Figure 1 shows a plot of part of the curve with equation $y = ext{cos}(x)$ where $x$ is measured in radians. Diagram 1, on the opposite page, is a copy of Figure 1. (a) Use Diagram 1 to show why the equation $$ ext{cos}(x) - 2x - rac{1}{2} = 0$$ has only one real root, giving a reason for your answer. Given that the root of the equation is $\alpha$, and that $\alpha$ is small, (b) use the small angle approximation for $\text{cos}(x)$ to estimate the value of $\alpha$ to 3 decimal places.

A-Level Edexcel Maths Pure Binomial Expansion: Figure 1 shows a plot of part of

Figure 1 shows a plot of part of the curve with equation $y = ext{cos}(x)$ where $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 4 - 2019 - Paper 1

Question 4

Figure 1 shows a plot of part of the curve with equation $y = ext{cos}(x)$ where $x$ is measured in radians. Diagram 1, on the opposite page, is a copy of Figure 1.... show full transcript

Worked Solution & Example Answer:Figure 1 shows a plot of part of the curve with equation $y = ext{cos}(x)$ where $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 4 - 2019 - Paper 1

Step 1

Use Diagram 1 to show why the equation $ \cos(x) - 2x - \frac{1}{2} = 0 $ has only one real root

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Answer

To determine the number of real roots of the equation ( \cos(x) - 2x - \frac{1}{2} = 0 ), we first need to analyze the graphs of ( y = \cos(x) ) and ( y = 2x + \frac{1}{2} ) using Diagram 1. The curve ( y = \cos(x) ) oscillates between -1 and 1, while the linear function ( y = 2x + \frac{1}{2} ) has a positive slope of 2 and intersects the y-axis at ( y = \frac{1}{2} ).

By examining the intersection points, we observe that the linear function ascends steeply and crosses the cosine curve only once in the range shown in Diagram 1. Since the cosine function oscillates and the linear function continuously increases, there will be no other points of intersection. Hence, the equation has only one real root.

Step 2

Estimate the value of $ \alpha $ using the small angle approximation for $ \text{cos}(x) $

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Answer

For small values of ( \alpha ), we can use the small angle approximation:

$\cos(x) \approx 1 - \frac{x^2}{2}$

Substituting this into our equation gives:

$1 - \frac{\alpha^2}{2} - 2\alpha - \frac{1}{2} = 0$

Simplifying, we obtain:

$\frac{\alpha^2}{2} + 2\alpha - \frac{1}{2} = 0$

Multiplying through by 2 to eliminate the fraction results in:

$\alpha^2 + 4\alpha - 1 = 0$

Using the quadratic formula, ( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), we find:

$\alpha = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$

By evaluating ( -2 + \sqrt{5} \approx -2 + 2.236 \approx 0.236 ) (selecting the positive root since ( \alpha ) must be small and positive), we estimate that:

$\alpha \approx 0.236$

Thus, rounding to three decimal places, we find:

$\alpha \approx 0.236$ .

Figure 1 shows a plot of part of the curve with equation $y = ext{cos}(x)$ where $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 4 - 2019 - Paper 1

Worked Solution & Example Answer:Figure 1 shows a plot of part of the curve with equation $y = ext{cos}(x)$ where $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 4 - 2019 - Paper 1

Use Diagram 1 to show why the equation \( \cos(x) - 2x - \frac{1}{2} = 0 \) has only one real root

Estimate the value of \( \alpha \) using the small angle approximation for \( \text{cos}(x) \)

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