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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 $x > 0$ - AQA - A-Level Maths: Mechanics - 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 $x > 0$ - AQA - A-Level Maths: Mechanics - Question 7 - 2019 - Paper 1
Step 1
By sketching the graphs of $y = \frac{1}{x}$ and $y = \sec 2x$
Answer
To sketch the graphs of ( y = \frac{1}{x} ) and ( y = \sec 2x ), we start by noting the key features of each function. ( y = \frac{1}{x} ) has a vertical asymptote at and decreases from infinity to 0 as increases, never touching the x-axis. The function ( y = \sec 2x ) has vertical asymptotes where ( \cos 2x = 0 ), occurring at ( x = \frac{(2n+1)\pi}{4} ), and oscillates between its peaks of ±1.
When we overlay these two graphs, they intersect at only one point in the first quadrant, showing that there is exactly one solution for ( x > 0 ).
Step 2
By considering a suitable change of sign, show that the solution lies between 0.4 and 0.6
Answer
Let ( f(x) = \sec 2x - \frac{1}{x} ). We evaluate ( f(0.4) ) and ( f(0.6) ) to find:
[ f(0.4) = \sec(0.8) - 2.5 \approx 1.515 - 2.5 = -0.985 ] [ f(0.6) = \sec(1.2) - 1.6667 \approx 3.077 - 1.6667 = 1.4103 ]
Since ( f(0.4) < 0 ) and ( f(0.6) > 0 ), by the Intermediate Value Theorem, there exists at least one solution in the interval ( (0.4, 0.6) ).
Step 3
Show that the equation can be rearranged to give $x = \frac{1}{2} \cos^{-1} x$
Answer
Starting from the original equation ( \frac{1}{x} = \sec 2x ), we can multiply both sides by ( x ) to obtain ( 1 = x \sec 2x ). Using the identity ( \sec 2x = \frac{1}{\cos 2x} ), we rearrange it to:
[ x = \frac{1}{\cos 2x} ]
Then we use the cosine inverse:
[ \cos 2x = \frac{1}{x} \implies 2x = \cos^{-1}(\frac{1}{x}) \implies x = \frac{1}{2} \cos^{-1}(\frac{1}{x}). ]
Step 4
Use the iterative formula $x_{n+1} = \frac{1}{2} \cos^{-1} x_n$
Answer
Using the initial value ( x_1 = 0.4 ):
- Calculate ( x_2 = \frac{1}{2} \cos^{-1}(0.4) \approx 0.5794 )
- Calculate ( x_3 = \frac{1}{2} \cos^{-1}(0.5794) \approx 0.4763 )
- Calculate ( x_4 = \frac{1}{2} \cos^{-1}(0.4763) \approx 0.5372 )
Thus, the values found are:
- .
Step 5
On the graph below, draw a cobweb or staircase diagram
Answer
The cobweb diagram should start at , where the first point is plotted. Then, move vertically to the line ( y = \frac{1}{2} \cos^{-1}(x) ) and then horizontally to the line ( y = x ). Repeat this for , , and to visually demonstrate the convergence towards the solution. Make sure to annotate the positions of , , and accurately on the diagram.