A/AS level Mathematics year one work booklet and mark scheme

MARK SCHEME
A-Level Mathematics Year 1 / AS — Edexcel
Master Revision Paper


119 questions · 604 marks · full worked solutions


How to mark yourself

The mark codes are the ones real Edexcel examiners use:
ˆ M1 — a method mark. Awarded for a correct method even if the arithmetic that follows is
wrong. This is why you write down the method before you use your calculator.
ˆ A1 — an accuracy mark. Only available if the relevant M mark has been earned. Lost for
wrong values, wrong rounding, or missing units.
ˆ B1 — an independent mark, awarded on sight of a correct statement or answer.
ˆ ft — “follow through”. Your answer is marked as correct given your earlier (wrong) value.
Mark strictly. If you would not have written that line in the exam, you do not get the mark.
Nearly all of the difference between a C and an A at this level is: (i) not showing method, (ii)
rounding too early, (iii) not answering in context, (iv) losing solutions in the second half of a
range. The red Watch out notes flag exactly where each of those bites — read them even on
questions you got right, because those are the marks you are currently leaving on the table.

Note on accuracy: unless a question says otherwise, non-exact answers are given to 3 significant figures. Keep
full accuracy in your calculator throughout and round only at the end ; rounding mid-question is the single most
common way to lose an A mark.

,MARK SCHEME — A-Level Maths Year 1 (Edexcel) 2



Section A — Pure Mathematics (331 marks)

Topic 1 — Indices and Surds 15 marks



4
1. (a) 2x3 y −2 = 24 x12 y −8 = 16x12 y −8 M1
16x12 y −8 × 4x−2 y 5 = 64x12−2 y −8+5 = 64x10 y −3 A1
64x10
= A1 (3)
y3

4
▶ Watch out: 2x3 is 16x12 , not 2x12 — the coefficient is raised to the power too. This single slip is
worth roughly a mark a paper.

3 3 3 3
(b) 25a6 b−4 2
= 25 2 · a6× 2 · b−4× 2 = 125a9 b−6 M1
125a9
= A1 (2)
b6
√
2. Multiply numerator and denominator by the conjugate 2 + 3: M1
√ √ √ √ √
(5 + 3)(2 + 3) 10 + 5 3 + 2 3 + 3 13 + 7 3
√ √ = = A1
(2 − 3)(2 + 3) 4 − 3 1
√
= 13 + 7 3 so a = 13, b = 7 A1 (3)
√ √ √ √ √ √
3. 50 = 5 2, 18 = 3 2, 8=2 2 M1
√ √ √ √
5 2 + 3 2 − 2 2 = 6 2, so k = 6 A1 (2)

2x+1
x−3
4. Write both sides in base 2: 42x+1 = 22 = 24x+2 and 8x−3 = 23 = 23x−9 M1
Equating indices: 4x + 2 = 3x − 9 M1
x = −11 A1 (3)
√ √
√

12 7+1 12 7 + 1 12 7 + 1
5. √ ×√ = = M1
7√−1 7+1 7−1 6
=2 7+2 A1 (2)

Topic 2 — Quadratics 21 marks



6. 6 × (−15) = −90; two numbers multiplying to −90 and summing to +1 are 10 and −9: 6x2 + 10x −
9x − 15 = 2x(3x + 5) − 3(3x + 5) = (3x + 5)(2x − 3) M1 A1
5 3
x=− or x = A1 (3)
3 2 √
−1 ± 1 + 360 −1 ± 19
▶ Alternative: Quadratic formula: x = = , giving the same two values. Full marks
12 12
either way.

7. (a) 3x2 − 12x + 5 = 3 x2 − 4x + 5


M1
= 3 (x − 2)2 − 4 + 5 = 3(x − 2)2 − 12 + 5
 
M1
= 3(x − 2)2 − 7 so a = 3, b = −2, c = −7 A1 (3)
▶ Watch out: After taking out the factor 3, the −4 inside the bracket must also be multiplied by 3.
Forgetting this gives 3(x − 2)2 − 4 + 5 and costs both A marks.

(b) Minimum at (2, −7) B1 (1)
(c) x = 2 B1 (1)

, MARK SCHEME — A-Level Maths Year 1 (Edexcel) 3


8. For kx2 + (k + 3)x + 4 = 0 to be a quadratic, k ̸= 0. B1
Two distinct real roots ⇒ b2 − 4ac > 0: (k + 3)2 − 4(k)(4) > 0 M1
k 2 + 6k + 9 − 16k > 0 ⇒ k 2 − 10k + 9 > 0 A1
(k − 1)(k − 9) > 0; positive quadratic, so the solution is outside the roots M1
k < 1 or k > 9, together with k ̸= 0 A1 (5)
▶ Watch out: Two classic losses here. (1) Forgetting k ̸= 0 — if k = 0 the equation is linear, not quadratic,
so it cannot have two roots. (2) Writing 1 < k < 9: for a positive quadratic, “> 0” means outside the critical
values. Always sketch the parabola.
√
9. Let u = x, so x = u2 and u ⩾ 0: u2 − 7u + 12 = 0 M1
(u − 3)(u − 4) = 0 ⇒ u = 3 or u = 4 A1
√ √
x = 3 ⇒ x = 9; x = 4 ⇒ x = 16 M1 A1 (4)
▶ Watch out: You must square back to find x. Leaving the answer as x = 3 or 4 throws away the last two
marks.

10. Let u = 3x , noting 32x = (3x )2 = u2 : u2 − 10u + 9 = 0 M1
(u − 1)(u − 9) = 0 ⇒ u = 1 or u = 9 A1
3x = 1 ⇒ x = 0 A1
3x = 9 = 32 ⇒ x = 2 A1 (4)
x 2
▶ Watch out: 32x is (3 ) , not 2 × 3x and not 9x treated as unrelated. Also, 3x = 1 does have a solution:
x = 0. Discarding it is a very common way to drop a mark.


Topic 3 — Equations and Inequalities 24 marks



11. Substitute y = 2x − 3 into x2 + y 2 = 5: M1
x2 + (2x − 3)2 = 5 ⇒ x2 + 4x2 − 12x + 9 = 5 M1
5x2 − 12x + 4 = 0 ⇒ (5x − 2)(x − 2) = 0 A1
2
x= or x = 2 A1
5
2 11
x = ⇒ y = 2 25 − 3 = − ; x = 2 ⇒ y = 1


A1 (5)
5 5
Solutions: 52 , − 11


5 and (2, 1).
2
▶ Watch out: Pair your answers up. Writing “x = 5 or 2, y = − 11
5 or 1” without linking them loses the
final A mark.

12. 2x2 + 3x − 9 = (2x − 3)(x + 3) M1
3
Critical values: x = and x = −3 A1
2
Positive quadratic, and we want ⩾ 0, so the solution is outside the critical values M1
{x : x ⩽ −3} ∪ x : x ⩾ 32

A1 (4)
3
▶ Watch out: Set notation must use ∪ (union), not ∩. Writing −3 ⩾ x ⩾ 2 is meaningless and scores zero
for the final mark.

13. 4x − 3 > 9 ⇒ 4x > 12 ⇒ x > 3 B1
x2 − 7x + 10 < 0 ⇒ (x − 2)(x − 5) < 0 M1
Positive quadratic, < 0 means between the roots: 2 < x < 5 A1
Both conditions must hold, so take the overlap of x > 3 and 2 < x < 5 M1
3<x<5 A1 (5)
▶ Watch out: Draw both solution sets on one number line. “Both” means intersection (overlap); “either”