AQA A-level MATHEMATICS Paper 1, Exams of Mathematics

A complete solution with detail of all relevant formulas and skills to create in the learner a problem solving strategy.Each solution has been verified using AI. Moreover,most of the solutions have been explained graphically to create visual learning. These are not merely solutions of problems but a problem solving strategy to solve similar issues

Typology: Exams

2023/2024

Available from 10/06/2024

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Solution
AQA A-level MATHEMATICS Paper 1
7357/1
PB/KL/Jun23/E7
shahbaz ahmed
October 2024
Relevant formula and skills
0! = 1
nCr=n!
r!(nr)!
d
dx cxn=cd
dx xn=nxn1
limn→∞ (1 + 1
n)n=e
Where e is an irrational number such that 2 <e<3
If ax=b logab=x
logex= ln x
1
pf3
pf4
pf5

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Solution

AQA A-level MATHEMATICS Paper 1

PB/KL/Jun23/E

shahbaz ahmed October 2024

Relevant formula and skills

0! = 1 nCr = (^) r!(nn−!r)! dx^ d cxn^ =^ c^ dxd xn^ =^ nxn−^1 limn−→∞(1 + (^1) n )n^ = e Where e is an irrational number such that 2 < e < 3 If ax^ = b ⇐⇒ loga b = x loge x = ln x

To expand cos(θ) using Taylor’s theorem up to three terms around x = 0, we can use the formula:

f (x) = f (0) + f ′(0)x + f^

2! x (^2) + f^ ′′′(0) 3! x (^3) + R 3 (x) where R 3 (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:

cos(θ) ≈ 1 + 0 · θ + −2!^1 θ^2 + 0 · θ 3 3! Simplifying this gives:

cos(θ) ≈ 1 − θ 2 2 Thus, the Taylor series expansion of cos(θ) up to three terms is:

log m + log n = log mn log mn = log m − log n log mn^ = n log m ........................................................................... Integration by parts Z u dv = uv −

Z

v du

Q1. Find the coefficient of x^7 in the expansion of (2x − 3)^7 Circle your answer.

− 2187 − 128 2

Solution Using the formula

Tk+1 =

n k

an−kbk

Putting a = 2x, b = − 3 , n = 7

Tk+1 =

k

(2x)^7 −k(−3)k

Tk+1 =

k

27 −kx^7 −k(−3)k

Since x^7 = x^7 −k 7 = 7 − k =⇒ k = 0 Putting k=

T0+1 =

27 x^7 (−3)^0 T 1 =

27 x^7 × 1 Also 7 0

70 ^ = 1 ×7!(7)! = 1

T 1 = 128x^7

Find the equation of the transformed curve. Circle your answer. To transform the curve y = ln x by a stretch parallel to the x-axis with a scale factor of 2, we replace x with x 2 :

y = ln

x 2