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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation
$(-2+\sqrt{3})\left(|\sqrt{x}-3|\right) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0,\quad x \ge 0,$
then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:
Solution
$(x-6\sqrt{x}+9) - (2-\sqrt{3})|\sqrt{x}-3| - 2\sqrt{3} = 0$
$\Rightarrow |\sqrt{x}-3|^2 - (2-\sqrt{3})|\sqrt{x}-3| - 2\sqrt{3} = 0$
$\Rightarrow |\sqrt{x}-3| = 2$ or $|\sqrt{x}-3| = -\sqrt{3}$ (not possible)
$\Rightarrow \sqrt{x} = 1$ or $5$
$\Rightarrow x = 1$ or $25$
$\Rightarrow \alpha = 1$ and $\beta = 25$
Aliter:
Let $x \ge 9$, let $\sqrt{x} = t \Rightarrow t \ge 3$
$(\sqrt{3}-2)(t-3) + (t-3)^2 - 2\sqrt{3} = 0$
let $t-3 = u$
$u^2 + (\sqrt{3}-2)u - 2\sqrt{3} = 0$
$u = 2$, or $u = -\sqrt{3}$
$\Rightarrow t-3 = 2$ or $t-3 = -\sqrt{3}$
$\Rightarrow t=5$ or $t=3-\sqrt{3}$ (rejected)
$\Rightarrow x=25$
Now let $0 < x < 9$,
$-(\sqrt{3}-2)(t-3) + (t-3)^2 - 2\sqrt{3} = 0$
let $t-3 = u$
$u^2 - (\sqrt{3}-2)u - 2\sqrt{3} = 0$
$\Rightarrow u = \sqrt{3}$ or $u=-2$
$\Rightarrow t=3+\sqrt{3}$ (rejected) or $t=1$
$\Rightarrow x=1$
$\alpha=1,; \beta=25$
Now $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 10$
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