AP EAPCET 08-Jul-2022 Shift 1 Paper

Given

x

(x−1)(x2+1)2

=

1

4

[

1

x−1

−

x+1

x2+1

]+y
Applying partial fraction in the expression

x

(x−1)(x2+1)2

=

A

x−1

+

Bx+C

x2+1

+

Dx+B

(x2+1)2

⇒ x=A(x2+1)2+(Bx+C)(x2+1)(x−1)+(Dx+E)(x−1)
⇒ x=Ax4+A2Ax2+(Bx+C)(x3+x−1)+Dx2+Ex−Dx−E
⇒ x=Ax4+A2Ax2+Bx4+Bx2−Bx3−Bx
+Cx3+Cx−Cx2−C+Dx2+Ex−DxE
Comparing the coefficients on both sides,
A+B=0⇒A=−B‌or‌B=−A .....(i)
⇒ −B+C=0⇒B=C=−A ......(ii)
⇒ 2A+B−C+D=0⇒2A+D=0 .....(iii)
⇒ −B+C+E−D=1⇒E=D+1 ....(iv)
⇒ A−C−E=0⇒2A−E=0 ........(v)
⇒ −D−(D+1)=0 [From eqs. (iii), (iv) and (v)]
−2D−1=0
D=−

1

2

E=D+1⇒E=−

1

2

+1=

1

2

⇒ 2A=

1

2

[From eq.(v)]
⇒ A=

1

4

⇒B=C=−

1

4

∴

x

(x−1)(x2+1)2

=

1

4(x−1)

−

1

4

(x+1)

(x2+1)

−

1

2

[

x−1

(x2+1)2

]
Compare it with

x

(x1)(x2+1)2

=

1

4

[

1

x−1

−

x+1

x2+1

]+y
⇒ y=

1

2

[

1−x

(x2+1)2

]