8 (a) Determine a sequence of transformations which maps the graph of $y = ext{sin} x$ onto the graph of $y = rac{ ext{√}3}{2} ext{sin} x - 3 ext{cos} x + 4$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 2
Question 8
8 (a) Determine a sequence of transformations which maps the graph of $y = ext{sin} x$ onto the graph of $y = rac{ ext{√}3}{2} ext{sin} x - 3 ext{cos} x + 4$.
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Worked Solution & Example Answer:8 (a) Determine a sequence of transformations which maps the graph of $y = ext{sin} x$ onto the graph of $y = rac{ ext{√}3}{2} ext{sin} x - 3 ext{cos} x + 4$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 2
Step 1
Determine a sequence of transformations which maps the graph of $y = sin x$ onto the graph of $y = \frac{\sqrt{3}}{2} \sin x - 3 \cos x + 4$
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Answer
To determine the transformations, we compare the given functions:
The original function is (y = \text{sin} x).
The transformed function can be expressed as (y = R \sin(x + \alpha) + d), where (R = \sqrt{(\frac{\sqrt{3}}{2})^2 + (-3)^2} = \sqrt{\frac{3}{4} + 9} = \sqrt{\frac{39}{4}} = \frac{\sqrt{39}}{2}).
The angle (\alpha) is determined from (R \cos \alpha = \frac{\sqrt{3}}{2}) and (R \sin \alpha = -3). Hence, (\tan \alpha = \frac{-3}{\frac{\sqrt{3}}{2}} = -\frac{6}{\sqrt{3}}).
The transformations include:
A vertical stretch by a factor of (\frac{\sqrt{39}}{2}) (scale in the y-direction).
A translation up by 4 units.
A translation left by the angle (\alpha).
Step 2
Show that the least value of \( \frac{1}{\text{√}3 \text{sin} x - 3 \text{cos} x + 4} \) is \( \frac{2 - \text{√}3}{2} \)
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Answer
We need to analyze the function (f(x) = \text{√}3 \text{sin} x - 3 \text{cos} x + 4).
To find the least value, we first find the critical points:
The least value occurs when (\sin(x + \frac{\pi}{3}) = -1), so we compute:
The minimum value is (\text{√}3(-1) - 3(0) + 4 = -\text{√}3 + 4 = 4 - \text{√}3).
Thus, the least value of (\frac{1}{f(x)}) is derived from:
(\frac{1}{4 - \text{√}3}) which simplifies to the required expression.
Step 3
Find the greatest value of \( \frac{1}{\text{√}3 \text{sin} x - 3 \text{cos} x + 4} \)
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Answer
The greatest value occurs when (\text{√}3 \text{sin} x - 3 \text{cos} x + 4) is minimized, leading to:
The maximum value can be calculated or inspected using the earlier minimum found.
So, we evaluate the parameter substitution analogous to the previous part leading us to derive the final answer as (\frac{2 + \text{√}3}{2}).