AIEEE 2005 Math Solved Paper

General term of [ax2+(‌

1

bx

)]11 is Tr+1.
‌Tr+1=‌11Cr(ax2)11−r(‌

1

bx

)r
‌=‌11Cr(a)11−r(b)−r(x)22−3r
For the coefficient of x7,
⇒22−3r=7
⇒r=5
So coefficient of x7=‌11C5(a)6(b)−5...(1)
Now General term of [ax−(‌

1

bx2

)]11 is Tr+1.
Tr+1=‌11Cr(ax)11−r(−‌

1

bx

)r
=‌11Cr(a)11−r(−1)r(b)−r(x)11−r(x)−2r
For the coefficient of x−7,
‌11−3r=−7
‌⇒r=6
∴ Coefficient of x−7=‌11C6(a)5(−1)6(b)−6
According to question,
Coefficient of x7= Coefficient of x−7
‌⇒‌11C5(a)6(b)−5=‌11C6(a)5(−1)6(b)−6
‌⇒ab=1