Differentiation: IGCSE Mathematics questions with answers
Free Differentiation questions for IGCSE Mathematics (0580), Extended tier (Supplement content), part of Algebra & graphs. Each question has the answer and the marking points an examiner would look for.
Find the coordinates of the two stationary points and determine the nature of each.
[6]
(b)
Find the set of values of k for which the equation f(x) = k, where f(x) is the expression above, has exactly three solutions.
[2]
(c)
Find the equation of the tangent to the curve at the point where x = 4.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
(−1, 1) maximum; (3, −31) minimum
dydx = 3x2 − 6x − 9 M1 set equal to 0 and solve M1 x = −1 and x = 3 A1 y = 1 and y = −31 A1 second derivative or gradient either side M1 (−1, 1) maximum, (3, −31) minimum A1
6
(b)
−31 < k < 1
uses the y-values of the stationary points M1 −31 < k < 1 A1
2
(c)
y = 15x − 84
substitute into dy/dx M1 gradient 15 A1 y = 15x − 84 A1
3
Question 2Structured · 8 marksA* challenge
A square sheet of card has side 18 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 − 72x2 + 324x.
[2]
(b)
Find the value of x that gives the maximum volume, and the maximum volume.
[5]
(c)
State the set of possible values of x for a box to exist.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
V = x(18 − 2x)2
base side 18 − 2x, height x M1 expanded correctly A1
2
(b)
x = 3, V = 432 cm³
dVdx = 12x2 − 144x + 324 M1 set to 0 and solve M1 x = 3 (x = 9 gives no box) A1 shows it is a maximum (second derivative or values either side) M1 V = 432 A1
5
(c)
0 < x < 9
0 < x < 9 B1
1
Question 3Structured · 5 marks
y = 2x3 − 15x2 + 36x + 4
(a)
Find dydx.
[2]
(b)
Find the coordinates of the two turning points.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
6x2 − 30x + 36
6x2 − 30x + 36 B2 B1 for two terms correct
2
(b)
(2, 32) and (3, 31)
Their dydx = 0 M1 x = 2, x = 3 A1 Both y-coordinates A1
3
Question 4Structured · 10 marksA* challenge
A curve has equation y = x3 − 9x2 + 24x − 3.
(a)
Find dydx.
[2]
(b)
Find the coordinates of the two turning points and determine whether each is a maximum or a minimum. Show your working.
[5]
(c)
Find the equation of the tangent to the curve at the point where x = 5. Give your answer in the form y = mx + c.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
3x2 − 18x + 24
two terms correct B1 all correct B1
2
(b)
(2, 17) maximum, (4, 13) minimum
their dydx = 0 M1 x = 2, x = 4 A1 y = 17, y = 13 A1 valid method to determine nature (e.g. sign of gradient either side, or shape of cubic) M1 (2, 17) maximum and (4, 13) minimum A1
5
(c)
y = 9x − 28
gradient = 9 B1 point (5, 17) B1 y = 9x − 28 B1
3
Question 5Structured · 11 marksA* challenge
A farmer has 120 m of fencing. He builds a rectangular enclosure against a long straight wall, so the wall forms one long side. The enclosure is divided into three equal pens by two fences parallel to the short sides. No fencing is needed along the wall.
(a)
Find the dimensions of the enclosure with the greatest possible area, and this area.
[5]
(b)
The wall is only 30 m long. The long side of the enclosure may extend past the end of the wall, but any part of that side not against the wall must be fenced. Find the greatest possible area now, with full justification.
[6]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
15 m by 60 m; 900 m²
L + 4w = 120 B1 A = w(120 − 4w) M1 dA/dw = 120 − 8w = 0 (or completing the square / symmetry) M1 w = 15, L = 60 A1 900 m² A1
5
(b)
703.125 m² (L = 37.5 m, w = 18.75 m)
for L ≤ 30: L + 4w = 120, A = L(120 − L)4, increasing, so best 675 at L = 30 M1 675 m² A1 for L > 30: extra fence L − 30, so 2L + 4w = 150 M1 A = L(150 − 2L)4A1 maximum at L = 37.5 (inside L > 30) M1 703.125 m², greater than 675, so this is the maximum A1
6
Question 6Structured · 11 marksA* challenge
A curve has equation y = x3 − 12x2 + 21x + 4.
(a)
Find the coordinates of the two stationary points and determine the nature of each.
[6]
(b)
Find the set of values of k for which the equation f(x) = k, where f(x) is the expression above, has exactly three solutions.
[2]
(c)
Find the equation of the tangent to the curve at the point where x = 8.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
(1, 14) maximum; (7, −94) minimum
dydx = 3x2 − 24x + 21 M1 set equal to 0 and solve M1 x = 1 and x = 7 A1 y = 14 and y = −94 A1 second derivative or gradient either side M1 (1, 14) maximum, (7, −94) minimum A1
6
(b)
−94 < k < 14
uses the y-values of the stationary points M1 −94 < k < 14 A1
2
(c)
y = 21x − 252
substitute into dy/dx M1 gradient 21 A1 y = 21x − 252 A1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in