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BITSAT 2021 Paper

Section: Mathematics

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Question : 120 of 150

Marks: +1, -0

The coefficient of the term independent of

x

in the expansion of

\left(\sqrt{\frac{x}{3}} + \frac{3}{2x^{2}}\right)^{10}

is

$\frac{5}{4}$
$\frac{7}{4}$
$\frac{9}{4}$
None of these

Solution:

The

(r+1)

th term in the expansion of

\left(\sqrt{\frac{x}{3}} + \frac{3}{2x^{2}}\right)^{10}

is given by

T_{r+1} = \binom{10}{r} \left(\sqrt{\frac{x}{3}}\right)^{10-r} \left(\frac{3}{2x^{2}}\right)^{r}

= \binom{10}{r} \frac{x^{5-(r/2)}}{3^{5-(r/2)}} \cdot \frac{3^{r}}{2^{r} x^{2r}}

= \binom{10}{r} \frac{3^{(3r/2)-5}}{2^{r}} x^{5-(5r/2)}

For

T_{r+1}

to be independent of

x

, we must have

5 - \frac{5r}{2} = 0

or

r = 2

.

Thus, the 3rd term is independent of

x

and is equal to

\binom{10}{2} \frac{3^{3-5}}{2^{2}} = \frac{10 \times 9}{2} \times \frac{3^{-2}}{4} = \frac{5}{4}

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