f(x) = 5 \, ext{cos} \, x + 12 \, ext{sin} \, x Given that f(x) = R \, ext{cos}(x - \alpha), \, ext{where} \, R > 0 \, ext{and} \, 0 < \alpha < \frac{\pi}{2}. (a) find the value of R and the value of \alpha \text{ to 3 decimal places.} (b) Hence solve the equation 5 \, ext{cos} \, x + 12 \, ext{sin} \, x = 6 \text{ for } 0 \leq x < 2\pi. (c) (i) Write down the maximum value of 5 \, ext{cos} \, x + 12 \, ext{sin} \, x. (ii) Find the smallest positive value of x for which this maximum value occurs.

A-Level Edexcel Maths Pure Functions: f(x) = 5 \, ext{cos} \, x + 12

f(x) = 5 \, ext{cos} \, x + 12 \, ext{sin} \, x Given that f(x) = R \, ext{cos}(x - \alpha), \, ext{where} \, R > 0 \, ext{and} \, 0 < \alpha < \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

Question 4

$f(x)-=-5-\,--ext{cos}-\,-x-+-12-\,--ext{sin}-\,-x-Given-that-f(x)-=-R-\,--ext{cos}(x---\alpha),-\,--ext{where}-\,-R->-0-\,--ext{and}-\,-0-<-\alpha-<-\frac{\pi}{2}-Edexcel-A-Level Maths Pure-Question 4-2008-Paper 5.png$

f(x) = 5 \, ext{cos} \, x + 12 \, ext{sin} \, x Given that f(x) = R \, ext{cos}(x - \alpha), \, ext{where} \, R > 0 \, ext{and} \, 0 < \alpha < \frac{\pi}{2}. ... show full transcript

Worked Solution & Example Answer:f(x) = 5 \, ext{cos} \, x + 12 \, ext{sin} \, x Given that f(x) = R \, ext{cos}(x - \alpha), \, ext{where} \, R > 0 \, ext{and} \, 0 < \alpha < \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

Step 1

find the value of R and the value of α to 3 decimal places.

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Answer

To find the values of R and α, we start by equating the functions: $R^2 = 5^2 + 12^2$ Calculating this, we get: $R^2 = 25 + 144 = 169$ Thus, $R = \sqrt{169} = 13.$

Next, we need to find α using: $\tan(\alpha) = \frac{12}{5}$ Calculating α, we find: $\alpha = \arctan\left(\frac{12}{5}\right) \approx 1.176.$

Step 2

Hence solve the equation 5 cos x + 12 sin x = 6 for 0 ≤ x < 2π.

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Answer

Starting from the rearranged equation: $5 \, \text{cos} \, x + 12 \, \text{sin} \, x - 6 = 0.$ Using the form: $\cos(x - \alpha) = \frac{6}{R} = \frac{6}{13},$ we find: $\cos(x - \alpha) = \frac{6}{13} \, \Rightarrow \, x - \alpha = \arccos\left(\frac{6}{13}\right) \approx 1.091.$

Therefore, solving for x, we add α: $x \approx 1.091 + 1.176 \approx 2.267.$

Step 3

Write down the maximum value of 5 cos x + 12 sin x.

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Answer

The maximum value of the function 5 cos x + 12 sin x is equal to R, which we calculated earlier: $\text{Maximum value} = R = 13.$

Step 4

Find the smallest positive value of x for which this maximum value occurs.

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Answer

The maximum occurs when: $\cos(x - \alpha) = 1 \, \text{or} \, \cos(x - \alpha) = 0.$ \nThese give us:

$x - \alpha = 0 \Rightarrow x = \alpha \approx 1.176.$

So, the smallest positive value of x for which this maximum occurs is approximately: $x \approx 1.176.$

f(x) = 5 \, ext{cos} \, x + 12 \, ext{sin} \, x Given that f(x) = R \, ext{cos}(x - \alpha), \, ext{where} \, R > 0 \, ext{and} \, 0 < \alpha < \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

Worked Solution & Example Answer:f(x) = 5 \, ext{cos} \, x + 12 \, ext{sin} \, x Given that f(x) = R \, ext{cos}(x - \alpha), \, ext{where} \, R > 0 \, ext{and} \, 0 < \alpha < \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

find the value of R and the value of α to 3 decimal places.

Hence solve the equation 5 cos x + 12 sin x = 6 for 0 ≤ x < 2π.

Write down the maximum value of 5 cos x + 12 sin x.

Find the smallest positive value of x for which this maximum value occurs.

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