When |x|

Question

(B) We have the expression $f(x) = \frac{x}{(x-2)(x-3)}$.Using partial fractions,$\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A

Vedclass · Accepted Answer

$\frac{5}{36}$

Vedclass · Answer

(B) We have the expression $f(x) = \frac{x}{(x-2)(x-3)}$.Using partial fractions,$\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$,we get $x = A(x-3) + B(x-2)$.For $x=2$,$2 = A(-1) \implies A = -2$.For $x=3$,$3 = B(1) \implies B = 3$.So,$f(x) = \frac{3}{x-3} - \frac{2}{x-2} = \frac{3}{-3(1 - \frac{x}{3})} - \frac{2}{-2(1 - \frac{x}{2})} = (1 - \frac{x}{2})^{-1} - (1 - \frac{x}{3})^{-1}$.Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:$(1 - \frac{x}{2})^{-1} = 1 + (\frac{x}{2}) + (\frac{x}{2})^2 + \dots = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$$(1 - \frac{x}{3})^{-1} = 1 + (\frac{x}{3}) + (\frac{x}{3})^2 + \dots = 1 + \frac{x}{3} + \frac{x^2}{9} + \dots$Subtracting these,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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