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 Modelling with Trigonometric Functions: 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 2 - 2019 - Paper 1

Question 2

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 2 - 2019 - Paper 1

Step 1

Use Diagram 1 to show why the equation cos x - 2x - 1/2 = 0 has only one real root, giving a reason for your answer.

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Answer

To demonstrate that the equation has only one real root, we can analyze the graphical representation of the function $y = ext{cos}(x) - 2x - rac{1}{2}$ in relation to the x-axis.

From Figure 1, we see that the curve $y = ext{cos}(x)$ oscillates between 1 and -1 while the line $y = 2x + rac{1}{2}$ is a straight line with a positive slope intersecting the y-axis above the x-axis.

At $x = 0$ , the value of $ext{cos}(0) = 1$ , giving us: $y = 1 - 0 - rac{1}{2} = rac{1}{2} > 0$ As $x$ increases towards $rac{1}{2}$ , the value of $y$ decreases. The line continues to rise, meaning it will intersect the oscillating curve at only one point because

The slope of the line is greater than the maximum decrease of the cosine function and,
The function $y = 2x + rac{1}{2}$ will eventually surpass the peaks of $y = ext{cos}(x)$ since it is linear in nature.

Thus, the equation only has one point of intersection, indicating a single real root.

Step 2

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

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Answer

To estimate the value of $\alpha$ using the small angle approximation, we start with: $\text{cos}(x) \approx 1 - \frac{x^2}{2}$ Given the equation: $\text{cos}(x) - 2x - \frac{1}{2} = 0$ Substituting the approximation: $1 - \frac{x^2}{2} - 2x - \frac{1}{2} = 0$ This simplifies to: $\frac{1}{2} - \frac{x^2}{2} - 2x = 0$ Multiplying through by 2 to eliminate the fraction gives: $1 - x^2 - 4x = 0$ Rearranging results in: $x^2 + 4x - 1 = 0$ Using the quadratic formula, we find: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \frac{-4 \pm \sqrt{16 + 4}}{2}= \frac{-4 \pm \sqrt{20}}{2}= -2 \pm \sqrt{5}$ Choosing the smaller root, since $\alpha$ is small: $\alpha = -2 + \sqrt{5}\approx -2 + 2.236 = 0.236$ Therefore, the estimate for $\alpha$ to three decimal places is: $\alpha \approx 0.236$ .

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 2 - 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 2 - 2019 - Paper 1

Use Diagram 1 to show why the equation cos x - 2x - 1/2 = 0 has only one real root, giving a reason for your answer.

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

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