BITSAT 2019 Slot 2 Solved Paper

We have,
(1+ax+bx2)(1−2x)18
(1+ax+bx2)(1−‌18C12x+‌18C2(2x2)− ‌18C3(2x)3+‌18C4(2x)4.........)
Coefficient of x3 is −‌18C3(2)3+a.‌18C2(2)2−b18C1(2)
and coefficient of x4 is ‌18C4(2)4−‌18C3(2)3a+‌18C2(2)2b
Coefficient of x3 and x4 are zero.
∴−‌18C3(2)3+‌18C2(2)2(a)−‌18C1(2)b=0
⇒

4×17×16

3×2

−17a+b=0 .........(i)
and 80−

32

3

a+b=0 ............(ii)
Solving Eqs. (i) and (ii), we get
a=16,b=

272

3