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JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Binomial Theorem

If the term independent of

x

in the expansion of

(\frac{3}{2} x^{2} - \frac{1}{3 x})^{9}

is

k

, then

18 k

is equal to:

(A)9
(B)5
(C)7✓ Correct
(D)11

Answer: (C)

Solution

The general term is

T_{r + 1} = (r 9) (\frac{3}{2} x^{2})^{9 - r} (- \frac{1}{3 x})^{r} = (r 9) (\frac{3}{2})^{9 - r} (- \frac{1}{3})^{r} x^{18 - 3 r}

. The term free of

x

needs

18 - 3 r = 0

, so

r = 6

:

T_{7} = (6 9) (\frac{3}{2})^{3} (\frac{1}{3})^{6} = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{7}{18}

. So

k = \frac{7}{18}

and

18 k = 7

.

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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