If x is numerically so small that x^2 and hig | vedclass

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

Question

(A) Given expression: $E = (1 + \frac{2x}{3})^{3/2} \cdot (32 + 5x)^{-1/5}$ Using the binomial approximation $(1+u)^n \approx 1+nu$ for small $u$: $E

Vedclass · Accepted Answer

$\frac{32+31x}{64}$

Vedclass · Answer

(A) Given expression: $E = (1 + \frac{2x}{3})^{3/2} \cdot (32 + 5x)^{-1/5}$ Using the binomial approximation $(1+u)^n \approx 1+nu$ for small $u$: $E \approx (1 + \frac{3}{2} \cdot \frac{2x}{3}) \cdot (32)^{-1/5} (1 + \frac{5x}{32})^{-1/5}$ $E \approx (1 + x) \cdot \frac{1}{2} \cdot (1 - \frac{1}{5} \cdot \frac{5x}{32})$ $E \approx \frac{1}{2} (1 + x) (1 - \frac{x}{32})$ Neglecting $x^2$ terms: $E \approx \frac{1}{2} (1 + x - \frac{x}{32}) = \frac{1}{2} (1 + \frac{31x}{32}) = \frac{32 + 31x}{64}$

If $x$ is numerically so small that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2x}{3}\right)^{3/2} \cdot (32+5x)^{-1/5}$ is approximately equal to

Similar Questions

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

If the coefficient of $x^{13}$ in the expansion of $\frac{(1+x)^2}{(1-2x)^3}$ is $A \times 2^{10}$,then $A=$

If $x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \dots \infty$,then $3x^2 + 6x$ is equal to

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

For $|x| < \frac{4}{3}$, the approximate value of $\frac{1}{(4-3 x)^{\frac{1}{2}}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module