When |x|

Question

(B) Given $f(x) = \frac{x}{(x-2)(x-3)}$.Using partial fractions: $\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$: $x =

Vedclass · Accepted Answer

$\frac{5}{36}$

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(B) Given $f(x) = \frac{x}{(x-2)(x-3)}$.Using partial fractions: $\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$: $x = A(x-3) + B(x-2)$.For $x=2$, $2 = A(-1) \implies A = -2$.For $x=3$, $3 = B(1) \implies B = 3$.Thus, $f(x) = \frac{3}{x-3} - \frac{2}{x-2} = \frac{3}{-3(1 - \frac{x}{3})} - \frac{2}{-2(1 - \frac{x}{2})} = (1 - \frac{x}{2})^{-1} - (1 - \frac{x}{3})^{-1}$.Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + \dots$ for $|z| $(1 - \frac{x}{2})^{-1} = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$$(1 - \frac{x}{3})^{-1} = 1 + \frac{x}{3} + \frac{x^2}{9} + \dots$Subtracting these: $f(x) = (1 - 1) + (\frac{1}{2} - \frac{1}{3})x + (\frac{1}{4} - \frac{1}{9})x^2 + \dots$The coefficient of $x^2$ is $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.

When $|x| < 2$, the coefficient of $x^2$ in the power series expansion of $\frac{x}{(x-2)(x-3)}$ is

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