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Q.

Let P(x) be a polynomial of degree n with leading 

coefficient 1 . Let v(x) be any function and v1(x)=v(x)dx

v2(x)=v1(x)dxvn+1(x)=vn(x)dx then P(x)v(x)dx is equal to 

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a

P(x)v1(x)+P(x)v2(x)2!+P′′(x)v3(x)3!+vn+1(x)(n+1)!

b

P(x)v1(x)P(x)v2(x)+P′′(x)v3(x)+(1)nn!vn+1(x)

c

P(x)v1(x)+P(x)v2(x)+P′′(x)v3(x)+nvn+1(x)

d

P(x)v1(x)P(x)v2(x)2!+P′′(x)v3(x)3!+(1)nvn+1(x)(n+1)!

answer is B.

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Detailed Solution

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Integrating by part repeatedly , we get 

P(x)v(x)dx=P(x)v1(x)P(x)v1(x)dx=P(x)v1(x)P(x)v2(x)+P′′(x)v2(x)dx=P(x)v1(x)P(x)v2(x)+P′′(x)v3(x).+(1)nP(n)(x)vn(x)dx

Now , use P(n)(x)=n!

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