If frac{3x + a}{x^2 - 3x + 2} = frac{A}{x - 2 | vedclass

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

Question

(D) Given the equation: $\frac{3x + a}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{10}{x - 1}$.Multiplying both sides by $(x - 2)(x - 1)$,we get:$3x + a

Vedclass · Accepted Answer

Both $A$ and $B$

Vedclass · Answer

(D) Given the equation: $\frac{3x + a}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{10}{x - 1}$.Multiplying both sides by $(x - 2)(x - 1)$,we get:$3x + a = A(x - 1) - 10(x - 2)$.Expanding the right side:$3x + a = Ax - A - 10x + 20$.$3x + a = (A - 10)x + (20 - A)$.Comparing the coefficients of $x$ and the constant terms on both sides:$3 = A - 10 \implies A = 13$.$a = 20 - A \implies a = 20 - 13 = 7$.Thus,both $a = 7$ and $A = 13$ are correct.

If $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$,then:

Similar Questions

Resolve the following expression into partial fractions: $\frac{x + 1}{(x - 1)(x - 2)(x - 3)}$

The coefficient of $x^n$ in the expansion of $\frac{5x + 6}{(2 + x)(1 - x)}$ in ascending powers of $x$ 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:

The partial fraction of $\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{1 + 5x + 6x^2} = $

$\frac{4x^2+5}{(x-2)^4} = \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4}$,then $\sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module