If x is so small that all terms containing x^ | vedclass

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

Question

(D) Given expression: $f(x) = \frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$Neglecting $x^2$ and higher powers, we use the binomial approximat

Vedclass · Accepted Answer

$\frac{1}{14}$

Vedclass · Answer

(D) Given expression: $f(x) = \frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$Neglecting $x^2$ and higher powers, we use the binomial approximation $(1+u)^n \approx 1+nu$.$f(x) = (1+\frac{2x}{3})^{-4} 	imes (4(1+\frac{5x}{4}))^{1/2} 	imes (9(1+\frac{x}{9}))^{-3/2}$$f(x) = (1+\frac{2x}{3})^{-4} 	imes 2(1+\frac{5x}{4})^{1/2} 	imes \frac{1}{27}(1+\frac{x}{9})^{-3/2}$$f(x) \approx \frac{2}{27} (1 - 4 \cdot \frac{2x}{3}) (1 + \frac{1}{2} \cdot \frac{5x}{4}) (1 - \frac{3}{2} \cdot \frac{x}{9})$$f(x) \approx \frac{2}{27} (1 - \frac{8x}{3}) (1 + \frac{5x}{8}) (1 - \frac{x}{6})$$f(x) \approx \frac{2}{27} (1 - \frac{8x}{3} + \frac{5x}{8} - \frac{x}{6}) = \frac{2}{27} (1 - \frac{64x - 15x + 4x}{24}) = \frac{2}{27} (1 - \frac{53x}{24})$Substituting $x = \frac{6}{371}$:$f(\frac{6}{371}) \approx \frac{2}{27} (1 - \frac{53}{24} \cdot \frac{6}{371}) = \frac{2}{27} (1 - \frac{53}{4 \cdot 371}) = \frac{2}{27} (1 - \frac{53}{1484})$$f(\frac{6}{371}) \approx \frac{2}{27} (\frac{1431}{1484}) = \frac{2}{27} \cdot \frac{53 \cdot 27}{1484} = \frac{106}{1484} = \frac{1}{14}$

If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$ when $x=\frac{6}{371}$, is

Similar Questions

If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected,then $\frac{(1 + x)^{3/2} - (1 + \frac{1}{2}x)^3}{(1 - x)^{1/2}}$ may be approximated as

The value of the expression $\frac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}$ at $x = 2$ is:

The sum of the coefficients in the expansion of $\left(1+\frac{x}{2}\right)^{12}$ is

The value of $^nC_1 \sum_{r=0}^1 {^1C_r} + ^nC_2 \left( \sum_{r=0}^2 {^2C_r} \right) + ^nC_3 \left( \sum_{r=0}^3 {^3C_r} \right) + \dots + ^nC_n \left( \sum_{r=0}^n {^nC_r} \right)$ is equal to

The coefficient of $x^4$ in the expansion of $(1+x-x^2-x^3)^{11}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module