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Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − (3/2)x The point P with x coordinate −2 lies on C - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4
Question 7
Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − (3/2)x The point P with x coordinate −2 lies on C. (a) Find an equation of the norm... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − (3/2)x The point P with x coordinate −2 lies on C - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4
Step 1
Find an equation of the normal to C at P.
Answer
To find the equation of the normal at point P, we first determine the coordinates of P when x = -2:
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Calculate y:
y = 2ln(2(-2) + 5) − (3/2)(-2) = 2ln(1) + 3 = 3.
Thus, point P is (-2, 3).
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Calculate the derivative of the curve:
rac{dy}{dx} = rac{4}{2x + 5} - rac{3}{2}.
Substituting x = -2 gives:
rac{dy}{dx} |_{x=-2} = rac{4}{1} - rac{3}{2} = 4 - 1.5 = 2.5.
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The slope of the normal is the negative reciprocal of the tangent slope:
m_{normal} = - rac{1}{2.5} = - rac{2}{5}.
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Using point-slope form:
y - 3 = - rac{2}{5}(x + 2).
Rearranging gives:
2x + 5y = 11 (or in the form ax + by = c where a = 2, b = 5, c = 11).
Step 2
Show that the x coordinate of Q is a solution of the equation x = 20/11 ln(2x + 5) − 2.
Answer
To show this, we derive the equation from the normal's intersection with the curve:
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Start with the equation of the normal:
5y + 2x = 11.
Substitute for y from the curve equation:
5(2ln(2x + 5) − (3/2)x) + 2x = 11.
Simplifying gives:
10ln(2x + 5) - (15/2)x + 2x = 11.
Combining terms yields:
10ln(2x + 5) = 11 + rac{15}{2}x - 2x.
Rearranging results in:
x = rac{20}{11}ln(2x + 5) - 2.
Step 3
Taking x_{1} = 2, find the values of x_{n} and x_{n+1}, giving each answer to 4 decimal places.
Answer
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Start with x_{1} = 2:
Calculate x_{2}:
x_{2} = rac{20}{11}ln(2(2) + 5) - 2 = rac{20}{11}ln(9) - 2 ≈ 1.9950.
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Calculate x_{3}:
Using x_{2}, find x_{3}:
x_{3} = rac{20}{11}ln(2(1.9950) + 5) - 2 ≈ 1.9920.
Thus, the approximate values are:
x_{2} ≈ 1.9950 x_{3} ≈ 1.9920.