Edexcel GCE Core Maths C4 January 2012
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C4 Edexcel Core Mathematics January 2012 Question 6
6. Figure 3 shows a sketch of the curve with equation y = 2 sin2x/(1 + cos x), 0 ≤ x ≤ π/2
The finite region R, shown shaded in Figure 3, is bounded by the curve and the x-axis.
The table below shows corresponding values of x and y for y = 2 sin2x/(1 + cos x)
(a) Complete the table above giving the missing value of y to 5 decimal places.
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an estimate for the area of R, giving your answer to 4 decimal places.
(c) Using the substitution u = 1 + cos x , or otherwise, show that
∫2 sin2x/(1 + cos x) dx = 4ln(1 + cos x) - 4cosx + k
where k is a constant.
(d) Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.
6 (a)(b) Trapezium Rule7. Relative to a fixed origin O, the point A has position vector ( 2i – j + 5k ), the point B has position vector ( 5i + 2j + 10k ), and the point D has position vector ( –i + j + 4k ).
The line l passes through the points A and B.
(a) Find the vector AB.
(b) Find a vector equation for the line l.
(c) Show that the size of the angle BAD is 109°, to the nearest degree.
The points A, B and D, together with a point C, are the vertices of the parallelogram ABCD, where AB = DC.
(d) Find the position vector of C.
(e) Find the area of the parallelogram ABCD, giving your answer to 3 significant figures.
(f) Find the shortest distance from the point D to the line l, giving your answer to 3 significant figures.
7 (a) Vectors7 (b) Vector Equation
7 (d)
7 (f)
8. (a) Express 1/[P(5-P)] in partial fractions.
A team of conservationists is studying the population of meerkats on a nature reserve. The population is modelled by the differential equation
dP/dt = 1/15 P(5 - P), t ≥ 0
where P, in thousands, is the population of meerkats and t is the time measured in years since the study began.
Given that when t = 0, P = 1,
(b) solve the differential equation, giving your answer in the form,
P = /(b + ce-1/3t) where a, b and c are integers.
(c) Hence show that the population cannot exceed 5000.
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