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  • If the value of x is so small that  and greater powers can be neglected, then <img alt="\frac{\sqrt{1+x}+\sqrt[3]{(

If the value of x is so small that x^2 and greater powers can be neglected, then \frac{\sqrt{1+x}+\sqrt[3]{(1-x)^2}}{1+x+\sqrt{1+x}} is equal to 

Option: 1

1+\frac{5}{6} x


Option: 2

1-\frac{5}{6} x


Option: 3

1+\frac{2}{3} x


Option: 4

1-\frac{2}{3} x


Answers (1)

best_answer

Given expression can be written as 

\begin{aligned} & =\frac{(1+x)^{1 / 2}+(1-x)^{2 / 3}}{1+x+(1+x)^{1 / 2}} \\ & \frac{\left[1+\frac{1}{2} x-\frac{1}{8} x^2 \ldots\right]+\left[1-\frac{2}{3} x-\frac{1}{9} x^2-\ldots\right]}{1+x+\left[1+\frac{1}{2} x-\frac{1}{8} x^2+\ldots\right]} \\ & \frac{\left[1-\frac{1}{12} x-\frac{1}{144} x^2 \ldots\right]}{\left[1+\frac{3}{4} x-\frac{1}{16} x^2 \ldots\right]}=1-\frac{5}{6} x+\ldots=1-\frac{5}{6} x \end{aligned}

when x^2, x^3, \dots are neglected

Posted by

HARSH KANKARIA

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