If sum of the coefficients of x7 and x4 in the expansion of x2a−bx11 is zero, then

Tr+1=11Crx2a11−r−bxr=11Cr(−1)rar−11brx22−3rFor coefficient of x 4 and x7 , we put 22 – 3r = 4 and 22 – 3r = 7 ⇒r=6  and r=5We are given  11C6(−1)6a−5b6+11C5(−1)5a−6b5=0⇒b6a5=b5a6⇒ab=1