The coefficient of x ^ 5 in the expansion of (2 x ^ 3 - 1 3 x ^ 2 ) ^ 5 is

Correct answer: A. (80)/(9) General term of the binomial expansion of (2 x ^ 3 - 1 3 x ^ 2 ) ^ 5 is T _ r + 1 = 5 _ C _ r ( 2 x ^ 3 ) ^ 5 - r ( - 1 3 x ^ 2 ) ^ r ⇒ T _ r + 1 = Cr 5 (2) ^ 5 - r (- (1)/(3)) ^ r x ^ 15 - 5r For coefficient of x ^ 5 , we must have 15 - 5r = 5 ⇒ r = 2 So, required coefficient is = C2 5 ( 2 ) ^ 3 ((1)/(9)) = 10 × 8 × (1)/(9) = (80)/(9)

JEE Main 202313 AprilEvening Shift

Exam: JEE Main
Year: 2023
Date: 13 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2023 · 13 April · Evening Shift · Mathematics · Binomial Theorem

The coefficient of x ^ 5 in the expansion of (2 x ^ 3 - 1 3 x ^ 2 )...

The coefficient of

{x}^{5} in the expansion of

{\left(2{x}^{3} - \dfrac{1}{3{x}^{2}}\right)}^{5} is

Options

A $\dfrac{80}{9}$
B $9$
C $8$
D $\dfrac{26}{3}$

Correct answer

A. $\dfrac{80}{9}$

Step-by-step solution

General term of the binomial expansion of ${\left(2{x}^{3} - \dfrac{1}{3{x}^{2}}\right)}^{5} is$ {T}_{r + 1} = {5}_{{C}_{r}} ( 2{x}^{3}{)}^{5 - r}{\left(\dfrac{- 1}{3{x}^{2}}\right)}^{r} $$ ⇒ {T}_{r + 1} = Cr 5 {\left(2\right)}^{5 - r}{\left(- \dfrac{1}{3}\right)}^{r}{x}^{15 - 5r} $For coefficient of$ {x}^{5} $, we must have$ 15 - 5r = 5 ⇒ r = 2 $So, required coefficient is$ = C2 5 ( 2{)}^{3}\left(\dfrac{1}{9}\right) = 10 \times 8 \times \dfrac{1}{9} = \dfrac{80}{9}

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