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 $$ \cos x - 2x - \frac{1}{2} = 0 $$ 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 $\cos x$ to estimate the value of $\alpha$ to 3 decimal places.

A-Level Edexcel Maths Pure Parametric Equations: 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 2

Question 4

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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 ... 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 2

Step 1

Use Diagram 1 to show why the equation cos x - 2x - 1/2 = 0 has only one real root

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Answer

To show that the equation has only one real root, we can analyze the graphs of the functions involved. The function $y = \cos x$ oscillates between 1 and -1, while the line $y = 2x + \frac{1}{2}$ is linear and increasing. Observing Diagram 1, we see that the curve of $y = \cos x$ intersects the line $y = 2x + \frac{1}{2}$ only once. As $x$ increases, the value of $2x + \frac{1}{2}$ becomes greater than 1 (the maximum value of $\cos x$ ), indicating that there are no further intersections. Thus, there is only one real root for the equation.

Step 2

use the small angle approximation for cos x to estimate the value of α to 3 decimal places

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Answer

Using the small angle approximation $\cos x \approx 1 - \frac{x^2}{2}$ , we can rewrite the equation as follows:

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

This simplifies to:

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

Multiplying through by -2 for simplicity, we have:

$x^2 + 4x - 1 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=1$ , $b=4$ , and $c=-1$ :

$x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$

Since $\alpha$ is small and within a reasonable range, we take $\alpha = -2 + \sqrt{5}$ . Calculating this gives:

$\sqrt{5} \approx 2.236$ . Thus:

$\alpha \approx -2 + 2.236 \approx 0.236$

Therefore, the estimated value of $\alpha$ to three decimal places is approximately 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 2

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 2

Use Diagram 1 to show why the equation cos x - 2x - 1/2 = 0 has only one real root

use the small angle approximation for cos x to estimate the value of α to 3 decimal places

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