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2026 O-Level A-Math Paper 1 — Full Worked Solutions

GCE O-Level · Additional Mathematics (4049/01) · October/November 2026

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1 Differentiation & integration 4 marks
a Find d/dx (x²sinx). [2]
Show working
Concept: Product rule
Using the product rule with u=x², v=sinx: d/dx(x²sinx) = x²·cosx + sinx·2x
= x²cosx + 2xsinx
b Find ∫ 3/(2x+1) dx. [2]
Show working
Concept: Integration of 1/(ax+b)
∫ 3/(2x+1) dx = (3/2) ∫ 2/(2x+1) dx
= (3/2)ln(2x+1) + C
2 Surds 6 marks

A calculator must not be used in this question.

a(i) Given that x = (−5+√57)/2 and x² + 5x = k, find the value of the constant k. [2]
Show working
Concept: Surds & substitution
k = ((−5+√57)/2)² + 5((−5+√57)/2)
= (25 − 10√57 + 57)/4 + (5√57 − 25)/2
= (41 − 5√57)/2 + (5√57 − 25)/2
k = 8
a(ii) State the other solution to the equation x² + 5x = k, giving your answer in surd form. [1]
Show working
Concept: Conjugate roots of a quadratic
Since x = (−5+√57)/2 is one root of x²+5x−8=0, the other root is its conjugate.
x = (−5−√57)/2
b Find the gradient of the straight line joining the points whose coordinates are (1, √2) and (√2, 4√2). Give your answer in the form a+b√2, where a and b are integers. [3]
Show working
Concept: Gradient & rationalising surds
m = (4√2 − √2)/(√2 − 1) = 3√2/(√2−1)
Rationalise by multiplying by (√2+1)/(√2+1): m = 3√2(√2+1)/(2−1) = (6+3√2)/1
m = 6 + 3√2
3 Quadratics — discriminant 6 marks
The curve y = (k−10)x² + (k−8)x − (2k+3) intersects the line y = 2x+k at two distinct points. Find the set of possible values for the constant k. [6]
Show working
Concept: Discriminant > 0
(k−10)x² + (k−8)x − (2k+3) = 2x+k
(k−10)x² + (k−10)x − (3k+3) = 0
Two distinct points → discriminant > 0: (k−10)² − 4(k−10)(−3k−3) > 0
(k−10)[(k−10) + 4(3k+3)] > 0 → (k−10)(13k+2) > 0
k < −2/13 or k > 10
4 Geometric progressions & logarithms 7 marks

A person sends an email to two friends. Each friend then forwards the email to two other friends. This process continues, each new person forwarding the email on to two new people. You may assume that no person receives the email more than once. This process is represented in the diagram using levels. Level 1 is the original sender.

Tree diagram showing the email forwarding process: Level 1 is the original sender (1 node), Level 2 is 2 new people, Level 3 is 4 new people, continuing to double at each level
a This process stops as soon as there are over 1 000 000 new people introduced to a level. Find the level at which the process is stopped. [3]
Show working
Concept: Geometric progression & logarithms
Let n be the level number. The number of new people at level n is 2n−1.
2n−1 = 1 000 000 → n−1 = lg(1 000 000)/lg2 = 6/lg2
n = 6/lg2 + 1 = 20.93 (2 d.p.)
Level = 21
b Find the level at which the total number of new people for this level and the one immediately before it first exceeds 8000. [4]
Show working
Concept: Geometric progression & logarithms
2n−1 + 2(n−1)−1 > 8000
2n−1(1 + 2⁻¹) > 8000 → 2n−1 > 16000/3
n−1 > lg(16000/3)/lg2 → n > 13.38 (2 d.p.)
Level = 14
5 Simultaneous equations 6 marks
Find the coordinates of the points of intersection of the curve x/y + 2/x = 1 with the line y = 2x−3. [6]
Show working
Concept: Simultaneous equations
x/y + 2/x = 1 …(1); y = 2x−3 …(2)
Sub (2) into (1): x/(2x−3) + 2/x = 1
Multiply through by x(2x−3): x² + 2(2x−3) = x(2x−3)
x² + 4x − 6 = 2x² − 3x → 0 = x² − 7x + 6
0 = (x−1)(x−6) → x = 1 or x = 6
When x=1, y=2(1)−3=−1. When x=6, y=2(6)−3=9
Intersections: (1, −1) or (6, 9)
6 Quadratics — completing the square 8 marks
a Express 5x² − 10x + k in the form A(x+B)² + C, where A and B are constants and C is an expression in terms of the constant k. [3]
Show working
Concept: Completing the square
5x²−10x+k = 5(x²−2x) + k = 5[(x−1)²−1] + k
= 5(x−1)² − 5 + k
b(i) State the coordinates of the turning point on the curve y = 5x² − 10x + k, giving your answer in terms of k. [2]
Show working
Concept: Turning point from completed square
Turning point = (1, k−5)
b(ii) Explain why the turning point is a minimum point. [1]
Show working
Since the coefficient of x² is 5 > 0, the curve is a u-shaped parabola, so the turning point is a minimum.
b(iii) Given that the curve y = 5x² − 10x + k is never below the line y = 3, find the set of possible values of k. [2]
Show working
Concept: Minimum value ≥ line
The minimum value of y is k−5 (at x=1). For the curve to never be below y=3: k−5 ≥ 3
k ≥ 8
7 Circles — coordinate geometry 7 marks

The diagram shows a circular target used in a game. The game involves throwing a dart at the target which has an inner circle and an outer circle with the same centre. The first throw hits a point on the outer circle of the target which has coordinates (5, −4). The second throw hits the centre of the target which has coordinates (2, 3).

A circular target made of two concentric circles (an outer circle and an inner circle) labelled with their centre
a Find the equation of the outer circle of the target. [3]
Show working
Concept: Circle equation from centre & a point
r = √[(5−2)² + (−4−3)²] = √58
(x−2)² + (y−3)² = 58
b The equation of the inner circle is x² + y² − px − qy − 23 = 0, where p and q are constants. Find the value of p and the value of q, and determine the radius of the inner circle. [4]
Show working
Concept: General circle equation & completing the square
(x−p/2)² + (y−q/2)² = 23 + (p/2)² + (q/2)². Centre of both circles is (2,3), so p/2=2, q/2=3
p = 4, q = 6
r² = 23 + 2² + 3² = 36
p = 4, q = 6, radius = 6
8 Applications of differentiation 8 marks

[The surface area, A, of a sphere, radius r, is given by A = 4πr².]
[The volume, V, of a sphere, radius r, is given by V = (4/3)πr³.]
The diagram shows a large storage tank. The tank consists of a cylinder, open at the upper end, with a hemisphere on the top. The hemisphere has the same radius, r m, as the radius of the cylinder. The height of the cylinder is h m. The tank is hollow and is made out of thin metal. The total surface area of the metal used to form the tank (including the circular base) is 80π m².

A storage tank made of a cylinder of radius r m and height h m, open at the top, with a hemisphere of radius r m on top
a Show that the volume, V m³, of the tank is given by V = 40πr − (5/6)πr³. [4]
Show working
Concept: Surface area & volume of composite solids
V = (2/3)πr³ + πr²h
Surface area (hemisphere curved surface + cylinder curved surface + circular base) = 2πr² + 2πrh + πr² = 80π
3r² + 2rh = 80 → h = (80−3r²)/(2r)
V = (2/3)πr³ + πr²·(80−3r²)/(2r) = (2/3)πr³ + 40πr − (3/2)πr³
V = 40πr − (5/6)πr³ (shown)
b The radius of this type of tank is allowed to vary. A tank needs to hold as much as possible. Find the radius and the height of the cylinder in this case. You are not required to justify that these values give a maximum. [4]
Show working
Concept: Stationary value by differentiation
dV/dr = 40π − (5/2)πr². At a stationary value, dV/dr = 0
(5/2)r² = 40 → r² = 16 → r = 4 (r = −4 rejected since r>0)
h = (80−3(4)²)/(2(4)) = (80−48)/8 = 4
r = 4 m, h = 4 m
9 Trigonometric identities & equations 10 marks
a Prove the following. tanx(sinx+cosx)(cosecx+secx) = (tanx+1)² [4]
Show working
Concept: Trigonometric identities
LHS = tanx(sinx+cosx)(cosecx+secx)
= tanx[sinx·cosecx + sinx·secx + cosx·cosecx + cosx·secx] = tanx[1 + tanx + cotx + 1]
= tanx[tanx + cotx + 2] = tan²x + 2tanx + 1 (since tanx·cotx = 1)
= (tanx)² + 2(1)(tanx) + (1)² = (tanx+1)² = RHS (shown)
b Hence solve the equation (sinx+cosx)(cosecx+secx) = 3(tanx+1) for 0° < x < 360°. [6]
Show working
Concept: Hence, using part (a)
Multiplying both sides by tanx (using part (a)): tanx(sinx+cosx)(cosecx+secx) = 3tanx(tanx+1)
(tanx+1)² = 3tanx(tanx+1) → (tanx+1)² − 3tanx(tanx+1) = 0
(tanx+1)[(tanx+1) − 3tanx] = 0 → (tanx+1)(1−2tanx) = 0
tanx = −1 (2nd & 4th quadrants): x = 135°, 315°
tanx = 1/2 (1st & 3rd quadrants): basic angle = tan⁻¹(1/2) = 26.565° → x = 26.565°, 206.565°
x = 26.6°, 135°, 206.6° or 315°
10 Trigonometric graphs & integration 10 marks

The diagram shows part of the curve y = 2sin3x + 3cos3x. The curve cuts the x-axis at the points A and B. The diagram shows two shaded regions, P (between O and A, above the x-axis) and Q (between A and B, below the x-axis).

Graph of y equals 2 sin 3x plus 3 cos 3x, with region P shaded between the origin and point A above the x-axis, and region Q shaded between points A and B below the x-axis
a Find, in radians, the x-coordinate of the point B. Give your answer correct to 4 significant figures. [3]
Show working
Concept: Trigonometric equations in radians
At the x-axis: 2sin3x + 3cos3x = 0 → tan3x = −3/2
Basic angle α = tan⁻¹(3/2) = 0.98279 rad. tan negative → 2nd or 4th quadrant
Point B is the second (further) root, in the 4th-quadrant branch: 3x = 2π − α = 5.3004
x = 1.767 rad (4 s.f.)
b When finding P−Q using calculus, it is not necessary to use the x-coordinate of A. Explain why. [2]
Show working
Concept: Signed area under a curve
P = ∫0xA y dx. Since Q lies below the x-axis, Q = −∫xAxB y dx (a negative area)
P − Q = ∫0xA y dx + ∫xAxB y dx = ∫0xB y dx
Since the integral for P−Q combines into a single integral from 0 to xB that does not involve xA, it is not necessary to use the x-coordinate of A.
c Find the value of P−Q. [4]
Show working
Concept: Definite integration
P−Q = ∫01.7668 (2sin3x+3cos3x) dx = [−(2/3)cos3x + sin3x]01.7668
At x=1.7668 (3x=5.3004 rad): −(2/3)cos(5.3004) + sin(5.3004) = −(2/3)(0.5547) + (−0.8321) = −1.2019
At x=0: −(2/3)cos0 + sin0 = −2/3 = −0.6667
P−Q = −1.2019 − (−0.6667) = −0.535 (3 s.f.)
d Explain what the answer to (c) tells you about the relative sizes of P and Q. [1]
Show working
Since P−Q < 0, P < Q, so Q is larger than P.
11 Coordinate geometry 10 marks

The diagram shows a trapezium ABCD with a line of symmetry L. The side AB is parallel to the x-axis and the side CD is parallel to the y-axis.

Trapezium ABCD on the coordinate plane with A and B near the top, a dashed line of symmetry L through the middle rising to the upper right, and C and D in the lower right
a Explain why the value of the gradient of L must be 1. [1]
Show working
Concept: Line of symmetry as perpendicular bisector
AB extended and CD extended meet L at points (say R and C's line) with RL = LC, and the angle between AB extended and L (∠BLC) is 90°, so triangle BLC is an isosceles right triangle with ∠BCL = 45°.
Since AB is horizontal and CD is vertical, BC therefore has gradient −1, and L is the perpendicular bisector of BC.
A line perpendicular to a line of gradient −1 has gradient 1, so L has gradient 1.

It is now given that the equation of the line L is y = x + 8.

b(i) Point A has coordinates (−3, 13). Find the equation of the line AD and hence find the coordinates of D. [6]
Show working
Concept: Perpendicular lines & midpoints
AD ⊥ L (since AD is a slanted side and L bisects the trapezium perpendicular to the parallel sides... AD has gradient −1, the negative reciprocal of L's gradient 1)
Line AD: y−13 = −(x+3) → y = −x+10 …(1)
L and AD intersect at the midpoint of AD: y = x+8 …(2)
x+8 = −x+10 → 2x=2 → x=1, y=9. Midpoint of AD = (1,9)
(1,9) = ((−3+xD)/2, (13+yD)/2) → xD=5, yD=5
Line AD: y = −x + 10; D = (5, 5)
b(ii) Given that the length of AB is 2 units, find the area of the trapezium. [3]
Show working
Concept: Shoelace formula
A = (−3,13), so B = (−1,13) (AB=2, parallel to x-axis)
Reflecting B in L (y=x+8) gives C = (5,7); D was found to be (5,5)
Using the shoelace formula on D(5,5), C(5,7), B(−1,13), A(−3,13):
Area = (1/2)|(5·7+5·13+(−1)·13+(−3)·5) − (5·5+7·(−1)+13·(−3)+13·5)| = (1/2)|28|
Area = 14 unit²
12 Applications of differentiation — stationary points 8 marks

The equation of a curve is given by y = 3x⁶ − 9x⁴ + 9x² + 2.

a Show that the x-coordinates of the stationary points on this curve satisfy the equation x(x²−1)² = 0. [3]
Show working
Concept: Stationary points
dy/dx = 18x⁵ − 36x³ + 18x. At a stationary point, dy/dx = 0
18x⁵ − 36x³ + 18x = 0 → divide by 18: x⁵ − 2x³ + x = 0
x(x⁴ − 2x² + 1) = 0 → x((x²)² − 2(x²) + 1) = 0
x(x²−1)² = 0 (shown)
b Solve the equation in (a) to find the x-coordinates of the three stationary points on the curve. [1]
Show working
x(x²−1)²=0 → x=0 or x²−1=0 → (x−1)(x+1)=0
x = 0, x = 1 or x = −1

This curve has exactly 1 turning point, has exactly 2 stationary points of inflexion, and is symmetrical about the y-axis.

c(i) Find the coordinates of the 2 stationary points of inflexion and justify that these are stationary points of inflexion. [2]
Show working
Concept: First derivative test
Since the curve is symmetrical about the y-axis, the 2 points of inflexion (from part (b)) are at x = ±1.
y(1) = 3(1)−9(1)+9(1)+2 = 5, and by symmetry y(−1) = 5 too
Checking dy/dx either side of x=1: at x=1⁻ and x=1⁺, dy/dx > 0 both sides (slope +, 0, +) — sign does not change, so it is a point of inflexion. Similarly at x=−1, dy/dx < 0 both sides (slope −, 0, −)
Points of inflexion: (1, 5) and (−1, 5) — both are stationary points of inflexion, since dy/dx does not change sign across them.
c(ii) Determine the nature of the turning point on the curve. [2]
Show working
Concept: Second derivative test
d²y/dx² = 90x⁴ − 108x² + 18 = 18(5x⁴−6x²+1) = 18(5x²−1)(x²−1)
At x=0: d²y/dx² = 18(−1)(−1) = 18 > 0
Since d²y/dx² > 0 at x=0, the turning point (0, 2) is a minimum point.

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Common questions about this paper

What topics are tested in the 2026 O-Level A-Math Paper 1?

This paper covers differentiation and integration, surds, the discriminant and nature of roots, geometric progressions, logarithms, simultaneous equations, completing the square and curve modelling, circle equations, applications of differentiation to optimisation problems, trigonometric identities and equations, and coordinate geometry involving lines of symmetry and area — a broad sweep of the Additional Mathematics (4049) syllabus.

How do you find the range of values of k for which a line and a curve do not intersect at two distinct points?

Set the curve's equation equal to the line's equation and rearrange into ax² + bx + c = 0 form. For the two graphs to intersect at two distinct points, the discriminant b² − 4ac must be greater than zero; solving the resulting inequality gives the range of the unknown constant.

How do you solve a trigonometric equation involving cosec and sec?

Rewrite cosec x and sec x as 1/sin x and 1/cos x, then simplify the equation so it is expressed only in terms of tan x (using identities like sec²x = tan²x + 1). Solve the resulting equation for tan x, then use the CAST rule to find every angle in the given range, not just the principal value.

How do you tell whether a stationary point is a turning point or a stationary point of inflexion?

Examine the sign of dy/dx just before and just after the stationary point (the first derivative test). If the sign changes from positive to negative or negative to positive, it is a turning point (maximum or minimum). If the sign stays the same on both sides, it is a stationary point of inflexion. The second derivative test can confirm a turning point's nature when it is non-zero there.

Is this Additional Mathematics or O-Level E-Math?

This paper is Additional Mathematics (Syllabus 4049), sat by students taking the A-Math elective — not the standard O-Level E-Math (Syllabus 4052). Topics like partial fractions, the remainder theorem, and calculus-based optimisation are unique to the A-Math syllabus.