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

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

Question

(D) We have $\frac{x^2+x+1}{x^2+2x+1} = \frac{(x^2+2x+1) - x}{x^2+2x+1} = 1 - \frac{x}{(x+1)^2}$. Now,express $\frac{x}{(x+1)^2}$ in partial fractions

Vedclass · Accepted Answer

$2C$

Vedclass · Answer

(D) We have $\frac{x^2+x+1}{x^2+2x+1} = \frac{(x^2+2x+1) - x}{x^2+2x+1} = 1 - \frac{x}{(x+1)^2}$. Now,express $\frac{x}{(x+1)^2}$ in partial fractions: $\frac{x}{(x+1)^2} = \frac{A'}{x+1} + \frac{B'}{(x+1)^2} = \frac{A'(x+1) + B'}{(x+1)^2}$. Comparing numerators,$x = A'x + (A'+B')$. Thus,$A' = 1$ and $A'+B' = 0$,which gives $B' = -1$. So,$\frac{x^2+x+1}{x^2+2x+1} = 1 - (\frac{1}{x+1} - \frac{1}{(x+1)^2}) = 1 - \frac{1}{x+1} + \frac{1}{(x+1)^2}$. Comparing this with $A + \frac{B}{x+1} + \frac{C}{(x+1)^2}$,we get $A=1$,$B=-1$,and $C=1$. Therefore,$A-B = 1 - (-1) = 2$. Since $C=1$,$2 = 2C$. Hence,$A-B = 2C$.

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

Similar Questions

Which among the following reactions will take place spontaneously?

If the circle $x^2 + y^2 + 6x - 2y + k = 0$ bisects the circumference of the circle $x^2 + y^2 + 2x - 6y - 15 = 0$,then $k =$

Calculate the amount of heat (in calories) required to convert $5\,g$ of ice at $0\,^{\circ}C$ to steam at $100\,^{\circ}C$.

$A$ circular disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is

$A$ plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module