If m and n are +ve integers and m>n and if (1+x)m+n(1−x)m−n is expanded as a polynomial in x, then the coefficient of x2 is
A
2m2−n
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B
2n2−m
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C
2m2+n
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D
2n2+m
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Solution
The correct option is D2n2−m This is a polynomial. Hence we differentiate it twice, the constant of the expression on differentiating twice, would be 2a2 where a2 is the coefficient of x2. f(x)=(1+x)m+n(1−x)×m−n f′(x)=(m+n)(1+x)m+n−1(1−x)×m−n−(m−n)(1+x)m+n(1−x)×m−n−1 f′′(x)=(m+n)(m+n−1)(1+x)m+n−2(1−x)×m−n−2(m+n)(m−n)(1+x)m+n−1(1−x)×m−n−1+(m−n)(m−n−1)(1+x)m+n(1−x)×m−n−2 Putting x=0 to get constant, Constant =2a2=(m+n)(m+n−1)−2(m+n)(m−n)+(m−n)(m−n−1)