See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(Q.48 – 49): Equation of line PM is y – 1 = tan  (x – 1) Coordinates of Q = 3 tan 1 3tan , 1 tan 1 tan               Area of AMQ = 1 tan 1 tan     Area of quadrilateral BMQC 1 1 tan 3 5 tan A 4 2 2 1 tan 1 tan              x 3 sin x 1 3 tan lim 2cosx 1 3             