Edexcel GCE A-Level Mathematics June 2026 Paper 1 Mark Scheme

⚠️ Unofficial Reconstructed Mark Scheme Disclaimer:
This is a reconstruction of the official Edexcel A Level Paper 1 2026 based on anecdotal data. Both exact question phrasing and specific mark distributions are professional estimates and may not fully match the official Pearson Edexcel materials. All questions are adjusted to sum exactly to 100 marks.

General Marking Guidance

M (Method) mark: Awarded for knowing a method and attempting to apply it.
A (Accuracy) mark: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.
B (Unconditional) mark: Independent accuracy marks, awarded regardless of method.
dM / ddM: Dependent method marks. The preceding step must have been completed correctly to award these.
ft (Follow Through): Marks awarded for correct calculations following an incorrect value from a previous part.
cso: Correct solution only. There must be absolutely no errors in this part of the question.

Question 1 Total: 4 Marks

Part	Scheme	Marks	AOs
(a)	x-transformation: −4 → −4 × ½ = −2 ; y-transformation: 5 → 5 × 4 = 20	M1	1.1b
(a)	Image point: (−2, 20)	A1	1.1b
(b)	x-transformation: −4 → −4 − 3 = −7 ; y-transformation: 5 → 5 − 2 = 3	M1	1.1b
(b)	Image point: (−7, 3)	A1	1.1b

Notes

(a) M1: Attempts both coordinate transformations correctly (halving the x-coordinate and multiplying the y-coordinate by 4).
(a) A1: Correct coordinate point only. Accept written separately as x = −2, y = 20.
(b) M1: Correctly shifts left by 3 units (x − 3) and down by 2 units (y − 2).
(b) A1: Correct coordinate point only. Accept written separately.

Question 2 Total: 5 Marks

Part	Scheme	Marks	AOs
(a)	f(1.1) = e^{2(1.1) − 1} + 3(1.1) − 7 = e^1.2 + 3.3 − 7 = −0.380... < 0 f(1.2) = e^{2(1.2) − 1} + 3(1.2) − 7 = e^1.4 + 3.6 − 7 = 0.655... > 0	M1	1.1b
(a)	States there is a change of sign and the function f(x) is continuous on [1.1, 1.2], concluding that there is a root α in the interval.	A1*	2.4
(b)	x₂ = ½(1 + ln(7 − 3(1.1))) = ½(1 + ln(3.7)) = 1.15423... ➔ awrt 1.1542	M1	1.1b
(b)	x₃ = 1.1293..., x₄ = 1.1415... ➔ continues iterations to obtain α = 1.1383 (correct to 4 d.p.)	A1	1.1b

Notes

(a) M1: Attempts values of f(1.1) and f(1.2) in radians. At least one must be correct to 1 s.f. rounded or truncated. Calculations done in degrees score M0.
(a) A1*: Both values correct to 1 s.f., notes change of sign, explicitly references continuity of f(x), and provides a clear conclusion.
(b) M1: Substitutes x₁ = 1.1 correctly into the formula and evaluates x₂. Award for x₂ = awrt 1.1542.
(b) A1: Correctly identifies the root α = 1.1383 to 4 d.p. from consecutive converging iterations.

Question 3 Total: 5 Marks

Part	Scheme	Marks	AOs
(a)	dydx = 3x² − 4x + 5	B1 B1	1.1b 1.1b
(b)	Substitute x = 2 into dydx: m = 3(2)² − 4(2) + 5 = 9	M1	1.1b
	Find the y-coordinate at x = 2: y = 2³ − 2(2)² + 5(2) − 7 = 3 ➔ P(2, 3)	B1	1.1b
	Equation of tangent: y − 3 = 9(x − 2) ➔ y = 9x − 15	A1	1.1b

Notes

(a) B1 B1: One mark for at least two correct terms differentiated. Second mark for a fully correct expression.
(b) M1: Substitutes x = 2 into their derivative and uses the result as their gradient in the straight line equation.
(b) B1: Correctly finds y = 3 when x = 2.
(b) A1: Correct tangent equation in the requested form.

Question 4 Total: 6 Marks

Part	Scheme	Marks	AOs
(a)	f(x) ≥ −2 (or y ≥ −2)	B1	1.1b
(b)	Evaluate f(2) = 3(2)² − 2 = 10	M1	1.1b
(b)	Substitute x = 10 into g(x): g(10) = 5(10) − 110 − 4 = 496	A1	1.1b
(c)	Let y = 5x − 1x − 4 ➔ y(x − 4) = 5x − 1 ➔ xy − 4y = 5x − 1	M1	1.1b
	xy − 5x = 4y − 1 ➔ x(y − 5) = 4y − 1 ➔ x = 4y − 1y − 5	A1	2.1
	g⁻¹(x) = 4x − 1x − 5, Domain: x ∈ ℝ, x ≠ 5	A1	2.5

Notes

(a) B1: Correct range. Accept any equivalent interval notation like [−2, ∞). Do not accept x ≥ −2.
(b) M1: Correct order of composition: finds f(2) first, then substitutes that value into g.
(b) A1: 49/6 or exact equivalent (like 8⅙).
(c) M1: Multiplies through to clear the fraction and collects all x-terms on one side of the equation.
(c) A1: Correctly factorises and isolates x.
(c) A1: Correct expression for the inverse in terms of x, accompanied by the correct domain constraint.

Question 5 Total: 5 Marks

Part	Scheme	Marks	AOs
(a)	Maximum height is 10 m (since the maximum value of sin(θ) is 1)	B1	3.4
(b)	Set y = 5: 10 sin(t²4) = 5 ➔ sin(t²4) = 0.5	M1	3.1b
	Finds second positive principal solution: t²4 = π − π6 = 5π6	M1	2.1
	Solve for t: t² = 20π6 = 10π3 ➔ t = √10π3 ≈ 3.236... seconds	dM1	1.1b
	Substitute t into x: x = 100 × √3.236... = 179.889... m ➔ awrt 179.9 m	A1	1.1b

Notes

(a) B1: Correct height of 10 m using the range of the sine wave.
(b) M1: Equates y to 5 and divides by 10 to isolate the trigonometric part.
(b) M1: Identifies the second boundary angle for the second time height is met: 5π6. Do not accept π6.
(b) dM1: Reaches a positive value for t and substitutes it into the equation for x.
(b) A1: Correct horizontal distance to 1 d.p. (awrt 179.9 m).

Question 6 Total: 6 Marks

Part	Scheme	Marks	AOs
(a)	Attempts integration of the form k ln(7x + 1)	M1	1.1b
(a)	∫ 37x + 1 dx = 37 ln(7x + 1) (condone missing constant)	A1	1.1b
(b)	Apply boundaries: 37 ln(7p + 1) − 37 ln(15) = 67	M1	1.1b
	Divide by 37: ln(7p + 1) − ln(15) = 2	M1	1.1b
	Merge logarithms: ln(7p + 115) = 2 ➔ 7p + 115 = e²	dM1	1.1b
	Solve for p: 7p + 1 = 15e² ➔ p = 15e² − 17	A1	2.1

Notes

(a) M1: Recognises reciprocal integration matches natural logarithm structure.
(a) A1: Correct coefficient of 3/7.
(b) M1: Evaluates boundaries p and 2 correctly and subtracts.
(b) M1: Rearranges the coefficients to simplify logarithmic terms.
(b) dM1: Correct exponential translation to undo logs. Dependent on previous M1.
(b) A1: Exact final value of p only.

Question 7 Total: 7 Marks

Part	Scheme	Marks	AOs
(a)	Area of minor sector BOC = ½ r² θ ; Area of triangle BOC = ½ r² sin θ Area of region R₁ (segment BC) = ½ r² (θ − sin θ)	M1	2.1
	Area of region R₂ = Semicircle − Sector BOC − Triangle AOB Area R₂ = ½ π r² − ½ r² θ − ½ r² sin(π − θ)	M1	3.1a
	Simplify using sin(π − θ) = sin θ: R₂ = ½ r² (π − θ − sin θ)	B1	1.1b
	Set R₁ = 2R₂: ½ r² (θ − sin θ) = 2 [ ½ r² (π − θ − sin θ) ] θ − sin θ = 2(π − θ − sin θ) ➔ θ − sin θ = 2π − 2θ − 2 sin θ	dM1	1.1b
	Collects terms to show: sin θ + 3θ − 2π = 0 ★	A1*	2.1
(b)	Let f(θ) = sin θ + 3θ − 2π ➔ f′(θ) = cos θ + 3 Evaluate at θ₁ = 1.3: f(1.3) = sin(1.3) + 3(1.3) − 2π ≈ −1.420... f′(1.3) = cos(1.3) + 3 ≈ 3.267...	M1	1.1b
(b)	θ₂ = 1.3 − −1.420...3.267... ≈ 1.7348... ➔ awrt 1.735	A1	1.1b

Notes

(a) M1: Employs correct area formulas for segments and sectors.
(a) M1: Complete geometric plan for region R₂.
(a) B1: Correct identity swap for supplementary angle sine.
(a) dM1: Sets up ratio, cancels ½ r² factor. Dependent on first M1.
(a) A1*: Pristine proof.
(b) M1: Differentiates f(θ) correctly and evaluates both values using radians.
(b) A1: Correct second value (awrt 1.735).

Question 8 Total: 5 Marks

Part	Scheme	Marks	AOs
(a)	lim_{δx → 0} ∑_{x = 0}³ x e^2x δx = 30∫ x e^2x dx	B1	1.1b
(b)	Integration by parts: u = x ➔ du = dx ; dv = e^2x dx ➔ v = ½ e^2x ∫ x e^2x dx = ½ x e^2x − ∫ ½ e^2x dx = ½ x e^2x − ¼ e^2x	M1 A1	2.1 1.1b
	Substitute boundaries: [ ½ x e^2x − ¼ e^2x ]₀³ = ( ³/₂ e⁶ − ¼ e⁶ ) − ( 0 − ¼ e⁰ )	dM1	1.1b
	Result = ⁵/₄ e⁶ + ¼ ➔ (A = ⁵/₄, B = ¼)	A1	1.1b

Notes

(a) B1: Correct translation of sum limit into definite integral notation.
(b) M1: Integrates by parts correctly.
(b) A1: Fully accurate indefinite integral.
(b) dM1: Applies limits correctly. Note that the lower limit evaluation must yield the constant term + ¼ (since e⁰ = 1).
(b) A1: Exact rational coefficients.

Question 9 Total: 6 Marks

Part	Scheme	Marks	AOs
(a)	(cos θ + sin θ)(cosec θ − sec θ) = (cos θ + sin θ)(¹/_{sin θ} − ¹/_{cos θ}) = (cos θ + sin θ) cos θ − sin θsin θ cos θ = cos² θ − sin² θsin θ cos θ	M1	2.1
(a)	= cos 2θ½ sin 2θ = 2 cot 2θ ➔ k = 2	A1	1.1b
(b)	Apply identity: 5(2 cot 2x) = 4 cosec²(2x) ➔ 10 cos 2xsin 2x = 4sin² 2x	M1	1.1b
	Multiply across: 10 cos 2x sin 2x = 4 ➔ 5(2 sin 2x cos 2x) = 4 ➔ 5 sin 4x = 4	M1	1.1b
	sin 4x = 0.8 ➔ 4x = 53.13°, 126.87°, −233.13°, −306.87°	dM1	1.1b
	Solve for x: x = −76.7°, −58.3°, 13.3°, 31.7°	A1 A1	1.1b 1.1b

Notes

(a) M1: Expresses terms as reciprocal functions, constructs common denominator, and multiplies numerator brackets.
(a) A1: Identifies double-angle forms correctly to show k = 2.
(b) M1: Integrates result of (a) into equation.
(b) M1: Condenses trigonometric formula to single term 5 sin 4x = 4.
(b) dM1: Computes standard solutions for 4x. Dependent on previous M1.
(b) A1 A1: Two correct solutions gets first accuracy mark; all four correct (awrt 1 d.p.) with no extras gets the second.

Question 10 Total: 10 Marks

Part	Scheme	Marks	AOs
(a)	At x = 2: y = 7(2)√3(2)² + 4 = 14√16 = 3.5 ➔ P(2, 3.5)	B1	1.1b
	Differentiate y = 7x(3x² + 4)^−½ using derivative rules: dydx = 7√3x² + 4 − 7x(½)(3x² + 4)^−½(6x)3x² + 4 = 28(3x² + 4)^1.5	M1 A1	1.1b 1.1b
	At x = 2: dydx = 2816^1.5 = 2864 = 716	M1	2.1
	Normal gradient = −167	M1	1.1b
	Normal line: y − 3.5 = −167(x − 2) ➔ 32x + 14y − 113 = 0 ★	A1*	1.1b
(b)	Set x = 0 ➔ 14y = 113 ➔ y = 11314 (or \(8.07\))	B1	1.1b
(c)	Perform integration via reverse chain rule / substitution: ∫ 7x(3x² + 4)^−½ dx = 73 (3x² + 4)^½	M1 A1	1.1b 1.1b
	Calculate Curve Area: [ 73 √3x² + 4 ]₀² = 283 − 143 = 143	B1	1.1b
	Calculate Trapezium Area: 20∫ (−167 x + 11314) dx = 817	dM1	3.1a
	Combined Area of R = 817 − 143 = 14521	A1	1.1b

Notes

(a) B1: Finds the coordinate P(2, 3.5).
(a) M1 A1: Differentiation completed with correct execution of the chain rule.
(a) M1: Computes the numerical gradient at x = 2.
(a) M1: Uses negative reciprocal gradient rule.
(a) A1*: Elegant algebraic reduction to the printed line equation.
(b) B1: Finds the correct y-intercept.
(c) M1 A1: Correctly integrates the reciprocal square root function.
(c) B1: Finds the definite integral bounds of the curve.
(c) dM1: Computes the linear trapezium bounds and subtracts the curve area.
(c) A1: Gives exact final fraction of 145/21.

Question 11 Total: 9 Marks

Part	Scheme	Marks	AOs
(a)	R cos(θ + α) = R cos θ cos α − R sin θ sin α R cos α = 4 ; R sin α = 13 ➔ R = √4² + 13² = √185	B1	1.1b
	tan α = 134 ➔ α = awrt 1.272	M1	1.1b
	Result: √185 cos(θ + 1.272)	A1	1.1b
(b)	Minimum daily consumption is D = 30 − √185 = awrt 16.4 GWh	B1ft	3.4
(c)	Minimum consumption occurs when cos(π t12 + 0.2 + 1.272) = −1	M1	3.4
	π t12 + 1.472 = π ➔ π t12 = π − 1.472 = 1.670... ➔ t = 6.377 hours	dM1	2.1
	Convert to clock time: 6.377 hours = 6 hours 23 mins after midnight ➔ 06:23	A1 A1	1.1b 3.2a
(d)	Energy usage significantly varies between weekdays and weekends (or due to seasonal temperature shifts) which are not accounted for by this simplified 24-hour model.	B1	3.5b

Notes

(a) B1: Exact surd value for R required.
(a) M1 A1: Correct division and evaluation of α.
(b) B1ft: Subtracts their exact R from 30.
(c) M1: Recognises cosine wave minimum is met when the cosine term is −1.
(c) dM1: Correct linear transposition to isolate t.
(c) A1 A1: Correct value of t is converted to accurate 24h clock time to the nearest minute.
(d) B1: Provides any valid energy grid physical limitation of the model.

Question 12 Total: 8 Marks

Part	Scheme	Marks	AOs
(a)	Draws reciprocal curve branches in quadrants 1 and 3 correctly.	B1	1.1b
	Draws V-shaped modulus curve with vertex on the positive x-axis.	B1	1.1b
	Coordinates written clearly as (q/p, 0) and (0, q).	A1	1.1b
(b)	Right-hand branch equation: kx = px − q ➔ px² − qx − k = 0	M1	3.1a
	Right discriminant: b² − 4ac = (−q)² − 4(p)(−k) = q² + 4pk Since p, k > 0, discriminant is always positive, yielding exactly 1 positive root intersection.	A1	2.4
	Left-hand branch equation: kx = q − px ➔ px² − qx + k = 0	dM1	1.1b
	Two roots required for exactly 3 total intersections: b² − 4ac > 0	A1	1.1b
	Deduces correct inequality: q² − 4pk > 0 ➔ 0 < k < q²4p	A1	2.2a

Notes

(a) B1: Draws both curves cleanly.
(a) B1: Modulus vertex lies correctly in quadrant 1.
(a) A1: Correct intercepts clearly labelled.
(b) M1: Formulates the quadratic from the positive branch of the modulus.
(b) A1: Proves that the positive branch always yields 1 intersection.
(b) dM1: Formulates the quadratic from the negative branch.
(b) A1: Uses discriminant inequality correctly.
(b) A1: Bounds the inequality with 0 < k to ensure three positive solutions.

Question 13 Total: 9 Marks

Part	Scheme	Marks	AOs
(a)	Write out ratio: r = √2 cos θsin θ and r = √3 cos θ cot θ√2 cos θ = √3 cot θ√2	M1	2.1
	Equate both ratios: √2 cos θsin θ = √3 cos θ√2 sin θ	dM1	1.1b
	Simplify equation: 2 = √3 cos θ ➔ cos θ = √32 ➔ θ = π6 ★	A1*	1.1b
	Evaluate ratio: r = √2 cos(π/6)sin(π/6) = √6	A1 A1	1.1b 1.1b
(b)	Find terms: u₃ = sin(π/6) = 0.5 u₂ = u₃r = 12√6 = √612 ; u₁ = u₂r = 112	M1 A1	1.1b 1.1b
	Add terms: Sum = 112 + √612 + 612	dM1	1.1b
	Sum = 7 + √612	A1	1.1b

Notes

(a) M1: Formulates two distinct geometric ratios.
(a) dM1: Equates and simplifies ratios correctly.
(a) A1*: Shows angle is exactly \(\pi/6\).
(a) A1 A1: Award first accuracy mark for getting a correct intermediate trigonometric relation, and second accuracy mark for finding r = √6 exactly.
(b) M1 A1: Divides to find previous terms correctly.
(b) dM1: Sums terms with a common denominator of 12.
(b) A1: Correct exact final term.

Question 14 Total: 11 Marks

Part	Scheme	Marks	AOs
(a)	Separate variables: ∫ 1x dx = ∫ (A − t) dt	M1	2.1
	Integrate: ln(x) = At − t²/2 + c	A1	1.1b
	Exponentiate: x = D e^{(At − t²/2)}	dM1	1.1b
	Substitute boundaries: t = 0, x = 0.3 ➔ D = 0.3	A1	3.4
	Substitute boundary: t = 4, x = 21 ➔ 21 = 0.3 e^{(4A − 8)} ➔ 70 = e^{(4A − 8)}	dM1	3.4
	Solve for A: 4A − 8 = ln 70 ➔ A = 2 + ¼ ln 70 ★	A1*	1.1b
(b)	Using model: Maximum occurs when dx/dt = 0 ➔ t = A	M1	3.1b
	Evaluate: x = 0.3 e^{(A² − A²/2)} = 0.3 e^(A²/2)	A1	1.1b
	Substitute A ≈ 3.062... ➔ x ≈ 32.607... ➔ awrt 32.6	A1	1.1b
(c)	Set x = 0.1 ➔ 0.3 e^{(At − t²/2)} = 0.1 ➔ At − t²/2 = −ln 3	M1	3.4
(c)	Rearrange to quadratic: t² − 2At − 2 ln 3 = 0 ➔ T = A + √A² + 2 ln 3 ≈ 6.464 hours	A1	1.1b

Notes

(a) M1 A1: Correctly separates variable parameters and integrates.
(a) dM1 A1: Translates to exponential form and uses first coordinate.
(a) dM1 A1*: Uses second coordinate boundary correctly to show value for A.
(b) M1: Sets first derivative equal to 0.
(b) A1: Isolates maximum concentration term in terms of A.
(b) A1: Substitutes numerical value of A and rounds correctly to 1 d.p.
(c) M1: Equates model to 0.1 and sets up a standard quadratic in t.
(c) A1: Solves for T to 3 d.p. (6.464). Reject negative root.

Question 15 Total: 4 Marks

Part	Scheme	Marks	AOs
-	Assume that a and b are both odd positive integers. Let a = 2m + 1 and b = 2n + 1, where m, n are non-negative integers.	M1	2.1
	Expand and sum squares: a² + b² = (2m + 1)² + (2n + 1)² = 4m² + 4m + 1 + 4n² + 4n + 1	M1	2.2a
	Factor out 2: a² + b² = 2(2m² + 2m + 2n² + 2n + 1) This demonstrates a² + b² is even, so c² must be even, which implies c is even.	dM1	1.1b
	Let c = 2k ➔ c² = (2k)² = 4k², which is a multiple of 4. However, a² + b² is of the form 4(m² + m + n² + n) + 2, leaving a remainder of 2 when divided by 4. Therefore, a² + b² ≠ c². This is a contradiction, hence a and b cannot both be odd.	A1	2.4

Notes

M1: Starts the contradiction proof by assuming a and b are both odd, expressing them algebraically as 2m + 1 and 2n + 1.
M1: Squares and adds both terms together.
dM1: Demonstrates that a² + b² is even, implying c is even. Dependent on previous M1.
A1: Establishes that c² must be a multiple of 4 while a² + b² is not, yielding a contradiction and completing the proof.

GCE Advanced Level Mathematics (9MA0/01)

Pure Mathematics 1 - June 2026 GCE exam

General Marking Guidance