JEE-Mains 2014 (7/30)

Note that $\displaystyle (1+ax+bx^2)(1-2x)^{18} = (1+ax+bx^2) \sum_{k=0}^{18}(-1)^k \dbinom {18}k (2x)^k$ .

Therefore the coefficient of $x^3$ is given by:

$\begin{aligned} (-1)^3 \binom {18}3 (2^3)(1) + (-1)^2 \binom {18}2 (2^2)a + (-1)^1 \binom {18}1 (2^1)b & = 0 \\ -6528 + 612a -36b & = 0 \\ 51a - 3b & = 544 & ...(1) \end{aligned}$

Similarly, the coefficient of $x^4$ is given by:

$\begin{aligned} (-1)^4 \binom {18}4 (2^4)(1) + (-1)^3 \binom {18}3 (2^3)a + (-1)^2 \binom {18}2 (2^2)b & = 0 \\ 48960 -6528a + 612b & = 0 \\ 32a - 3b & = 240 & ...(2) \end{aligned}$

$\begin{aligned} (1)-(2): \quad 19a & = 304 \\ \implies a & = 16 \end{aligned}$

$\begin{aligned} (2): \quad 32(16) - 3b & = 240 \\ \implies b & = \frac {512-240}3 = \frac {272}3 \end{aligned}$

Therefore, $(a, b) = \boxed{\left(16, \dfrac {273}3 \right)}$ .

18C3(-2)^3+a×18C16(-2)^2+b×18C17(-2)=0
-6528+612a-36b=0 put the options

Wrong question Actually in jee mains it was(1+ax+bx^2)(1-2x)^18

This question that is posted can have infinite (a,b) pairs and none of the options satisfy