If frac{9}{(x - 1)(x + 2)^2} = frac{A}{x - 1} | vedclass

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

Question

(C) Given the partial fraction decomposition: $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$.Multiplying both

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the partial fraction decomposition: $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$.Multiplying both sides by $(x - 1)(x + 2)^2$,we get: $9 = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1)$.To find $A$,put $x = 1$: $9 = A(1 + 2)^2 \implies 9 = 9A \implies A = 1$.To find $C$,put $x = -2$: $9 = C(-2 - 1) \implies 9 = -3C \implies C = -3$.To find $B$,compare the coefficients of $x^2$: $0 = A + B \implies B = -A = -1$.Now,calculate $A - B - C = 1 - (-1) - (-3) = 1 + 1 + 3 = 5$.

If $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$,then $A - B - C = $

Similar Questions

In a $\triangle ABC$,$a^2 \sin 2C + c^2 \sin 2A$ is equal to (in $\Delta$)

$A$ light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $\frac{g}{8}$,then the ratio of the masses is

$A$ parachutist after bailing out falls $50\, m$ without friction. When the parachute opens,it decelerates at $2\, m/s^2$. He reaches the ground with a speed of $3\, m/s$. At what height did he bail out?

The number of four-letter words that can be formed using the letters of the word $BARRACK$ is:

If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$,then which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module