If frac{x-4}{x^2-5x+6} can be expanded in the | vedclass

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

Question

(A) First,we decompose the expression into partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$,we get

Vedclass · Accepted Answer

$\frac{-73}{648}$

Vedclass · Answer

(A) First,we decompose the expression into partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$,we get $\frac{2}{x-2} - \frac{1}{x-3}$.Rewrite the expression as: $2(x-2)^{-1} - (x-3)^{-1} = 2(-2)^{-1}(1-\frac{x}{2})^{-1} - (-3)^{-1}(1-\frac{x}{3})^{-1}$.$= -\left[1 + \frac{x}{2} + (\frac{x}{2})^2 + (\frac{x}{2})^3 + \dots\right] + \frac{1}{3}\left[1 + \frac{x}{3} + (\frac{x}{3})^2 + (\frac{x}{3})^3 + \dots\right]$.The coefficient of $x^3$ is $-\left(\frac{1}{2}\right)^3 + \frac{1}{3}\left(\frac{1}{3}\right)^3$.$= -\frac{1}{8} + \frac{1}{81} = \frac{-81 + 8}{648} = -\frac{73}{648}$.

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

Similar Questions

If $K_{sp}$ of $CaF_2$ at $25\,^{\circ}C$ is $1.7 \times 10^{-10}$,the combination amongst the following which gives a precipitate of $CaF_2$ is

For positive numbers $x, y, \text{ and } z$,the numerical value of the determinant $\left| \begin{array}{ccc} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{array} \right|$ is

In which of the following compounds does the transition metal have a zero oxidation state?

Zero growth means:

The standard e.m.f. of a cell,involving one electron change is found to be $0.591 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction is ($F = 96,500 \ C \ mol^{-1}$; $R = 8.314 \ JK^{-1} \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module