If sum of the coefficients of x7 and x4 in the expansion of x2a−bx11 is zero, then
ab = 1
a = b
ab = –1
a + b = 0
Tr+1=11Crx2a11−r−bxr=11Cr(−1)rar−11brx22−3r
For coefficient of x 4 and x7 , we put 22 – 3r = 4 and 22 – 3r = 7
⇒r=6 and r=5
We are given
11C6(−1)6a−5b6+11C5(−1)5a−6b5=0
⇒b6a5=b5a6⇒ab=1