In the diagram below, the graphs of $f(x) = an x$ and $g(x) = 2 ext{sin} 2x$ are drawn for the interval $x \in [-180^{\circ} ; 180^{\circ}]$ - NSC Mathematics - Question 6 - 2022 - Paper 2

To determine the value of $k$, we need to evaluate $g(60^{\circ})$:

$g(60^{\circ}) = 2\text{sin}(2 \cdot 60^{\circ}) = 2\text{sin}(120^{\circ}) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

Thus, the value of $k$ is:

$k = \sqrt{3}$.

Let's find the coordinate of point B. We know that A is at $(60^{\circ}; k)$. To find the intersection points, we need to evaluate where $f(x) = g(x)$, particularly at $x = -120^{\circ}$:

Substituting into $f(x)$ and $g(x)$ gives:

$f(-120^{\circ}) = \tan(-120^{\circ}) \quad \text{and} \quad g(-120^{\circ}) = 2\text{sin}(-240^{\circ})$

Given that:

$f(-120^{\circ}) = \tan(-120^{\circ}) = \tan(60^{\circ}) = \sqrt{3}$ $g(-120^{\circ}) = 2\text{sin}(120^{\circ}) = \sqrt{3}$

Therefore, the coordinates of B are $(-120^{\circ}, \sqrt{3})$.

The range of the function $g(x) = 2\text{sin} 2x$ is obtained by analyzing the values of the sine function, which ranges from $-1$ to $1$. Hence, the range of $g(x)$ is:

$-2 \leq g(x) \leq 2$

Doubling this range for $2g(x)$ results in:

$-4 \leq 2g(x) \leq 4$

To identify the values of $x$ where:

$g(x + 5^{\circ}) - f(x + 5^{\circ}) \leq 0$

We need to analyze the function over the interval $x \in [-90^{\circ}; 0^{\circ}]$. This will require plotting or checking points, but the critical s points are:

$-65^{\circ} \leq x \leq -5^{\circ}$.

We start from:

$ext{sin} x \cdot ext{cos} x = p$

Using the double angle identity, we rewrite this as:

rac{1}{2} \text{sin}(2x) = p

To have exactly two real roots in the interval $x \in [-180^{\circ}; 180^{\circ}]$, we need:

$-1 \leq p \leq \frac{1}{2}$.