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 Share with your friends Share 0 Aarushi Mishra answered this 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 point⇒f'0=00+0+c=0c=0also x=0 is not an extremum point for fx i.e. it is neither local maxima nor local minima⇒f''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-7∫xx2+7+∫1x2+7dx=∫xx2+7dx-7∫xx2+7dx+∫1x2+72dxUsing ∫1x2+a2dx=ln x+x2+a2+cI=∫xx2+7dx-7∫xx2+7dx+ln x+x2+7+cLet x2+7=tdt=2x dxI=12∫tdt-72∫1tdt+ln x+x2+7+c=12∫t12dt-72∫t-12dt+ln x+x2+7+c∫xndx=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 0 View Full Answer