Edexcel A-level Maths coverage

Algebra and functions

Section 2
11 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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2.1

Understand and use the laws of indices for all rational exponents.

  • For a non-zero base, aman=am+na^m a^n=a^{m+n}, am/an=amna^m/a^n=a^{m-n} and (am)n=amn(a^m)^n=a^{mn}.
  • Interpret rational powers through roots: ap/q=apq=(aq)pa^{p/q}=\sqrt[q]{a^p}=(\sqrt[q]{a})^p whenever the real-valued expression is defined.
  • For example, 163/4=(164)3=23=816^{3/4}=(\sqrt[4]{16})^3=2^3=8.
  • A negative exponent means reciprocal, not a negative value: ar=1/ara^{-r}=1/a^r; also avoid applying index laws across addition.

Tier 1 · Easy

2 marks
ORIGINAL

Evaluate 813/481^{3/4} without using a calculator.

Tier 2 · Standard

2 marks
ORIGINAL

Given x>0x>0, simplify x5/2x1/2\dfrac{x^{5/2}}{x^{-1/2}} to a single power of xx.

Tier 3 · Hard

4 marks
ORIGINAL

Solve 82x1=4x+28^{2x-1}=4^{x+2}, giving the exact value of xx.

2.2

Use and manipulate surds, including rationalising the denominator.

  • A surd is an exact irrational root; simplify it by extracting the largest square factor, as in 50=52\sqrt{50}=5\sqrt2.
  • Combine only like surds, and rationalise a binomial denominator by multiplying numerator and denominator by its conjugate.
  • For example, 1/(3+2)=(32)/(92)=(32)/71/(3+\sqrt2)=(3-\sqrt2)/(9-2)=(3-\sqrt2)/7.
  • Do not split roots across addition: a+b\sqrt{a+b} is not generally a+b\sqrt a+\sqrt b; after squaring an equation, check every candidate in the original.

Tier 1 · Easy

2 marks
ORIGINAL

Write 72\sqrt{72} in the form a2a\sqrt2, where aa is an integer.

Tier 2 · Standard

3 marks
ORIGINAL

Rationalise and simplify 52+3\dfrac{5}{2+\sqrt3}.

Tier 3 · Hard

5 marks
ORIGINAL

Solve x+6x3=1\sqrt{x+6}-\sqrt{x-3}=1.

2.3

Work with quadratic functions and their graphs; the discriminant, including conditions for real and repeated roots; completing the square; solution of quadratic equations, including solving quadratic equations in a function of the unknown.

  • For ax2+bx+cax^2+bx+c, the discriminant b24acb^2-4ac is positive for two distinct real roots, zero for a repeated root and negative for no real roots.
  • Complete the square to expose the turning point and range, or use factorisation and the quadratic formula to solve equations.
  • For example, x26x+11=(x3)2+2x^2-6x+11=(x-3)^2+2, so its minimum point is (3,2)(3,2) and it has no real roots.
  • When the quadratic is in a function of the unknown, substitute a new variable, solve the quadratic in that variable, then solve every resulting equation and reject invalid roots.

Tier 1 · Easy

2 marks
ORIGINAL

Write x28x+5x^2-8x+5 in completed-square form.

Tier 2 · Standard

3 marks
ORIGINAL

Find the two values of kk for which 2x2+(k1)x+8=02x^2+(k-1)x+8=0 has a repeated root.

Tier 3 · Hard

6 marks
ORIGINAL

Solve (x25x)25(x25x)14=0(x^2-5x)^2-5(x^2-5x)-14=0, giving exact answers.

2.4

Solve simultaneous equations in two variables by elimination and by substitution, including one linear and one quadratic equation.

  • A simultaneous solution is an ordered pair satisfying every equation; graphically, solutions are intersection points.
  • Use elimination when coefficients align conveniently, or substitute an expression from the linear equation into the nonlinear equation.
  • For example, substituting y=5xy=5-x into xy=6xy=6 gives x(5x)=6x(5-x)=6, whose roots generate the two intersection pairs.
  • After finding one coordinate, substitute each possible value back separately; losing the second quadratic root or mismatching coordinates are common errors.

Tier 1 · Easy

2 marks
ORIGINAL

Solve x+y=9x+y=9 and xy=1x-y=1 simultaneously.

Tier 2 · Standard

4 marks
ORIGINAL

Find every solution of y=x+1y=x+1 and x2+y2=25x^2+y^2=25.

Tier 3 · Hard

5 marks
ORIGINAL

Solve the system x+2y=7x+2y=7 and x2+y2=13x^2+y^2=13.

2.5

Solve linear and quadratic inequalities in one variable and interpret them graphically, including with brackets and fractions; express solutions using 'and'/'or' or set notation; represent linear and quadratic inequalities graphically.

  • An inequality solution is a set of values; endpoints are included for \leq or \geq and excluded for << or >>.
  • For a quadratic or rational inequality, locate every zero and undefined value, then use a sign diagram or graph on the resulting intervals.
  • For example, (x2)(x+3)<0(x-2)(x+3)<0 between its roots, so 3<x<2-3<x<2.
  • Multiplying by an expression of unknown sign can reverse the inequality unpredictably; bring a fractional inequality to one side and analyse signs instead.

Tier 1 · Easy

2 marks
ORIGINAL

Solve 52x<115-2x<11.

Tier 2 · Standard

3 marks
ORIGINAL

Solve (x4)(x+1)0(x-4)(x+1)\leq0.

Tier 3 · Hard

4 marks
ORIGINAL

Solve x+2x3>2\dfrac{x+2}{x-3}>2 and give the result in set notation.

2.6

Manipulate polynomials algebraically, including expanding brackets, collecting like terms, factorisation and simple algebraic division; use the factor theorem; simplify rational expressions by factorising, cancelling and algebraic division.

  • Polynomial manipulation uses expansion, collection and factorisation while preserving equality; choose the form that exposes the required structure.
  • The factor theorem states that xax-a is a factor of f(x)f(x) exactly when f(a)=0f(a)=0; after finding a factor, divide to obtain the remaining polynomial.
  • For example, f(2)=0f(2)=0 for f(x)=x33x24x+12f(x)=x^3-3x^2-4x+12, and division by x2x-2 gives x2x6x^2-x-6.
  • Cancel factors, not separate terms, in rational expressions, and retain exclusions from the original denominator even when a factor cancels.

Tier 1 · Easy

2 marks
ORIGINAL

Factorise x29x+20x^2-9x+20 completely.

Tier 2 · Standard

4 marks
ORIGINAL

Use the factor theorem to factorise x34x2+x+6x^3-4x^2+x+6 fully.

Tier 3 · Hard

5 marks
ORIGINAL

Simplify x34x2+x+6x24x+4\dfrac{x^3-4x^2+x+6}{x^2-4x+4} as far as possible, then use algebraic division. State any excluded value.

2.7

Understand and use graphs of functions; sketch curves including polynomials, the modulus of a linear function, y = a/x and y = a/x² with asymptotes; use graph intersections to solve equations; understand proportional relationships.

  • A useful sketch records intercepts, turning points, end behaviour, symmetry and asymptotes rather than relying on a table of isolated points.
  • For y=a/xy=a/x the coordinate axes are asymptotes and the relation is inverse proportionality; y=a/x2y=a/x^2 is inverse-square, symmetric about the yy-axis and has the same sign as aa.
  • An equation f(x)=g(x)f(x)=g(x) is solved graphically at the xx-coordinates where the two graphs intersect; modulus reflects negative outputs above the xx-axis.
  • Do not draw a reciprocal curve crossing an asymptote, and distinguish direct proportionality y=kxy=kx from a relation that merely has positive correlation.

Tier 1 · Easy

2 marks
ORIGINAL

For the curve y=6xy=-\dfrac{6}{x}, state both asymptotes and the two quadrants containing its branches.

Tier 2 · Standard

4 marks
ORIGINAL

Find the intersection values of xx for the graphs y=2x3y=|2x-3| and y=x+4y=x+4.

Tier 3 · Hard

4 marks
ORIGINAL

The curves y=4xy=\dfrac4x and y=x+3y=x+3 meet twice. Determine the exact coordinates of both intersections.

2.8

Understand and use composite functions; inverse functions and their graphs.

  • The composite fg(x)fg(x) means f(g(x))f(g(x)): apply the right-hand function first and keep its whole output inside the outer function.
  • To find an inverse, write y=f(x)y=f(x), rearrange for xx, then exchange the variable labels; a restricted domain may be needed to make ff one-to-one.
  • A function and its inverse have graphs reflected in y=xy=x, with domain and range interchanged.
  • In general fggffg\ne gf; for inverse quadratics, use the stated domain to choose the correct square-root branch.

Tier 1 · Easy

2 marks
ORIGINAL

Let f(x)=3x1f(x)=3x-1 and g(x)=x2+2g(x)=x^2+2. Calculate fg(2)fg(2).

Tier 2 · Standard

4 marks
ORIGINAL

For f(x)=3x+2x1f(x)=\dfrac{3x+2}{x-1}, find f1(x)f^{-1}(x) and state its domain.

Tier 3 · Hard

6 marks
ORIGINAL

The function f(x)=(x3)2+1f(x)=(x-3)^2+1 has domain x3x\geq3. Find f1(x)f^{-1}(x), state its domain, and solve f1(2x+1)=x+3f^{-1}(2x+1)=x+3.

2.9

Understand the effect of simple transformations on the graph of y = f(x), including sketching associated graphs: y = af(x), y = f(x) + a, y = f(x + a), y = f(ax) and combinations of these transformations.

  • Transformations outside ff act on output coordinates: y=af(x)y=af(x) scales vertically by factor a|a| and reflects in the xx-axis when a<0a<0.
  • Transformations inside ff act oppositely on inputs: f(x+a)f(x+a) moves left by aa, while f(ax)f(ax) scales horizontally by factor 1/a1/|a| and may reflect in the yy-axis.
  • For example, a point (u,v)(u,v) on y=f(x)y=f(x) maps to (u/2,v+3)(u/2,v+3) on y=f(2x)+3y=f(2x)+3.
  • Apply combined transformations to coordinates or rewrite the complete formula; describing f(x+a)f(x+a) as a shift right is the standard sign error.

Tier 1 · Easy

2 marks
ORIGINAL

Describe fully the transformation from y=f(x)y=f(x) to y=f(x4)+2y=f(x-4)+2.

Tier 2 · Standard

3 marks
ORIGINAL

The point (6,1)(6,-1) lies on y=f(x)y=f(x). Find the corresponding point on y=2f(3x)+1y=-2f(3x)+1.

Tier 3 · Hard

5 marks
ORIGINAL

Let f(x)=x24x+1f(x)=x^2-4x+1. For g(x)=f(2x+2)+4g(x)=-f(2x+2)+4, find the turning point and both xx-intercepts, then state whether it is a maximum or minimum.

2.10

Decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerators constant or linear).

  • Factor the denominator first and assign one partial-fraction term to each linear factor.
  • A repeated factor needs a term for every power: (xa)2(x-a)^2 requires A/(xa)+B/(xa)2A/(x-a)+B/(x-a)^2.
  • After multiplying through by the common denominator, substitute convenient roots or compare coefficients to determine the constants.
  • If the numerator degree is at least the denominator degree, divide first; omitting a repeated-factor term produces an identity that cannot hold.

Tier 1 · Easy

3 marks
ORIGINAL

Express 7(x+1)(x+2)\dfrac7{(x+1)(x+2)} as partial fractions.

Tier 2 · Standard

4 marks
ORIGINAL

Decompose 5x+1(x1)(x+2)\dfrac{5x+1}{(x-1)(x+2)} into partial fractions.

Tier 3 · Hard

5 marks
ORIGINAL

Express 5x+1(x1)2(x+2)\dfrac{5x+1}{(x-1)^2(x+2)} as three partial-fraction terms.

2.11

Use of functions in modelling, including consideration of limitations and refinements of the models.

  • A function model links clearly defined variables over a stated domain; parameters should have units or contextual meaning where possible.
  • Use the given data to determine parameters, then calculate with the model and interpret the result in context with appropriate rounding.
  • For example, in P(t)=P0rtP(t)=P_0r^t, P0P_0 is the value at t=0t=0 and rr is the multiplier per time interval.
  • A model is an approximation: check whether extrapolation, a fixed growth rate, ignored constraints or measurement variation makes a prediction unreliable, and suggest a refinement tied to that weakness.

Tier 1 · Easy

2 marks
ORIGINAL

A colony is modelled by P(t)=120(1.08)tP(t)=120(1.08)^t, where tt is measured in days. Estimate the number of organisms after 33 days.

Tier 2 · Standard

5 marks
ORIGINAL

A cooling model is T(t)=18+62e0.15tT(t)=18+62e^{-0.15t}, where TT is in degrees Celsius and tt is in minutes. Find the first whole minute for which the model gives T<30T<30, and state one limitation.

Tier 3 · Hard

7 marks
ORIGINAL

The cross-section of a greenhouse roof is modelled by h(x)=kx(30x)h(x)=kx(30-x) for 0x300\leq x\leq30, with distances in metres. The point (5,10)(5,10) lies on the model. Find kk, the maximum modelled height, and the horizontal width over which h(x)12h(x)\geq12. Give one limitation of the model.