. . . (1) Now, according to the question, (−1)r‌
‌9Cr
29−r
4r=−84 =(−1)r‌9Cr23r−9=21×4 Only natural value of r possible if 3r−9=0r=3 and ‌9C3=84 ∴1=5 from equation (1) Now, coefficient of x−31=x‌
45
2
−‌
5r
2
−lr at 1=5, gives ‌r=5 ‌∴‌9c5(−1)‌
45
24
=2α×β ‌=−63×27 ‌⇒α=7,β=−63 ‌∴‌‌‌ value of ‌|αℓ−β|=98