AP EAMCET Engineering 2012 Solved Paper

Given expansion is (

x2

a

−

b

x

)11.
∴ The general term is
Tr+1=‌11Cr(

x2

a

)11−r(−

b

x

)r
=‌11Cr(x)22−3r(−b)r(

1

a

)11−r
∴ For coefficient x7, put 22−3r=7
⇒ 3r=15
⇒ r=5
and for coefficient of x4, put 22−3r=4
⇒3r=18⇒r=6
∴T6=‌11C5(

1

a

)6(−b)5
and T7=‌11C6(

1

a

)6(−b)6
According to the given condition,
T6+T7=0
∴ ‌11C5(

1

a

)6(−b)5+‌11C6(

1

a

)5(−b)6=0
⇒ ‌11C5(

1

a

)5(−b)5(

1

a

−b)=0
⇒

1

a

−b=0
⇒ ab=1