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EAMCET Engineering 2005 Solved Paper

Section: Mathematics

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Question : 91 of 160

Marks: +1, -0

If

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

=A+\;\frac{B}{2x-1}+\;\frac{C}{x+2}+\;\frac{D}{x-3}

A

$\;\frac{1}{2}$
$-\;\frac{1}{50}$
$-\;\frac{8}{25}$
$\;\frac{27}{25}$

Solution:

Given that

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

A+\;\frac{B}{2x-1}+\;\frac{C}{x+2}+\;\frac{D}{x-3}

Let

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

=\;\frac{x^3}{(2x-1)(x^2-x-6)}

\;=\;\frac{x^3}{2x^3-3x^2-11x+6}

Here we see that the power of

x

will be same in

\text{Nr}

and Dr.

\therefore

First we divide the numerator by denominator

2x^3-3x^2-11x+6 \sqrt{x^3}

x^3-\;\frac{3}{2}x^2-\;\frac{11}{2}x+3

\;\frac{-\frac{1}{2}}{\frac{3}{2}x^2+\;\frac{11}{2}x-3}

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

\Rightarrow \;\; \;\frac{1}{2}+\;\frac{\;\frac{3}{2}x^2+\frac{11}{2}x-3}{(2x-1)(x+2)(x-3)}

A=\;\frac{1}{2}

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