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Question: If the value of *x* is so small that \(x^{2}\) and higher power can be neglected, then \(\frac{\sqrt...

If the value of x is so small that x2x^{2} and higher power can be neglected, then 1+x+(1−x)231+x+1+x\frac{\sqrt{1 + x} + \sqrt[3]{(1 - x)^{2}}}{1 + x + \sqrt{1 + x}} is equal to

A

1+56x1 + \frac{5}{6}x

B

1−56x1 - \frac{5}{6}x

C

1+23x1 + \frac{2}{3}x

D

1−23x1 - \frac{2}{3}x

Answer

1−56x1 - \frac{5}{6}x

Explanation

Solution

Given expression can be written as

= (1+x)1/2+(1−x)2/31+x+(1+x)1/2\frac{(1 + x)^{1/2} + (1 - x)^{2/3}}{1 + x + (1 + x)^{1/2}}

= (1+12x+(−18)x2+.....)+(1−23x−19x2−....)1+x+[1+12x−18x2+.....]\frac{\left( 1 + \frac{1}{2}x + \left( - \frac{1}{8} \right)x^{2} + ..... \right) + \left( 1 - \frac{2}{3}x - \frac{1}{9}x^{2} - .... \right)}{1 + x + \left\lbrack 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + ..... \right\rbrack}

= 1−112x−1144x2+.....1+34x−116x2+.....=1−56x+.....\frac{1 - \frac{1}{12}x - \frac{1}{144}x^{2} + .....}{1 + \frac{3}{4}x - \frac{1}{16}x^{2} + .....} = 1 - \frac{5}{6}x + .....=1−56x1 - \frac{5}{6}x,

when x2,x3....x^{2},x^{3}.... are neglected.