The sketch below shows the graphs of $f(x) = x^2 - 2x - 3$ and $g(x) = x - 3$ - NSC Mathematics - Question 5 - 2017 - Paper 1

To find the length of line segment AB, we first determine the x-intercepts of the function $f$. The x-intercepts can be found by solving:

$f(x) = 0 \\ x^2 - 2x - 3 = 0$

Factoring, we get:

$(x - 3)(x + 1) = 0$

Hence, the x-intercepts are at $x = 3$ and $x = -1$. The length of AB is the difference between these intercepts:

$$AB = |3 - (-1)| = 4 ext{ units}.$

The turning point D of the parabola $f(x) = x^2 - 2x - 3$ can be found by calculating the vertex. The x-coordinate of the vertex for a parabola in the form $ax^2 + bx + c$ is given by:

$x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$ Substituting $x = 1$ back into $f(x)$ to find the y-coordinate:

$f(1) = (1)^2 - 2(1) - 3 = -4$ Thus, the coordinates of D are $(1, -4)$.

The average gradient (slope) between two points can be calculated using the formula:

$\text{Average Gradient} = \frac{f(y_2) - f(y_1)}{x_2 - x_1}$ Here, we use points C $(3, 0)$ and D $(1, -4)$:

$\text{Average Gradient} = \frac{-4 - 0}{1 - 3} = \frac{-4}{-2} = 2$

The triangle OCB is an isosceles triangle where OC = OB = 3 units (as both are the distance from the origin to point B and C respectively). Since both legs are equal, we can find the angle $heta$ at O using trigonometric properties. The angle $heta$ at O (angle OCB) can be calculated knowing that the triangle is isosceles:

$\angle OCB = 45^\circ$

For $f(x) = k$ to have two unequal positive real roots, the discriminant of the quadratic equation must be positive. The equation can be rewritten as:

$x^2 - 2x - (3 + k) = 0$ The discriminant $D$ is given by:

$D = b^2 - 4ac = (-2)^2 - 4(1)(-3 - k) = 4 + 12 + 4k = 16 + 4k$ Setting this greater than zero for two real roots gives:

$16 + 4k > 0$ Solving, we find: $k > -4$

First, we calculate the derivatives: f'(x) = $2x - 2$ and f''(x) = $2$. The product of the derivatives is:

$f'(x) imes f''(x) = (2x - 2) imes 2$ Setting this greater than zero:

$(2x - 2) imes 2 > 0 \\ 2x - 2 > 0 \\ 2x > 2 \\ x > 1$ Thus, $f'(x) imes f''(x) > 0$ for $x > 1$.