Pearson Edexcel International Advanced Level Mathematics: Differentiating and integrating powers of x: gradients, tangents, normals and indefinite integrals (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 P1: Pure Mathematics 1, sections 4.1 to 4.3 and 5.1 to 5.2 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). P1 is compulsory 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. Differentiate f(x) = 2x² − 3x from first principles. [3]

2. Given that y = (5x − 2)(x + 3), find dy/dx and its value when x = −1. [3]

3. Given that y = (x⁴ − 3x² + 8)/(4x²), find dy/dx. [4]

4. The function f is defined for x > 0 by f(x) = (x² − 6x + 5)/√x.

(a) Write f(x) as a sum of powers of x and find f′(x). [3] (b) Find the exact value of f′(4). [2]

5. A curve has equation y = 2x³ + 6/x for x ≠ 0.

(a) Find dy/dx and d²y/dx². [4] (b) Find the values of x for which dy/dx = 0. [2]

6. The curve C has equation y = x³ − 4x² + 3. The point P on C has x-coordinate 2.

(a) Find an equation of the tangent to C at P. [3] (b) Find an equation of the normal to C at P, as a linear equation with all terms on one side and integer coefficients. [3]

7. The curve with equation y = x³ − 6x² + 5x + 2 has tangents at two points that are parallel to the line y = 7 − 4x. Find the coordinates of these two points. [5]

8. Find ∫(8x³ − 3√x + 5/x²) dx. [4]

9. Find ∫(2x + 3)²/√x dx. [4]

10. A curve has gradient dy/dx = 12√x − 3 and passes through the point (4, 30). Find the equation of the curve in the form y = f(x). [5]

11. A curve C has gradient dy/dx = 3x² − 4/x² and passes through the point (2, 11).

(a) Find the equation of C. [4] (b) The point P on C has x-coordinate 1. Find an equation of the normal to C at P. [4] (c) Find the coordinates of the point where this normal meets the x-axis. [1]

12. The number of bacteria in a culture after t hours is P = 500 + 40t^(3/2), for t ≥ 0. Find dP/dt when t = 9 and interpret your answer, giving units. [3]

Answers

1. f(x + h) = 2(x + h)² − 3(x + h) = 2x² + 4xh + 2h² − 3x − 3h [1] [f(x + h) − f(x)]/h = (4xh + 2h² − 3h)/h = 4x + 2h − 3 [1] As h → 0, f′(x) = 4x − 3 [1] [3] Examiner insight: A first-principles answer should show the limit notation and cancel h before letting h tend to 0. A bare answer of 4x − 3 does not show the method the question demands.

2. Expand: y = 5x² + 13x − 6 [1] dy/dx = 10x + 13 [1] When x = −1: 10(−1) + 13 = 3 [1] [3] Examiner insight: Differentiating each bracket separately gives 5 × 1 = 5, which is a classic wrong route. Expanding first is the only method available in this unit.

3. Divide each term by 4x²: y = x²/4 − 3/4 + 2x⁻² [2] dy/dx = x/2 + 0 − 4x⁻³ [1] dy/dx = x/2 − 4/x³ [1] [4] Examiner insight: The constant −3/4 differentiates to zero, so it must not reappear. Check each division of powers (x⁴ ÷ x² = x²) before differentiating.

4. (a) f(x) = x^(3/2) − 6x^(1/2) + 5x^(−1/2) [1] f′(x) = (3/2)x^(1/2) − 3x^(−1/2) − (5/2)x^(−3/2) [2] (b) f′(4) = (3/2)(2) − 3(1/2) − (5/2)(1/8) [1] = 3 − 3/2 − 5/16 = 19/16 [1] [5] Examiner insight: When an exact value is asked for, leave a fraction such as 19/16 rather than a rounded decimal. Evaluate x^(−3/2) as 1/(√x)³, which is 1/8 at x = 4.

5. (a) y = 2x³ + 6x⁻¹ [1] dy/dx = 6x² − 6x⁻² [1] d²y/dx² = 12x + 12x⁻³ [2] [4] (b) 6x² − 6/x² = 0, so x⁴ = 1 [1] x = 1 or x = −1 [1] [2] Examiner insight: x⁴ = 1 has two real roots, and the curve is defined at both. Stating only x = 1 loses the second solution.

6. (a) At x = 2: y = 8 − 16 + 3 = −5 [1] dy/dx = 3x² − 8x = 12 − 16 = −4 [1] Tangent: y + 5 = −4(x − 2), so y = −4x + 3 [1] [3] (b) Normal gradient = 1/4 [1] y + 5 = (1/4)(x − 2), so 4y + 20 = x − 2 [1] x − 4y − 22 = 0 [1] [3] Examiner insight: The question asks for a specific form with integer coefficients, so clear the fraction and move every term to one side. Leaving y = x/4 − 11/2 does not answer the question as set.

7. dy/dx = 3x² − 12x + 5 [1] Parallel lines have equal gradients, so 3x² − 12x + 5 = −4 [1] 3x² − 12x + 9 = 0, so x² − 4x + 3 = 0 and x = 1 or x = 3 [1] x = 1: y = 1 − 6 + 5 + 2 = 2; x = 3: y = 27 − 54 + 15 + 2 = −10 [1] (1, 2) and (3, −10) [1] [5] Examiner insight: The y-coordinates come from the original equation, not from dy/dx. Reading the gradient of y = 7 − 4x as 7 is a frequent slip.

8. 8x³ → 2x⁴ [1] −3x^(1/2) → −3 · x^(3/2)/(3/2) = −2x^(3/2) [1] 5x⁻² → −5x⁻¹ [1] 2x⁴ − 2x^(3/2) − 5/x + c [1] [4] Examiner insight: An indefinite integral without + c is incomplete. Differentiate your answer as a quick check on each term.

9. (2x + 3)² = 4x² + 12x + 9 [1] Divide by x^(1/2): 4x^(3/2) + 12x^(1/2) + 9x^(−1/2) [1] Integrate: 4 · (2/5)x^(5/2) + 12 · (2/3)x^(3/2) + 9 · 2x^(1/2) [1] (8/5)x^(5/2) + 8x^(3/2) + 18x^(1/2) + c [1] [4] Examiner insight: Integrating the numerator and denominator separately is not valid. Expand and split first, and take care that dividing by a fraction means multiplying by its reciprocal.

10. y = 12 · x^(3/2)/(3/2) − 3x + c [1] = 8x^(3/2) − 3x + c [1] Substitute (4, 30): 8(8) − 12 + c = 30 [1] c = −22 [1] y = 8x^(3/2) − 3x − 22 [1] [5] Examiner insight: The final line should be a full equation, not just the value of c. Substituting the point before integrating is a common source of error.

11. (a) ∫3x² dx = x³ [1] ∫−4x⁻² dx = 4x⁻¹, so y = x³ + 4/x + c [1] At (2, 11): 8 + 2 + c = 11 [1] y = x³ + 4/x + 1 [1] [4] (b) At x = 1: y = 1 + 4 + 1 = 6, and dy/dx = 3 − 4 = −1 [1] Normal gradient = 1 [1] Normal: y − 6 = 1(x − 1) [1] y = x + 5 [1] [4] (c) y = 0 gives (−5, 0) [1] [1] Examiner insight: In a multi-part question, an equation found in an earlier part is needed later. Say clearly which equation you use, and check that your answer to (a) fits the given point before moving on.

12. dP/dt = 40 · (3/2)t^(1/2) = 60t^(1/2) [1] When t = 9: dP/dt = 60 × 3 = 180 [1] After 9 hours the number of bacteria is increasing at a rate of 180 bacteria per hour [1] [3] Examiner insight: Interpretation questions need the units and the direction (increasing). The constant 500 differentiates to zero, so it plays no part in the rate.

Where marks are usually lost

• Differentiating or integrating a product or fraction without expanding or splitting first.
• Treating 1/x² or 1/√x as if they were positive powers.
• Dividing by n instead of n + 1 when integrating.
• Dropping + c, or forgetting to use the given point to find it.
• Slips with fractional arithmetic, such as (3/2) × 2 or dividing by 2/5.
• Using the gradient of the tangent as the gradient of the normal.
• Giving a line in the wrong form when a form is specified.
• Setting h = 0 before dividing by h in a first-principles argument.
• Omitting units and direction in a rate-of-change interpretation.
• Reporting only one of two roots when solving dy/dx = 0.

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 P1 sections 4.1 to 4.3 and 5.1 to 5.2.