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Question

# If m and n are +ve integers and m>n and if (1+x)m+n(1âˆ’x)mâˆ’n is expanded as a polynomial in x, then the coefficient of x2 is

A
2m2n
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B
2n2m
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C
2m2+n
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D
2n2+m
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Solution

## The correct option is D 2n2−mThis 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−nf′(x)=(m+n)(1+x)m+n−1(1−x)×m−n−(m−n)(1+x)m+n(1−x)×m−n−1f′′(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−2Putting x=0 to get constant,Constant =2a2=(m+n)(m+n−1)−2(m+n)(m−n)+(m−n)(m−n−1)Hence,a2=(m+n)(m+n−1)−2(m+n)(m−n)+(m−n)(m−n−1)=2n2−m

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