OCR A Level Further Maths B (MEI) Further Pure with Technology (minor option - choose one with a major, or three alone) June 2025 question paper Y436

A family of curves is given by the cartesian equation

where a, b and m are real numbers with a and b non-zero.

In this part of the question a = 2 and b = 1.

On the axes in the Printed Answer Booklet, sketch the curve in each of these cases.

• m = 0

• m = 1

• m = 2

State one feature of the curve for the case m = 0 that is not a feature of the curve in the

cases m = 1 and m = 2.

For the remainder of this question m = 0.

Verify that the parametric equations of the curve are

x () t = a cos () t, y () t = b sin () t,

where 0 ⩽ t < 2 π is a parameter.

Show that the equation of the normal to the curve at the point with parameter t is

Show that the parametric equations of the envelope of the normal to the curve are

where 0 ⩽ t < 2 π is a parameter.

In this part of the question a = 2 and b = 1.

On the axes in the Printed Answer Booklet, sketch the envelope of the normal to the curve.

By considering the expression (ax())t + (by())t or otherwise, determine a cartesian

equation of the envelope of the normal to the curve.

Write down 7¹⁰(mod 1000).

Fermat’s little theorem states that if p is a prime and x is an integer which is co-prime to p then x^p-1 ≡ 1(mod p).

Explain why Fermat’s little theorem implies 2¹² ≡ 1(mod 13).

Determine 2^(12q+1)(mod 13), where q is a positive integer.

In the rest of this question the highest common factor of positive integers m and n is denoted by (m, n).

Determine 7^(400r+10)(mod 1000), where r is a positive integer.

Using part (b)(iv), determine the tens digit of 7²⁰¹⁰.

Suppose that p ⩾ 3 is a prime number and that x and y are positive integers.

By considering the equation p^2x + 1 = 2^2y modulo 4, or otherwise, prove there are no integer

solutions to the equation p^2x + 1 = 2^2y.

This question concerns the family of differential equations

where a is a constant.

In this part of the question a = 0.

Verify that y = 2e^-x + x - 1 is the particular solution of (**) that satisfies y = 1 when x = 0

The solution y = 2e^-x + x - 1 has a minimum value at the point (m, n) where 0 < m < 1.

Find the exact value of m.

Sketch the particular solution of (**) given in part (a)(i) for 0 ⩽ x ⩽ 3 on the axes in the

Printed Answer Booklet.

The figure below shows the tangent field for an unspecified value of a. A sketch of the solution curve y = g(x) which passes through the point (0, 1) is shown for 0 ⩽ x ⩽

Continue the sketch of the solution curve for ⩽ x ⩽ 3 on the axes in the Printed Answer

Booklet.

The standard Runge-Kutta method of order 4 for the solution of the differential equation

Construct a spreadsheet to solve (**) so that the value of a and the value of h can be varied in the case x₀ = 0 and y₀ = 4. State the formulae you have used in your

spreadsheet.

In this part of the question a = 1, x₀ = 0 and y₀ = 4.

Use your spreadsheet with h = 0.05 to determine an approximation of the value of y for

the solution to (**) when x = 1, giving your answer correct to 2 decimal places.

In this part of the question a = 0.5, x₀ = 0 and y₀ = 4.

The solution to (**) has a minimum point (r, s) with 1 < r < 2.

Use your spreadsheet with suitable values of h to determine the value of r correct to

2 decimal places.

In this part of the question h = 0.05, x₀ = 0 and y₀ = 4.

There is a value of a > 0 such that y (1) is 2.47 to 2 decimal places.

By varying the value of a in your spreadsheet, find a suitable value of a correct to

1 decimal place.