If the constant term in the expansion of (3 x ^ 3 - 2 x ^ 2 + 5 x ^ 5 ) ^ 10 is 2 ^ k . l , where l is an odd integer, then the value of k is equal to

Correct answer: D. 9 For (3 x ^ 3 - 2 x ^ 2 + 5 x ^ 5 ) ^ 10 the general term is given by T _ r + 1 = 10 ! r _ 1 ! r _ 2 ! r _ 3 ! (3) ^ r _ 1 (- 2) ^ r _ 2 (5) ^ r _ 3 (x) ^ 3 r _ 1 + 2 r _ 2 - 5 r _ 3 For term independent of x the exponent 3 r _ 1 + 2 r _ 2 - 5 r _ 3 = 0 (1) Also we know r _ 1 + r _ 2 + r _ 3 = 10 ⋯ (2) From (1) and (2), we get r _ 1 + 2(10 - r _ 3 ) - 5 r _ 3 = 0 i.e. r _ 1 + 20 = 7 r _ 3 So r _ 1 , r _ 2 , r _ 3 = 1, 6, 3 Hence the constant term =(10 !)/(1 ! 6 ! 3 !) (3) ^ 1 (- 2) ^ 6 (5) ^ 3 = 2 ^ 9 · 3 ^ 2 · 5 ^ 4 · 7 ^ 1 ⇒ k = 9

JEE Main 202229 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2022 · 29 June · Morning Shift · Mathematics · Binomial Theorem

If the constant term in the expansion of (3 x ^ 3 - 2 x ^ 2 + 5 x ^...

If the constant term in the expansion of

{\left(3{x}^{3} - 2{x}^{2} + \dfrac{5}{{x}^{5}}\right)}^{10} is

{2}^{k} . l

, where

is an odd integer, then the value of

k is equal to

Options

A $6$
B $7$
C $8$
D $9$

Correct answer

D. $9$

Step-by-step solution

For ${\left(3{x}^{3} - 2{x}^{2} + \dfrac{5}{{x}^{5}}\right)}$ ^{10} $the general term is given by$ {T}_{r + 1} = \dfrac{10 !}{{r}_{1} ! {r}_{2} ! {r}_{3} !}{\left(3\right)}^{{r}_{1}}{\left(- 2\right)}^{{r}_{2}}{\left(5\right)}^{{r}_{3}}{\left(x\right)}^{3{r}_{1} + 2{r}_{2} - 5{r}_{3}} $For term independent of$ x $the exponent$ 3{r}_{1} + 2{r}_{2} - 5{r}_{3} = 0 \ldots $\left(1\right)$

Also we know ${r}_{1} + {r}_{2} + {r}_{3} = 10 ⋯ \left(2\right)From \left(1\right)$ and $\left(2\right), we get $ {r}_{1} + 2\left(10 - {r}_{3}\right) $- 5{r}_{3} = 0$

i.e. ${r}_{1} + 20 = 7{r}_{3}$

So ${r}_{1}, {r}_{2}, {r}_{3} = 1, 6, 3$

Hence the constant term =\dfrac{10 !}{1 ! 6 ! 3 !} ${\left(3\right)}$ ^{1}{\left(- 2\right) $}^{6}{\left(5\right)$ }^{3} $$ = {2}^{9} \cdot {3}^{2} \cdot {5}^{4} \cdot {7}^{1} $$ ⇒ k = 9 $$

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