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Given $$f(x) = an \left( \frac{1}{2} x \right)$$ and $$g(x) = \sin \left( x - 30^{\circ} \right) \text{ for } x \in [-90^{\circ}; 180^{\circ}]$$ 6.1 On the same set of axes draw the graphs of $f$ and $g$ - NSC Mathematics - Question 6 - 2017 - Paper 2

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Given---$$f(x)-=--an-\left(-\frac{1}{2}-x-\right)$$--and---$$g(x)-=-\sin-\left(-x---30^{\circ}-\right)-\text{-for-}-x-\in-[-90^{\circ};-180^{\circ}]$$--6.1-On-the-same-set-of-axes-draw-the-graphs-of-$f$-and-$g$-NSC Mathematics-Question 6-2017-Paper 2.png

Given $$f(x) = an \left( \frac{1}{2} x \right)$$ and $$g(x) = \sin \left( x - 30^{\circ} \right) \text{ for } x \in [-90^{\circ}; 180^{\circ}]$$ 6.1 On the sa... show full transcript

Worked Solution & Example Answer:Given $$f(x) = an \left( \frac{1}{2} x \right)$$ and $$g(x) = \sin \left( x - 30^{\circ} \right) \text{ for } x \in [-90^{\circ}; 180^{\circ}]$$ 6.1 On the same set of axes draw the graphs of $f$ and $g$ - NSC Mathematics - Question 6 - 2017 - Paper 2

Step 1

6.1 On the same set of axes draw the graphs of f and g.

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Answer

To draw the graphs of the functions, we start with the graph of f(x)=tan(12x)f(x) = \tan \left( \frac{1}{2} x \right). The function has vertical asymptotes at the points where the tangent function is undefined, which occur at:

x=(2n+1)90x = (2n + 1) \cdot 90^{\circ}

for integer values of nn. For our interval, the asymptotes occur at x=90x = -90^{\circ} and x=90x = 90^{\circ}. The graph will oscillate between these points, with turning points occurring within these intervals.

For the function g(x)=sin(x30)g(x) = \sin(x - 30^{\circ}), the graph will oscillate between -1 and 1, with a phase shift of 3030^{\circ}. We can plot this on the same axes, showing how both functions intersect, their turning points, and the asymptotes for f(x)f(x).

Step 2

6.2 Write down the period of f.

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Answer

The period of the function f(x)=tan(12x)f(x) = \tan \left( \frac{1}{2} x \right) can be calculated as:

Period(f)=18012=360.\text{Period}(f) = \frac{180^{\circ}}{\frac{1}{2}} = 360^{\circ}.

Step 3

6.3 For what values of x is f(x).g(x) < 0 for x \in [-90^{\circ}; 120^{\circ}]?

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Answer

To find the values of xx for which f(x)g(x)<0f(x) \cdot g(x) < 0, we analyze the signs of both functions over the interval [90;120][-90^{\circ}; 120^{\circ}].

  • For xx in the interval [90,0][-90^{\circ}, 0^{\circ}], f(x)f(x) is negative and g(x)g(x) is negative.
  • For xx in the interval (0,30)(0^{\circ}, 30^{\circ}), f(x)f(x) is positive and g(x)g(x) is negative.

Thus, the solution for when f(x)g(x)<0f(x) \cdot g(x) < 0 occurs in:

x(0,30).x \in (0^{\circ}, 30^{\circ}).

Step 4

6.4 Write down the equation(s) of the asymptotes of h if h(x) = f(x + 10^{\circ}) for x \in [-90^{\circ}; 180^{\circ}].

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Answer

The function h(x)=f(x+10)=tan(12(x+10))h(x) = f(x + 10^{\circ}) = \tan \left( \frac{1}{2} (x + 10^{\circ}) \right) will have asymptotes where:

12(x+10)=(2n+1)90\frac{1}{2} (x + 10^{\circ}) = (2n + 1) \cdot 90^{\circ}

Solving for xx gives:

x+10=(2n+1)180x + 10^{\circ} = (2n + 1) \cdot 180^{\circ}

x=(2n+1)18010.x = (2n + 1) \cdot 180^{\circ} - 10^{\circ}.

The specific asymptotes in the interval [90;180][-90^{\circ}; 180^{\circ}] will be determined by selecting integer values for nn.

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