If x is so small that x^2 and higher powers o | vedclass

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

Question

(D) Given expression: $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ Neglecting $x^2$ and higher powers,we use the binomial approximation

Vedclass · Accepted Answer

$\frac{1}{16}(1+7x)$

Vedclass · Answer

(D) Given expression: $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ Neglecting $x^2$ and higher powers,we use the binomial approximation $(1+nx) \approx 1+nx$. $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{2^4(1-\frac{3}{2}x)^4}$ $E = \frac{1}{16} (1+\frac{2}{3}x)^{-3} (1-15x)^{-1/5} (1-\frac{3}{2}x)^{-4}$ Applying the approximation $(1+ax)^n \approx 1+nax$: $(1+\frac{2}{3}x)^{-3} \approx 1 + (-3)(\frac{2}{3}x) = 1-2x$ $(1-15x)^{-1/5} \approx 1 + (-\frac{1}{5})(-15x) = 1+3x$ $(1-\frac{3}{2}x)^{-4} \approx 1 + (-4)(-\frac{3}{2}x) = 1+6x$ Multiplying these: $E \approx \frac{1}{16} (1-2x)(1+3x)(1+6x)$ $E \approx \frac{1}{16} (1+3x-2x)(1+6x) = \frac{1}{16} (1+x)(1+6x)$ $E \approx \frac{1}{16} (1+6x+x) = \frac{1}{16}(1+7x)$

If $x$ is so small that $x^2$ and higher powers of $x$ may be neglected,then the approximate value of $\frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ is:

Similar Questions

If $f(x)=(2011+x)^n$,where $x$ is a real variable and $n$ is a positive integer,then the value of $f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\ldots+\frac{f^{(n-1)}(0)}{(n-1) !}$ is

The number of non-zero terms in the expansion of $(1 + 3\sqrt{2}x)^9 + (1 - 3\sqrt{2}x)^9$ is

If the first three terms in the binomial expansion of $(1+bx)^n$ in ascending powers of $x$ are $1, 6x$ and $6x^2$ respectively,then $b+n=$

$\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \dots + 2^n\binom{n}{n}$ is equal to

Expand the expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module