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Find the value of \sqrt{\frac{y}{x+y}} \cdot \sqrt{\frac{y}{y-x}}, if x is very small as compared to y.

Option: 1

1-\frac{1}{2} \cdot \frac{y^{2}}{x^{2}}


Option: 2

1+\frac{1}{2} \cdot \frac{y^{2}}{x^{2}}


Option: 3

1-\frac{1}{2} \cdot \frac{x^{2}}{y^{2}}


Option: 4

1+\frac{1}{2} \cdot \frac{x^{2}}{y^{2}}


Answers (1)

best_answer

Binomial Theorem for any  Index

\\\text{For negative or fractional Index and } |x|<1,\\\\\mathrm{(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+......+\frac{n(n-1)(n-2).....(n-r+1)}{r!}x^r....\infty}

 

Now,

\begin{aligned} \sqrt{\frac{y}{x+y}} \sqrt{\frac{y}{y-x}} &=\left(\frac{1}{1+\frac{x}{y}}\right)^{1 / 2}\left(\frac{1}{1-\frac{x}{y}}\right)^{1 / 2} \\ &=\left(1-\frac{x^{2}}{y^{2}}\right)^{-1 / 2}=1+\frac{1}{2} \cdot \frac{x^{2}}{y^{2}} \\ &\text { (ignoring higher powers of } x / y \text { as } x / y \,\,is\,\,small) \end{aligned}

Option D is correct.

Posted by

Ritika Harsh

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