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

To find the value of $k$ at the point A where $x = 60^ ext{o}$, we substitute $x = 60^ ext{o}$ into the equation of $g(x)$: g(60^ ext{o}) = 2 ext{sin}(2 imes 60^ ext{o}) = 2 ext{sin}(120^ ext{o}) = 2 imes rac{ ext{√3}}{2} = ext{√3}. Thus, $k = ext{√3} ext{ or approximately } 1.73.$

The coordinates of point B can be determined from the intersection points. From the graph and the interval, we determine: $B(-120^ ext{o}; - ext{√3})$.

The range of $g(x)$ is $[-2; 2]$. Thus, the range of $2g(x)$ is: $[-4; 4].$

To solve for $x$ where $g(x + 59^ ext{o}) - f(x + 59^ ext{o}) ext{ ≤ } 0$, we analyze the functions within the interval: $-65^ ext{o} ≤ x ≤ -5^ ext{o}.$

We start from the equation: $ext{sin} x ext{cos} x = p$ Using the identity, we get: rac{1}{2} ext{sin}(2x) = p. For this to have exactly two real roots in the given interval, we find the values of $p$ such that: p = rac{1}{2} ext{ or } -rac{1}{2}.