A-Level AQA Maths Pure Trigonometric Functions: By sketching the graphs of $y =

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 $x > 0$ - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 1

Question 7

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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 ... 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 $x > 0$ - 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 $x > 0$.

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Answer

To analyze this graphically, we can sketch the two functions. The function $y = \frac{1}{x}$ is a hyperbola that approaches the x-axis as $x$ increases and has a vertical asymptote at $x = 0$ . The function $y = \sec 2x$ oscillates and has vertical asymptotes where $ext{cos}(2x) = 0$ , specifically at $x = \frac{(2n+1)\pi}{4}$ for integer $n$ . In the interval $0 < x < \frac{\pi}{2}$ , the two graphs will intersect once because $\sec 2x$ approaches positive infinity near the asymptotes while $\frac{1}{x}$ decreases to zero. Thus, there is exactly one intersection point for $x > 0$ .

Step 2

By considering a suitable change of sign, show that the solution to the equation lies between 0.4 and 0.6.

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Answer

Define the function $f(x) = \sec(2x) - \frac{1}{x}$ . We will evaluate $f$ at the endpoints $0.4$ and $0.6$ .

Calculating $f(0.4)$ : [ f(0.4) = \sec(0.8) - \frac{1}{0.4} \approx 1.152 - 2.5 < 0 ]

Calculating $f(0.6)$ : [ f(0.6) = \sec(1.2) - \frac{1}{0.6} \approx 1.32 - 1.6667 > 0 ]

Thus, since $f(0.4) < 0$ and $f(0.6) > 0$ , by the Intermediate Value Theorem, there is at least one root in the interval $(0.4, 0.6)$ .

Step 3

Show that the equation can be rearranged to give $ x = \frac{1}{2} \cos^{-1} x $.

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Answer

Starting from the original equation ( \frac{1}{x} = \sec(2x) ), we multiply by $x$ yielding: [ 1 = x \cdot \sec(2x) = \frac{x}{\cos(2x)} ] Rearranging gives: [ x\cos(2x) = 1 ] Now apply $\cos^{-1}$ : [ 2x = \cos^{-1}(1/x) ] Finally, we divide by 2: [ x = \frac{1}{2} \cos^{-1}(1/x) ] Thus achieving the desired rearrangement.

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.

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Answer

Start with $x_1 = 0.4$ : [ x_2 = \frac{1}{2} \cos^{-1}(0.4) \approx 0.5796 ]
Now calculate $x_3$ : [ x_3 = \frac{1}{2} \cos^{-1}(0.5796) \approx 0.4763 ]
Then calculate $x_4$ : [ x_4 = \frac{1}{2} \cos^{-1}(0.4763) \approx 0.5372 ] The final values are:

$x_2 \approx 0.5796$
$x_3 \approx 0.4763$
$x_4 \approx 0.5372$ .

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 $x_4$.

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Answer

In this diagram, we reflect points $(x_n, y = x_n)$ and the function $y = \frac{1}{2} \cos^{-1}(x)$ .

Start at $x_1 = 0.4$ and plot it on the $x$ -axis.
Move vertically to the curve $y = \frac{1}{2} \cos^{-1}(0.4)$ to find $x_2$ .
Draw a horizontal line to the $x$ -axis to locate $x_2$ .
Repeat the process from $x_2$ to find $x_3$ and from $x_3$ to find $x_4$ .
The intersections and points will show the convergence clearly.

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 $x > 0$ - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 1

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 $x > 0$ - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 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 $x > 0$.

By considering a suitable change of sign, show that the solution to the equation lies between 0.4 and 0.6.

Show that the equation can be rearranged to give \( x = \frac{1}{2} \cos^{-1} x \).

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.

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 $x_4$.

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