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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

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=1
General term Tr+1 in the expansion of (a+b)n is given by
Tr+1=nCranrbr
Applying to (ax2+1bx)11, we get
Tr+1=11Crx223ra11(ab)r ...(i)
Therefore 223r=7 for coefficient of x7
r=5
Thus, (i) reduces to
T6=11C5x7a11(ab)5
Similarly applying to (ax1bx2)11, we get
Tr+1=(1)r11Crx113ra11(ab)r ...(ii)

Therefore 113r=7 for coefficient of x7
r=6
Equation (ii) reduces to
T7=11C6x7a11ab6 ...(b)
Hence applying the given condition we get
11C6a11(ab)6=11C5a11(ab)5 from(a,b)
ab=1
Hence answer is C

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