Question

Marks

Answer / Key Steps

Q1 — Translation of Co-ordinates

P(−4, 5)

a)  4f(2x)

b)  f(x+3) − 2

4

a)  (−2, 20)  →  2 marks

b)  (−7, 3)   →  2 marks

Q2 — Approximation / Root Finding

f(x) = c·x³ + 3x − 2

A)  Show root lies in [1.1, 1.2]

B)  Find x₂

Bi) Find A

a)2
bi)1
bii)2

Total 5

A)  f(1.1) and f(1.2) change sign → continuous → root in interval

B)  x₂ =1.158(apply Newton–Raphson or iteration)

Bi) a =1.138

 (value from iteration)

Q3 — Differentiation

f(x) = (given function)

Find dy/dx, then use coordinate (2, 3)

5d

dy/dx = (differentiated expression)

Tangent: y = 9x − 15

Q4 — Functions

g(x) = (5x − 1)/(x − 4)

f(x)=3x^2 -2

a)  Minimum point of f(x)

b)  Find gf(x) when x=2?

c)  Find g⁻¹(x)

5

a)  f≥ −2

b)  49/6

c)  g⁻¹(x) = (4x − 1)/(x − 5)

Q5 — Helicopter Parametric

x = 100√t,   y = 10 sin(t²/4)

a)  Maximum vertical height

b)  Horizontal distance at max height of 5 m

5

-someone said 6

a)  10 m

b)  180 m

Q6 — Integration

∫₂ᵖ [3 / (7x + 1)] dx = 6/7,   p > 2

5

P = (15e² − 1) / 7

Q7 — Area of Semicircle Segment

R₁ = 2R₂  (given)

a)  Show:  p·sinθ + q·θ − 2π = 0

b)  1st approx α = 1.3; find 2nd approx via Newton–Raphson

7

a)  p = 1,  q = 3  →  sinθ + 3θ − 2π = 0

   Area of segment: A = ½r²(θ − sinθ)

   R₁ = 2R₂ leads to θ − sinθ = 2(π − θ) − 2sin(π − θ)

   sin(π − θ) = sinθ  →  simplifies to 3θ + sinθ − 2π = 0  ✓

b)  f(a) = 3a + sin(a) − 2π

   f′(a) = 3 + cos(a)

   Apply NR:  aₙ₊₁ = aₙ − f(aₙ)/f′(aₙ)

Q8 — Integration (Area)

x-coord of P = 2  (given)

5

Someone said 6

Answer: 5/4 · e⁶ + ¼

Q9 — Trig Identity

(sinθ + cosθ)(cosecθ − secθ) = k·cot2θ

a)  Find k

b)  Solve (sinx + cosx)(cscx − secx) = 4csc²(2x)

   for −90° < x < 90°

a)3 or 4?
b)5

2 people said total 8

a)  k = 2

   cosecθ − secθ = (cosθ − sinθ)/(sinθ cosθ)

   = 2(cosθ − sinθ)/sin2θ

   Full product = 2cos2θ/sin2θ = 2cot2θ  ✓

b)  Note: equation reduces to sin4x = 4 (no solution)

   Likely intended: 2csc²(2x) → check original paper

Q10 — Modulus Graph

a)  Draw modulus graph

b)  Find range of k using y = k/x

7

Intercept = q,  vertex = q/p

k/x intersects negative-gradient region twice:

0 < k < q²/(4p)

Q11 — Curve & Area

y_C = 7x / √(3x² + 4)

a)  Equation of line L:  ax + by + c = 0

b)  y-coord of point Q

c)  Area of bounded region R

10

a)  14y + 32x − 133 = 0  (intercept 113/14)

c)  ∫₀² (−16x/7 + 113/14 − 7x/√(3x²+4)) dx = 145/21

Q12 — Harmonic Form

a)  4cosx − 13sinx = R·cos(x + α)

b)  D = 30 + 4cos(πt/12 + 0.2) − 13sin(πt/12 + 0.2)

bi)  Find minimum value of D

bii) Find time T at minimum

c)  Why can't this model be used year-round?

a:3  b:1  c:4  d:1

Someone said 7

a)  R = √185,   α = 1.272

bi) Minimum D = 30 − √185 ≈ 16.4 (stated as ~2.7 in notes)

bii) t ≈ 6:23

c)  Model is periodic / doesn't account for seasonal variation

Q13 — Geometric Series

u₃ = sinθ,  u₄ = √2·cosθ,  u₅ = √3·cotθ

a)  Find common ratio r (exact)

b)  Find sum of terms 1, 2 and 3

7

a)4
b)3

a)  θ = π/6,   r = √6

b)  uₙ = (1/12)(√6)ⁿ⁻¹

   S = u₁ + u₂ + u₃ = 1/12(1 + √6 + 6) = (7 + √6)/12

Q14 — Differential Equation

dx/dt = x(A − t)

a)  Solve (exact value)

b)  Find maximum value of x

c)  Find time T when x returns to start

11  (4+3+3 approx)

a)  x = 0.3·e^(2t + t/4·ln70 − t²/2)

b)  x_max ≈ 32.6  when t = A

c)  T ≈ 6.46

Q15 — Proof by Contradiction

Given a² + b² = c²

Prove a and b cannot both be odd

4

Let a = 2m+1,  b = 2n+1  (both odd)

a² + b² = 4m²+4m+1 + 4n²+4n+1 = 4m²+4m+4n²+4n+2

= 4(m²+m+n²+n) + 2  →  even but ≢ 0 (mod 4)

So c² ≡ 2 (mod 4).  But squares are 0 or 1 (mod 4).

Contradiction  →  a and b cannot both be odd  ✓

question

marks

answer

1 Translation of co-ordinates

P(-4,5)

1. 4f(2x)
2. f(x+3) - 2
   

4

(-2,20) → 2 marks

(-7,3) → 2 marks

2 Approximation

A)Show root lies in 1.1-1.2

B)FIND X2

Bi) FIND A

f(1.1) f(1.2) change in sign continuous

x2=

a=

3 differentiation question

f(x)=

dy/dx=

 coordinate(2, 3)

 Y = 9x-15

5

4  Function

f(x) =

g(x) =
a)

b)

c) find g^-1(x)

5

a)F>-2

b) 49/6

C)Y=(4x-1)/x-5  

5 Helicopter parametric

X = 100√ t   y = 10sin(t2/4)

1. Maximum vertical height
2. Horizontal distance at a maximum height of 5m

5x

1. 10m

b)180m

6

  p

∫ from 2 to p [3/(7x + 1)] dx =  

 2
= 6/7 (haha edexcel’s a comedian)
p>2

5

P = (15e^2 - 1)/7

7 Area of semicircle segment

R1 = 2R2 (given)

a “Show psinθ + qθ - 2π = 0

b “1st approximation for α = 1.3. Use newton raphson method to find 2nd approximation for α “

7

a)

1sinθ + 3θ  - 2π = 0

Aka p = 1 ,  q = 3

To solve, use areausearea of segment A = (θ-sinθ)r2/2

R1 = (θ-sinθ)r2/2 = 2R2

R2 = (π - θ - sin(π - θ))r2/2

Therefore (θ-sinθ)r2/2 = 2(π - θ - sin(π - θ))r2/2

This simplifies to (θ-sinθ) = 2π - 2θ - 2sin(π - θ)

sin(-θ + π) = sin(θ)

Therefore θ-sinθ = 2π - 2θ - 2sinθ

Rearrange to 3θ + sinθ - 2π = 0

b)

For Newton Raphson, f(x) = 0

Since 3a + sin(a) - 2π = 0

We can say:

f(a) = 3a + sin(a) - 2π

f’(a) = 3+cos(a)

Apply NR method to

8

 Integration

x of P = 2 (given)

Wrong q thats q 11 aight

5

5/4 ( e^6) + ¼

9 Trig equation (sinθ+cosθ)(cosecθ-secθ)=kcot2θ

2nd Part of Question:

(sinx + cosx)(cscx - secx) = 4csc2 (2x)

For this u made 4cosec2^(2x)=2cot(2x) then make the coesec^2(2x) cot^2x from the identity sin^2 + cos^2=1 by dividing by sin^2
Make it 4(cot2(2x) +1) = 2cot(2x)

^^

I think this function is wrong, because it gives the result sin4x = 4

Might have been 2csc squared then

The values in the student room ai markscheme is what i got

-90o < x < 90o

Solve for x

k=2

Cscθ = 1/sinθ

Secθ = 1/cosθ

Multiply top and bottom, so the left bracket =

((cosθ - sinθ)/(sinθcosθ))

= 2(cosθ - sinθ)/(2sinθcosθ)

= 2(cosθ - sinθ)/(sin2θ)

Therefore total equation is:

( (cosθ + sinθ) ⨉ 2(cosθ - sinθ) ) / (sin2θ)

= 2(cos2θ - sin2θ) / (sin2θ)

= 2(cos2θ) / (sin2θ)

= 2 cot2θ

Therefore k = 2

[error somewhere]

Second part of the Question:

(sinx + cosx)(cscx - secx) = 4csc22x [given]

(sinx + cosx)(cscx - secx) = 2cot2x [substitution from part a]

Therefore, 4csc22x = 2cot2x

Csc22x = 1/sin22x

Cot2x = cos2x/sin2x

Therefore

4 = 2sin2(2x)cos(2x)/sin2x

4 = 2sin(2x)cos(2x)

4 = sin4x ??

10 (a) Drawing modulus graph

(b) range of k using y=k/x

7

1. Intercept is q. Vertex is q/p
2. Pretty sure k/x intersects the negative gradient area twice → 0<k<q2/4p

11

yC = 7x / root(3x2 + 4)

a) find equation of line L in form ax + by + c = 0

b) find y co-ordinate of point Q

b) find area of bounded region R

10

14y-32x-113

Was it not

14y+32x-133=0

Intercept 113/14

∫₀² (−16x/7 + 113/14 − 7x/√(3x² + 4)) dx  = 145/21

(6.904…) was the answer not

12) Harmonic Form

a)

4cosx - 13sinx = Rcos(x+a)

b)

D = 30 + 4cos(πt/12 + 0.2) - 13sin(πt/12 + 0.2)

bi) find MINIMUM

bii) find time T @ minimum

6?

How many marks was this

Part a, 3 marks

B,1

C, 4

D,1

Was c just dy/dx=0 i forgot  wha  I equated it to -1 so i got shook when u said dy/dx

1. R = root(185)

a= 1.272

2. 30GW it asked for the max/min no?B)6:23 It was 6:23

C)2.7

13 Geometric series

u3 = sinθ

u4 = √root2 cosθ

u5 = √root3 cotθ

a) find the common ratio (r) exact value.

b) find the sum of term 1, 2, and 3

7 6? 3 and 3

a=pi/6
R = sqrt 6
7+root6 over 3 It’s over 12 for sure not over 3 I got over 12

Where did 7 come from???

u3 = ar2

u4 = ar3

u5 ar4

Solution to part a)

ar2 = sinθ

ar3 = √2 cosθ

ar4 = root3 cotθ

ar3 = rsinθ = root2 cosθ

Ar4 = r root2 cosθ = root3 cotθ

r = root(6)

Solution to part b)

14. Differential equation
a) dx/dt = x(A-t)

11

A

x = 0.3e^(2t+t/4 ln70 - t^2/2)

B max is like 32.6 when t = A = 32.600522346
C
T = 6.46

15 Proof
Given
a2 + b2 = c2
 Prove that both a and b can’t be odd (or smth like that)
Let a = 2m + 1
Let b = 2n +1

4

Person 1’s Answer:

Expand out a^2 and b^2 using 2m+1 and 2n+1 and sum. This gives you 4m^2 + 4m + 4n^2 + 4n + 2 = c^2.

c^2 is even, therefore c is even, therefore c = 2p where p is an integer. Therefore c^2 = 4p^2.

Factor out 2 on both sides, you get 2m^2 + 2m + 2n^2 + 2n + 1 = 2p^2. Therefore even = odd, therefore contradiction.

Person 2’s Answer:

Expand out a^2 and b^2 using 2m+1 and 2n+1 and sum. Root this answer. Take out factor of root two. Show inside other root is always odd so can never equal any multiple of root two therefore c is always irrational.