Assuming x to be so small that x^2 and higher | vedclass

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

Question

(D) Given expression: $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$$= \frac{1}{2} (1-x)^{1/3} (1-\frac{x}{16})^{-1/4} + \frac{1}{2} (1-5x)^2 (1-\frac{x}

Vedclass · Accepted Answer

$-\frac{989}{192}$

Vedclass · Answer

(D) Given expression: $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$$= \frac{1}{2} (1-x)^{1/3} (1-\frac{x}{16})^{-1/4} + \frac{1}{2} (1-5x)^2 (1-\frac{x}{16})^{-1/4}$Using binomial expansion $(1+z)^n \approx 1+nz$ for small $z$:$= \frac{1}{2} (1-\frac{1}{3}x)(1+\frac{x}{64}) + \frac{1}{2} (1-10x)(1+\frac{x}{64})$$= \frac{1}{2} [ (1 - \frac{1}{3}x + \frac{x}{64}) + (1 - 10x + \frac{x}{64}) ]$$= \frac{1}{2} [ 2 - x(\frac{1}{3} + 10 - \frac{2}{64}) ]$$= 1 - \frac{x}{2} (\frac{1}{3} + 10 - \frac{1}{32})$$= 1 - \frac{x}{2} (\frac{32 + 960 - 3}{96}) = 1 - \frac{989}{192}x$Therefore,the coefficient of $x$ is $-\frac{989}{192}$.

Assuming $x$ to be so small that $x^2$ and higher powers of $x$ can be neglected,the coefficient of $x$ in $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$ is equal to

Similar Questions

The sum of the series $1 + \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \dots$ is equal to

If $x = \frac{2 \cdot 5}{2! \cdot 3} + \frac{2 \cdot 5 \cdot 7}{3! \cdot 3^2} + \frac{2 \cdot 5 \cdot 7 \cdot 9}{4! \cdot 3^3} + \ldots$,then $x^2 + 8x + 8 = $

$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :

The sum to infinite terms of the series $\frac{3}{10}+\frac{3 \cdot 7}{10 \cdot 15}+\frac{3 \cdot 7 \cdot 11}{10 \cdot 15 \cdot 20}+\ldots$ to $\infty$ is

If $|x| < 1$,then in the expansion of $(1 + 2x + 3x^2 + 4x^3 + ....)^{1/2}$,the coefficient of $x^n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module