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AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

\frac{x^{2}+1}{x^{4}+4} = \frac{A x+B}{x^{2}-2 x+2} + \frac{C x+D}{x^{2}+2 x+2} \Rightarrow 3 A+2 B+3 C =

$-D$
$D$
$2D$
$-2D$

Solution:

\frac{x^2+1}{x^4+4} = \frac{Ax+B}{x^4-2x+2} + \frac{Cx+D}{x^2+2x+2}

\Rightarrow \frac{x^2+1}{x^4+4}

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

\Rightarrow x^2+1 = (A+C)x^2 + (2A+B-2C+D)x^2

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

On Comparing both sides:

A+C=0 \Rightarrow A=-C, 2B+2D=1

and

2A+B-2C+D=1

\Rightarrow 2(A-C)+B+D=1

\Rightarrow 4A + \frac{1}{2} = 1

\Rightarrow 4A = 1 - \frac{1}{2} = \frac{1}{2}

\Rightarrow A = \frac{1}{8}

and

C = -\frac{1}{8}

Now,

2A+2B+2C-2D=0

\Rightarrow 2(A+C) + 2(B-D) = 0

\Rightarrow B = D

and

2B+2D=1

\Rightarrow 4B=1 \Rightarrow B=D = \frac{1}{4}

Now,

3A+2B+3C=2B

\because 3A+3C=0

=2D

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