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

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

Question

(B) We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. Comparing this with $(x^2+ax+1)(x^2-ax+1) = x^4+(2-a^2)x^2+1$,we get $2-a^2=1$,which implies $a^2=1

Vedclass · Accepted Answer

$2a$

Vedclass · Answer

(B) We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. Comparing this with $(x^2+ax+1)(x^2-ax+1) = x^4+(2-a^2)x^2+1$,we get $2-a^2=1$,which implies $a^2=1$. Thus,$a=1$.Given $\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+ax+1} + \frac{Cx+D}{x^2-ax+1}$.Multiplying both sides by $(x^2+ax+1)(x^2-ax+1)$,we get $1 = (Ax+B)(x^2-ax+1) + (Cx+D)(x^2+ax+1)$.Expanding the $RHS$: $1 = (A+C)x^3 + (-aA+B+aC+D)x^2 + (A-aB+C+aD)x + (B+D)$.Comparing coefficients:$1$) $A+C=0 \Rightarrow C=-A$$2$) $B+D=1$$3$) $-aA+B+aC+D = 0 \Rightarrow -aA+B-aA+D = 0 \Rightarrow -2aA + (B+D) = 0 \Rightarrow -2aA+1=0 \Rightarrow A = \frac{1}{2a}$. Thus $C = -\frac{1}{2a}$.$4$) $A-aB+C+aD = 0 \Rightarrow (A+C) - a(B-D) = 0 \Rightarrow 0 - a(B-D) = 0 \Rightarrow B=D$.Since $B+D=1$ and $B=D$,we have $B=D=\frac{1}{2}$.Now,$A+B-C+D = \frac{1}{2a} + \frac{1}{2} - (-\frac{1}{2a}) + \frac{1}{2} = \frac{1}{a} + 1$.Since $a=1$,$A+B-C+D = 1+1 = 2 = 2a$.

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

Similar Questions

If $\frac{e^x + 2}{(e^x - 1)(2e^x - 3)} = -\frac{3}{e^x - 1} + \frac{B}{2e^x - 3}$,then $B = $

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

If $F_1$ and $F_2$ are irreducible factors of $x^4+x^2+1$ with real coefficients and $\frac{x^3-2x^2+3x-4}{x^4+x^2+1}=\frac{Ax+B}{F_1}+\frac{Cx+D}{F_2}$,then $A+B+C+D=$

If $\frac{2x^4-x^3+3x^2-x+4}{x^2-3x+2} = f(x) + \frac{A}{x-1} + \frac{B}{x-2}$,then:

$\frac{x^2 + 1}{(2x - 1)(x^2 - 1)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module