Basic Of Mathematics Question 1

Question 1 - 01 February - Shift 1

Let

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

Then $n(S)$ is equal to

(1) 2

(2) 4

(3) 6

(4) 0

Show Answer

Answer: (2)

Solution:

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

$t+\frac{1}{t}=10$

$\Rightarrow \quad t=5+2 \sqrt{6}, 5-2 \sqrt{6}$

for $t=5+2 \sqrt{6}$

$\Rightarrow \quad(\sqrt{3}+\sqrt{2})^{x^{2}-4}=5+2 \sqrt{6}= (\sqrt{3}+\sqrt{2})^{2}$

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

$\Rightarrow x= \pm \sqrt{6}$

and

$t+\frac{1}{t}=10$

for $t=5-2 \sqrt{6}$

$\Rightarrow \quad(\sqrt{3}+\sqrt{2})^{x^{2}-4}=5-2 \sqrt{6}= (\sqrt{3}+\sqrt{2})^{-2}$

$\Rightarrow \quad x^{2}-4=-2 \quad$ or $x^{2}=2$

$\Rightarrow x= \pm \sqrt{2}$

Hence $ x= \pm \sqrt{6}, \pm \sqrt{2}$

$S = \lbrace \pm \sqrt{6}, \pm \sqrt{2} \rbrace $

$n(S) = 4$