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f(x) = x² - 3x + 2 cos(\frac{x}{2}), \, 0 \leq x \leq \pi - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 6
Question 7
f(x) = x² - 3x + 2 cos(\frac{x}{2}), \, 0 \leq x \leq \pi. (a) Show that the equation f(x) = 0 has a solution in the interval 0.8 < x < 0.9. (b) The curve with... show full transcript
Worked Solution & Example Answer:f(x) = x² - 3x + 2 cos(\frac{x}{2}), \, 0 \leq x \leq \pi - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 6
Step 1
Show that the equation f(x) = 0 has a solution in the interval 0.8 < x < 0.9.
Answer
To find a solution for f(x) = 0 in the interval (0.8, 0.9), we first evaluate f at the endpoints:
-
Calculate f(0.8):
[ f(0.8) = (0.8)^2 - 3(0.8) + 2 \cos\left(\frac{0.8}{2}\right) ]
[ = 0.64 - 2.4 + 2 \cos(0.4) \approx 0.64 - 2.4 + 1.920 = 0.16 ] -
Calculate f(0.9):
[ f(0.9) = (0.9)^2 - 3(0.9) + 2 \cos\left(\frac{0.9}{2}\right) ]
[ = 0.81 - 2.7 + 2 \cos(0.45) \approx 0.81 - 2.7 + 1.870 = -0.02 ]
Since f(0.8) > 0 and f(0.9) < 0, and because f is continuous, by the Intermediate Value Theorem, there exists at least one solution to f(x) = 0 in the interval (0.8, 0.9).
Step 2
Show that the x-coordinate of P is the solution of the equation x = \frac{3 + sin(\frac{x}{2})}{2}.
Answer
The minimum point P occurs where the derivative f'(x) equals zero.
We begin by finding f'(x):
[ f'(x) = 2x - 3 - \frac{\sin(\frac{x}{2})}{2} ]
Setting f'(x) = 0 gives:
[ 0 = 2x - 3 - \frac{\sin(\frac{x}{2})}{2} \implies 2x = 3 + \frac{\sin(\frac{x}{2})}{2} \implies x = \frac{3 + sin(\frac{x}{2})}{2} ]
This establishes that the x-coordinate of point P is indeed described by the given equation.
Step 3
Using the iteration formula x_{n+1} = \frac{3 + sin(\frac{x_n}{2})}{2}, x_0 = 2, find the values of x_1, x_2, and x_3, giving your answers to 3 decimal places.
Answer
-
Start with x_0 = 2:
[ x_{1} = \frac{3 + \sin(1)}{2} \approx 1.921 ] -
Next, compute x_2 using x_1:
[ x_{2} = \frac{3 + \sin(\frac{1.921}{2})}{2} \approx 1.910 ] -
Finally, compute x_3 using x_2:
[ x_{3} = \frac{3 + \sin(\frac{1.910}{2})}{2} \approx 1.908 ]
Thus, the values are:
- x_1 = 1.921
- x_2 = 1.910
- x_3 = 1.908.
Step 4
By choosing a suitable interval, show that the x-coordinate of P is 1.9078 correct to 4 decimal places.
Answer
We will narrow down the solution with higher accuracy.
We previously found f(1.907) and f(1.908):
- f(1.907) gives a positive value,
- f(1.908) gives a negative value, hence f changes sign.
Evaluating further in the interval (1.907, 1.908):
- Check f(1.9075):
If it is found to be positive, then the solution lies between (1.9075, 1.908) which approximates 1.9078 correct to 4 decimal places.