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Question

# Let f(x) be a polynomial of degree three such that f(0)=1,f(1)=2 and f(x) has a critical point at x=0 where f(x) does not have a local extremum, then ∫f(x)x2+1dx is equal to

A
xlog(x2+1)+tan1x+c
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B
x+12log(x2+1)tan1x+c
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C
12x2+12log(x2+1)tan1x+c
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D
12(x2log(x2+1))+tan1x+c
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Solution

## The correct option is B 12(x2−log(x2+1))+tan−1x+cGiven that f(x) is a polynomial of degree 3. Let f(x)=ax3+bx2+cx+d.Given that f(0)=1⇒d=1Also given that f(x) has a critical point at x=0 where it does not have a local extremum.⇒f′(x)=0 and f′′(x)=0 at x=0....(1)f′(x)=3ax2+2bx+c=0, f′′(x)=6ax+2b....(2)Using (1) and (2), we get c=0;b=0⇒f(x)=ax3+1Also given that f(1)=2⇒a+1=2⇒a=1⇒f(x)=x3+1So, ∫x3+1x2+1dx.=∫x(x2+1)+(1−x)x2+1=∫[x+1−x1+x2]dx=∫x+11+x2−122x1+x2.dx=x22+tan−1x−12log(1+x2)+C=12[x2−log(1+x2)]+tan−1x+C

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