If the coefficients of x ^ 2 and x ^ 3 , are both zero, in the expansion of the expression (1 + ax + b x ^ 2 ) (1 - 3x) ^ 15 , in powers of x , then the ordered pair (a, b) is equal to

Correct answer: A. (28, 315) (1 + ax + b x ^ 2 ) (1 - 3x) ^ 15 = (1 + ax + b x ^ 2 )(C0 15 - ^ 15 C _ 1 3x + ^ 15 C _ 2 (3x) ^ 2 - ^ 15 C _ 3 (3x) ^ 3 + - ^ 15 C _ 15 (3x) ^ 15 ) ∴ Coefficient of x ^ 2 is: b . ^ 15 C _ 0 - 3 . a ^ 15 C _ 1 + 3 ^ 2 . . ^ 15 C _ 2 = 0 (Given) ⇒ b - 45a + 945 = 0 . . . . . . (i) Coefficient of x ^ 3 is: - b . 3 . ^ 15 C _ 1 + 9a . ^ 15 C _ 2 - 27 . ^ 15 C _ 3 = 0 ⇒ - 45b + 9a × (15)/(2) × (14)/(1) - 27 × (15)/(3) × (14)/(2) × (13)/(1) = 0 ⇒ - b + 21a - 273 = 0 ⇒ b - 21a + 273 = 0 . . . . . . . (ii) From (i) and (ii) a = 28 and b = 315

JEE Main 201910 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Binomial Theorem

If the coefficients of x ^ 2 and x ^ 3 , are both zero, in the...

If the coefficients of

{x}^{2} and {x}^{3}

, are both zero, in the expansion of the expression \left(1 + ax + b{x}^{2}\right){\left(1 - 3x\right)}^{15}

, in powers of

, then the ordered pair \left(a, b\right)

is equal to

Options

A $\left(28, 315\right)$
B $\left(- 21, 714\right)$
C $\left(28, 861\right)$
D $\left(- 54, 315\right)$

Correct answer

A. $\left(28, 315\right)$

Step-by-step solution

$\left(1 + ax + b{x}^{2}\right){\left(1 - 3x\right)}$ ^{15} = \left(1 + ax + b{x}^{2}\right)\left(C0 15 {-}^{15}{C}_{1}3x{+}^{15}{C}_{2}{\left(3x\right) $}^{2}{-}^{15}{C}_{3}{\left(3x\right)$ }^{3} + \ldots {-}^{15}{C}_{15}{\left(3x\right)}^{15}\right) $$ ∴ $Coefficient of$ {x}^{2} $is:$ {b .}^{15}{C}_{0} - 3 . {a}^{15}{C}_{1} + {3}^{2 .}{.}^{15}{C}_{2} = 0 $(Given) $ ⇒ b - 45a + 945 = 0 . . . . . . \left(i\right)

Coefficient $of {x}^{3}$ is:

$- b . 3{.}^{15}{C}_{1} + 9a{.}^{15}{C}_{2} - 27{.}^{15}{C}_{3} = 0$

$⇒ - 45b + 9a \times$ \dfrac{15}{2} $\times$ \dfrac{14}{1} - $27 \times$ \dfrac{15}{3} $\times$ \dfrac{14}{2} $\times$ \dfrac{13}{1} = $0$

$⇒ - b + 21a - 273 = 0$

$⇒ b - 21a + 273 = 0 . . . . . . . \left(ii\right)$

From \left(i\right) $and \left(ii\right)$

$a = 28 and b = 315$

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