If frac{x^2-7 x+2}{x^4+3 x^2+4}=frac{A x+B}{x | vedclass

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

Question

(B) Given $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$  ...$(i)$Factorize the denominator: $x^4+3 x^2+4 = (x^2+2)^2

Vedclass · Accepted Answer

$2 a+b$

Vedclass · Answer

(B) Given $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$  ...$(i)$Factorize the denominator: $x^4+3 x^2+4 = (x^2+2)^2 - x^2 = (x^2+x+2)(x^2-x+2)$.Thus,$\frac{x^2-7 x+2}{(x^2+x+2)(x^2-x+2)} = \frac{Px+Q}{x^2+x+2} + \frac{Rx+S}{x^2-x+2}$.Equating numerators: $x^2-7x+2 = (Px+Q)(x^2-x+2) + (Rx+S)(x^2+x+2)$.Expanding the right side: $x^2-7x+2 = x^3(P+R) + x^2(-P+Q+R+S) + x(2P-Q+2R+S) + (2Q+2S)$.Comparing coefficients:$P+R=0$$-P+Q+R+S=1$$2P-Q+2R+S=-7$$2Q+2S=2 \Rightarrow Q+S=1$Solving these,we get $P=0, Q=4, R=0, S=-3$.So,$\frac{x^2-7x+2}{x^4+3x^2+4} = \frac{4}{x^2+x+2} + \frac{-3}{x^2-x+2}$.Comparing with $(i)$,we have $a=1, b=-1$ (since $a>b$),$A=0, B=4, C=0, D=-3$.Then $B+D = 4-3 = 1$.Checking options: $2a+b = 2(1) + (-1) = 1$.Therefore,$B+D = 2a+b$.

If $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$ and $a>b$ then $B+D=$

Similar Questions

The partial fraction of $\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{1 + 5x + 6x^2} = $

If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$,then $a-b+c=$

For $a > 0$,let $\frac{1}{a(a+1)(a+2) \ldots(a+20)}=\sum_{k=0}^{20} \frac{A_{k}}{a+k}$. Then the value of $100\left(\frac{A_{14}+A_{15}}{A_{13}}\right)^{2}$ is equal to $....$

If $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{2x-1}+\frac{B}{3x+1}$,then $16A+9B$ is equal to

Resolve $\frac{2x}{x^4 + x^2 + 1}$ into partial fractions.

Vedclass Test Series

Exam Paper Generator

Online Exam Module