)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