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TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

If

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

A+2B+C+2D=

0
1
$-1$
2

Solution:

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

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

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

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

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

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

On comparing the corresponding coefficients

A+C=0

......(i)

A+B+D=1

......(ii)

A+B+C=-1

......(iii)

B+D=1

......(iv)

From Eqs. (i)and (iii), we have

B=-1

From Eq. (iv), we have

-1+D=1

\Rightarrow D=2

From Eq. (ii), we have

A+(-1)+2=1

A=0

From Eq. (i), we have

A+C=0

0+C=0

C=0

\therefore A+2B+C+2D

=0+2(-1)+0+2(2)

=-2+4=2

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