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

A-Level Edexcel Maths Pure Rational Expressions: 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 3 - 2019 - Paper 1

Question 3

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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 3 - 2019 - Paper 1

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 demonstrate that the equation has only one real root, we analyze the graph of $y = ext{cos} \, x$ and the line $y = 2x + \frac{1}{2}$ . Noticeably, the cosine function oscillates between -1 and 1, while the linear function $y = 2x + \frac{1}{2}$ is a straight line with a positive slope.

From the graph, we find that the two graphs intersect only once. The height of the cosine function (maximum 1) compared to the increasing linear function indicates that after this intersection point, $y = 2x + \frac{1}{2}$ will always be above the cosine curve. Therefore, we conclude that there is only one intersection point, hence 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

For small angles, we use the approximation: $\text{cos} \, x \approx 1 - \frac{x^2}{2}$ Substituting into the equation: $1 - \frac{x^2}{2} - 2x - \frac{1}{2} = 0$ This simplifies to: $-\frac{x^2}{2} - 2x + \frac{1}{2} = 0$ Multiplying through by -2 gives: $x^2 + 4x - 1 = 0$ Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$ , $b = 4$ , $c = -1$ : $x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$ Since $\alpha$ is small, we take the negative root: $\alpha = -2 + \sqrt{5}\approx -2 + 2.236 \approx 0.236$ Thus, the estimated value of $\alpha$ to 3 decimal places is $0.236$ .

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 3 - 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 3 - 2019 - Paper 1

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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