Explanation

(b)
Given equation of curve is
$y = 2x^{3} + ax^{2} + bx + c$ …(i)
$\Rightarrow \ \ \ \ \ \ 0 = 2(0) + a(0) + b(0) + c$ [passes through (0, 0)]
$\Rightarrow \ \ \ \ c = 0$ ...(ii)
On differentiating Eq. (i), we get
$\frac{dy}{dx} = 6x^{2} + 2ax + b$
Since, the tangent 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. (iii) and (iv), we get
$a = - 3,\ \ \ \ \ b = - 12$ and $c = 0$