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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

\frac{1}{(3-5x)(2+3x)}

= \frac{A}{3-5x} + \frac{B}{2+3x}

A+B=

$\frac{7}{19}$
$\frac{8}{19}$
$\frac{9}{19}$
$\frac{10}{19}$

Solution:

Given,

\frac{1}{(3-5x)(2+3x)}

= \frac{A}{3-5x} + \frac{B}{2+3x}

\Rightarrow \frac{1}{(3-5x)(2+3x)}

= \frac{2A+3Ax+3B-5Bx}{(3-5x)(2+3x)}

\Rightarrow 1 = x(3A-5B) + (2A+3B)

On comparing the coefficients of

x

and constant.

3A - 5B = 0

...(i)

and

2A + 3B = 1

...(ii)

Eq. (i)

\times 3 +

Eq. (ii)

\times 5

, we get

19A = 5

A = \frac{5}{19}, B = \left(\frac{1-2A}{3}\right)

\Rightarrow B = \frac{3}{19}

\therefore A+B = \frac{5}{19} + \frac{3}{19} = \frac{8}{19}

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