JEE Advanced 2021 Paper 2 Q02 Mathematics Trigonometry Trigonometric Identities Hard

JEE Advanced 2021 Paper 2 · Q02 · Trigonometric Identities

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$, respectively. Then which of the following statements is (are) TRUE?

  1. A. $\cos P \ge 1 - \dfrac{p^2}{2qr}$
  2. B. $\cos R \ge \left(\dfrac{q-r}{p+q}\right)\cos P + \left(\dfrac{p-r}{p+q}\right)\cos Q$
  3. C. $\dfrac{q+r}{p} < 2\dfrac{\sqrt{\sin Q \sin R}}{\sin P}$
  4. D. If $p < q$ and $p < r$, then $\cos Q > \dfrac{p}{r}$ and $\cos R > \dfrac{p}{q}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B

Solution

(A) By cosine rule $\cos P = \dfrac{q^2+r^2-p^2}{2qr}$. Since $q^2+r^2 \ge 2qr$ by AM-GM, $\cos P \ge \dfrac{2qr - p^2}{2qr} = 1 - \dfrac{p^2}{2qr}$. TRUE. (B) By triangle inequality $p+q > r$. Using projection formula $r = p\cos Q + q\cos P$ etc, rearranging gives the claimed inequality (strict, as triangle inequality is strict). TRUE. (C) By AM-GM, $q+r \ge 2\sqrt{qr}$, so $\dfrac{q+r}{p} \ge \dfrac{2\sqrt{qr}}{p} = 2\dfrac{\sqrt{\sin Q\sin R}}{\sin P}$ by sine rule. Hence the $<$ is FALSE. (D) If true, $r\cos Q > p$ and $q\cos R > p$. Adding and using projection $p = r\cos Q + q\cos R$ gives $p > 2p$, i.e. $p < 0$, contradiction. FALSE.

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