The partial fraction of frac{x^2 - 5}{x^2 - 3 | vedclass

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

Question

(A) Given expression is $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.Since the degree of the numerator is equal to the degree of the denominator,we perform l

Vedclass · Accepted Answer

$1 + \frac{4}{x - 1} - \frac{1}{x - 2}$

Vedclass · Answer

(A) Given expression is $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long division:$x^2 - 5 = 1(x^2 - 3x + 2) + (3x - 7)$.So,$f(x) = 1 + \frac{3x - 7}{(x - 1)(x - 2)}$.Now,decompose $\frac{3x - 7}{(x - 1)(x - 2)}$ into partial fractions:$\frac{3x - 7}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$.$3x - 7 = A(x - 2) + B(x - 1)$.For $x = 1$: $3(1) - 7 = A(1 - 2)$ $\Rightarrow -4 = -A$ $\Rightarrow A = 4$.For $x = 2$: $3(2) - 7 = B(2 - 1)$ $\Rightarrow -1 = B$ $\Rightarrow B = -1$.Therefore,$f(x) = 1 + \frac{4}{x - 1} - \frac{1}{x - 2}$.

The partial fraction of $\frac{x^2 - 5}{x^2 - 3x + 2}$ is:

Similar Questions

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

The fraction $\frac{x^2}{(x-a)(x-b)}$ is

If $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \left[ \frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} \right] + y$,then $y =$

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module