If |x| is so small that x^3 and higher powers | vedclass

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

Question

(D) We have $\frac{1}{\sqrt{4-x}(2+x)^3} = (4-x)^{-1/2} (2+x)^{-3}$.$= 4^{-1/2} \left(1-\frac{x}{4}\right)^{-1/2} \cdot 2^{-3} \left(1+\frac{x}{2}\rig

Vedclass · Accepted Answer

$\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)$

Vedclass · Answer

(D) We have $\frac{1}{\sqrt{4-x}(2+x)^3} = (4-x)^{-1/2} (2+x)^{-3}$.$= 4^{-1/2} \left(1-\frac{x}{4}\right)^{-1/2} \cdot 2^{-3} \left(1+\frac{x}{2}\right)^{-3}$$= \frac{1}{2} \cdot \frac{1}{8} \left(1 - \frac{1}{2}\left(-\frac{x}{4}\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} \left(-\frac{x}{4}\right)^2\right) \left(1 + (-3)\left(\frac{x}{2}\right) + \frac{(-3)(-4)}{2} \left(\frac{x}{2}\right)^2\right)$$= \frac{1}{16} \left(1 + \frac{x}{8} + \frac{3}{128} x^2\right) \left(1 - \frac{3x}{2} + \frac{3x^2}{2}\right)$$= \frac{1}{16} \left(1 - \frac{3x}{2} + \frac{3x^2}{2} + \frac{x}{8} - \frac{3x^2}{16} + \frac{3x^2}{128}\right)$$= \frac{1}{16} \left(1 - \frac{11x}{8} + \frac{171x^2}{128}\right)$.

If $|x|$ is so small that $x^3$ and higher powers of $x$ can be neglected,then an approximate value of $\frac{1}{\sqrt{4-x}(2+x)^3}$ is

Similar Questions

In the expansion of $\left(1+\frac{3x}{2}\right)^{-5}$,the coefficient of $x^{10}$ is equal to the coefficient of $x^{10}$ in $(1+ax)^n$,where $n \in N$. Then $na$ is equal to:

The set of all real values of $x$ for which the expansion of $\left(125 x^2-\frac{27}{x}\right)^{-2/3}$ is valid,is

If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$,then

If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$

If $5|b| < 2|a|$,then the $4^{th}$ term in the expansion of $(2a + 5b)^{-4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module