IPMAT Indore 2022 · Quantitative Ability · Question 31

Everything turns on the signs and relative sizes of the two denominators. Note \sqrt{2}\approx 1.414 and \sqrt{3}\approx 1.732 radians — both in the second quadrant region \left( \pi /2\approx 1.571 \right)? Let us be careful: \sqrt{2}<\pi /2<\sqrt{3}, so \sqrt{2} is in the first quadrant and \sqrt{3} in the second.

Step 1 — the {x}^{2} denominator. Sine increases on \left( 0,\pi /2 \right) and decreases on \left( \pi /2,\pi \right). Use the sum-to-product form instead, which is safer:

\sin \sqrt{2}-\sin \sqrt{3}=2\cos \left( \frac{\sqrt{2}+\sqrt{3}}{2}\right)\sin \left( \frac{\sqrt{2}-\sqrt{3}}{2}\right)

Here \frac{\sqrt{2}+\sqrt{3}}{2}\approx 1.573>\frac{\pi}{2}, so the cosine is negative; and \frac{\sqrt{2}-\sqrt{3}}{2}\approx -0.159, so the sine is negative. Negative × negative gives

\sin \sqrt{2}-\sin \sqrt{3}>0

Numerically: \sin \left( 1.414 \right)\approx 0.9878, \sin \left( 1.732 \right)\approx 0.9870, difference \approx +0.0008 — positive but tiny.

Step 2 — the {y}^{2} denominator. Cosine is strictly decreasing on \left( 0,\pi \right) and \sqrt{2}<\sqrt{3}, so

\cos \sqrt{2}>\cos \sqrt{3}\Rightarrow \cos \sqrt{2}-\cos \sqrt{3}>0

Numerically: \cos \left( 1.414 \right)\approx 0.1559, \cos \left( 1.732 \right)\approx -0.1609, difference \approx +0.3168.

Step 3 — classify. Both denominators are positive, so the curve is an ellipse.

Step 4 — locate the foci. For \frac{{x}^{2}}{A}+\frac{{y}^{2}}{B}=1 with A,B>0, the foci lie on the axis belonging to the larger denominator. Since

B\approx 0.3168 \gg A\approx 0.0008

the major axis is vertical and the foci lie on the y-axis.

Answer: Option 1 — an ellipse with the foci on the y-axis

Shortcut. Two sign checks decide everything, and neither needs a calculator.

Cosine is decreasing on \left( 0,\pi \right), and \sqrt{2}<\sqrt{3}, so \cos \sqrt{2}-\cos \sqrt{3}>0 immediately.

Both \sqrt{2} and \sqrt{3} sit very close to \pi /2\approx 1.571, where sine is at its maximum and nearly flat. So \sin \sqrt{2} and \sin \sqrt{3} are both \approx 0.987 and their difference is positive but minuscule.

Both denominators positive \Rightarrow ellipse, which kills (c) and (d). And the y denominator is hundreds of times larger \Rightarrow foci on the y-axis.

The design of the question is the flatness of sine near its peak — that is what makes one denominator tiny and settles the orientation without any precise computation.