If the coefficient of x7 in ax2+1bx11 equals the coefficient of x−7 in ax−1bx211, then a and b satisfy the relation

For ax2+1bx11,     Tr+1=11Crax211−r1bxr=11Cra11−r1brx22−3rFor x7,     22−3r=7or 3r=15or r=5⇒T6=11C5a61b5x7⇒ Coefficient of x7 is 11C5a6b5similarly, coefficient of x-7 in ax−1bx211 is 11C6a5b6.Given that  11C5a6b5=11C6a5b6⇒ a=1bor ab = 1