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TS EAMCET 7 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

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

\frac{A}{x+1}

\frac{B}{(x+1)^2}

\frac{C}{x+3}

, then A + B + C =

$\frac{11}{4}$
$\frac{5}{2}$
- $\frac{5}{2}$
- $\frac{11}{4}$

Solution:

Given:

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

=

\frac{A}{x+1}

+

\frac{B}{(x+1)^2}

+

\frac{C}{x+3}

3x-2=A(x+1)(x+3)+B(x+3)

+C(x+1)^2

Substitute

x=-1

in above equation,

3(-1)-2=B(-1+3)

-3-2=2B

B=\frac{-5}{2}

Substitute

x=-3

in above equation,

3(-3)-2=C(-3+1)^2

-9-2=4C

C=\frac{-11}{4}

Substitute

x=0

in above equation,

3\times0-2=A(1)(3)+B(3)

+C(1)^2

-2=3A+3B+C

-2=3A+3\left(\frac{-5}{2}\right)-\frac{11}{4}

3A=\frac{15}{2}+\frac{11}{4}-2

A=\frac{11}{4}

The sum of

A, B

, and

C

A+B+C=\frac{11}{4}-\frac{5}{2}-\frac{11}{4}

=-\frac{5}{2}

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