Figure 1 shows a plot of part of the curve with equation $y = ext{cos}\, x$ where $x$ is measured in radians. (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. (2) 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. (3)

A-Level Edexcel Maths Pure Circles: 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. (a) Use Diagram 1 to show why the equation $$ ext{cos... 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 $ ext{cos}\, x - 2x - \frac{1}{2} = 0$ has only one real root

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Answer

To demonstrate that the equation has only one real root, we start by analyzing the graph presented in Diagram 1. The function $y = ext{cos}\, x$ oscillates between -1 and 1, while the line $y = 2x + \frac{1}{2}$ is a linear function with a positive slope.

As $x$ increases, $ext{cos}\, x$ decreases and the line slowly rises. Notably, the graph shows that they intersect at most once because:

The cosine function is periodic and continuous, while the line is not oscillatory.
Since $ext{cos}\, x$ has a maximum of 1 at $x=0$ , the linear function is only able to bridge the variability of the cosine curve at one point before $ext{cos}\, x$ dips below the line's height due to this intersection condition.

Thus, we conclude that there is only one point where $y = ext{cos}\, x$ intersects with $y = 2x + \frac{1}{2}$ .

Step 2

use the small angle approximation for $\text{cos}\, x$

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Answer

To estimate the value of $\alpha$ using the small angle approximation, we utilize the approximation $\text{cos}\, x \approx 1 - \frac{x^2}{2}$ for small values of $x$ .

Substituting this into our equation:

$\text{cos}\, \alpha - 2\alpha - \frac{1}{2} = 0 \implies \left(1 - \frac{\alpha^2}{2}\right) - 2\alpha - \frac{1}{2} = 0$

This simplifies to:

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

Rearranging gives us:

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

To solve for $\alpha$ , we can use the quadratic formula:

$\alpha = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-4 \pm \sqrt{{16 + 4}}}}{2} = \frac{{-4 \pm \sqrt{{20}}}}{2} = -2 \pm \sqrt{5}$

Given that $\alpha$ is small, we take the positive root:

$\alpha = -2 + \sqrt{5}$

Finally, estimating $\sqrt{5} \approx 2.236$ , we find:

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

Rounding this to three decimal places yields $\alpha \approx 0.236$ .

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 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 $ ext{cos}\, x - 2x - \frac{1}{2} = 0$ has only one real root

use the small angle approximation for $\text{cos}\, x$

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