If the coefficient of x7 in [ax+(1bx)]11is55a11, then a and b satisfy the relation
A
a+b=1
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B
a−b=1
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C
ab=1
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D
ab=1
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Solution
The correct option is Cab=1 General term of [ax+(1bx)]11 is, Tr+1=11Cr(ax)r⋅(1bx)11−r=11Crarbr−11x2r−11 Let Tr+1 posses x7⇒2r−11=7=>r=9 Hence coefficient of x7 is =11C9a9b−2=55a9b−2=55a11 (given) ∴a2b2=1⇒ab=±1