In the expansion of (9x - frac 1 3√(x) ) 18 , x 0, if the term independent of x is 221k , then k is equal to:

Mathematics · Binomial Theorem

JEE Main 2026 — 4 April, Evening Shift — Question 33

In the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x > 0,$ if the term independent of $x$ is $221k$ , then k is equal to:

Option A:
84
Correct
Option B:
78
Option C:
168
Option D:
198

Answer: A

Step-by-step solution

General term in above expression is $\mathrm{T}_{\mathrm{r}+1}={ }^{18} \mathrm{C}_{\mathrm{r}}(9 \mathrm{x})^{18-\mathrm{r}}\left(-\frac{1}{3 \sqrt{\mathrm{x}}}\right)^{\mathrm{r}}$ $=\left(-\frac{1}{3}\right)^{\mathrm{r}}{ }^{18} \mathrm{C}_{\mathrm{r}} 9^{18-\mathrm{r}} \cdot \mathrm{x}^{18-\frac{3 \mathrm{r}}{2}}$ For term independent of $\mathrm{x}, 18-\frac{3 \mathrm{r}}{2}=0 \Rightarrow \mathrm{r}=12$ ∴ Coefficient of term independent of x $=\left(-\frac{1}{3}\right)^{12}{ }^{18} \mathrm{C}_{12} 9^{18-12}$ $=18564$ $=221 \mathrm{k}$ (given) $\Rightarrow \mathrm{k}=84$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Introduction to Binomial Theorem