The term independent of x in {left[ {sqrt{fra | vedclass

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(B) The general term $T_{r+1}$ in the expansion of ${\left[ {\sqrt{\frac{x}{3}} + \frac{{\sqrt{3}}}{{{x^2}}}} \right]^{10}}$ is given by:$T_{r+1} = {^

Vedclass · Accepted Answer

$\frac{5}{3}$

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(B) The general term $T_{r+1}$ in the expansion of ${\left[ {\sqrt{\frac{x}{3}} + \frac{{\sqrt{3}}}{{{x^2}}}} ight]^{10}}$ is given by:$T_{r+1} = {^{10}C_r} {\left( \sqrt{\frac{x}{3}} ight)^{10-r}} {\left( \frac{\sqrt{3}}{x^2} ight)^r}$$T_{r+1} = {^{10}C_r} {\left( \frac{x^{1/2}}{3^{1/2}} ight)^{10-r}} {\left( \frac{3^{1/2}}{x^2} ight)^r}$$T_{r+1} = {^{10}C_r} \cdot \frac{x^{(10-r)/2}}{3^{(10-r)/2}} \cdot \frac{3^{r/2}}{x^{2r}}$$T_{r+1} = {^{10}C_r} \cdot \frac{3^{r/2}}{3^{(10-r)/2}} \cdot x^{\frac{10-r}{2} - 2r}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$\frac{10-r}{2} - 2r = 0$ $\Rightarrow 10 - r - 4r = 0$ $\Rightarrow 5r = 10$ $\Rightarrow r = 2$Substituting $r = 2$ into the expression:$T_{2+1} = {^{10}C_2} \cdot \frac{3^{2/2}}{3^{(10-2)/2}} = {^{10}C_2} \cdot \frac{3^1}{3^4} = \frac{10 	imes 9}{2 	imes 1} \cdot \frac{1}{3^3} = 45 \cdot \frac{1}{27} = \frac{45}{27} = \frac{5}{3}$.

The term independent of $x$ in ${\left[ {\sqrt{\frac{x}{3}} + \frac{{\sqrt{3}}}{{{x^2}}}} \right]^{10}}$ is

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