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 $$x \in [-90^{\circ}; 240^{\circ}]$$. A and B are the y-intercepts of f and g respectively. 7.1 Determine the length of AB. 7.2 Write down the range of 3f(x) + 2. 7.3 Read off from the graphs a value of x for which g(x) - f(x) = \frac{\sqrt{3}}{2}. 7.4 For which values of x, in the interval $$x \in [-90^{\circ}; 240^{\circ}]$$, will: 7.4.1 f(x)g(x) > 0 7.4.2 g(x - 5^{\circ}) > 0

NSC Mathematics Functions: In the diagram below, the graphs

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

Question 7

$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.png$

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... show full transcript

Worked Solution & Example Answer: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

Step 1

Determine the length of AB.

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Answer

To find the length of AB, we first identify the y-intercepts of both functions. From the graph, the point A is located at (0, 1/2) and point B at (0, -1/2). The length of AB can be calculated as follows:

$\text{Length of AB} = |y_A - y_B| = \left|\frac{1}{2} - \left(-\frac{1}{2}\right)\right| = |1| = 1\text{ unit}.$

Step 2

Write down the range of 3f(x) + 2.

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Answer

The range of f(x) is given as $y \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$ . By multiplying f(x) by 3 and adding 2, we have:

$3f(x) + 2 \Rightarrow y \in \left[ 3 \cdot -\frac{1}{2} + 2, 3 \cdot \frac{1}{2} + 2 \right] = \left[ -\frac{3}{2} + 2, \frac{3}{2} + 2 \right] = \left[ \frac{1}{2}, \frac{7}{2} \right].$

Thus, the range of 3f(x) + 2 is: $\left[ \frac{1}{2}, \frac{7}{2} \right]$ .

Step 3

Read off from the graphs a value of x for which g(x) - f(x) = \frac{\sqrt{3}}{2}.

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Answer

By examining the graph of g(x) and f(x), we need to find the x-coordinate where the difference g(x) - f(x) equals \frac{\sqrt{3}}{2}. Observing the graph, this occurs at:

$x = 90^{\circ}$ .

Step 4

For which values of x, in the interval x in [-90^{\circ}; 240^{\circ}], will:

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Answer

7.4.1 f(x)g(x) > 0
From the given interval, f(x) and g(x) are both positive in the ranges:

$30^{\circ} < x < 90^{\circ}$
or
$210^{\circ} < x < 240^{\circ}.$

7.4.2 g(x - 5^{\circ}) > 0
To determine where this condition holds, we check:

$-55^{\circ} < x < 125^{\circ}$ .

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

Worked Solution & Example Answer: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

Determine the length of AB.

Write down the range of 3f(x) + 2.

Read off from the graphs a value of x for which g(x) - f(x) = \frac{\sqrt{3}}{2}.

For which values of x, in the interval x in [-90^{\circ}; 240^{\circ}], will:

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