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AQA A-level MATHEMATICS Paper 1, 7357/1.PB/KL/Jun23/E7,Solution. $7.99   Add to cart
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AQA A-level MATHEMATICS Paper 1, 7357/1.PB/KL/Jun23/E7,Solution.
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It is a solution of AQA A-level MATHEMATICS Paper 1.Introduction consists of all the relevant formulas and skills to help the learner .Most of the solutions have been explained graphically to make learning visual. Document is not merely solution of problems but a problem solving strategy to create...

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Document information

  • September 30, 2024
  • 59
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

Subjects

  • 73571
  • pbkljun23e7
  • calculus
  • the trapezium rule
  • aqa a level mathematics paper 1

Written for

  • Mathematics
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Solution

AQA A-level MATHEMATICS Paper 1

7357/1

PB/KL/Jun23/E7

shahbaz ahmed

September 2024



Relevant formula and skills

0! = 1
n n!
Cr = r!(n−r)!
d n d n
dx cx = c dx x = nxn−1

limn−→∞ (1 + n1 )n = e

Where e is an irrational number such that 2 < e < 3

If ax = b ⇐⇒ loga b = x

loge x = ln x

................................................................

1

, To expand cos(θ) using Taylor’s theorem up to three terms around

x = 0, we can use the formula:


f ′′ (0) 2 f ′′′ (0) 3
f (x) = f (0) + f ′ (0)x + x + x + R3 (x)
2! 3!

where R3 (x) is the remainder term. For cos(θ):

1. **Function and its derivatives**: - f (θ) = cos(θ) - f ′ (x) =

− sin(x) - f ′′ (x) = − cos(x) - f ′′′ (x) = sin(x) - f (4) (x) = cos(x) (and it

repeats)

2. **Evaluating at x = 0**: - f (0) = cos(0) = 1 - f ′ (0) = − sin(0) =

0 - f ′′ (0) = − cos(0) = −1 - f ′′′ (0) = sin(0) = 0

3. **Constructing the expansion**: Using the derivatives:

−1 2 θ3
cos(θ) ≈ 1 + 0 · θ + θ +0·
2! 3!

Simplifying this gives:

θ2
cos(θ) ≈ 1 −
2

Thus, the Taylor series expansion of cos(θ) up to three terms is:




2

, θ2
cos(θ) ≈ 1 − + 0 · θ3
2

For practical purposes, we can stop at the quadratic term, resulting

in:


θ2
cos(θ) ≈ 1 −
2

............................................................................

cos(α + β) = cos α cos β − sin α sin β

Put α = β = x

cos 2x = cos2 x − sin2 x

From equation

cos2 x + sin2 x = 1 ⇐⇒ cos2 x = 1 − sin2 x

=⇒ cos 2x = cos2 x − sin2 x = 1 − sin2 x − sin2 x = 1 − 2 sin2 x

Similarly

From equation

cos2 x + sin2 x = 1 ⇐⇒ sin2 x = 1 − cos2 x

=⇒

cos 2x = cos2 x − sin2 x = cos2 x − (1 − cos2 x) = 2 cos2 x − 1


3

, .....................................................................

log m + log n = log mn

log m
n = log m − log n

log mn = n log m

...........................................................................

Integration by parts

Z Z
u dv = uv − v du




..........................................................................



Q1. Find the coefficient of x7 in the expansion of

(2x − 3)7

Circle your answer.
128
−2187 − 128 2

Solution

Using the formula


4

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