Question

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

• A. $1+\frac{5}{6}x$
• B. $1-\frac{5}{6}x$
• C. $1+\frac{2}{3}x$
• D. $1-\frac{2}{3}x$

Answer 1

The correct option is B $1-\frac{5}{6}x$

Given expression can be written as
$=\frac{(1+x{)}^{\frac{1}{2}}+(1-x{)}^{\frac{2}{3}}}{1+x+(1+x{)}^{\frac{1}{2}}}$

$=\frac{[1+\frac{1}{2}x-\frac{1}{8}{x}^2\dots ]+[1-\frac{2}{3}x-\frac{1}{9}{x}^2-\dots ]}{1+x+[1+\frac{1}{2}x-\frac{1}{8}{x}^2+\dots ]}$

$=\frac{[1-\frac{1}{12}x-\frac{1}{144}{x}^2\dots ]}{[1+\frac{3}{4}x-\frac{1}{16}{x}^2\dots ]}=1-\frac{5}{6}x+\dots =1-\frac{5}{6}x$
when $x^2,x^3\dots$ are neglected