IPMAT Indore 2023 · Quantitative Ability · Question 20

Step 1 — parametrise by one variable. Put \cos \alpha =t. The constraint gives \cos \beta =1-t. For both cosines to be legitimate we need

-1\le t\le 1 \text{and}-1\le 1-t\le 1 \Rightarrow 0\le t\le 1

Step 2 — maximise the sines. We have \sin \alpha =\pm \sqrt{1-{t}^{2}} and \sin \beta =\pm \sqrt{1-{\left( 1-t \right)}^{2}}. To make \sin \alpha -\sin \beta as large as possible, take \sin \alpha positive and \sin \beta negative:

f\left( t \right)=\sqrt{1-{t}^{2}}+\sqrt{1-{\left( 1-t \right)}^{2}}=\sqrt{1-{t}^{2}}+\sqrt{2t-{t}^{2}}

Step 3 — maximise by symmetry. Swapping t\leftrightarrow 1-t exchanges the two square roots, so f is symmetric about t=1/2 — which is where the maximum sits.

f\left( 1/2 \right)=\sqrt{1-1/4}+\sqrt{1-1/4}=2\sqrt{3/4}=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}

Step 4 — interpret. At t=1/2 both cosines equal 1/2, i.e. \alpha ={60}^{∘} and \beta =-{60}^{∘}, giving

\sin{60}^{∘}-\sin \left( -{60}^{∘}\right)=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}✓

Answer: Option 4 — \sqrt{3}

Shortcut. Guess the symmetric point and verify. The constraint \cos \alpha +\cos \beta =1 is symmetric in the two angles, so the extremum of a difference of sines occurs at the symmetric split \cos \alpha =\cos \beta =1/2, i.e. \alpha ={60}^{∘}, \beta =-{60}^{∘}. That gives \sqrt{3} in one line.

The trap is the sign of \sin \beta. If you take both angles in \left[ 0,\pi \right] (as a calculator’s \arccos would), you get a maximum of only 1 — which is option (c), sitting there waiting. Nothing says \beta is in the first or second quadrant, and letting \sin \beta go negative is exactly what the word maximum demands.

Option (a) 2 is impossible for a different reason: \sin \alpha -\sin \beta =2 would force \sin \alpha =1 and \sin \beta =-1, making both cosines 0 and their sum 0, not 1.