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 \[x=2\] \[6{{(2)}^{2}}+2a(2)+b=0\] On solving Eqs. (iii) and (iv), we get \[a=-3,\text{ }b=-12\]and\[c=0\]You need to login to perform this action.
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