When x is so small that its square and its hi | vedclass

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

Question

(A) Given that $x^n \approx 0$ for $n \ge 2$. Let $y = \frac{(1+\frac{3}{4}x)^{-4} \sqrt{3+x}}{\sqrt{(3-x)^3}}$. $y = \frac{(1+\frac{3}{4}x)^{-4} \sqr

Vedclass · Accepted Answer

$\frac{1}{3}-\frac{7 x}{9}$

Vedclass · Answer

(A) Given that $x^n \approx 0$ for $n \ge 2$. Let $y = \frac{(1+\frac{3}{4}x)^{-4} \sqrt{3+x}}{\sqrt{(3-x)^3}}$. $y = \frac{(1+\frac{3}{4}x)^{-4} \sqrt{3}(1+\frac{x}{3})^{1/2}}{3^{3/2}(1-\frac{x}{3})^{3/2}}$. $y = \frac{1}{3} (1+\frac{3}{4}x)^{-4} (1+\frac{x}{3})^{1/2} (1-\frac{x}{3})^{-3/2}$. Using the binomial expansion $(1+z)^n \approx 1+nz$ for small $z$: $y \approx \frac{1}{3} (1 - 4 \cdot \frac{3}{4}x) (1 + \frac{1}{2} \cdot \frac{x}{3}) (1 - (-\frac{3}{2}) \cdot \frac{x}{3})$. $y \approx \frac{1}{3} (1 - 3x) (1 + \frac{x}{6}) (1 + \frac{x}{2})$. $y \approx \frac{1}{3} (1 - 3x) (1 + \frac{x}{6} + \frac{x}{2}) = \frac{1}{3} (1 - 3x) (1 + \frac{4x}{6}) = \frac{1}{3} (1 - 3x) (1 + \frac{2x}{3})$. $y \approx \frac{1}{3} (1 + \frac{2x}{3} - 3x - 2x^2) \approx \frac{1}{3} (1 - \frac{7x}{3})$. $y \approx \frac{1}{3} - \frac{7x}{9}$.

When $x$ is so small that its square and its higher powers may be neglected,then the value of $\frac{\left(1+\frac{3}{4} x\right)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}$ is approximately equal to

Similar Questions

The sum of coefficients in $(1 + x - 3x^2)^{2134}$ is

If $f(x) = 1 + nx + \frac{n(n-1)}{2} x^{2} + \frac{n(n-1)(n-2)}{6} x^{3} + \ldots + x^{n}$,then $f^{\prime \prime}(1)$ is equal to

The coefficient of $x^8$ in the expansion of $(x-1)(x-2)(x-3)\ldots(x-10)$ is:

If $\{ p \}$ denotes the fractional part of the number $p$,then $\left\{\frac{3^{200}}{8}\right\}$ is equal to

If $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i$ for all $x$ in $\mathbb{R},$ then $a_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module