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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Parametric differentiation · Core · Paper 3 style · 7 marks
(a)A curve is defined by the parametric equations x = t² + 3t, y = t³ − 2t. Find dy/dx in terms of t.[3]
(b)Find the equation of the tangent to the curve at the point where t = 1, giving your answer in the form ax + by + c = 0.[4]
B1 dx/dt = 1/tB1 dy/dt = 2t − 2/t²A1 divide to obtain the given result
(b) (1 + ln 2, 3)
M1 2t² − 2/t = 0 so t³ = 1, t = 1A1 x = 1 + ln 2, y = 3
Q3Parametric differentiation · Standard · Paper 3 style · 8 marks
(a)The parametric equations of a curve are x = e²ᵗ − 1, y = t eᵗ. Show that dy/dx = (1 + t)/(2eᵗ).[4]
(b)Find the exact coordinates of the point on the curve at which the tangent is parallel to the x-axis.[2]
(c)Find the equation of the normal to the curve at the origin.[2]
Show answer
(a) dy/dx = (eᵗ + t eᵗ)/(2e²ᵗ) = (1 + t)/(2eᵗ)
B1 dx/dt = 2e²ᵗB1 dy/dt = eᵗ + t eᵗ (product rule)M1 divide and cancel eᵗ, A1 given answer
(b) (e⁻² − 1, −e⁻¹)
M1 1 + t = 0 so t = −1A1 x = e⁻² − 1, y = −e⁻¹
(c) y = −2x
B1 origin when t = 0, gradient 1/2A1 normal gradient −2, y = −2x
Q4Parametric differentiation · Standard · Paper 3 style · 8 marks
(a)A curve has parametric equations x = 4 sin θ, y = 3 cos 2θ, for −½π < θ < ½π. Show that dy/dx = −3 sin θ.[3]
(b)Find the equation of the tangent to the curve at the point where θ = π/6, giving your answer in the form ax + by = c where a, b and c are integers.[3]
(c)Find the Cartesian equation of the curve in the form y = f(x).[2]
Show answer
(a) dy/dx = −6 sin 2θ ÷ 4 cos θ = −12 sin θ cos θ ÷ 4 cos θ = −3 sin θ
B1 dx/dθ = 4 cos θ, B1 dy/dθ = −6 sin 2θM1 use sin 2θ = 2 sin θ cos θ and simplify to the given answer
(b) 3x + 2y = 9
B1 point (2, 3/2) and gradient −3/2M1 y − 3/2 = −3/2(x − 2)A1 3x + 2y = 9
(c) y = 3 − 3x²/8
M1 cos 2θ = 1 − 2 sin²θ = 1 − 2(x/4)²A1 y = 3 − 3x²/8
Q5Parametric differentiation · Challenge · Paper 3 style · 9 marks
(a)A spiral curve is given by x = eᵗ cos t, y = eᵗ sin t, for 0 ≤ t ≤ π. Show that dy/dx = (sin t + cos t)/(cos t − sin t).[3]
(b)Find the exact coordinates of the point on the curve where the tangent is parallel to the y-axis.[3]
(c)Find the exact value of t at which the tangent is parallel to the x-axis, and find the equation of the tangent at the point where t = 0.[3]
Show answer
(a) dx/dt = eᵗ(cos t − sin t), dy/dt = eᵗ(sin t + cos t); divide
B1 both derivatives by the product ruleM1 divide and cancel eᵗA1 given answer
(b) t = π/4; (e^(π/4)/√2, e^(π/4)/√2)
M1 cos t − sin t = 0A1 t = π/4A1 x = y = e^(π/4)/√2 (or ½√2 e^(π/4))
(c) t = 3π/4; tangent at t = 0: y = x − 1
B1 sin t + cos t = 0 gives t = 3π/4B1 point (1, 0), gradient 1B1 y = x − 1
Q6Implicit differentiation · Core · Paper 2 style · 5 marks
(a)The equation of a curve is x² + 3xy + 2y² = 6. Show that the point (1, 1) lies on the curve and find dy/dx in terms of x and y.[3]
(b)Find the equation of the tangent to the curve at (1, 1), giving your answer in the form ax + by = c.[2]
Show answer
(a) 1 + 3 + 2 = 6; dy/dx = −(2x + 3y)/(3x + 4y)
B1 verifyM1 differentiate 3xy as a product: 3y + 3x dy/dx, and 2y² as 4y dy/dxA1 dy/dx = −(2x + 3y)/(3x + 4y)
(b) 5x + 7y = 12
M1 gradient −5/7A1 5x + 7y = 12
Q7Implicit differentiation · Standard · Paper 3 style · 6 marks
(a)A curve has equation y³ + x²y − 2x = 8. Find dy/dx in terms of x and y.[3]
(b)Find the equation of the normal to the curve at the point where it crosses the y-axis.[3]
Show answer
(a) dy/dx = (2 − 2xy)/(3y² + x²)
M1 implicit: 3y² dy/dx + 2xy + x² dy/dx − 2 = 0A1 correct derivative of each termA1 rearranged
(b) y = 2 − 6x
B1 x = 0 gives y³ = 8, y = 2M1 gradient at (0, 2) = 2/12 = 1/6, so normal gradient −6A1 y = 2 − 6x
Q8Implicit differentiation · Standard · Paper 3 style · 7 marks
(a)The equation of a curve is 2x² − xy + y² = 28. Show that dy/dx = (y − 4x)/(2y − x).[3]
(b)Find the exact coordinates of the two points on the curve at which the tangent is parallel to the x-axis.[4]
Show answer
(a) 4x − y − x dy/dx + 2y dy/dx = 0, so dy/dx = (y − 4x)/(2y − x)
M1 implicit differentiation including product rule on xyA1 4x − (y + x dy/dx) + 2y dy/dx = 0A1 given answer
(b) (√2, 4√2) and (−√2, −4√2)
M1 y = 4xM1 substitute: 2x² − 4x² + 16x² = 28A1 x² = 2A1 both points
Q9Implicit differentiation · Challenge · Paper 3 style · 8 marks
(a)A curve has equation x³ + y³ = 6xy. Find dy/dx in terms of x and y.[3]
(b)Find the exact coordinates of the point, other than the origin, at which the tangent to the curve is parallel to the x-axis. Give your answer in the form (p∛2, q∛4) where p and q are integers.[4]
(c)Use the symmetry of the equation to state the coordinates of the point where the tangent is parallel to the y-axis.[1]
M1 ln x = 0 or ln x = 2A1 (1, 0)A1 (e², 4/e²)M1 sign of dy/dx either side (or second derivative)A1 (1, 0) minimum and (e², 4e⁻²) maximum
Q18Tangents & normals · Core · Paper 1 style · 6 marks
(a)A curve has equation y = 2x + 8/x. Find the equation of the normal to the curve at the point where x = 4, giving your answer in the form ax + by = c.[4]
(b)Find the coordinates of the stationary point of the curve for which x > 0.[2]
Show answer
(a) 2x + 3y = 38
B1 point (4, 10)M1 dy/dx = 2 − 8/x², = 3/2 at x = 4M1 normal gradient −2/3, y − 10 = −2/3(x − 4)A1 2x + 3y = 38
(b) (2, 8)
M1 2 − 8/x² = 0A1 (2, 8)
Q19Tangents & normals · Standard · Paper 1 style · 7 marks
(a)The curve y = √(4x + 1) passes through the point P(2, 3). Find the equation of the tangent to the curve at P.[3]
(b)The tangent and the normal at P meet the x-axis at A and B respectively. Find the area of triangle PAB.[4]
B1 A(−5/2, 0)M1 normal: y − 3 = −3/2(x − 2)A1 B(4, 0)B1 area ½ × 6.5 × 3 = 9.75
Q20Tangents & normals · Challenge · Paper 1 style · 6 marks
(a)A curve has equation y = x² − 4x + 7. The tangent to the curve at the point where x = a passes through the origin O. Show that a² = 7.[3]
(b)Hence find the equations of the two tangents to the curve that pass through O, giving the gradients in exact form.[3]
Show answer
(a) Tangent at x = a: y − (a² − 4a + 7) = (2a − 4)(x − a); through O: −(a² − 4a + 7) = −a(2a − 4) gives a² = 7
B1 gradient 2a − 4M1 substitute (0, 0) into the tangentA1 a² = 7
(b) y = (2√7 − 4)x and y = (−2√7 − 4)x
M1 a = ±√7A1 y = (2√7 − 4)xA1 y = −(2√7 + 4)x
Q21Stationary points & nature · Core · Paper 1 style · 5 marks
(a)Find the coordinates of the stationary points of the curve y = x³ − 6x² + 9x + 2.[3]
(b)Use the second derivative to determine the nature of each stationary point.[2]
Show answer
(a) (1, 6) and (3, 2)
M1 dy/dx = 3x² − 12x + 9 = 0A1 x = 1, 3A1 y = 6, 2
(b) (1, 6) maximum, (3, 2) minimum
M1 d²y/dx² = 6x − 12: −6 at x = 1, 6 at x = 3A1 both correct
Q22Stationary points & nature · Standard · Paper 1 style · 5 marks
(a)A curve has equation y = x² + 16/x, x ≠ 0. Find the coordinates of the stationary point.[3]
(b)Find d²y/dx² and determine the nature of the stationary point.[2]
Show answer
(a) (2, 12)
M1 dy/dx = 2x − 16/x²A1 x³ = 8, x = 2A1 y = 12
(b) d²y/dx² = 2 + 32/x³ = 6 > 0, minimum
B1 2 + 32/x³B1 6 > 0 so minimum
Q23Stationary points & nature · Challenge · Paper 1 style · 8 marks
(a)A workshop in Sialkot makes open boxes from rectangular metal sheets 30 cm by 16 cm. A square of side x cm is cut from each corner and the sides are folded up. Show that the volume V cm³ of the box is given by V = 4x³ − 92x² + 480x.[2]
(b)Find the value of x for which V is stationary, and find this value of V. Give V correct to 1 decimal place.[4]
(c)Determine whether this value of V is a maximum or a minimum.[2]
Show answer
(a) V = x(30 − 2x)(16 − 2x) = 4x³ − 92x² + 480x
M1 V = x(30 − 2x)(16 − 2x)A1 expanded to the given result
M1 d²V/dx² = 24x − 184 = −104 at x = 10/3A1 negative, so maximum
Q24Rates of change · Standard · Paper 1 style · 4 marks
(a)A water tank in a village is in the shape of an inverted cone. When the depth of water is h cm, the radius of the water surface is ½h cm. Show that the volume of water is V = πh³/12 cm³.[1]
(b)Water is poured in at a constant rate of 5 cm³ per second. Find the rate at which the depth is increasing when h = 10. Give your answer correct to 3 significant figures.[3]
Show answer
(a) V = ⅓π(½h)²h = πh³/12
B1 substitute r = h/2
(b) 1/(5π) ≈ 0.0637 cm/s
B1 dV/dh = πh²/4 = 25πM1 dh/dt = (dV/dt) ÷ (dV/dh)A1 0.0637 cm per second
Q25Rates of change · Challenge · Paper 1 style · 7 marks
(a)A point P moves along the curve y = 6/(3x − 1) for x > ⅓, so that its x-coordinate increases at a constant rate of 0.04 units per second. Find the rate of change of the y-coordinate of P when x = 1.[3]
(b)Find the rate of change of the distance OP, where O is the origin, at the instant when x = 1. Give your answer correct to 3 significant figures.[4]
Show answer
(a) −0.18 units per second
M1 dy/dx = −18/(3x − 1)²A1 −4.5 at x = 1A1 dy/dt = −4.5 × 0.04 = −0.18
(b) −0.158 units per second
B1 P(1, 3), OP = √10M1 d(OP)/dt = (x dx/dt + y dy/dt)/OPM1 (0.04 − 0.54)/√10A1 −0.158
Q26Increasing / decreasing · Core · Paper 1 style · 3 marks
(a)The function f is defined by f(x) = x³ − 3x² − 9x + 4. Find the set of values of x for which f is decreasing.[3]
Show answer
(a) −1 < x < 3
M1 f′(x) = 3x² − 6x − 9A1 critical values −1 and 3A1 −1 < x < 3 (or ≤)
Q27Increasing / decreasing · Challenge · Paper 1 style · 6 marks
(a)The function f is defined by f(x) = x³ + kx² + 12x for all real x, where k is a constant. Find the set of values of k for which f is an increasing function.[3]
(b)For the case k = −7.5, find the set of values of x for which f is decreasing.[3]
Show answer
(a) −6 ≤ k ≤ 6
M1 f′(x) = 3x² + 2kx + 12 ≥ 0 for all xM1 discriminant 4k² − 144 ≤ 0A1 −6 ≤ k ≤ 6
