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Question

If in the expansion of (1+x)m (1−x)n, the coefficients of x and x2 are 3 and −6 respectively, then m is

A
6
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B
9
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C
12
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D
24
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Solution

The correct option is C 12
(1+x)m(1x)n=(1+mx+m(m1)x22!)(1nx+n(n1)2!x2)
=1+(mn)x+[n2n2mn+(m2m)2]x2
Given mn=3 or n=m3
Hence n2n2mn+m2m2=6
(m3)(m4)2m(m3)+m2m2=6
m27m+122m2+6m+m2m+12=0
2m+24=0m=12.

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