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Question

If the coefficient of x7 in (ax2+1bx)11 is equal to the coefficient of x7 in (ax1bx2)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)11r(1bx)r = 11Cr(a)11r(1br)x223r

For x7, we must have 22 - 3r = 7 ⇒ r = 5, and

the coefficient of x7 = 11C5.a11515) = 11C5 a6b5

Similarly, in the expansion of (ax1bx2)11, the

general term is Tr+1 = 11Cr(1)r a11rbr.x113r

For x7 we must have, 11 - 3r = -7 ⇒ r = 6, and

the coefficient of x7 is 11C6 a5b6 = 11C5 a5b6.

As given, 11C5 a6b5 = 11C5 a5b6 ⇒ ab = 1.


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