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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 ab =

A

1

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B

1/2

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C

2

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D

3

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Solution

## The correct option is A 1 In the expansion of (ax2+1bx)11, the general term is Tr+1 = 11Cr(ax2)11−r(1bx)r = 11Cr(a)11−r(1br)x22−3r For x7, we must have 22 - 3r = 7 ⇒ r = 5, and the coefficient of x7 = 11C5.a11−515) = 11C5 a6b5 Similarly, in the expansion of (ax−1bx2)11, the general term is Tr+1 = 11Cr(−1)r a11−rbr.x11−3r For x−7 we must have, 11 - 3r = -7 ⇒ r = 6, and the coefficient of x−7 is 11C6 a5b6 = 11C5 a5b6. As given, 11C5 a6b5 = 11C5 a5b6 ⇒ ab = 1.

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