If frac{1}{x^4+1}=frac{A x+B}{x^2+sqrt{2} x+1 | vedclass

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

Question

(A) Given that $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$.Multiplying both sides by $(x^4+1)$,we get $(A x+B)(x^2

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(A) Given that $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$.Multiplying both sides by $(x^4+1)$,we get $(A x+B)(x^2-\sqrt{2} x+1)+(C x+D)(x^2+\sqrt{2} x+1)=1$.Comparing coefficients:For $x^3$: $A+C=0 \Rightarrow C=-A$.For $x^0$ (constant term): $B+D=1$.For $x^2$: $B-\sqrt{2} A+D+\sqrt{2} C=0 \Rightarrow (B+D)-\sqrt{2}(A-C)=0$.Since $B+D=1$ and $C=-A$,we have $1-\sqrt{2}(2A)=0 \Rightarrow A=\frac{1}{2\sqrt{2}}$ and $C=-\frac{1}{2\sqrt{2}}$.For $x^1$: $A-\sqrt{2} B+C+\sqrt{2} D=0 \Rightarrow (A+C)+\sqrt{2}(D-B)=0$.Since $A+C=0$,we have $\sqrt{2}(D-B)=0 \Rightarrow D=B$.Since $B+D=1$,we get $2B=1 \Rightarrow B=\frac{1}{2}$ and $D=\frac{1}{2}$.Now,$B D-A C = (\frac{1}{2})(\frac{1}{2}) - (\frac{1}{2\sqrt{2}})(-\frac{1}{2\sqrt{2}}) = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$.

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

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 $\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = f(x) + \frac{A}{x - 2} + \frac{B}{x - 3}$,then $f(x) = $

If $\frac{x+1}{\left(x^2+1\right)(x-1)^2}=\frac{A x+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$, then $A+B+C+D=$

Let $x$ be a real number and $-2 < x < 2$. When $\frac{x+1}{(x+3)(x-2)}$ is expanded in powers of $x$,then the coefficient of $x^3$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module