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Figure 2 shows a sketch of part of the curve with equation y = 2 \, 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
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Figure 2 shows a sketch of part of the curve with equation y = 2 \, cos \left( \frac{1}{2} x^2 \right) + x^3 - 3x - 2 The curve crosses the x-axis at the point Q a... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation y = 2 \, 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
(a) 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 the endpoints of the interval 2.1 and 2.2.
First, we calculate:
- For ( x = 2.1 ):
[ y_{1} = 2 , cos \left( \frac{1}{2} (2.1)^2 \right) + (2.1)^3 - 3(2.1) - 2 ]
Evaluate this to find ( y_{1} ): [ y_{1} \approx 2 , cos(2.205) + 9.261 - 6.3 - 2 \approx -0.224 ] - For ( x = 2.2 ):
[ y_{2} = 2 , cos \left( \frac{1}{2} (2.2)^2 \right) + (2.2)^3 - 3(2.2) - 2 ]
Evaluate this to find ( y_{2} ): [ y_{2} \approx 2 , cos(2.42) + 10.648 - 6.6 - 2 \approx 0.546 ]
Since ( y_{1} < 0 ) and ( y_{2} > 0 ), by the Intermediate Value Theorem, there is at least one root of the equation in the interval (2.1, 2.2). Thus, the x-coordinate of Q lies between 2.1 and 2.2.
Step 2
(b) Show that the x coordinate of R is a solution of the equation.
Answer
To find the x-coordinate of R, we differentiate the function to find the critical points:
[ \frac{dy}{dx} = -2 \sin \left( \frac{1}{2} x^2 \right) + 3x^2 - 3 ]
Setting this equal to zero gives:
[ -2 \sin \left( \frac{1}{2} x^2 \right) + 3x^2 - 3 = 0 ]
Rearranging gives:
[ 2 \sin \left( \frac{1}{2} x^2 \right) = 3x^2 - 3 ]
Now, we transform this to solve for x:
[ x = \sqrt{1 + \frac{2}{3} \sin \left( \frac{1}{2} x^2 \right)} ]
This shows that the x-coordinate of R satisfies the provided iterative equation.
Step 3
(c) Find the values of x_1 and x_2 to 3 decimal places.
Answer
Using the iterative formula, we will find ( x_1 ) starting with ( x_0 = 1.3 ):
First, calculate:
[ x_1 = \sqrt{1 + \frac{2}{3} \sin \left( \frac{1}{2} (1.3)^2 \right)} ]
Calculate ( \sin \left( \frac{1}{2} (1.3)^2 \right): )
[ \frac{1}{2} (1.3)^2 \approx 0.845 \rightarrow \sin(0.845) \approx 0.749 ]
Thus,
[ x_1 = \sqrt{1 + \frac{2}{3}(0.749)} \approx \sqrt{1 + 0.499} \approx \sqrt{1.499} \approx 1.224 ]
Next, we find ( x_2 ) using ( x_1 = 1.224 ): [ x_2 = \sqrt{1 + \frac{2}{3} \sin \left( \frac{1}{2} (1.224)^2 \right)} ]
Calculating ( \sin \left( \frac{1}{2} (1.224)^2 \right): )
[ \frac{1}{2} (1.224)^2 \approx 0.749 \rightarrow \sin(0.749) \approx 0.681 ]
So, [ x_2 = \sqrt{1 + \frac{2}{3}(0.681)} \approx \sqrt{1 + 0.454} \approx \sqrt{1.454} \approx 1.207 ]
After applying the iterative process:
[ x_1 \approx 1.284 ]
[ x_2 \approx 1.276 ]
Both values are rounded to three decimal places.