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June 2010 p32 q4
1837
The diagram shows the curve \(y = \frac{\sin x}{x}\) for \(0 < x \leq 2\pi\), and its minimum point \(M\).
(i) Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \tan x\).
(ii) The iterative formula \(x_{n+1} = \arctan(x_n) + \pi\) can be used to determine the \(x\)-coordinate of \(M\). Use this formula to determine the \(x\)-coordinate of \(M\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
Solution
(i) To find the \(x\)-coordinate of the minimum point \(M\), we need to find the derivative of \(y = \frac{\sin x}{x}\) and set it to zero.
Using the quotient rule, the derivative is:
\(y' = \frac{x \cos x - \sin x}{x^2}\).
Setting \(y' = 0\), we have:
\(x \cos x - \sin x = 0\).
Rearranging gives:
\(x \cos x = \sin x\).
Dividing both sides by \(\cos x\), we get:
\(x = \tan x\).
(ii) Using the iterative formula \(x_{n+1} = \arctan(x_n) + \pi\), we start with an initial guess and iterate:
1. \(x_1 = \arctan(x_0) + \pi\)
2. \(x_2 = \arctan(x_1) + \pi\)
Continue iterating until the value stabilizes to 4 decimal places.
The final \(x\)-coordinate of \(M\) is 4.49, correct to 2 decimal places.