The absolute value of the difference of the c | vedclass

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

Question

(A) Given the expression $\frac{2 x^2}{(x^2+1)(x^2+2)}$.Using partial fractions,$\frac{2 x^2}{(x^2+1)(x^2+2)} = \frac{A}{x^2+1} + \frac{B}{x^2+2}$.Sol

Vedclass · Accepted Answer

$\frac{13}{4}$

Vedclass · Answer

(A) Given the expression $\frac{2 x^2}{(x^2+1)(x^2+2)}$.Using partial fractions,$\frac{2 x^2}{(x^2+1)(x^2+2)} = \frac{A}{x^2+1} + \frac{B}{x^2+2}$.Solving for $A$ and $B$,we get $2x^2 = A(x^2+2) + B(x^2+1)$.For $x^2 = -1$,$A = -2$. For $x^2 = -2$,$B = 4$.So,the expression is $-2(1+x^2)^{-1} + 2(1+\frac{x^2}{2})^{-1}$.Expanding using the binomial series $(1+y)^{-1} = 1 - y + y^2 - y^3 + \dots$:$-2(1 - x^2 + x^4 - x^6 + \dots) + 2(1 - \frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{8} + \dots)$.Coefficient of $x^4 = -2(1) + 2(\frac{1}{4}) = -2 + \frac{1}{2} = -\frac{3}{2}$.Coefficient of $x^6 = -2(-1) + 2(-\frac{1}{8}) = 2 - \frac{1}{4} = \frac{7}{4}$.The absolute value of the difference is $|-\frac{3}{2} - \frac{7}{4}| = |-\frac{6}{4} - \frac{7}{4}| = |-\frac{13}{4}| = \frac{13}{4}$.

The absolute value of the difference of the coefficients of $x^4$ and $x^6$ in the expansion of $\frac{2 x^2}{(x^2+1)(x^2+2)}$ is

Similar Questions

If $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$,then $C$ is

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

If $\frac{17x-2}{12x^2-x-20}=\frac{A}{ax+5}+\frac{B}{3x+b}$ then $a \cdot A+b \cdot B=$

The partial fraction of $\frac{x^2}{x^2+3x-4}$ is

If $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$,then $A+B = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module