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

In the expansion of the expression $1 + (1 + x) + (1 + x)^2 + \dots + (1 + x)^n$,the coefficient of $x^k$ $(0 \le k \le n)$ is

The expression $\frac{1}{{\sqrt {4x + 1} }}\left[ {{{\left[ {\frac{{1 + \sqrt {4x + 1} }}{2}} \right]}^7} - {{\left[ {\frac{{1 - \sqrt {4x + 1} }}{2}} \right]}^7}} \right]$ is a polynomial in $x$ of degree

Expand the expression $\left(x+\frac{1}{x}\right)^{6}$.

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

If in the expansion of $(1 + x)^m(1 - x)^n$,the coefficients of $x$ and $x^2$ are $3$ and $-6$ respectively,then $m$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module