XAT 2011 — QA & DI Question 13

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As S is not the circumcentre, PS ≠ ST and QS ≠ SR

∵ PT and QR are chords of the circle intersecting at S, PS × ST = QS × SR … (i)

We know that Arithmetic mean ≥ Geometric mean

$\therefore \frac{PS+ST}{2}$ ≥ $\sqrt{PS\times ST}$

But as PS ≠ ST,

$\frac{PS+ST}{2}$ > $\sqrt{PS\times ST}$

$\therefore \frac{PS+ST}{2}$ > $\sqrt{QS\times SR}$

$\therefore \frac{PS+ST}{2}$ > $2\sqrt{QS\times SR}$

$\therefore \frac{PS+ST}{PS\times ST}$ > $\frac{2\sqrt{QS+SR}}{QS\times SR}$

$\therefore \frac{1}{PS}$ + $\frac{1}{ST}$ > $\frac{2}{\sqrt{QS\times SR}}$      ...(i)

∴ Option 1 is false.

Also,

$\frac{QS+SR}{2}$ > $\sqrt{QS\times SR}$

$\therefore \frac{2}{QR}$ < $\frac{1}{\sqrt{QS\times SR}}$

$\therefore \frac{4}{QR}$ < $\frac{2}{\sqrt{QS\times SR}}$

$\therefore \frac{1}{PS}$ + $\frac{1}{ST}$ > $\frac{2}{\sqrt{QS\times SR}}$ > $\frac{4}{QR}$    ...From (i)

Hence, option (d).

Note: As the result is a general one, we can, without loss of generality, consider an equilateral triangle PQR with point S being the mid-point of QR and verify all options using numbers.