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 $
$n(S) = 4$
