The coefficient of x^3 in the expansion of fr | vedclass

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

Question

(C) Let $f(x) = \frac{1-2x}{(2x+1)(2-x)}$. Using partial fractions,we write: $\frac{1-2x}{(2x+1)(2-x)} = \frac{A}{2x+1} + \frac{B}{2-x}$.Solving for $

Vedclass · Accepted Answer

$-\frac{103}{16}$

Vedclass · Answer

(C) Let $f(x) = \frac{1-2x}{(2x+1)(2-x)}$. Using partial fractions,we write: $\frac{1-2x}{(2x+1)(2-x)} = \frac{A}{2x+1} + \frac{B}{2-x}$.Solving for $A$ and $B$: $1-2x = A(2-x) + B(2x+1)$.For $x = 2$,$1-4 = B(4+1) \implies -3 = 5B \implies B = -\frac{3}{5}$.For $x = -\frac{1}{2}$,$1+1 = A(2+\frac{1}{2}) \implies 2 = A(\frac{5}{2}) \implies A = \frac{4}{5}$.So,$f(x) = \frac{4}{5(2x+1)} - \frac{3}{5(2-x)} = \frac{4}{5}(1+2x)^{-1} - \frac{3}{10}(1-\frac{x}{2})^{-1}$.Using the binomial expansion $(1+u)^{-1} = 1-u+u^2-u^3+\dots$:$f(x) = \frac{4}{5}(1 - 2x + 4x^2 - 8x^3) - \frac{3}{10}(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8})$.The coefficient of $x^3$ is $\frac{4}{5}(-8) - \frac{3}{10}(\frac{1}{8}) = -\frac{32}{5} - \frac{3}{80} = -\frac{512+3}{80} = -\frac{515}{80} = -\frac{103}{16}$.

The coefficient of $x^3$ in the expansion of $\frac{1-2x}{(2x+1)(2-x)}$ is

Similar Questions

If $\frac{x-4}{x^2-5x+6}$ can be expanded in the ascending powers of $x$,then the coefficient of $x^3$ is

If $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$,then $A+B = $

The partial fraction decomposition of $\frac{9x-7}{(x+3)(x^2+1)}$ is

$\begin{aligned} & \text{If } \frac{x^3}{(2x-1)(x-1)^2} = A + \frac{B}{2x-1} + \frac{C}{x-1} \\ & + \frac{D}{(x-1)^2}, \text{ then } 2A - 3B + 4C + 5D = \end{aligned}$

If $\frac{x^{4}}{(x - 1)(x - 2)} = f(x) + \frac{A}{x - 1} + \frac{B}{x - 2}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module