Xavier Aptitude Test 2011 Paper

(d)We know by geemetry PS $\times \mathit{S}\mathit{T}=\mathrm{}\mathit{Q}\mathit{S}\times \mathit{S}\mathit{R}\mathrm{}......$ (1)
$\because \mathrm{}\mathit{S}$ is not the centre of circulm circle,
$\mathrm{}\mathit{P}\mathit{S}\ne \mathit{S}\mathit{T}$
And we know that for two unequal real numbers.
H.M.

$\Rightarrow 1\mathrm{}\mathit{P}\mathit{S}+\frac{1}{\mathrm{}\mathit{S}\mathit{T}}>\mathrm{}\frac{2}{\sqrt{\mathrm{}\mathit{P}\mathit{S}\times \mathit{S}\mathit{T}}}$
$\Rightarrow \mathrm{}\frac{1}{\mathrm{}\mathit{P}\mathit{S}}+\mathrm{}\frac{1}{\mathit{S}\mathit{T}}>\mathrm{}2\sqrt{\mathrm{}\mathit{Q}\mathit{S}\times \mathit{S}\mathit{R}}\mathrm{}($ using eqn $\left(1\right)......\left(2\right)$
Also
$\Rightarrow \frac{1}{\sqrt{\mathit{Q}\mathit{S}\times \mathit{S}\mathit{R}}}>\mathrm{}\frac{2}{\mathit{Q}\mathit{R}}$
$\Rightarrow \frac{2}{\sqrt{\mathit{Q}\mathit{S}\times \mathit{S}\mathit{R}}}>\frac{4}{\mathit{Q}\mathit{R}}\mathrm{}........\left(3\right)$
From equations $\left(2\right)$ and $\left(3\right),$ we get $\mathrm{}\frac{1}{\mathit{P}\mathit{S}}+\mathrm{}\frac{1}{\mathit{S}\mathit{T}}>\frac{4}{\mathit{Q}\mathit{R}}$
$\therefore \mathrm{}\left(\mathit{d}\right)$ is the correct option.