Let x be a real number and -2

Question

(A) Using partial fractions,we write: $\frac{x+1}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$.Solving for $A$ and $B$,we get $A = \frac{2}{5}$ and $B

Vedclass · Accepted Answer

$-\frac{55}{1296}$

Vedclass · Answer

(A) Using partial fractions,we write: $\frac{x+1}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$.Solving for $A$ and $B$,we get $A = \frac{2}{5}$ and $B = \frac{3}{5}$.Thus,$\frac{x+1}{(x+3)(x-2)} = \frac{2}{5(x+3)} + \frac{3}{5(x-2)} = \frac{2}{15(1 + x/3)} - \frac{3}{10(1 - x/2)}$.Using the binomial expansion $(1+u)^{-1} = 1 - u + u^2 - u^3 + \dots$ and $(1-u)^{-1} = 1 + u + u^2 + u^3 + \dots$,we have:$= \frac{2}{15} \left(1 - \frac{x}{3} + \frac{x^2}{9} - \frac{x^3}{27} + \dots ight) - \frac{3}{10} \left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots ight)$.The coefficient of $x^3$ is $\frac{2}{15} 	imes (-\frac{1}{27}) - \frac{3}{10} 	imes \frac{1}{8} = -\frac{2}{405} - \frac{3}{80} = -\frac{32 + 243}{6480} = -\frac{275}{6480} = -\frac{55}{1296}$.

Let $x$ be a real number and $-2 < x < 2$. When $\frac{x+1}{(x+3)(x-2)}$ is expanded in powers of $x$,then the coefficient of $x^3$ is

Similar Questions

If $\frac{e^x + 2}{(e^x - 1)(2e^x - 3)} = -\frac{3}{e^x - 1} + \frac{B}{2e^x - 3}$,then $B = $

If $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A + \frac{B}{2 x-1} + \frac{C}{x+2} + \frac{D}{x-3}$,then $A$ is equal to

If $\frac{3x^2+x+1}{(x-1)^4} = \frac{a}{(x-1)} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4}$,then $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is equal to

If $\frac{\sin^2 x + 1}{2\sin^2 x - 5\sin x + 3} = \frac{A}{2\sin x - 3} + \frac{B}{\sin x - 1} + C$,then:

The partial fraction of $\frac{x^2 - 5}{x^2 - 3x + 2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module