OxfordAQA International AS and A-level Mathematics: Differentiation: gradients, tangents, normals and stationary points (9660) – 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 cover specification section P1.3 (Differentiation) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The content is part of International AS Unit P1 and also counts towards the International A-level. Learn the method first from the study guide or the revision notes. See also the course hub, the printable checklist and the free diagnostics. Mixed Unit P1 work is in the Unit P1 practice set.

Questions

1. (a) The chord joining the points with x = 3 and x = 3.01 on the curve y = x² − 2x has gradient 4.01. State what value the chord gradient approaches as the second point moves towards the first, and what that value represents. [2] (b) Given f(x) = 7x³ − 12√x + 5/x, find f’(x). [3]

2. Find the gradient of the curve y = (6x² − 4√x)/x at the point where x = 4. [4]

3. The curve y = x³ − 2x² + 4 passes through the point where x = −1. Find the equation of the tangent and the equation of the normal at this point. Give each answer in the form ax + by + c = 0 with integer coefficients. [6]

4. The curve y = x³ − 6x² + 5x has tangents parallel to the line y = 9 − 4x at two points. Find the coordinates of both points. [4]

5. (a) The function f is given by f(x) = 2x³ + 3x² − 72x + 1. Find the set of values of x for which f is decreasing. [4] (b) Show that g(x) = x³ + 3x² + 7x + 4 is increasing for every real value of x. [3]

6. Find the coordinates of the stationary points of y = 2x + 18/x. Determine the nature of each. [7]

7. The volume of water, V litres, in a tank t minutes after filling starts is modelled by V = 200 + 12t^(3/2), for 0 ≤ t ≤ 25. (a) Find dV/dt. [2] (b) Find the value of dV/dt when t = 16 and state its units. [2] (c) Find the time at which the tank is filling at 45 litres per minute. [2]

8. A curve has equation y = x√x − 3x + 4, for x > 0. Find the coordinates of its stationary point and show that it is a minimum. [5]

9. A sheet of card measures 24 cm by 15 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box. (a) Show that the volume of the box is V = 4x³ − 78x² + 360x cm³. [2] (b) Find the value of x that maximises V and find this maximum volume. Prove that it is a maximum. [5]

10. The running cost of a delivery van is modelled as C pence per kilometre, where C = 2880/v + 0.2v and v is its steady speed in kilometres per hour, v > 0. (a) Find dC/dv. [2] (b) Find the speed at which the cost per kilometre is stationary. [3] (c) Show that this speed gives the minimum cost and find that minimum cost. [2]

11. The curve C has equation y = √x(12 − x), for x ≥ 0. (a) Show that y = 12x^(1/2) − x^(3/2), and find the coordinates of the points where C meets the x-axis. [2] (b) Find dy/dx. [2] (c) Find the coordinates of the maximum point of C, and justify that it is a maximum. [4] (d) Find the equation of the normal to C at the point where x = 9, in the form ax + by + c = 0 with integer coefficients. [3]

Answers

1. (a) The chord gradient tends to the gradient of the tangent to the curve at x = 3 [1]. Since dy/dx = 2x − 2, this is 4 [1]. (b) Rewrite as f(x) = 7x³ − 12x^(1/2) + 5x⁻¹. Then f’(x) = 21x² [1] − 6x^(−1/2) [1] − 5x⁻² [1]. f’(x) = 21x² − 6/√x − 5/x² [3] Examiner insight: in (a), say that the value is the tangent gradient, not just a number close to 4.01. In (b), rewrite the root and reciprocal as powers on the page.

2. Divide each term by x: y = 6x − 4x^(−1/2) [1]. Then dy/dx = 6 [1] + 2x^(−3/2) [1]. At x = 4, x^(−3/2) = 1/8, so dy/dx = 6 + 2/8 = 6.25 [1]. [4] Examiner insight: differentiating the top and bottom separately is a common slip on this kind of fraction. Splitting it first keeps every step to the power rule and gives a clean exact answer.

3. At x = −1, y = −1 − 2 + 4 = 1, so the point is (−1, 1) [1]. dy/dx = 3x² − 4x [1] = 3 + 4 = 7 [1]. Tangent: y − 1 = 7(x + 1), so 7x − y + 8 = 0 [1]. The normal gradient is −1/7 [1], so y − 1 = −(1/7)(x + 1), giving 7y − 7 = −x − 1, so x + 7y − 6 = 0 [1]. [6] Examiner insight: show the y-coordinate being calculated, because the point is needed for both lines. Clear the fraction and check that the coefficients are integers, as the question asks for that form.

4. The line has gradient −4 [1]. dy/dx = 3x² − 12x + 5, and setting this equal to −4 gives 3x² − 12x + 9 = 0 [1], so x² − 4x + 3 = 0 and x = 1 or x = 3 [1]. Substituting gives the points (1, 0) and (3, −12) [1]. [4] Examiner insight: the final step is easy to forget. Parallel tangents are found from the x-values, but the question asks for coordinates, so each x must be substituted back into the curve.

5. (a) f’(x) = 6x² + 6x − 72 [1] = 6(x − 3)(x + 4) [1]. f’(x) = 0 at x = −4 and x = 3 [1]. The quadratic is negative between its roots, so f is decreasing for −4 < x < 3 [1]. (b) g’(x) = 3x² + 6x + 7 [1] = 3(x + 1)² + 4 [1]. A square is at least 0, so g’(x) ≥ 4 > 0 for every x, and g is increasing everywhere [1]. [7] Examiner insight: for a “show that” write the completed-square form explicitly and then state the inequality in words. Saying “the discriminant is negative” can also work, but only if you show the value.

6. dy/dx = 2 − 18x⁻² [1]. Setting this to 0 gives x² = 9 [1], so x = 3 or x = −3 [1]. Then y = 12 and y = −12, giving the points (3, 12) and (−3, −12) [1]. d²y/dx² = 36x⁻³ [1]. At x = 3 this is 4/3 > 0, so (3, 12) is a minimum [1]. At x = −3 it is −4/3 < 0, so (−3, −12) is a maximum [1]. [7] Examiner insight: the maximum has a lower y-value than the minimum, which is possible because these are local turning points. Supporting each classification with its own second-derivative value gives a more convincing answer than a bare statement.

7. (a) dV/dt = 18 [1] t^(1/2) [1]. dV/dt = 18√t (b) At t = 16, dV/dt = 18 × 4 = 72 [1], in litres per minute [1]. (c) 18√t = 45, so √t = 2.5 [1] and t = 6.25 minutes [1]. [6] Examiner insight: units are part of the answer when a rate is interpreted. Writing “litres per minute” shows you know that dV/dt is the change in V per unit of t.

8. y = x^(3/2) − 3x + 4 [1]. dy/dx = (3/2)x^(1/2) − 3 [1]. Setting this to 0 gives x^(1/2) = 2, so x = 4 [1]. Then y = 8 − 12 + 4 = 0, so the point is (4, 0) [1]. d²y/dx² = (3/4)x^(−1/2) = 3/8 > 0 at x = 4, so it is a minimum [1]. [5] Examiner insight: rewriting x√x as x^(3/2) comes first, and the working should show it. A “show that” answer needs the second derivative’s value or a sign check, not only the word minimum.

9. (a) The base has sides (24 − 2x) and (15 − 2x) and the height is x, so V = x(24 − 2x)(15 − 2x) [1]. Expanding: (24 − 2x)(15 − 2x) = 360 − 78x + 4x², so V = 4x³ − 78x² + 360x [1]. (b) dV/dx = 12x² − 156x + 360 [1] = 12(x − 3)(x − 10), so x = 3 or x = 10 [1]. The card is only 15 cm wide, so x < 7.5 and x = 10 is rejected [1]. d²V/dx² = 24x − 156 = −84 < 0 at x = 3, so x = 3 gives a maximum [1]. V = 3 × 18 × 9 = 486 cm³ [1]. [7] Examiner insight: the domain argument is part of the solution. If you leave in x = 10, you have not used the physical limit on x.

10. (a) C = 2880v⁻¹ + 0.2v [1], so dC/dv = 0.2 − 2880v⁻² [1]. (b) Setting dC/dv = 0 gives 0.2 = 2880/v² [1], so v² = 14 400 [1] and v = 120 km/h (v > 0) [1]. (c) d²C/dv² = 5760v⁻³, which is positive for v > 0, so this is a minimum [1]. C = 2880/120 + 0.2 × 120 = 24 + 24 = 48 pence per kilometre [1]. [7] Examiner insight: the second derivative is positive for every v > 0, so the argument covers the whole domain at once. Give the final cost with its unit.

11. (a) √x(12 − x) = 12x^(1/2) − x × x^(1/2) = 12x^(1/2) − x^(3/2) [1]. It meets the x-axis when y = 0, at (0, 0) and (12, 0) [1]. (b) dy/dx = 6x^(−1/2) [1] − (3/2)x^(1/2) [1]. (c) Setting dy/dx = 0: 6x^(−1/2) = (3/2)x^(1/2) [1]. Multiply by x^(1/2): 6 = (3/2)x, so x = 4 [1]. Then y = 12(2) − 8 = 16, giving (4, 16) [1]. d²y/dx² = −3x^(−3/2) − (3/4)x^(−1/2) = −3/8 − 3/8 = −3/4 < 0, so it is a maximum [1]. (d) At x = 9: y = 3(12 − 9) = 9 [1]. dy/dx = 6/3 − (3/2)(3) = −5/2, so the normal gradient is 2/5 [1]. The normal is y − 9 = (2/5)(x − 9), so 5y − 45 = 2x − 18, giving 2x − 5y + 27 = 0 [1]. [11] Examiner insight: in a longer question, label each part and keep exact values throughout. Parts (c) and (d) both use the derivative from (b), so check it at one easy value before moving on.

Where marks are usually lost

• Differentiating before rewriting a root or reciprocal as a power.
• Losing a negative sign when differentiating x⁻² or x^(−1/2).
• Forgetting to subtract 1 from a fractional power.
• Writing a tangent or normal without finding the y-coordinate of the point.
• Using the tangent gradient for the normal, or taking the reciprocal without the minus sign.
• Giving only the x-values of stationary points when coordinates were requested.
• Stating “maximum” or “minimum” without a second-derivative value or a gradient test.
• Leaving a rejected value, such as one outside the physical domain, in the final answer.
• Dropping units on a rate of change.

Next steps

• Revise the method with the revision notes.
• Re-read the explanations in the study guide.
• Return to the OxfordAQA Mathematics hub and tick off the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: section 3.1, International AS Unit P1 (Pure Maths), content P1.3 Differentiation.