A) 12,-3 and 0
B) - 3,-12 and 0
C) - 3,12 and 0
D) 3, -12 and 0
Correct Answer: B
Solution :
Given equation of curve is \[y=2{{x}^{3}}+a{{x}^{2}}+bx+c\] ...(i) Since it is passes through \[(0,0)\] \[\Rightarrow \] \[0=2(0)+a(0)+b(0)+c\] \[\Rightarrow \] \[c=0\] ?.(ii) On differentiating Eq. (i), we get \[\frac{dy}{dx}=6{{x}^{2}}+2ax+b\] Since the tangents at \[x=-1\]and \[x=2\]are parallel to x-axis. \[\therefore \] \[\frac{dy}{dx}=0\] At \[x=-1\] \[6{{(-1)}^{2}}+2a(-1)+b=0\] \[\Rightarrow \] \[6-2a+b=0\] ??(iii) At \[x=2\] \[6{{(2)}^{2}}+2a(2)+b=0\] \[\Rightarrow \] \[24+4a+b=0\] ?..(iv) On solving Eqs. (in) and (iv), we get \[a=-3,\]\[b=-12\] and \[c=0\]
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