Q32 / 75 · JEE Main 2026 (05 April Shift 2) · Mathematics
The coefficient of $x^2$ in the expansion of $\left(2x^2 + \dfrac{1}{x}\right)^{10}$, $x \neq 0$, is :
- A$3240$
- Correct answerB$3360$
- C$3480$
- D$3600$
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: $3360$
$T_{r+1} = ^{10}C_{r} (2x^2)^{10-r} \left(\dfrac{1}{x}\right)^r$
$T_{r+1} = ^{10}C_{r} 2^{10-r} x^{20-2r} x^{-r}$
$T_{r+1} = ^{10}C_{r} 2^{10-r} x^{20-3r}$
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: $3360$
Hence, the correct answer is option: 'B'.