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

(C) Given that $x$ is very small,we can use the binomial approximation $(1+nx) \approx (1+x)^n$ for $|x| < 1$. Applying this to the given expression:

Vedclass · Accepted Answer

$\frac{24+41 x}{24}$

Vedclass · Answer

(C) Given that $x$ is very small,we can use the binomial approximation $(1+nx) \approx (1+x)^n$ for $|x| Applying this to the given expression: $\left(1+\frac{3}{4} x\right)^{\frac{1}{2}} \approx 1 + \frac{1}{2} \cdot \frac{3}{4} x = 1 + \frac{3}{8} x$ $\left(1-\frac{2}{3} x\right)^{-2} \approx 1 + (-2) \cdot \left(-\frac{2}{3} x\right) = 1 + \frac{4}{3} x$ Multiplying these two approximations and neglecting $x^2$ terms: $\left(1+\frac{3}{8} x\right)\left(1+\frac{4}{3} x\right) \approx 1 + \frac{3}{8} x + \frac{4}{3} x + \frac{12}{24} x^2$ Neglecting the $x^2$ term: $1 + \left(\frac{9+32}{24}\right) x = 1 + \frac{41}{24} x = \frac{24+41 x}{24}$ Thus,the correct option is $C$.

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

Similar Questions

$1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots$ is equal to

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

The $13^{\text{th}}$ term in the expansion of $(1-4x)^{-4}$ is

The expansion of $(1+x+x^2)^{-3/2}$ in powers of $x$ is valid if

If $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$,then $(a, b) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module