In the diagram below, the graphs of $f(x) = \frac{1}{2} \cos x$ and $g(x) = \sin(x - 30^\circ)$ are drawn for the interval $x \in [-90^\circ; 240^\circ]$ - NSC Mathematics - Question 7 - 2022 - Paper 2

First, we need to determine the range of $f(x)$. From the function $f(x) = \frac{1}{2} \cos x$, the maximum and minimum values occur at:

• Maximum: $\frac{1}{2}$ (when $\cos x = 1$)
• Minimum: $-\frac{1}{2}$ (when $\cos x = -1$)

Thus, the range of $f(x)$ is: $[-\frac{1}{2}, \frac{1}{2}]$

To find the range of $3f(x) + 2$, we apply the transformation:

• Maximum of $3f(x) + 2 = 3(\frac{1}{2}) + 2 = 3.5$
• Minimum of $3f(x) + 2 = 3(-\frac{1}{2}) + 2 = 0.5$

So, the range is: $[0.5, 3.5]$

To find the value of $x$ such that $g(x) - f(x) = \frac{\sqrt{3}}{2}$, we need to look at the graph and find where the difference between the two functions equals $\frac{\sqrt{3}}{2}$. Upon inspection, this occurs at:

$x = 90^\circ$ (when the value of $g(x)$ meets the requirement relative to $f(x)$).

To determine where $f(x)g(x) > 0$, we find when both functions are positive or both are negative:

• $f(x)$ is positive when $\cos x > 0$ (i.e. between $-90^\circ < x < 90^\circ$).
• $g(x)$ is positive when $\sin(x - 30^\circ) > 0$. This occurs:
  • When $30^\circ < x < 210^\circ$ (which simplifies to two ranges)

Combining the findings gives us:

• $30^\circ < x < 90^\circ$
• $210^\circ < x < 240^\circ$

So, the final intervals are: $30^\circ < x < 90^\circ$ and $210^\circ < x < 240^\circ$.

To solve where $g(x - 5^\circ) > 0$, we need to make the transformation:
$\sin((x - 5^\circ) - 30^\circ) > 0$
from which we derive that:

• This inequality holds true between the critical values related to the sine function.
• We can solve this by determining: $-90^\circ < x - 5^\circ < 90^\circ$
  Which implies:
  $-85^\circ < x < 95^\circ$. Also we find: $210^\circ < x < 240^\circ$ does not alter these values, so we restrict to the first interval.

Thus, the answer is: $-85^\circ < x < 95^\circ$.