If frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + fr | vedclass

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

Question

(D) Given the equation: $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$.Multiply both sides by $(2x -

Vedclass · Accepted Answer

$6q - 3r + 2s = 3$

Vedclass · Answer

(D) Given the equation: $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$.Multiply both sides by $(2x - 1)(x + 2)(x - 3)$:$x^3 = p(2x - 1)(x + 2)(x - 3) + q(x + 2)(x - 3) + r(2x - 1)(x - 3) + s(2x - 1)(x + 2)$.Expanding the term with $p$: $p(2x - 1)(x^2 - x - 6) = p(2x^3 - 2x^2 - 12x - x^2 + x + 6) = p(2x^3 - 3x^2 - 11x + 6)$.Equating the coefficient of $x^3$ on both sides: $1 = 2p \Rightarrow p = \frac{1}{2}$.Now,to find the constant term,set $x = 0$ in the equation:$0 = p(-1)(2)(-3) + q(2)(-3) + r(-1)(-3) + s(-1)(2)$.$0 = 6p - 6q + 3r - 2s$.Substituting $p = \frac{1}{2}$:$0 = 6(\frac{1}{2}) - 6q + 3r - 2s$.$0 = 3 - 6q + 3r - 2s$.$6q - 3r + 2s = 3$.

If $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$,then which of the following is true?

Similar Questions

If $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$,then $k =$

$\begin{aligned} & \frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2} \\ & \Rightarrow 3 A+2 B+3 C=\end{aligned}$

When $|x| < \frac{1}{2}$,the coefficient of $x^4$ in the expansion of $\frac{3x^2-5x+3}{(x-1)(2x+1)(x+3)}$ is

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A + 2B + C = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module