begin{aligned} & frac{x^2+x+1}{(x-1)(x-2)(x-3 | vedclass

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

Question

(D) Given,$\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ Multiplying both sides by $(x-1)(x-2)(x-3)$,we get: $x^2+x+1 = A

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given,$\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ Multiplying both sides by $(x-1)(x-2)(x-3)$,we get: $x^2+x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$ To find $A$,put $x=1$: $1^2+1+1 = A(1-2)(1-3) \Rightarrow 3 = A(-1)(-2) \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}$ To find $C$,put $x=3$: $3^2+3+1 = C(3-1)(3-2) \Rightarrow 13 = C(2)(1) \Rightarrow 13 = 2C \Rightarrow C = \frac{13}{2}$ Therefore,$A+C = \frac{3}{2} + \frac{13}{2} = \frac{16}{2} = 8$

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C= \end{aligned}$

Similar Questions

The area (in sq unit) of the region bounded by the curves $2x = y^2 - 1$ and $x = 0$ is

If the volume of a sphere increases at the rate of $2 \pi \text{ cm}^3/\text{s}$, then the rate of increase of its radius (in $\text{cm/s}$), when the volume is $288 \pi \text{ cm}^3$, is

The mean of a set of observations is $\bar{x}$. If each observation is divided by $\alpha$ $(\alpha \neq 0)$ and then increased by $10$,what is the mean of the new set?

If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors having magnitudes $1, 2, 3$ respectively,then $[\vec{a}+\vec{b}+\vec{c} \quad \vec{b}-\vec{a} \quad \vec{c}]=$

Which one of the following pairs of organisms are exotic species introduced in India?

Vedclass Test Series

Exam Paper Generator

Online Exam Module