if the coefficients of  x7  in (ax2+1bx)11  and x−7  in (ax−1bx2)11  are equal, then ab=

Given expansion is  (ax2+1bx)11   We know that rth  term in the expansion of (x+a)n isTr+1=nCr xn−r ar Tr+1= 11Cr (ax2)11−r (1nx)r ................(1)  Tr+1= 11Cr (a)11−r (x2)11−r (1n)r (x−1)r Tr+1= 11Cr (a)11−r (x)22−3r (1n)r  And this will contain x7  compare with  (x)22−3r=x7 Only if 22−3r=7  i.e. if r=5 .These   value sub in Eqn (1)Hence T6 = 11C5(ax2)6(1bx)5                   =  11C5a6b5x7  contains x7  And given another expansion is  (ax−1bx2),  We know that rth  term in the expansion of (x+a)n isTr+1=nCr xn−r ar The general term is Tr+1= 11Cr (ax)11−r (−1bx2)r .................(2) Tr+1= 11Cr (a)11−r(x)11−r (−1b)r (x−2)r Tr+1= 11Cr (a)11−r(x)11−3r (−1b)r  And this contains x−7 compare with  (x)11−3r ⇒(x)11−3r=x−7 Only if 11−3r=−7  i.e. if r=6 .These value in Eqn (2) then we getHence T'7= 11C6(ax5(−1bx2)6   = 11C6a5b6x−7  contains x−7  . According to problem  given that 11C5a6b5=  11C6a5b6  ⇒  a6b6=a5b5⇒  ab  =  1