Assuming $|x|$ to be so small, that $x^2$ and higher powers of $x$ can be neglected, then $\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=$

Question

Assuming $|x|$ to be so small, that $x^2$ and higher powers of $x$ can be neglected, then $\frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=$, $1+\frac{5 x}{4}$, $1-\frac{5 x}{4}$, $1+\frac{4 x}{5}$, $1-\frac{4 x}{5}$

MARKS App · Accepted Answer

Given, $|x|$ is very small, $x^2$ is negligible $\begin{aligned} & \frac{\sqrt{1+x}+(1-x)^{3 / 2}}{(1+x)+\sqrt{1+x}}=\frac{1+\frac{1}{2} x+1-\frac{3}{2} x}{1+x+1+\frac{x}{2}} \ & =\frac{2-x}{2+\frac{3 x}{2}}=\frac{4-2 x}{4+3 x}=\frac{(4-2 x)(4-3 x)}{\left(16-9 x^2\right)} \ & =\frac{16-20 x+6 x^2}{16-9 x^2}=\frac{16-20 x}{16}\left\{\because x^2 \text { negligible }\right\} \ & =1-\frac{5 x}{4} \end{aligned}$

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