If the coefficients of x³ and x⁴ in (1+ax+bx²)(1-2x)¹⁸ are both zero, then (a,b) is (a) (14, 272 3 ) (b) (16, 272 3 ) (c) (16, 251 3 ) (d) (14, 251 3 )

Correct answer: (b) (16, 272 3 ) Setting the two coefficients to 0 gives 51a-3b=544 and -544a+51b=-4080, solving to a=16, b= 272 3. JEE Main 2014 · Binomial Theorem — verified solution by the Vidaara Team.

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[JEE Main 2014] if the coefficients of x 3 and x 4 in 1 ax bx 2

VAVidaara Admin Asked 2d ago 0 views 1 answer

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

(a) $\left(14,\dfrac{272}3\right)$
(b) $\left(16,\dfrac{272}3\right)$
(c) $\left(16,\dfrac{251}3\right)$
(d) $\left(14,\dfrac{251}3\right)$

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Correct answer: (b) $\left(16,\dfrac{272}3\right)$

Setting the two coefficients to $0$ gives $51a-3b=544$ and $-544a+51b=-4080$, solving to $a=16,\ b=\frac{272}3$.

JEE Main 2014 · Binomial Theorem — verified solution by the Vidaara Team.

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