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Figure 2 shows a sketch of part of the curve with equation $$y = 2 \, ext{cos} \left( \frac{1}{2} x^{2} \right) + x^{3} - 3x - 2$$ The curve crosses the x-axis at the point Q and has a minimum turning point at R - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 5
Question 6
Figure 2 shows a sketch of part of the curve with equation $$y = 2 \, ext{cos} \left( \frac{1}{2} x^{2} \right) + x^{3} - 3x - 2$$ The curve crosses the x-axis at... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $$y = 2 \, ext{cos} \left( \frac{1}{2} x^{2} \right) + x^{3} - 3x - 2$$ The curve crosses the x-axis at the point Q and has a minimum turning point at R - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 5
Step 1
Show that the x coordinate of Q lies between 2.1 and 2.2.
Answer
To determine the x-coordinate of point Q where the curve crosses the x-axis, we need to evaluate the function at x = 2.1 and x = 2.2:
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Calculate (y(2.1)) and (y(2.2)):
(y(2.1) = 2 \cos\left( \frac{1}{2} (2.1)^{2} \right) + (2.1)^{3} - 3(2.1) - 2)
Calculating this gives: (y(2.1) \approx -0.224), which is negative.
Now, for (y(2.2):)
(y(2.2) = 2 \cos\left( \frac{1}{2} (2.2)^{2} \right) + (2.2)^{3} - 3(2.2) - 2)
This yields: (y(2.2) \approx 0.546), which is positive.
Since (y(2.1) < 0) and (y(2.2) > 0), by the Intermediate Value Theorem, there exists a root between 2.1 and 2.2, confirming that the x-coordinate of Q lies within this range.
Step 2
Show that the x coordinate of R is a solution of the equation
Answer
To verify that the x coordinate of R satisfies the equation provided, we differentiate the original equation:
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Differentiate: [\frac{dy}{dx} = -2\sin\left( \frac{1}{2} x^{2} \right) + 3x^{2} - 3.]
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We need to find R where the derivative equals to zero, let’s set: [\frac{dy}{dx} = -2\sin\left( \frac{1}{2} x^{2} \right) + 3x^{2} - 3 = 0.]
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Rearranging leads us to: [x = \sqrt{1 + \frac{2}{3} \sin\left( \frac{1}{2} x^{2}\right)}.]
Thus, we confirm that the x-coordinate of R satisfies this equation.
Step 3
Using the iterative formula find the values of x1 and x2 to 3 decimal places.
Answer
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Start the iteration with (x_{0} = 1.3).
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Using the iterative formula: [x_{1} = \sqrt{1 + \frac{2}{3} \sin\left( \frac{1}{2} (1.3)^{2}\right)}.]
Calculate: (x_{1} \approx 1.284).
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Now calculating for (x_{2}): [x_{2} = \sqrt{1 + \frac{2}{3} \sin\left( \frac{1}{2} (x_{1})^{2}\right)}]
Substituting (x_{1}) gives: (x_{2} \approx 1.276).
Thus, (x_{1} \approx 1.284) and (x_{2} \approx 1.276).