4MA1: Sequences, functions and graphs. Version: Issue 2, November 2017; first assessment June 2018; linear Mathematics A.
Foundation teaching and Higher additions are labelled below. This reference packages the existing native-lesson crosswalk. It does not certify unreviewed specification rows or a whole qualification. Original diagnostics are separate and are not reproduced.
Arithmetic sequences and nth terms · Foundation
Find a constant difference for an arithmetic sequence. Its nth term is a+(n-1)d. A term-to-term rule describes how to reach the next term; a position-to-term rule gives a term directly.
For 5,8,11,14,... the common difference is 3. The nth term is 5+3(n-1)=3n+2. At n=8, u₈=26. To find the position of 62, solve 3n+2=62, giving n=20.
The first term has index 1, so the exponent is n-1. A sequence is a list; a series is a sum. A geometric sequence can alternate in sign and still converge.
Generate several terms and check a proposed nth-term rule. Infinite geometric series and advanced sum formulae are excluded from this Foundation/Core lesson.
Arithmetic sequences and finite sums · Higher
For an arithmetic sequence, u_n=a+(n-1)d. Pairing the first and last terms gives equal pair sums a+u_n. Therefore the first n terms have sum S_n=n(a+u_n)/2=n[2a+(n-1)d]/2. A sequence lists terms; a series adds them. State the first index before using a formula.
For a=5,d=3,n=8, u_8=5+7×3=26 and S_8=8(5+26)/2=124. If the second term is 7 and the fifth is 19, then 3d=19−7=12, so d=4 and a=7−4=3. The first four terms 3,7,11,15 sum to 36, also 4(3+15)/2=36. The sum formula depends on a constant difference.
A term is not a sum. The first term is indexed by 1 in these formulae. An arithmetic sum requires a constant difference; increasing terms alone do not establish this.
4MA1 Higher 3.1 A–C covers common difference, the arithmetic nth term and the sum of the first n arithmetic terms. Geometric finite/infinite sum formulae are excluded from this supported focus lesson.
Straight lines and gradients · Foundation
Gradient is change in y divided by change in x. A straight line has y=mx+c, where c is its y-intercept. Parallel lines have equal gradients.
Through (2,5) with gradient 3, substitute to get 5=3×2+c, so c=-1 and y=3x-1. Points (1,2) and (4,8) give gradient (8-2)/(4-1)=2.
A vertical line has no finite gradient; do not force it into y=mx+c. Read the signs of a circle's centre carefully. The radius to a tangent is perpendicular to the tangent.
Plot a straight line using two checked points and label its intercept. Perpendicular-gradient formulae and circle equations are not part of this Foundation/Core lesson.
Coordinate geometry and tangents · Higher
A line through (x₁,y₁) with gradient m has y-y₁=m(x-x₁). Parallel lines have equal gradients. Finite perpendicular gradients multiply to -1. A circle has (x-a)²+(y-b)²=r².
Through (2,5) with gradient 3, y-5=3(x-2), so y=3x-1. A perpendicular through the same point has y-5=-(x-2)/3. The circle (x-2)²+(y+1)²=25 has centre (2,-1) and radius 5.
A vertical line has no finite gradient; do not force it into y=mx+c. Read the signs of a circle's centre carefully. The radius to a tangent is perpendicular to the tangent.
Before solving a line-circle intersection, predict whether there are zero, one or two intersections. Substitution produces a quadratic whose discriminant checks the prediction.
Domains, inverses and composition · Higher
State the domain and range. For an inverse, first ensure the function is one-to-one on its domain. Composition fg means apply g first, then f; the intermediate output must be an allowed input to f.
For f(x)=√(x-2), x≥2 and the range is y≥0. From y=√(x-2), x=y²+2. Thus f inverse(x)=x²+2 with x≥0. For g(x)=x+3, fg(1)=f(4)=√2.
Squaring can introduce extraneous solutions. Restricting a parabola's domain is essential before claiming an inverse. A horizontal translation inside f has the opposite sign to the graph's movement.
Check f(f inverse(x))=x on the inverse domain. Use a sketch to test whether a horizontal line meets the original graph more than once.
Derivatives and stationary points · Higher
For a polynomial term ax^n with nonnegative integer n, the gradient term is anx^(n-1). Add the differentiated terms. A stationary point has zero gradient.
For y=x³-3x, dy/dx=3x²-3. At x=1, the gradient is 0 and y=-2. The gradient 3x²-3 changes from negative to positive at x=1, so this is a local minimum. At x=-1, y=2 and the gradient changes from positive to negative, giving a local maximum. Use gradient signs and the graph shape, within this specification.
A zero derivative does not always mean a maximum or minimum: y=x³ is stationary at 0 but continues increasing. An endpoint can also produce an extreme value on a restricted domain.
This advanced-tier IGCSE lesson is limited to polynomial differentiation, tangent gradients and stationary points. Chain, product, quotient and implicit differentiation are excluded.
