If frac{1}{x^4+x^2+1}=frac{Ax+B}{x^2+x+1}+fra | vedclass

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

Question

(B) Given,$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.Equating the numerators:$1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given,$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.Equating the numerators:$1 = (Ax+B)(x^2-x+1) + (Cx+D)(x^2+x+1)$$1 = Ax^3 - Ax^2 + Ax + Bx^2 - Bx + B + Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D$$1 = (A+C)x^3 + (B-A+C+D)x^2 + (A-B+C+D)x + (B+D)$Comparing coefficients on both sides:$A+C = 0 \Rightarrow C = -A$$B-A+C+D = 0 \Rightarrow B+D = A-C = 2A$$A-B+C+D = 0$ $\Rightarrow A+C = B-D$ $\Rightarrow 0 = B-D$ $\Rightarrow B=D$$B+D = 1$ $\Rightarrow 2B = 1$ $\Rightarrow B = \frac{1}{2}, D = \frac{1}{2}$Since $B+D = 2A$,we have $1 = 2A \Rightarrow A = \frac{1}{2}$.Since $C = -A$,we have $C = -\frac{1}{2}$.Thus,$A+B+C+D = \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} = 1$.Therefore,$\cos^{-1}(A+B+C+D) = \cos^{-1}(1) = 0$.

If $\frac{1}{x^4+x^2+1}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}$,then $\cos^{-1}(A+B+C+D)=$

Similar Questions

If $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, then $A+B+C+D=$

If $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$,then $51apB=$ (in $bqA$)

If $\frac{x^2+5x+7}{(x-3)^3}=\frac{A}{(x-3)}+\frac{B}{(x-3)^2}+\frac{C}{(x-3)^3}$,then $9A-3B+C=$

$\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = $

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