Question

If the curve $y=2x^3+a{x}^2+bx+c$ passes through the origin and the tangents drawn to it at $x=-1$ and $x=2$ are parallel to the $x-$ axis, then the values of a, b and c are respectively

• A. $12,-3$ and 0
• B. $-3,-12$ and 0
• C. $-3,12$ and 0
• D. $3,-12$ and 0

Answer 1

Answer: (B)

Given equation of curve is $y=2x^3+a{x}^2+bx+c$ ...(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 $\frac{dy}{dx}=6x^2+2ax+b$ 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 $a=-3,b=-12$ and $c=0$