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JEE Main 20265 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2026 · 5 April · Evening Shift · Mathematics · Binomial Theorem

The coefficient of x^2 in the expansion of (2x^2 + (1)/(x))^ 10 , x...

The coefficient of

x^2

in the expansion of \left(2x^2 + \dfrac{1}{x}\right)^{10}

,

x \neq 0$, is :

Options

A $3240$
B $3360$
C $3480$
D $3600$

Correct answer

B. $3360$

Step-by-step solution

The general term in the expansion of

\left(2x^2 + \dfrac{1}{x}\right)^{10}

is given by:

T_{r+1} = ^{10}C_{r} (2x^2)^{10-r} \left(\dfrac{1}{x}\right)^r

T_{r+1} = ^{10}C_{r} 2^{10-r} x^{20-2r} x^{-r}

T_{r+1} = ^{10}C_{r} 2^{10-r} x^{20-3r}

For the coefficient of

x^2

, we equate the power of

x

to

2

:

20 - 3r = 2

3r = 18 \Rightarrow r = 6

Substituting

r = 6

in the coefficient part, we get:

Coefficient

= ^{10}C_{6} 2^{10-6}

= ^{10}C_{4} 2^4

= \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times 16

= 210 \times 16

= 3360

Answer:

3360

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