The function y = 2x3 + ax2 + bx has critical points at x=2 and x=1. Find the values of a and b, the values of x for which y is minimum, and the minimum value of y.

A minimum point of a function is a point where the function's value is less than or equal to its values at nearby points. At a minimum point, the first derivative of the function is zero, and the second derivative is positive.

The given function is $y = 2x^3 + ax^2 + bx$.
To find the critical points, we need to find the first derivative $\frac{dy}{dx}$ and set it to zero.

We are given that the critical points are at $x=2$ and $x=1$. This means $\frac{dy}{dx} = 0$ at these points.

For $x=2$:

For $x=1$:

Subtract equation $(*2)$ from equation $(*1)$:

Substitute $a = -9$ into equation $(*2)$:

a) (i) The values of $a$ and $b$:

The first derivative is $\frac{dy}{dx} = 6x^2 + 2ax + b$.
Substitute the values of $a$ and $b$:

Now, find the second derivative $\frac{d^2y}{dx^2}$:

For $x=2$:

Since $\frac{d^2y}{dx^2} > 0$ at $x=2$, this point is a minimum.

For $x=1$:

Since $\frac{d^2y}{dx^2} < 0$ at $x=1$, this point is a maximum.

a) (ii) The values of $x$ for which $y$ is minimum:

Substitute $x=2$, $a=-9$, and $b=12$ into the original function $y = 2x^3 + ax^2 + bx$.

a) (iii) The minimum value of $y$: