Q.44 plz answer fast!

​Q44. Let f (x) be a polynomial of degree three such that f (0) = 1, f (1) = 2. If 0 is a critical point of f (x) which doesn't have local extremum at x = 0 then f   x x 2   + 7 dx  =

    (A)  1 3 x 2   + 7 3 2   - 7 x 2   + 7 1 2   +   log   x + x 2   + 7 + c

    (B)  1 3 x 2   + 7 3 2   + 7 x 2   + 7 1 2   +   log   x + x 2   + 7 + c
    
     (C)  1 3 x 2   + 7 3 2   + 7 x 2   + 7 1 2   -   log   x + x 2   + 7 + c           
    
     (D) none of these
 

Let fx=ax3+bx2+cx+df0=10+0+0+d=1d=1fx=ax3+bx2+cx+1f'x=3ax2+2bx+cGiven x=0 is a critical pointf'0=00+0+c=0c=0also x=0 is not an extremum point for fx i.e. it is neither local maxima nor local minimaf''0=0f''x=6ax+2bf''0=2b=0b=0fx=ax3+1Givenf1=2a+1=2a=1fx=x3+1I=fxx2+7dx=x3+1x2+7dx=x3x2+7dx+1x2+7dx=x.x2+7-7x2+7dx+1x2+7dx=x.x2+7x2+7dx-7xx2+7+1x2+7dx=xx2+7dx-7xx2+7dx+1x2+72dxUsing 1x2+a2dx=ln x+x2+a2+cI=xx2+7dx-7xx2+7dx+ln x+x2+7+cLet x2+7=tdt=2x dxI=12tdt-721tdt+ln x+x2+7+c=12t12dt-72t-12dt+ln x+x2+7+cxndx=xn+1n+1+c=12×23t32-72×2t12+ln x+x2+7+c=13t32-7t12+ln x+x2+7+c=13x2+732-7x2+712+ln x+x2+7+c

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