JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Properties of Triangle

(A) By the cosine rule: $\cos \mathrm{P}=\frac{{\mathrm{q}}^2+{\mathrm{r}}^2-{\mathrm{p}}^2}{2\mathrm{qr}}=\frac{{\mathrm{q}}^2+{\mathrm{r}}^2}{2\mathrm{qr}}-\frac{{\mathrm{p}}^2}{2\mathrm{qr}}$. By AM-GM inequality, ${\mathrm{q}}^2+{\mathrm{r}}^2\ge 2\mathrm{qr}$, so $\frac{{\mathrm{q}}^2+{\mathrm{r}}^2}{2\mathrm{qr}}\ge 1$. Thus $\cos \mathrm{P}\ge 1-\frac{{\mathrm{p}}^2}{2\mathrm{qr}}$. (A) is correct.

(B) The inequality $(\mathrm{p}+\mathrm{q})\cos \mathrm{R}\ge (\mathrm{q}-\mathrm{r})\cos \mathrm{P}+(\mathrm{p}-\mathrm{r})\cos \mathrm{Q}$ can be rearranged using projection formulas to give $\mathrm{p}+\mathrm{q}\ge \mathrm{r}$, which is the triangle inequality. (B) is correct.

(C) By sine rule and AM-GM: $\frac{\mathrm{q}+\mathrm{r}}{\mathrm{p}}=\frac{\sin \mathrm{Q}+\sin \mathrm{R}}{\sin \mathrm{P}}\ge \frac{2\sqrt{\sin \mathrm{Q}\sin \mathrm{R}}}{\sin \mathrm{P}}$. The inequality is reversed. (C) is incorrect.

(D) The condition does not necessarily hold for all triangles with p < q and p < r. (D) is incorrect.

Correct answers: (A), (B).