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Question

If the coefficient of x7 in [ax2+(1bx)]11 equals the coefficient of x−7 in [ax−(1bx2)]11, then a and b satisfy the relation

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

The correct option is A ab=1
Let x7 is contained in (r+1)th term in the expansion of (ax2+1bx)11

Thus Tr+1=11Cr(ax2)11r(1bx)r

=11Cra11rbrx223r
For coefficient of x7,

223r=7

3r=15

r=5

Thus T6=11C5a6b5x7
Therefore, coefficient of x7 in the expansion of

(ax2+1bx)11=11C5a6b5

Similarly, coefficient of x7 in the expansion of

(ax1bx2)11=11C6a5b6

Given, 11C5a6b5=11C6a5b6

a6b5=a5b6 (11C5=11C6)

ab=1

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