It is given that $\mathrm{f}(x) = 2x^3 + ax^2 + x + b$.

1. Factorise the given quadratic factor to find its roots.

Let $2x^2 + x - 1 = 0$
$(2x - 1)(x + 1) = 0$
$x = \frac{1}{2}$ or $x = -1$
Since $2x^2 + x - 1$ is a factor of $\mathrm{f}(x)$, both $(2x - 1)$ and $(x + 1)$ are factors of $\mathrm{f}(x)$.

2. Apply the Factor Theorem using $x = -1$.

By the Factor Theorem, $\mathrm{f}(-1) = 0$:
$2(-1)^3 + a(-1)^2 + (-1) + b = 0$
$-2 + a - 1 + b = 0$
$a + b = 3 \quad \text{--- (1)}$

3. Apply the Factor Theorem using $x = \frac{1}{2}$.

By the Factor Theorem, $\mathrm{f}\left(\frac{1}{2}\right) = 0$:
$2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) + b = 0$
$2\left(\frac{1}{8}\right) + \frac{a}{4} + \frac{1}{2} + b = 0$
$\frac{1}{4} + \frac{a}{4} + \frac{1}{2} + b = 0$
Multiply by 4:
$1 + a + 2 + 4b = 0$
$a + 4b = -3 \quad \text{--- (2)}$

4. Solve the simultaneous equations to find $a$ and $b$.

Subtract (1) from (2):
$(a + 4b) - (a + b) = -3 - 3$
$3b = -6$
$b = -2$

Substitute $b = -2$ into (1):
$a + (-2) = 3$
$a = 5$

1. Express $\mathrm{f}(x)$ as a product of the given quadratic factor and a linear factor by comparing coefficients.

From (i), $2x^2 + x - 1$ is a factor of $\mathrm{f}(x)$.
Let $\mathrm{f}(x) = (2x^2 + x - 1)(cx + d)$.
Comparing the coefficient of $x^3$: $2c = 2 \implies c = 1$.
Comparing the coefficient of $x$: $d - 1 = 1 \implies d = 2$.
Thus, $\mathrm{f}(x) = (2x^2 + x - 1)(x + 2)$.

2. Solve $\mathrm{f}(x) = 0$ to find all roots.

$(2x^2 + x - 1)(x + 2) = 0$
$(2x - 1)(x + 1)(x + 2) = 0$
$x = \frac{1}{2}$, $x = -1$, or $x = -2$

1. Relate the new equation to the original function $f(x)$

The given equation is $\frac{1}{4}y^3 + \frac{a}{4}y^2 + \frac{1}{2}y + b = 0$.
Notice that substituting $x = \frac{y}{2}$ into $f(x) = 2x^3 + ax^2 + x + b$ gives:
$f\left(\frac{y}{2}\right) = 2\left(\frac{y}{2}\right)^3 + a\left(\frac{y}{2}\right)^2 + \left(\frac{y}{2}\right) + b$
$f\left(\frac{y}{2}\right) = \frac{1}{4}y^3 + \frac{a}{4}y^2 + \frac{1}{2}y + b$
Thus, the equation is equivalent to $f\left(\frac{y}{2}\right) = 0$.

2. Use the roots from part (ii) to find $y$

From part (ii), the roots of $f(x) = 0$ are $x = \frac{1}{2}, -1, -2$.
Therefore, we set $\frac{y}{2} = \frac{1}{2}, -1, -2$.
Solving for $y$ gives:
$y = 1, -2, -4$

1. Determine the amplitude and period of the function

For $y = 3 \sin 2x$, the amplitude is $3$.
The period is $\frac{2\pi}{2} = \pi$.

2. Identify key points and sketch the graph for $0 \le x \le \pi$

The graph completes one full cycle in the interval $[0, \pi]$.
Key points:

• Start: $(0, 0)$
• Maximum: $(\frac{\pi}{4}, 3)$
• $x$-intercept: $(\frac{\pi}{2}, 0)$
• Minimum: $(\frac{3\pi}{4}, -3)$
• End: $(\pi, 0)$