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EAMCET Engineering 2006 Solved Paper

Section: Mathematics

Question : 91 of 160

Marks: +1, -0

\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}

A+C-B

Solution:

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

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

\text{ On putting } x+1=0

\Rightarrow 3(-1)+2=A[2(-1)^2+3]

\Rightarrow -3+2=A(5)

\Rightarrow A=-\frac{1}{5}

Now on comparing the coefficients of like power of

x

Coefficient of

x^2

0=2A+B

\Rightarrow B=\frac{2}{5}

and coefficient of

x

3=B+C

\Rightarrow \;\; C=3-\frac{2}{5}=\frac{13}{5}

\therefore \;\; A+C-B=-\frac{1}{5}+\frac{13}{5}-\frac{2}{5}

=\frac{13-2-1}{5}=\frac{10}{5}=2

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