The partial fraction decomposition of 3x+1/(x-1)² (x+2) is

Correct Answer is : −59(1x+2)+591x−1+431(x−1)2-5/9≤ft(1/x+2)+5/91/x-1+4/31/(x-1)²9−5(x+21)+95x−11+34(x−1)21

-5/9≤ft(1/x+2)+4/31/(x-1)²+2/x-1

-5/9≤ft(1/x+2)+5/91/x-1+4/31/(x-1)²

-5/9≤ft(1/x+2)+5/9≤ft(1/x-1)+2/(x-1)²

Correct Answer is : −59(1x+2)+591x−1+431(x−1)2-5/9≤ft(1/x+2)+5/91/x-1+4/31/(x-1)²9−5(x+21)+95x−11+34(x−1)21

Test Index

TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

\frac{3x+1}{(x-1)^2 (x+2)}

$\frac{4}{3}\frac{1}{(x-1)^2}+\frac{5}{9}\frac{1}{x-1}+\frac{5}{9}\frac{1}{x+2}$
$\frac{-5}{9}\left(\frac{1}{x+2}\right)+\frac{4}{3}\frac{1}{(x-1)^2}+\frac{2}{x-1}$
$\frac{-5}{9}\left(\frac{1}{x+2}\right)+\frac{5}{9}\frac{1}{x-1}+\frac{4}{3}\frac{1}{(x-1)^2}$
$\frac{-5}{9}\left(\frac{1}{x+2}\right)+\frac{5}{9}\left(\frac{1}{x-1}\right)+\frac{2}{(x-1)^2}$

Solution:

\;\frac{3x+1}{(x+2)(x-1)^2}=\;\frac{A}{x+2}+\;\frac{B}{(x-1)}+\;\frac{C}{(x-1)^2}

\Rightarrow 3x+1=A(x-1)^2+B(x-1)(x+2)+C(x+2)

\Rightarrow 0\cdot x^{2}+3x+1=(A+B)x^{2}

+(-2A+B+C)x+(A-2B+2C)

On comparing, we get

$A+B=0$	$-2A+B+C=3$	$A-2B+2C=1$
$\Rightarrow A=-B \ldots$ (i)	$\Rightarrow 3B+C=3 \ldots$ (ii)	$\Rightarrow -3B+2C=1 \ldots$ (iii)

Adding Eqs. (ii) and (iii),

3C=4 \Rightarrow C=\frac{4}{3}

From Eq. (ii), we get

\;3B+\;\frac{4}{3}\;\;=3 \Rightarrow 3B=3-\;\frac{4}{3}=\;\frac{5}{3}

\Rightarrow \;\; B\;B

From Eq. (i), we get

A=-B=\;\frac{-5}{9}

Hence,

\;\frac{3x+1}{(x+2(x-1)^2)}=\;\frac{-5}{9}\left(\frac{1}{x+2}\right)+\;\frac{5}{9}\left(\frac{1}{x-1}\right)

+\;\frac{4}{3}\left(\frac{1}{(x-1)^2}\right)

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