) = 4$

Solution

Q.
Let $S = {x : x \in R, {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10} .$ Then $n (S)$ is equal to [2023]

1 2

2 4

3 0

4 6

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Solution

Q. Let S={x:x∈R, (3+2)x2-4+(3-2)x2-4=10}. Then n(S) is equal to [2023]

1 2

2 4

3 0

4 6

Ans. (2) S={x:x∈R and (3+2)x2-4+(3-2)x2-4=10} Let (3+2)x2-4=t and (3-2)x2-4=1t ∴ t+1t=10⇒t2-10t+1=0⇒t=10±962 ⇒t=5±26=(3±2)2 ⇒ (3+2)x2-4=(3+2)2 and (3+2)x2-4=(3-2)2=(3+2)-2 ⇒ x2-4=2 and x2-4=-2⇒ x2=6 and x2=2 ⇒ x=±6 and x=±2⇒ n(S)=4

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Q.
Let $S = {x : x \in R, {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10} .$ Then $n (S)$ is equal to [2023]