Let P(x)=53−6x−9x2 and Q(y)=−4y2+4y+132. If there exists unique pair of real numbers (x, y) such that P(x)Q(y)=20, thenthe value of 1xyis_____.

We have P(x)=53−6x−9x2=−(3x+1)2+83⇒Pmax=83Similarly, Q(y)=−4y2+4y+132=−(2y−1)2+152⇒ Qmax=152Now, Pmax×Qmax=83×152=20So, (x,y)≡−13,12Hence,  1xy=−6