Pearson Edexcel International Advanced Level Mathematics: Stationary points, definite integrals, areas and the trapezium rule (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow Unit P2: Pure Mathematics 2, sections 7.1 and 8.1 to 8.3 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). P1 and P2 are compulsory units for IAS and IAL Mathematics (and Pure Mathematics), as the specification states. Learn the content first from the study guide or the revision notes.

Course links: the Edexcel A Level Mathematics hub, the printable checklist and the free 10-minute diagnostics. Show full working throughout.

Questions

1. Find the coordinates of the stationary points of the curve y = x³ + 3x² − 45x + 2 and determine their nature. [4]

2. The function f is given by f(x) = 2x³ − 9x² − 24x. State the range of x over which f is increasing. [4]

3. A curve has equation y = 3x + 12/x for x ≠ 0. Find the coordinates of its stationary points and determine their nature. [6]

4. The curve C has equation y = (x + 1)²(x − 2).

(a) Find where C crosses or touches each axis, giving coordinates. [2] (b) Find the coordinates of the stationary points of C and determine their nature. [3] (c) Sketch C, labelling the points found in (a) and (b). [1]

5. A closed storage box has height h cm and a square base with sides of length x cm. Its six faces have a total area of 600 cm².

(a) Show that the volume of the box is V = 150x − x³/2 cm³. [3] (b) Find the value of x for which V is a maximum, justify that it is a maximum, and find this greatest volume. [4]

6. Evaluate ∫ from 1 to 9 of (x + 4)/√x dx. [4]

7. Given that k > 1 and ∫ from 1 to k of (4x − 3) dx = 10, find the value of k. [4]

8. The curve with equation y = 9x − x² and the line with equation y = 2x + 10 intersect at two points. Find the area of the finite region enclosed between them. [6]

9. Find the total area of the finite regions enclosed between the curve y = x(x + 1)(x − 3) and the x-axis. [6]

10. The curves y = 2x² − 6x and y = 6x − x² intersect at two points.

(a) Find the x-coordinates of the points of intersection. [2] (b) Find the area of the region enclosed between the curves. [4]

11. (a) Use the trapezium rule with 4 strips to estimate ∫ from 0 to 2 of x³ dx. [3]

(b) Find the exact value of the integral and state, with a reason, whether your answer to (a) is an over-estimate or an under-estimate. [2] (c) With 2 strips the trapezium rule gives 5. Compare the errors for 2 and 4 strips and say what happens to the error when the number of strips is doubled. [2]

12. The depth of a river, in metres, is measured at 1 m intervals across its 6 m width, starting at one bank.

Distance from bank (m)	0	1	2	3	4	5	6
Depth (m)	0	1.2	2.1	2.6	2.4	1.5	0

(a) Use the trapezium rule with 6 strips to estimate the cross-sectional area of the river. [3] (b) Use the trapezium rule with 3 strips, using only the depths at 0, 2, 4 and 6 m, to give a second estimate. [2] (c) Comment on the difference between your two estimates. [2] (d) Give one reason why the trapezium rule gives only an approximation here. [1]

Answers

1. dy/dx = 3x² + 6x − 45 = 3(x + 5)(x − 3) [1] dy/dx = 0 gives x = 3 or x = −5 [1] y = −79 when x = 3 and y = 177 when x = −5 [1] d²y/dx² = 6x + 6, which is 24 > 0 at x = 3 and −24 < 0 at x = −5. Minimum at (3, −79); maximum at (−5, 177) [1] [4] Examiner insight: Give both coordinates and say how you know the nature. A statement such as “it is a minimum” with no sign of d²y/dx² or of the gradient gives no evidence for the claim.

2. f′(x) = 6x² − 18x − 24 = 6(x − 4)(x + 1) [2] f is increasing where f′(x) > 0 [1] The quadratic opens upwards, so x < −1 or x > 4 [1] [4] Examiner insight: “Increasing” means f′(x) > 0, so say so. A single inequality such as −1 < x < 4 is the decreasing interval, and the answer must list two separate ranges.

3. y = 3x + 12x⁻¹, so dy/dx = 3 − 12x⁻² [1] 3 − 12/x² = 0 gives x² = 4, so x = ±2 [1] At x = 2, y = 12; at x = −2, y = −12 [1] d²y/dx² = 24x⁻³ = 24/x³ [1] At x = 2 it is 3 > 0, so (2, 12) is a minimum [1] At x = −2 it is −3 < 0, so (−2, −12) is a maximum [1] [6] Examiner insight: Both roots of x² = 4 are needed. Notice the “maximum” is lower than the “minimum”: these are local turning points, not the greatest and least values of y.

4. (a) y = 0 at x = −1 and x = 2 [1]; y = (1)(−2) = −2 when x = 0 [1] (b) y = x³ − 3x − 2, so dy/dx = 3x² − 3 = 3(x − 1)(x + 1) [1] Stationary points at x = 1, y = −4 and x = −1, y = 0 [1] d²y/dx² = 6x: −6 at x = −1 (maximum), 6 at x = 1 (minimum). Maximum (−1, 0); minimum (1, −4) [1] (c) Cubic with positive x³ coefficient; touches the x-axis at (−1, 0), crosses at (2, 0), passes through (0, −2) with minimum (1, −4) [1] [6] Examiner insight: A repeated root means the curve touches the axis at that point, and that point is also the turning point. In a sketch, shape and labelled features matter more than scale.

5. (a) Surface area: 2x² + 4xh = 600 [1] h = (300 − x²)/(2x) [1] V = x²h = x(300 − x²)/2 = 150x − x³/2 [1] (b) dV/dx = 150 − 3x²/2 [1] dV/dx = 0 gives x² = 100, so x = 10 (x > 0) [1] d²V/dx² = −3x = −30 < 0, so V is a maximum [1] V = 1500 − 500 = 1000 cm³ [1] [7] Examiner insight: A “show that” needs every step written, including where h comes from. In the second part, justifying the maximum is a separate requirement from solving dV/dx = 0.

6. (x + 4)/√x = x^(1/2) + 4x^(−1/2) [1] Antiderivative: (2/3)x^(3/2) + 8x^(1/2) [1] At 9: 18 + 24 = 42; at 1: 2/3 + 8 = 26/3 [1] 42 − 26/3 = 100/3 [1] [4] Examiner insight: Split the fraction before integrating. Substitute both limits, and leave the exact fraction unless a decimal is asked for.

7. [2x² − 3x] from 1 to k = (2k² − 3k) − (−1) [2] 2k² − 3k + 1 = 10 gives 2k² − 3k − 9 = 0 [1] (2k + 3)(k − 3) = 0, and k > 1, so k = 3 [1] [4] Examiner insight: The condition k > 1 is there to reject k = −3/2, so reject it explicitly. Leaving both roots usually loses the final credit.

8. 9x − x² = 2x + 10 gives x² − 7x + 10 = 0, so x = 2 and x = 5 [2] At x = 3 the curve gives 18 and the line gives 16, so the curve is above [1] Integrand: (9x − x²) − (2x + 10) = 7x − x² − 10 [1] [7x²/2 − x³/3 − 10x] from 2 to 5 = −25/6 − (−26/3) [1] Area = 9/2 [1] [6] Examiner insight: Check which graph is on top with a test value. Subtracting the wrong way gives a negative area, which signals the error but still has to be corrected.

9. Roots at x = −1, 0 and 3 [1] y = x³ − 2x² − 3x [1] F(x) = x⁴/4 − 2x³/3 − 3x²/2 [1] From −1 to 0: 0 − F(−1) = 0 − (−7/12) = 7/12, above the axis [1] From 0 to 3: F(3) − 0 = −45/4, below the axis, so area 45/4 [1] Total area = 7/12 + 135/12 = 71/6 [1] [6] Examiner insight: A single integral from −1 to 3 gives −32/3, which is the signed value and not an area. Treat each region separately so that the positive and negative parts do not cancel.

10. (a) 2x² − 6x = 6x − x² gives 3x² − 12x = 0, so x = 0 and x = 4 [2] (b) At x = 1 the curves give −4 and 5, so y = 6x − x² is the upper curve [1] Integrand: 12x − 3x² [1] [6x² − x³] from 0 to 4 = 96 − 64 [1] Area = 32 [1] [6] Examiner insight: Simplify (upper − lower) before integrating, because it halves the work. Make clear which curve you took as upper.

11. (a) h = 0.5, with y-values 0, 0.125, 1, 3.375, 8 [1] (0.5/2)[0 + 8 + 2(0.125 + 1 + 3.375)] [1] = 0.25 × 17 = 4.25 [1] (b) Exact: [x⁴/4] from 0 to 2 = 4 [1] 4.25 > 4, so over-estimate, because the curve bends upwards (y″ = 6x > 0) and the chords lie above it [1] (c) Error with 2 strips = 5 − 4 = 1; error with 4 strips = 0.25 [1] Doubling the strips cut the error to about a quarter [1] [7] Examiner insight: Give a reason tied to the shape of the curve. Saying “trapezia always overshoot” is wrong, because for a curve bending downwards the rule under-estimates.

12. (a) h = 1 [1] (1/2)[0 + 0 + 2(1.2 + 2.1 + 2.6 + 2.4 + 1.5)] [1] = 9.8 m² [1] (b) h = 2 and (2/2)[0 + 0 + 2(2.1 + 2.4)] [1] = 9.0 m² [1] (c) The estimates differ by 0.8 m² [1] The 6-strip value uses more measurements, so it is likely to be closer to the true area and the 3-strip value probably underestimates by about 0.8 m² [1] (d) The bed between measurements is curved, but the rule joins the points with straight lines. [1] [8] Examiner insight: A comment should refer to the numbers and to why one estimate is better. “The more accurate one is bigger” without a reason about strips does not use the information given.

Where marks are usually lost

• Giving x-values for stationary points and not the y-coordinates.
• Claiming a nature without a second derivative sign or a gradient check.
• Writing one inequality for an increasing function that has two ranges.
• Forgetting to rewrite 1/x or 1/√x as powers before differentiating or integrating.
• Substituting only the upper limit, or subtracting the limits the wrong way.
• Integrating over a root of the curve and quoting the signed value as an area.
• Omitting the intersection points, or integrating (lower − upper).
• Not justifying a minimum in an optimisation problem.
• Using the wrong number of y-values or the wrong h in the trapezium rule.
• Stating over- or under-estimate with no reason connected to the curve.

Next steps

• Go back to the revision notes for any rule you missed.
• Read the study guide for the full explanations.
• Browse the Edexcel A Level Mathematics hub and tick off topics on the printable checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, published by Pearson Education Limited. Unit P2 sections 7.1 and 8.1 to 8.3.