TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The given expression is
=1+‌

A

2x−1

+‌

B

x−1

+‌

C

x+1

‌‌ (By partial fraction)
On comparing, we get
−1+A+2B+2C=8....(i)
−2+B−3C=−2.....(ii)
and 1−A−B+C=−2 .....(iii)
On solving we have A=1,B=3 and C=1
∴ Given expression is
∴ Coefficient of x10 is
−210−3+1=−2−210=l
and constant term is 1−1−3+1=−2=m
∴‌‌lm=4(1+29)