Edexcel International GCSE Maths Differentiation Past Paper Questions and Solutions (4MA1 Higher)

Part (a)

1. Differentiate each term: multiply by the power, then reduce the power by 1.
2. 4x³ → 12x², x² → 2x, −20x → −20

Answer: 12x² + 2x − 20

Part (b)

1. Set the gradient equal to 4: 12x² + 2x − 20 = 4
2. 12x² + 2x − 24 = 0. Divide by 2: 6x² + x − 12 = 0
3. Factorise: (3x − 4)(2x + 3) = 0

Answer: x = ⁴⁄₃ or x = −³⁄₂

Part (a)

1. Differentiate each term; the constant 4 disappears.

Answer: 15x² − 2x − 6

Part (b)

1. Set the gradient equal to 2: 15x² − 2x − 6 = 2, so 15x² − 2x − 8 = 0
2. Factorise: (5x − 4)(3x + 2) = 0

Answer: x = ⁴⁄₅ and x = −²⁄₃

Part (a)

1. V = (2x + 5)(x + 1)(3 − x)
2. (2x + 5)(x + 1) = 2x² + 7x + 5
3. (2x² + 7x + 5)(3 − x) = 6x² + 21x + 15 − 2x³ − 7x² − 5x
4. Collect like terms: 15 + 16x − x² − 2x³

Answer: V = 15 + 16x − x² − 2x³

Part (b)

1. At a maximum, ^dV⁄_dx = 0: ^dV⁄_dx = 16 − 2x − 6x²
2. 16 − 2x − 6x² = 0, so 3x² + x − 8 = 0
3. Quadratic formula: x = ^{−1 ± √1 + 96}⁄₆ = ^{−1 ± √97}⁄₆
4. x = 1.4748… or x = −1.81…. The lengths must be positive, so (x + 1) > 0 means x > −1; reject the negative root.

Answer: x = 1.47

1. Volume of the cylinder = πx²(20 − 4x) = 20πx² − 4πx³
2. Volume of the hemisphere = ²⁄₃πx³
3. V = 20πx² − 4πx³ + ²⁄₃πx³ = 20πx² − ¹⁰⁄₃πx³
4. V = ⅓πy, so y = 3V ÷ π = 60x² − 10x³
5. For a maximum, ^dy⁄_dx = 120x − 30x² = 0, so 30x(4 − x) = 0 and x = 4 (the radius cannot be 0)
6. y = 60 × 16 − 10 × 64 = 960 − 640

Answer: 320

1. Velocity is the rate of change of displacement, so v = ^ds⁄_dt. Write ¹²⁵⁄_t as 125t⁻¹ to differentiate.
2. v = 8t − 125t⁻² = 8t − ¹²⁵⁄_t²
3. Set v = 0: 8t = ¹²⁵⁄_t², so 8t³ = 125 and t³ = ¹²⁵⁄₈
4. t = ∛(¹²⁵⁄₈) = ⁵⁄₂ = 2.5 seconds
5. Substitute into s: 4 × 2.5² + ¹²⁵⁄_2.5 = 25 + 50

Answer: 75 m

1. The gradient is given by the derivative. Differentiate each term: ^dy⁄_dx = 3x² − 40
2. Set the gradient equal to 8: 3x² − 40 = 8
3. 3x² = 48, so x² = 16 and x = 4 or x = −4
4. Substitute each x into the equation of the curve to find y.
5. x = 4: y = 64 − 160 + 1 = −95
6. x = −4: y = −64 + 160 + 1 = 97

Answer: (4, −95) and (−4, 97)

1. At a stationary point the gradient is 0, so differentiate: ^dy⁄_dx = 3x² − 16x − 12
2. Set it equal to 0: 3x² − 16x − 12 = 0
3. Factorise: (3x + 2)(x − 6) = 0, so x = −²⁄₃ or x = 6
4. A has the larger x coordinate, so x = 6.
5. y = 6³ − 8 × 6² − 12 × 6 + 5 = 216 − 288 − 72 + 5 = −139

Answer: (6, −139)

1. Velocity v = ^ds⁄_dt = 3t² − 18t + 33
2. The velocity is smallest when ^dv⁄_dt = 0: 6t − 18 = 0, so t = 3
3. v = 3(9) − 18(3) + 33 = 27 − 54 + 33 = 6
4. v is always positive here (3t² − 18t + 33 = 3(t − 3)² + 6), so the minimum speed is 6.

Answer: 6 m/s

1. V = πx² × (⁸⁰⁰⁄_πx − x) = 800x − πx³
2. At the maximum, ^dV⁄_dx = 800 − 3πx² = 0, so x² = ⁸⁰⁰⁄_3π and x = 9.213… cm
3. V = 800 × 9.213… − π × 9.213…³ = 7370.5… − 2456.8… = 4913.6…

Answer: 4914 cm³

Part (a)

1. Differentiate each term.

Answer: 3x² − 4x − 9

Part (b)

1. Positive gradient means ^dy⁄_dx > 0, so solve 3x² − 4x − 9 > 0.
2. First find where it equals 0 using the quadratic formula: x = ^{4 ± √16 + 108}⁄₆ = ^{4 ± √124}⁄₆
3. x = 2.5225… or x = −1.1892…
4. 3x² − 4x − 9 is a positive (U-shaped) quadratic, so it is positive outside its two roots.

Answer: x < −1.19 or x > 2.52

Part (a)

1. Write ^p⁄_x as px⁻¹ and differentiate: ^dy⁄_dx = 2x + px⁻² = 2x + ^p⁄_x²
2. At the turning point the gradient is 0. Substitute x = −3: 2(−3) + ^p⁄₉ = 0
3. ^p⁄₉ = 6

Answer: p = 54

Part (b)

1. y = k is a horizontal line, so it touches the curve at the turning point (where the gradient is 0).
2. ^dy⁄_dx = 2x + ¹⁶⁄_x² = 0, so 2x³ = −16, x³ = −8 and x = −2
3. y = (−2)² − ¹⁶⁄₋₂ = 4 + 8 = 12

Answer: k = 12

1. Velocity = ^ds⁄_dt = 3t² + 8t − 5
2. At time T: 3T² + 8T − 5 = V, so 3T² + 8T − (5 + V) = 0
3. Quadratic formula: T = ^{−8 ± √64 + 12(5 + V)}⁄₆ = ^{−8 ± √124 + 12V}⁄₆
4. √124 + 12V = √4(31 + 3V) = 2√31 + 3V, so T = ^{−8 ± 2√31 + 3V}⁄₆ = ^{−4 ± √31 + 3V}⁄₃
5. Time cannot be negative, so take the + sign.

Answer: T = ^{−4 + √31 + 3V}⁄₃, so k = 31 and m = 3

1. Gradient: ^dy⁄_dx = 3px² − m
2. Negative gradient: 3px² − m < 0, so x² < ^m⁄_3p
3. x² is less than a positive number when x lies between its negative and positive square roots.

Answer: −√^m⁄_3p < x < √^m⁄_3p

1. Velocity of P: ^ds⁄_dt = 3t² − 8t + 5 = (3t − 5)(t − 1). For t > 1 this is negative when t < ⁵⁄₃ and positive when t > ⁵⁄₃.
2. Velocity of Q: ^dx⁄_dt = 2t − 4. This is negative when t < 2 and positive when t > 2.
3. Same direction means both velocities have the same sign.
4. Both negative: 1 < t < ⁵⁄₃. Both positive: t > 2.

Answer: 1 < t < ⁵⁄₃ and t > 2

1. ^dy⁄_dx = 16x − 14 = 10, so x = 1.5
2. y = 8 × 1.5² − 14 × 1.5 − 6 = 18 − 21 − 6 = −9, so G = (1.5, −9)
3. The tangent has gradient 10, so Q has gradient −¹⁄₁₀
4. y + 9 = −¹⁄₁₀(x − 1.5). Multiply by 10: 10y + 90 = −x + 1.5
5. x + 10y + 88.5 = 0. Multiply by 2 to make the numbers integers.

Answer: 2x + 20y + 177 = 0

1. A lies on C: 8a + 4b − 24 + 6 = −6, so 8a + 4b = 12 and 2a + b = 3
2. Gradient: ^dy⁄_dx = 3ax² + 2bx − 12. At x = 2: 12a + 4b − 12 = 16, so 12a + 4b = 28 and 3a + b = 7
3. Subtract the two equations: a = 4, then b = 3 − 8 = −5
4. y = 4x³ − 5x² − 12x + 6. At x = 3: 108 − 45 − 36 + 6

Answer: 33