TS EAMCET 10 May 2024 Shift 2 Solved Paper

We know that
‌x4+x2+1=(x2+x+1)(x2−x+1)
‌{∴a=1}
‌‌

1

x4+x2+1

=‌

Ax+B

x2+x+1

+‌

Cx+D

x2−x+1

‌=‌

1(x2+x+1)(x2−x+1)

(x2+x+1)(x2−x+1)

‌1=(Ax+B)(x2−x+1)+(Cx+D)(x2+x+1)
‌1=Ax3−Ax2+Ax+Bx2−Bx+B
‌‌‌+Cx3+Cx2+Cx+Dx2+Dx+D
On comparing, we get
Coefficient of x3=0A+C=0
Coefficient of x2=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=‌

1

2

,C=−‌

1

2

So, A+B−C+D=1=2a