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TS EAMCET 10 May 2024 Shift 2 Solved Paper

Section: Mathematics

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Question : 21 of 160

Marks: +1, -0

If

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

A+B-C+D=

[TG EAPCET 10 May 2024 Shift 2]

$a$
$2 a$
$3 a$
$4 a$

Solution:

We know that

\; x^4+x^2+1 = (x^2+x+1)(x^2-x+1)

\; \{ \therefore a=1\}

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

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

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

\; 1 = A x^3 - A x^2 + A x + B x^2 - B x + B

\; \; \; + C x^3 + C x^2 + C x + D x^2 + D x + D

On comparing, we get

Coefficient of

x^3=0 A+C=0

Coefficient of

x^2=0

-A+B+C+D=0

Coefficient of

x=0

\; A-B+C=0

\; B+D=1

From Eqs. (i) and (iv), we get

A-C=1

and from Eqs. (i) and (v) we get,

and from Eqs. (i) and (v) we get,

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

So,

A+B-C+D=1=2 a

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