Find the set of values of $x$ for which the curve $y = -x^2 + 2x + 4$ lies below another curve $y = x^2 - 2x - 2$ and represent this set on a number line. The line $y = 2x + k$ is a normal to the curve $y = -x^2 + 2x + 4$ at the point A. The line $y = 2x + k$ is a normal to the curve $y = -x^2 + 2x + 4$ at the point A. Find the x-coordinate of A. Find the value of the constant $k$.

1. Set up the inequality based on the condition that the first curve lies below the second curve.

$-x^2 + 2x + 4 < x^2 - 2x - 2$

2. Rearrange the inequality into standard quadratic form.

$0 < 2x^2 - 4x - 6$
$2x^2 - 4x - 6 > 0$
$x^2 - 2x - 3 > 0$

3. Factorise the quadratic expression to find the critical values.

$(x - 3)(x + 1) > 0$
Critical values are $x = 3$ and $x = -1$.

4. Determine the solution set and represent it on a number line.

Since the quadratic is $> 0$, the solution is $x < -1$ or $x > 3$.

(Draw a number line with open circles at $-1$ and $3$, with an arrow pointing left from $-1$ and an arrow pointing right from $3$.)

1. Identify the gradient of the normal from the given equation.

The equation of the normal is $y = 2x + k$.
The gradient of the normal, $m_n = 2$.

2. Find the gradient of the tangent using the relationship $m_t \times m_n = -1$.

Let the gradient of the tangent be $m_t$.
$m_t \times 2 = -1 \implies m_t = -\frac{1}{2}$

3. Differentiate the curve equation and equate to the tangent gradient to find the x-coordinate.

The curve is $y = -x^2 + 2x + 4$.
$\frac{dy}{dx} = -2x + 2$
At point A, $\frac{dy}{dx} = m_t = -\frac{1}{2}$.
$-2x + 2 = -\frac{1}{2}$
$-2x = -\frac{5}{2}$
$x = \frac{5}{4}$

1. Find the $x$-coordinate of point A (from part (b)(i))

The gradient of the normal is $2$, so the gradient of the tangent is $-\frac{1}{2}$.
$\frac{dy}{dx} = -2x + 2$
$-2x + 2 = -\frac{1}{2}$
$2x = \frac{5}{2} \implies x = \frac{5}{4}$

2. Find the $y$-coordinate of point A

Substitute $x = \frac{5}{4}$ into the equation of the curve:
$y = -\left(\frac{5}{4}\right)^2 + 2\left(\frac{5}{4}\right) + 4$
$y = -\frac{25}{16} + \frac{10}{4} + 4$
$y = -\frac{25}{16} + \frac{40}{16} + \frac{64}{16} = \frac{79}{16}$

3. Substitute the coordinates of A into the equation of the normal to find $k$

Substitute $x = \frac{5}{4}$ and $y = \frac{79}{16}$ into $y = 2x + k$:
$\frac{79}{16} = 2\left(\frac{5}{4}\right) + k$
$\frac{79}{16} = \frac{10}{4} + k$
$\frac{79}{16} = \frac{40}{16} + k$
$k = \frac{39}{16}$