If the value of x is so small that x^2 and hi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression is $E = \frac{(1 + x)^{1/2} + (1 - x)^{2/3}}{1 + x + (1 + x)^{1/2}}$.Using binomial expansion $(1 + x)^n \approx 1 + nx$ for smal

Vedclass · Accepted Answer

$1 - \frac{5}{6}x$

Vedclass · Answer

(B) Given expression is $E = \frac{(1 + x)^{1/2} + (1 - x)^{2/3}}{1 + x + (1 + x)^{1/2}}$.Using binomial expansion $(1 + x)^n \approx 1 + nx$ for small $x$:$(1 + x)^{1/2} \approx 1 + \frac{1}{2}x$$(1 - x)^{2/3} \approx 1 - \frac{2}{3}x$Substituting these into the expression:$E \approx \frac{(1 + \frac{1}{2}x) + (1 - \frac{2}{3}x)}{1 + x + (1 + \frac{1}{2}x)}$$E \approx \frac{2 - \frac{1}{6}x}{2 + \frac{3}{2}x} = \frac{2(1 - \frac{1}{12}x)}{2(1 + \frac{3}{4}x)}$$E \approx (1 - \frac{1}{12}x)(1 + \frac{3}{4}x)^{-1}$Using $(1 + z)^{-1} \approx 1 - z$:$E \approx (1 - \frac{1}{12}x)(1 - \frac{3}{4}x)$$E \approx 1 - \frac{3}{4}x - \frac{1}{12}x = 1 - (\frac{9+1}{12})x = 1 - \frac{10}{12}x = 1 - \frac{5}{6}x$.

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

Similar Questions

The coefficient of $x^2$ in the expansion of $(1-3x)^{-1/4}$ is

If the set of all values of $x$ for which the expansion of $(7-5 x)^{-\frac{2}{3}}$ is valid is equal to $(-a, a)$,then $5 a+7$ is equal to

If $x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots$,then $x+\frac{1}{x}=$

For $n, p \in N-\{1\}$,the coefficient of $x^3$ in $\frac{(1-x)^{-1 / p}}{(1-x)^n}$ is:

Find the binomial expansion of $\frac{1}{\sqrt[3]{6 - 3x}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module