For |x|

Question

(B) Let $f(x) = \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}$.We can rewrite the expression as:$f(x) = (1 - 8x + 16x^2) (1 - 2x^2)^{1/2} \cdot 4^

Vedclass · Accepted Answer

$-\frac{61}{64}$

Vedclass · Answer

(B) Let $f(x) = \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}$.We can rewrite the expression as:$f(x) = (1 - 8x + 16x^2) (1 - 2x^2)^{1/2} \cdot 4^{-3/2} (1 - \frac{x}{4})^{-3/2}$.Since $4^{-3/2} = \frac{1}{8}$, we have:$f(x) = \frac{1}{8} (1 - 8x + 16x^2) (1 - 2x^2)^{1/2} (1 - \frac{x}{4})^{-3/2}$.Using the binomial expansion $(1 + u)^n = 1 + nu + \dots$, we get:$(1 - 2x^2)^{1/2} = 1 - x^2 + \dots$$(1 - \frac{x}{4})^{-3/2} = 1 + (-\frac{3}{2})(-\frac{x}{4}) + \dots = 1 + \frac{3x}{8} + \dots$Substituting these back:$f(x) = \frac{1}{8} (1 - 8x + 16x^2) (1 + \dots) (1 + \frac{3x}{8} + \dots)$.To find the coefficient of $x$, we multiply the terms:$f(x) = \frac{1}{8} [1 \cdot (\frac{3x}{8}) - 8x \cdot 1] + \dots$$f(x) = \frac{1}{8} (\frac{3}{8} - 8) x + \dots = \frac{1}{8} (\frac{3 - 64}{8}) x = -\frac{61}{64} x$.Thus, the coefficient of $x$ is $-\frac{61}{64}$.

For $|x| < \frac{1}{\sqrt{2}}$, the coefficient of $x$ in the expansion of $\frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}$ is

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