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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

\frac{2x+1}{(x-1)^{2}(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{Cx+D}{x^{2}+1}

A+B+C+D=

$1$
$2$
$\frac{3}{4}$
$\frac{1}{2}$

Solution:

If

\frac{2x+1}{(x-1)^{2}(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{Cx+D}{x^{2}+1}

\Rightarrow \;\; 2x+1 = A(x-1)(x^{2}+1) + B(x^{2}+1)

+ (Cx+D)(x-1)^{2}

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

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

On comparing coefficients of

x^{3}, x^{2}, x

and constant terms both sides we get.

A+C=0

. . . (i)

-A+B-2C+D=0

. . . (ii)

A+C-2D=2

. . . (iii)

and

\;\; -A+B+D=1

. . . (iv)

On solving Eqs. (i), (ii), (iii) and (iv), we get

A = -\frac{1}{2}, B = \frac{3}{2}, C = \frac{1}{2}

and

D=-1

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

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