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Question

# If the coefficient of x7 in (ax2+1bx)11 is equal to the coefficient of xâˆ’7 in (axâˆ’1bx2)11 then

A
ab=1
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B
a=b
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C
ab=1
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D
a+b=0
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Solution

## The correct option is A ab=1General term Tr+1 in the expansion of (a+b)n is given by Tr+1=nCran−rbrApplying to (ax2+1bx)11, we getTr+1=11Crx22−3ra11(ab)−r ...(i)Therefore 22−3r=7 for coefficient of x7⇒r=5Thus, (i) reduces toT6=11C5x7a11(ab)−5Similarly applying to (ax−1bx2)11, we getTr+1=(−1)r11Crx11−3ra11(ab)−r ...(ii)Therefore 11−3r=−7 for coefficient of x−7⇒r=6Equation (ii) reduces to T7=11C6x−7a11ab−6 ...(b)Hence applying the given condition we get11C6a11(ab)−6=11C5a11(ab)−5 from(a,b)ab=1Hence answer is C

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