JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q9

MathematicsSingle correct+4 / -1Question 9 of 75

In the expansion of $\left(9x-\dfrac1{3\sqrt x}\right)^{18}$ , $x>0$ , if the term independent of $x$ is $(221)k$ , then $k$ is equal to:

168

198

Answer and worked solution

Official answer: A

General term. $\binom{18}r(9x)^{18-r}\left(-\frac1{3\sqrt x}\right)^r$ , with power of $x$ equal to $18-\frac{3r}2$ , which is 0 at $r=12$ .

Value. $\binom{18}{12}\frac{9^6}{3^{12}}=\binom{18}6=18564=221\times84$ .

(A)

Topics:General and middle terms

JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 9