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By sketching the graphs of y = \frac{1}{x} and y = \sec{2x} on the axes below, show that the equation \frac{1}{x} = \sec{2x} has exactly one solution for x > 0 By considering a suitable change of sign, show that the solution to the equation lies between 0.4 and 0.6 - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 1
Question 7
By sketching the graphs of y = \frac{1}{x} and y = \sec{2x} on the axes below, show that the equation \frac{1}{x} = \sec{2x} has exactly one solution for x ... show full transcript
Worked Solution & Example Answer:By sketching the graphs of y = \frac{1}{x} and y = \sec{2x} on the axes below, show that the equation \frac{1}{x} = \sec{2x} has exactly one solution for x > 0 By considering a suitable change of sign, show that the solution to the equation lies between 0.4 and 0.6 - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 1
Step 1
By sketching the graphs of y = \frac{1}{x} and y = \sec{2x}, show that the equation \frac{1}{x} = \sec{2x} has exactly one solution for x > 0
Answer
To begin, sketch the graph of (y = \frac{1}{x}) for (x > 0). This graph is a hyperbola that approaches the axes but never touches them, with its curve in the first quadrant. Next, sketch (y = \sec(2x)), which oscillates and has vertical asymptotes at (x = \frac{\pi}{2}n) for odd integers n. Notably, in the interval (0, (\frac{\pi}{2})), these two graphs will intersect. Since (y = \sec(2x)) increases rapidly towards the vertical asymptotes, we can observe that (y = \frac{1}{x}) starts from infinity at (x = 0) and decreases towards 0 as (x) increases, while (y = \sec(2x)) has multiple peaks within this range but does not exceed 1. Hence, there is exactly one point of intersection where (x > 0), establishing that the equation has exactly one positive solution.
Step 2
By considering a suitable change of sign, show that the solution to the equation lies between 0.4 and 0.6.
Answer
Define a function (f(x) = \sec(2x) - \frac{1}{x}). We evaluate this function at two points:
- For (x = 0.4), we find (f(0.4) = \sec(0.8) - 2.5 \approx 0.698 - 2.5 < 0).
- For (x = 0.6), we get (f(0.6) = \sec(1.2) - \frac{5}{3} \approx 1.357 - 1.666 > 0). Thus, we have (f(0.4) < 0) and (f(0.6) > 0), indicating a change of sign. By the Intermediate Value Theorem, there exists at least one root between (0.4) and (0.6).
Step 3
Show that the equation can be rearranged to give x = \frac{1}{2} \cos^{-1}{x}
Answer
Starting with the original equation (\sec(2x) = \frac{1}{x}), we can rearrange it as follows:
- Take the reciprocal: (x = \cos(2x)).
- Rearranging gives ( \cos(2x) = x), and utilizing the double-angle identity for cosine results in: [x = \frac{1}{2} \cos^{-1}{x}.]
Step 4
Use the iterative formula x_{n+1} = \frac{1}{2} \cos^{-1}{x_n} with x_1 = 0.4 to find x_2, x_3, and x_4, giving your answers to four decimal places.
Answer
Using the formula with (x_1 = 0.4):
- Calculate: [x_2 = \frac{1}{2} \cos^{-1}{0.4} \approx 0.5796]
- Next, with (x_2 = 0.5796): [x_3 = \frac{1}{2} \cos^{-1}{0.5796} \approx 0.4763]
- Finally, with (x_3 = 0.4763): [x_4 = \frac{1}{2} \cos^{-1}{0.4763} \approx 0.5372] Thus, the results are: (x_2 \approx 0.5796), (x_3 \approx 0.4763), and (x_4 \approx 0.5372).
Step 5
On the graph below, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of x_2, x_3, and x_4.
Answer
In the cobweb or staircase diagram:
- Begin at the point (x_1 = 0.4) on the x-axis.
- Move vertically to (y = \frac{1}{2} \cos^{-1}(0.4)) to find (x_2) and mark this point.
- From (x_2), draw a horizontal line to the line (y = x) which intersects at (x_2).
- Repeat for (x_3) and (x_4) to continue the process, illustrating the convergence towards the solution on the diagram clearly.