The coefficient of x 2 in the expansion of (2 x 2 +frac 1 x ) 10 , x neq 0 , is :

Mathematics · Binomial Theorem

JEE Main 2026 — 5 April, Evening Shift — Question 31

The coefficient of $x^{2}$ in the expansion of $\left(2 \mathrm{x}^{2}+\frac{1}{\mathrm{x}}\right)^{10}, \mathrm{x} \neq 0$ , is :

Option A:
$3240$
Option B:
$3360$
Correct
Option C:
$3480$
Option D:
$3600$

Answer: B

Step-by-step solution

General term: $T_{r+1} = \binom{10}{r} (2x^2)^{10-r} \left(\frac{1}{x}\right)^r = \binom{10}{r} 2^{10-r} x^{20-2r-r} = \binom{10}{r} 2^{10-r} x^{20-3r}$ . For coefficient of $x^2$ , set $20-3r = 2 \Rightarrow r = 6$ . Term: $T_7 = \binom{10}{6} 2^{10-6} x^2 = \binom{10}{4} 2^4 x^2$ . $\binom{10}{4} = 210$ , $2^4 = 16$ . Coefficient = $210 \times 16 = 3360$ . Thus the coefficient is $3360.$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Introduction to Binomial Theorem