The term independent of

Question:

The term independent of $\mathrm{x}$ in the expansion of

$\left(\frac{1}{60}-\frac{x^{8}}{81}\right) \cdot\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$ is equal to :

  1. 36

  2. $-108$

  3. $-72$

  4. $-36$


Correct Option: , 4

Solution:

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

its general term

$\frac{1}{60}{ }^{6} \mathrm{C}_{\mathrm{r}} 2^{6-\mathrm{r}}(-3)^{\mathrm{r}} \mathrm{x}^{12-\mathrm{r}}-\frac{1}{81}{ }^{6} \mathrm{C}_{\mathrm{r}} 2^{6-\mathrm{r}}(-3)^{\mathrm{r}} 12^{20-4 \mathrm{r}}$

for term independent of $x, r$ for $I^{\text {st }}$ expression is 3 and $r$ for second expression is 5

$\therefore$ term independent of $\mathrm{x}=-36$

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