If frac{3x^2+x+1}{(x-1)^4} = frac{a}{(x-1)} + | vedclass

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

Question

(B) Let $x-1 = t$,so $x = t+1$. Substituting this into the numerator: $3(t+1)^2 + (t+1) + 1 = 3(t^2+2t+1) + t + 2 = 3t^2 + 7t + 5$.The expression beco

Vedclass · Accepted Answer

$\begin{bmatrix} 0 & 3 \ 7 & 5 \end{bmatrix}$

Vedclass · Answer

(B) Let $x-1 = t$,so $x = t+1$. Substituting this into the numerator: $3(t+1)^2 + (t+1) + 1 = 3(t^2+2t+1) + t + 2 = 3t^2 + 7t + 5$.The expression becomes $\frac{3t^2+7t+5}{t^4} = \frac{3}{t^2} + \frac{7}{t^3} + \frac{5}{t^4}$.Comparing this with $\frac{a}{t} + \frac{b}{t^2} + \frac{c}{t^3} + \frac{d}{t^4}$,we get $a=0, b=3, c=7, d=5$.Thus,$\begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} 0 & 3 \ 7 & 5 \end{bmatrix}$.

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 $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is equal to

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 $\frac{ax + b}{(3x + 4)^2} = \frac{1}{3x + 4} - \frac{3}{(3x + 4)^2}$,then:

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

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

The coefficient of $x^5$ in the expansion of $\frac{x^2 + 1}{(x^2 + 4)(x - 2)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module