JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

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

Solution

General term: $T_{r + 1} = (r 18) (9 x)^{18 - r} (- \frac{1}{3 x})^{r} = (r 18) (- \frac{1}{3})^{r} 9^{18 - r} x^{18 - r - r /2}$ .

Power of $x$ : $18 - \frac{3 r}{2}$ . Set this to zero: $r = 12$ .

Coefficient $= (12 18) (- \frac{1}{3})^{12} 9^{6} = (12 18) \frac{9 ^{6}}{3 ^{12}} = (12 18) \frac{3 ^{12}}{3 ^{12}} = (12 18)$ .

$(12 18) = (6 18) = 18564$ .

$18564 = 221 \times 84$ , so $k = 84$ .

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