OxfordAQA International GCSE Mathematics: Differentiation, tangents and stationary points (9260) – 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 OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), section 3.2.2, statements A13 to A15. Every question is Extension only. If you need the method first, read the study guide or the revision notes. Course pages: the OxfordAQA IGCSE Mathematics hub and the checklist. Want a quick check of your level? Try the free 10-minute diagnostics.

Questions

1. (Extension only)

(a) Differentiate y = 6x⁵ − 4x² + 9. [2] (b) Differentiate y = 3(x + 2)(x − 5). [3]

2. (Extension only) Given y = (8x⁵ + 12x³)/(4x²), find dy/dx and work out the gradient of the curve where x = −1. [4]

3. (Extension only) A curve has equation y = x³ − 5x².

(a) Find the gradient of the curve where x = 2. [2] (b) Find the x-coordinates of the points on the curve where the gradient is 8. [3]

4. (Extension only) Find the equation of the tangent to the curve y = 2x³ − 5x² + 4 at the point where x = 2. Give your answer in the form y = mx + c. [4]

5. (Extension only) A curve has equation y = x⁴ − 3x + 2.

(a) Find the equation of the tangent to the curve at the point where x = −1. [4] (b) This tangent crosses the x-axis at the point T. Find the coordinates of T. [2]

6. (Extension only) The curve C has equation y = x³ − 6x + 1. Two tangents to C are parallel to the line y = 6x + 3. Find the coordinates of the two points of contact and the equations of the two tangents. [7]

7. (Extension only) Find the coordinates of the stationary points of y = 2x³ − 9x² + 12x − 1 and decide whether each is a maximum or a minimum. [6]

8. (Extension only) Show that the curve y = x³ − 3x² + 3x + 5 has exactly one stationary point, and that it is a point of inflection. State its coordinates. [4]

9. (Extension only) A curve has equation y = x⁴ − 8x² + 3.

(a) Find the coordinates of the stationary points. [4] (b) Determine the nature of each stationary point. [3] (c) Describe a sketch of the curve, naming its key features. [1]

10. (Extension only) The curve y = x³ + px² + qx has stationary points where x = 1 and x = 3.

(a) Find the values of p and q. [4] (b) Find the coordinates of the two stationary points. [2] (c) Determine the nature of each stationary point. [2] (d) Describe a sketch of the curve that shows the position of the stationary points and the point where the curve meets the y-axis. [1]

11. (Extension only) A curve has equation y = x⁴ + 4x³.

(a) Find the coordinates of its stationary points. [3] (b) Determine the nature of each stationary point. [3] (c) Describe a sketch of the curve. [1]

Answers

1. (a) 6x⁵ → 30x⁴ [1]. −4x² → −8x, and the constant 9 disappears [1]. dy/dx = 30x⁴ − 8x (b) Expand: 3(x² − 3x − 10) = 3x² − 9x − 30 [2]. Differentiate: 6x − 9 [1]. [5] Examiner insight: Expanding the brackets first is the safe route and it shows your method. Differentiating each bracket separately and multiplying does not give the gradient function.

2. Divide each term by 4x²: y = 2x³ + 3x [2]. dy/dx = 6x² + 3 [1]. At x = −1: 6 × 1 + 3 = 9 [1]. [4] Examiner insight: Show the simplified form of y. If the division goes wrong, the line of working shows where, and an answer with no working has nothing to build on.

3. (a) dy/dx = 3x² − 10x [1]. At x = 2: 12 − 20 = −8 [1]. (b) 3x² − 10x = 8, so 3x² − 10x − 8 = 0 [1]. (3x + 2)(x − 4) = 0 [1]. x = 4 or x = −2/3 [1]. [5] Examiner insight: “Where the gradient is 8” means set dy/dx equal to 8. A common slip is to set y equal to 8, which uses the wrong equation.

4. At x = 2: y = 16 − 20 + 4 = 0 [1]. dy/dx = 6x² − 10x [1]. At x = 2: 24 − 20 = 4 [1]. y − 0 = 4(x − 2), so y = 4x − 8 [1]. [4] Examiner insight: Work out the point on the curve before the gradient. Here y is 0, so check that your final line also passes through (2, 0).

5. (a) At x = −1: y = 1 + 3 + 2 = 6 [1]. dy/dx = 4x³ − 3 [1]. At x = −1: −4 − 3 = −7 [1]. y − 6 = −7(x + 1), so y = −7x − 1 [1]. (b) On the x-axis y = 0, so −7x − 1 = 0 [1]. x = −1/7, so T = (−1/7, 0) [1]. [6] Examiner insight: Use the tangent from part (a) in part (b). Writing y = 0 and the coordinates of T as a pair makes the answer clear.

6. The tangents have gradient 6, the gradient of y = 6x + 3 [1]. dy/dx = 3x² − 6 [1]. 3x² − 6 = 6, so x² = 4 and x = 2 or x = −2 [1]. x = 2: y = 8 − 12 + 1 = −3. x = −2: y = −8 + 12 + 1 = 5. Points (2, −3) and (−2, 5) [2]. At (2, −3): y + 3 = 6(x − 2), so y = 6x − 15 [1]. At (−2, 5): y − 5 = 6(x + 2), so y = 6x + 17 [1]. [7] Examiner insight: Parallel lines have equal gradients. Check both roots of x² = 4, because stopping at x = 2 loses the second tangent.

7. dy/dx = 6x² − 18x + 12 = 6(x − 1)(x − 2) [1]. So x = 1 or x = 2 [1]. y at x = 1: 2 − 9 + 12 − 1 = 4. y at x = 2: 16 − 36 + 24 − 1 = 3. Points (1, 4) and (2, 3) [1]. Test x = 0, 1.5, 3: dy/dx = 12, −1.5, 12 [1]. Positive then negative at x = 1: (1, 4) is a maximum [1]. Negative then positive at x = 2: (2, 3) is a minimum [1]. [6] Examiner insight: Quote the gradient values you tested. A conclusion with no evidence is harder to credit than one backed by numbers.

8. dy/dx = 3x² − 6x + 3 = 3(x − 1)² [1]. This is zero only when x = 1, so there is one stationary point. y = 1 − 3 + 3 + 5 = 6, so it is (1, 6) [1]. At x = 0, dy/dx = 3. At x = 2, dy/dx = 3 [1]. The gradient is positive on both sides, so the point is a point of inflection, not a maximum or minimum [1]. [4] Examiner insight: On a “show that” question, state the reason as well as the result. Here that means saying the gradient does not change sign.

9. (a) dy/dx = 4x³ − 16x [1]. = 4x(x − 2)(x + 2) = 0, so x = 0, 2, −2 [1]. x = 0: y = 3 [1]. x = ±2: y = 16 − 32 + 3 = −13 [1]. Points (0, 3), (2, −13), (−2, −13). (b) Test x = −3, −1, 1, 3: dy/dx = −60, 12, −12, 60 [1]. At x = −2 the gradient goes − then +: minimum [1]. At x = 0 it goes + then −: maximum [1]. At x = 2 it goes − then +: minimum. (c) Symmetric about the y-axis, both ends rising, minima at (±2, −13), maximum at (0, 3), which is also the y-intercept [1]. [8] Examiner insight: Choose test values that sit between the stationary x-values. Testing x = 0.5 and x = 3 only would miss the sign change at x = −2.

10. (a) dy/dx = 3x² + 2px + q [1]. x = 1: 3 + 2p + q = 0 [1]. x = 3: 27 + 6p + q = 0 [1]. Subtracting gives 24 + 4p = 0, so p = −6 and q = 9 [1]. (b) The curve is y = x³ − 6x² + 9x. At x = 1: 1 − 6 + 9 = 4. At x = 3: 27 − 54 + 27 = 0. Points (1, 4) and (3, 0) [2]. (c) dy/dx = 3(x − 1)(x − 3). Test x = 0, 2, 4: dy/dx = 9, −3, 9 [1]. So (1, 4) is a maximum and (3, 0) is a minimum [1]. (d) The curve passes through the origin (y = x(x − 3)²), rises to the maximum at (1, 4), falls to touch the x-axis at the minimum (3, 0), then rises again [1]. [9] Examiner insight: In part (a), “stationary at x = 1” means dy/dx = 0 when x = 1. Substitute into dy/dx, never into y.

11. (a) dy/dx = 4x³ + 12x² = 4x²(x + 3) [1]. So x = 0 or x = −3 [1]. y at x = 0 is 0. y at x = −3 is 81 − 108 = −27. Points (0, 0) and (−3, −27) [1]. (b) Test x = −4, −1, 1: dy/dx = −64, 8, 16 [1]. At x = −3 the gradient goes − then +: (−3, −27) is a minimum [1]. At x = 0 the gradient is positive on both sides: (0, 0) is a point of inflection [1]. (c) Both ends rise. The curve falls to the minimum at (−3, −27), then rises through a flat spot at the origin and keeps rising [1]. [7] Examiner insight: A repeated factor such as x² in dy/dx often points to an inflection, but the sign test is what proves it.

Where marks are usually lost

• Differentiating brackets one at a time instead of expanding the product first.
• Dropping the multiplication by the power, so 6x⁵ becomes 6x⁴.
• Treating a constant as if it still contributes a term in the gradient.
• Stopping at the stationary x-values and not giving the y-coordinates.
• Putting x-values into dy/dx to find heights, when y = f(x) is needed.
• Declaring a maximum or minimum with no sign test, or testing values that straddle two stationary points.
• Missing the second solution when dy/dx = k gives a quadratic, so only one tangent is found.
• Forgetting that a stationary point of inflection exists: the gradient is zero but no sign change occurs.
• Sketches with sharp corners at stationary points, or ends that go the wrong way for the highest power.

Next steps

• Revise the key results in the revision notes.
• Re-read the methods in the study guide.
• Go to the OxfordAQA IGCSE Mathematics hub and tick off topics on the checklist.
• Take the free 10-minute diagnostics to find your weakest areas.
• Get help from a teacher: book a free trial class.

Official syllabus

These questions follow the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA: section 3.2.2, statements A13, A14 and A15.