CIE May 2020 9709 Pure Maths Paper 32 (pdf)
- Find the quotient and remainder when 6x4 + x3 − x2 + 5x − 6 is divided by 2x2 − x + 1
- The variables x and y satisfy the equation y2 = Aekx, where A and k are constants. The graph of ln y against x is a straight line passing through the points (1.5, 1.2) and (5.24, 2.7) as shown in the diagram.
Find the values of A and k correct to 2 decimal places.
- Find the exact value of
- A curve has equation y = cos x sin 2x.
Find the x-coordinate of the stationary point in the interval 0 < x < 1/2 π, giving your answer correct to 3 significant figures.
- (a) Express √2 cos x − √5 sin x in the form R (cosx + α), where R > 0 and 0° < α < 90°. Give the exact value of R and the value of α correct to 3 decimal places
(b) Hence solve the equation √2 cos 2θ − √5 sin 2θ = 1, for 0° < θ < 180°.
- The diagram shows the curve y , for x ≥ 0, and its maximum point M.
(a) Find the x-coordinate of M, giving your answer correct to 3 decimal places
(b) Using the substitution u = √3x2, find by integration the exact area of the shaded region bounded by the curve, the x-axis and the line x = 1.
- The variables x and y satisfy the differential equation
It is given that y = 2 when x = 0.
Solve the differential equation, obtaining an expression for y in terms of x.
- (a) Solve the equation (1 + 2i)w + iw* = 3 + 5i. Give your answer in the form x + iy, where x and y are real.
(b) (i) On a sketch of an Argand diagram, shade the region whose points represent complex
numbers z satisfying the inequalities |z − 2 − 2i| ≤ 1 and arg(z − 4i) ≥ −1/4 π
(ii) Find the least value of Im z for points in this region, giving your answer in an exact for
- The diagram shows the curves y = cos x and y = k/(1 + x), where k is a constant, for 0 ≤ x ≤ 1/2 π. The curves touch at the point where x = p.
(a) Show that p satisfies the equation tan p = 1/(1 + p).
(b) Use the iterative formula
to determine the value of p correct to 3 decimal
places. Give the result of each iteration to 5 decimal places.
(c) Hence find the value of k correct to 2 decimal places.
- With respect to the origin O, the points A and B have position vectors given by
OA = 6i + 2j and
OB = 2i + 2j + 3k. The midpoint of OA is M. The point N lying on AB, between A and B, is such
that AN = 2NB.
(a) Find a vector equation for the line through M and N.
The line through M and N intersects the line through O and B at the point P.
(b) Find the position vector of P
(c) Calculate angle OPM, giving your answer in degrees
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