JEE Advanced 2021 · Paper 2, Mathematics question 2

(A) By the cosine rule and $q^2+r^2\ge2qr$:
$$\cos P=\frac{q^2+r^2-p^2}{2qr}\ge\frac{2qr-p^2}{2qr}=1-\frac{p^2}{2qr}.$$
True.

(B) The projection formulas give $p=q\cos R+r\cos Q$, $q=p\cos R+r\cos P$ and $r=p\cos Q+q\cos P$.

• Adding the first two: $(p+q)\cos R=p+q-r(\cos P+\cos Q)$.
• Using the third: $(q-r)\cos P+(p-r)\cos Q=r-r(\cos P+\cos Q)$.

The difference between the two sides is $p+q-r>0$, so the inequality holds. True.

(C) By the sine rule, the claim becomes $q+r<2\sqrt{qr}$, which contradicts AM-GM. False.

(D) Counterexample: $p=1$, $q=r=10$. Then $\cos Q=\frac1{20}<\frac pr=\frac1{10}$. False.

(A), (B)