Question

Please send me the solution of this question​

Answer 1

Step-by-step explanation:

We have,

$\bigg( \sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x} \bigg) ^{12}$

$\tt{General \:\:term \:\:of\:\: the\:\: given \:\:expression\:\: is}$

$t_{r + 1} = \:^{12}C_{r} \bigg( \sqrt{ \frac{x}{3} } \bigg)^{ 12 - r}. \bigg( \frac{ \sqrt{3} }{2x} \bigg)^{r}$

$\implies \: t_{r + 1} = \:^{12}C_{r} \bigg(\frac{x}{3} \bigg)^{ \frac{12 - r}{2}}. \frac{ (\sqrt{3} )^{r} }{2 ^{r}. x^{r} }$

$\implies \: t_{r + 1} = \:^{12}C_{r} \bigg(\frac{x}{3} \bigg)^{ 6 - \frac{ r}{2}}. \frac{ (3 )^{ \frac{r}{2} } }{2 ^{r}. x^{r} }$

$\implies \: t_{r + 1} = \:^{12}C_{r} (x)^{ 6 - \frac{ r}{2}}. {3}^{ (\frac{r}{2} - 6)} . \frac{ (3 )^{ \frac{r}{2} } }{2 ^{r}. x^{r} }$

$\implies \: t_{r + 1} = \:^{12}C_{r} (x)^{ 6 - \frac{ r}{2} - r}. {3}^{ (\frac{r}{2} - 6 + \frac{r}{2} )} . 2 ^{ - r}$

$\implies \: t_{r + 1} = \:^{12}C_{r} (x)^{ 6 - \frac{ 3r}{2}}. {3}^{ (r - 6 )} . 2 ^{ - r}$

For term independent of x, we must have, $6-\frac{3r}{2}=0$

So, $r=4$

So, the required term is

$\implies \: t_{4 + 1} = \:^{12}C_{4} (x)^{ 6 - \frac{ 3 \times 4}{2}}. {3}^{ (4 - 6 )} . 2 ^{ - 4}$

$\implies \: t_{5} = \:^{12}C_{4} . {3}^{ - 2} . 2 ^{ - 4}$

$\implies \: t_{5} = \frac{ \:^{12}C_{4} }{ {3}^{ 2} . 2 ^{ 4} }$

$\implies \: t_{5} = \frac{ 12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1 \times 9 \times 16 }$

$\implies \: t_{5} = \frac{ 12 \times 11 \times 10 }{12 \times 2 \times 1 \times 16 }$

$\implies \: t_{5} = \frac{ 11 \times 10 }{ 2 \times 1 \times 16 }$

$\implies \: t_{5} = \frac{ 11 \times 5 }{ 1 \times 16 }$

$\implies \: t_{5} = \frac{ 55 }{ 16 }$