JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q67

MathematicsSingle correct+4 / -1Question 67 of 90

If the coefficients of $x^3$ and $x^4$ in the expansion of $(1+ax+bx^2)(1-2x)^{18}$ in powers of $x$ are both zero, then $(a,b)$ is equal to :

\left(14,\dfrac{272}{3}\right)

\left(16,\dfrac{272}{3}\right)

\left(16,\dfrac{251}{3}\right)

\left(14,\dfrac{251}{3}\right)

Answer and worked solution

Official answer: B

Let $c_k$ be the coefficient of $x^k$ in $(1-2x)^{18}$ : $c_1=-36$ , $c_2=612$ , $c_3=-6528$ , $c_4=48960$ .

Setting the two coefficients to zero:

Substituting: $3876a=62016$ , so $a=16$ and $b=\frac{272}3$ .

JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 67