Find OCR A Level Maths tutors

Across both full A Levels, the assessment objectives allocate about 50% to techniques (AO1), 25% to reasoning and communication (AO2), and 25% to problem solving and modelling (AO3), each with a two-percentage-point tolerance. Being able to carry out a calculation is only part of preparation.

The following constructed examples show the sort of difficulty a tutor can explore. They are illustrations, not OCR exam questions or accounts of a particular lesson.

Example: find and classify the stationary points of y = x³ − 3x² + 2. A stationary point has zero gradient. Differentiate to get dy/dx = 3x² − 6x = 3x(x − 2), then set this equal to zero: x = 0 or x = 2.

Stopping there misses the coordinates and the nature of the points. Substitute into the original function to get (0, 2) and (2, −2). The second derivative is 6x − 6: it is negative at x = 0, so (0, 2) is a local maximum; it is positive at x = 2, so (2, −2) is a local minimum.

For a follow-up, change the constant from +2 to +5. The derivative is unchanged, so the stationary points stay at the same x-values and keep their classification; their y-values rise by 3. Explaining that change shows understanding beyond repeating the method.

Example: test whether a coin favours heads. Before tossing it, choose a one-sided test at the 5% significance level. Let p be the probability of heads, assumed constant across 20 independent tosses. The null hypothesis is H₀: p = 0.5; the alternative is H₁: p > 0.5.

Suppose there are 15 heads. Under H₀, the number of heads X follows a binomial distribution with n = 20 and p = 0.5. Calculate P(X ≥ 15) = 1 − P(X ≤ 14) ≈ 0.0207. This includes results at least as extreme in the direction of the alternative; P(X = 15) alone is the wrong probability for this test.

Since 0.0207 is below 0.05, reject H₀. A suitable conclusion is: there is sufficient evidence at the 5% level that the coin favours heads. This does not prove bias, and 0.0207 is not the probability that the coin is fair.

With 14 heads, P(X ≥ 14) ≈ 0.0577, so the same test would not reject H₀. A useful check is whether the learner can choose the correct tail and explain that changed conclusion without prompting. OCR’s hypothesis-test guidance explains why the wording of the conclusion matters.

Example: a particle moves in a straight line with velocity v(t) = 6 − 2t metres per second, for 0 ≤ t ≤ 5 seconds. Positive velocity means motion in the chosen positive direction. The particle stops momentarily at t = 3 and then reverses.

Integrating velocity gives the signed displacement: an antiderivative is 6t − t², so from t = 0 to t = 5 the displacement is 30 − 25 = 5 m.

Distance counts the motion in both directions. Before the reversal, the particle travels 6(3) − 3² = 9 m. From t = 3 to t = 5, its displacement is 5 − 9 = −4 m, so it travels another 4 m. Total distance is therefore 13 m, although the final position is only 5 m from the starting position in the positive direction.

On a velocity–time graph, the area above the time axis is 9 and the area below has magnitude 4. Subtract them for displacement; add their magnitudes for distance. A tutor can use the graph to check why a learner who obtained 5 m for both quantities overlooked the reversal.

H640 Paper 3 includes an unseen passage about applications of pure mathematics. Mechanics and statistics knowledge is assumed, but is not the focus of that section. OCR recommends practising mathematical reading throughout the course, once the relevant content has been taught.

Consider this short invented passage: “A model gives the concentration of a dye as C = 80e^(−0.2t) mg/L, where t is time in hours. The model is intended for 0 ≤ t ≤ 5.” Start by identifying the quantity, units and stated time interval.

A learner might read −0.2 as a fixed amount lost each hour. Instead, each extra hour multiplies C by e^(−0.2) ≈ 0.819, a reduction of about 18.1%. Differentiating gives dC/dt = −16e^(−0.2t), so the initial rate is −16 mg/L per hour; that rate changes as the concentration falls.

At t = 5, the model gives C ≈ 29.4 mg/L. A calculator can also evaluate the formula at t = 10, but the passage supplies no basis for trusting the model beyond five hours. A stronger answer connects the calculation with the stated limitation.

This miniature example practises extracting meaning and explaining a result. Later practice should use full H640 comprehension questions, with unfamiliar wording and notation.

Use a free introductory meeting to discuss the learner’s goals, the tutor’s approach and practical arrangements before paid lessons. It is an introductory conversation, not normally a full teaching lesson. A recent attempted question is more useful than a general request to “get better at maths”.

Choose a rhythm that leaves time to use feedback. For example, a regular lesson with two short practice periods between sessions may be manageable; the right frequency depends on the learner’s starting point, deadline, workload and budget. A focused topic problem may need a different plan from gaps across the course.

Use the learner’s work to decide which kind of support comes next:

Latimer’s standard service is online one-to-one tuition. Microsoft Teams is the default platform, with another meeting method possible by agreement. Any in-person lessons need a separate arrangement with the tutor.

For maths, agree how to share handwritten work: a clear photo, a camera view of paper or a writing tablet may suit the learner. Have the question, calculator and relevant working ready. Check that both sides can read small details such as indices, minus signs and graph labels.

Agree lesson times and length with the tutor, along with how practice work and feedback will be handled. The how tutoring works guide explains the wider process.