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Question

If m and n are +ve integers and m>n and if (1+x)m+n(1x)mn 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 2n2m
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(1x)×mn
f(x)=(m+n)(1+x)m+n1(1x)×mn(mn)(1+x)m+n(1x)×mn1
f′′(x)=(m+n)(m+n1)(1+x)m+n2(1x)×mn2(m+n)(mn)(1+x)m+n1(1x)×mn1+(mn)(mn1)(1+x)m+n(1x)×mn2
Putting x=0 to get constant,
Constant =2a2=(m+n)(m+n1)2(m+n)(mn)+(mn)(mn1)
Hence,
a2=(m+n)(m+n1)2(m+n)(mn)+(mn)(mn1)=2n2m

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