If the coefficients x7 in ax2+1bx11and coefficient of x-7in ax−1bx211are equal, then the value of ab is _____.

Let x7 occurs in Tr+1term, then Tr+1=nCrax2n−r1bxr=11Cra11−rbr⋅x22−2r−rFor x7,22−3r=7 or r=5Hence, coefficients of x7 is 11C5a6b5.Let x−7 occur in Tr+1term then Tr+1=11Cr(ax)11−−1bx2r=11Cra11−r(−b)rx11−3rFor x−7⇒11−3r=−7⇒r=6Hence, coefficient of  x−7 is 11C6a5b6Now,  11C5a5b6=11C6a6b5⇒11C5a=11C6a5b6⇒11C5a=11C11−61b⇒11C5a=11C51b ⇒ab=1