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 C12 (1+x)m(1−x)n=(1+mx+m(m−1)x22!)(1−nx+n(n−1)2!x2) =1+(m−n)x+[n2−n2−mn+(m2−m)2]x2 Given m−n=3 or n=m−3 Hence n2−n2−mn+m2−m2=6 ⇒(m−3)(m−4)2−m(m−3)+m2−m2=−6 ⇒m2−7m+12−2m2+6m+m2−m+12=0 ⇒−2m+24=0⇒m=12.