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JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Binomial Theorem

The term independent of

x

in the expansion of

(\frac{1}{60} - \frac{x ^{8}}{81}) \cdot (2 x^{2} - \frac{3}{x ^{2}})^{6}

is equal to:

(A)36
(B)-36✓ Correct
(C)-108
(D)-72

Answer: (B)

Solution

Each term of

(2 x^{2} - \frac{3}{x ^{2}})^{6}

is

(r 6) 2^{6 - r} (- 3)^{r} x^{12 - 4 r}

. Multiplying by

\frac{1}{60}

needs

12 - 4 r = 0

, so

r = 3

, giving

\frac{1}{60} (3 6) 2^{3} (- 3)^{3} = \frac{1}{60} (20) (8) (- 27) = - 72

. Multiplying by

- \frac{x ^{8}}{81}

needs

8 + 12 - 4 r = 0

, so

r = 5

, giving

- \frac{1}{81} (5 6) 2^{1} (- 3)^{5} = - \frac{1}{81} (6) (2) (- 243) = 36

. Adding,

- 72 + 36 = - 36

.

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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