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f(x) = 2 \\sin(x^2) + x - 2, \\ 0 \\leq x \\lt 2\pi (a) Show that f(x) = 0 has a root \( \alpha \) between x = 0.75 and x = 0.85 - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 3
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f(x) = 2 \\sin(x^2) + x - 2, \\ 0 \\leq x \\lt 2\pi (a) Show that f(x) = 0 has a root \( \alpha \) between x = 0.75 and x = 0.85. The equation f(x) = 0 can be wri... show full transcript
Worked Solution & Example Answer:f(x) = 2 \\sin(x^2) + x - 2, \\ 0 \\leq x \\lt 2\pi (a) Show that f(x) = 0 has a root \( \alpha \) between x = 0.75 and x = 0.85 - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 3
Step 1
Show that f(x) = 0 has a root \( \alpha \) between x = 0.75 and x = 0.85
Answer
To check for the existence of a root between the intervals, we evaluate the function at the endpoints:
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Calculate ( f(0.75) ): [ f(0.75) = 2 \sin(0.75^2) + 0.75 - 2 ] [ f(0.75) \approx -0.18 ]
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Calculate ( f(0.85) ): [ f(0.85) = 2 \sin(0.85^2) + 0.85 - 2 ] [ f(0.85) \approx 0.17 ]
Since ( f(0.75) < 0 ) and ( f(0.85) > 0 ), there is a change of sign in the interval ( [0.75, 0.85] ), indicating a root exists by the Intermediate Value Theorem.
Step 2
Use the iterative formula to find the values of \( x_1, \ x_2, \ and \ x_3 \)
Answer
Starting with ( x_0 = 0.8 ):
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To find ( x_1 ): [ x_1 = [\arcsin(1 - 0.5 \cdot 0.8)]^{\frac{1}{2}} ] After calculating, ( x_1 \approx 0.80219 ).
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To find ( x_2 ): [ x_2 = [\arcsin(1 - 0.5 \cdot 0.80219)]^{\frac{1}{2}} ] After calculating, ( x_2 \approx 0.80133 ).
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To find ( x_3 ): [ x_3 = [\arcsin(1 - 0.5 \cdot 0.80133)]^{\frac{1}{2}} ] After calculating, ( x_3 \approx 0.80167 ).
Step 3
Show that \( \alpha = 0.80157 \) is correct to 5 decimal places
Answer
To verify that ( \alpha = 0.80157 ) is correct to 5 decimal places, we evaluate the function at this point:
- Calculate ( f(0.80157) ): [ f(0.80157) = 2 \sin((0.80157)^2) + 0.80157 - 2 \approx -2.7 \times 10^{-5} ]
- Perform a similar calculation for ( f(0.801575) ): [ f(0.801575) \approx +8.6 \times 10^{-6} ]
Both calculations show a change of sign, confirming that ( \alpha ) lies between ( 0.80157 ) and ( 0.801575 ), supporting the conclusion that ( \alpha = 0.80157 ) is correct to 5 decimal places.