Pure: Algebra and Functions Depth
In A-level Pure Mathematics, algebra and functions at depth covers partial fractions (distinct, repeated and irreducible-quadratic denominators, including cases needing algebraic division first), harder domain, range, composite and inverse function problems with restricted domains, modulus equations and inequalities, and sequences of graph transformations applied together. The marks reward checking the numerator's degree against the denominator's before choosing a partial-fraction form, finding a function's range correctly before treating it as another function's domain, and verifying every candidate solution of a modulus equation or inequality in the original statement rather than the squared or case-split version.
Before you start
Make sure you're comfortable with these topics first:
Method
- For an algebraic fraction to decompose into partial fractions, first compare the degree of the numerator with the degree of the denominator: if the numerator's degree is greater than or equal to the denominator's, divide first (by algebraic long division or by inspection) to leave a polynomial plus a proper fraction.
- Factorise the denominator fully, then match the partial-fraction form to it: a distinct linear factor (ax+b) gets a single constant-over-it term; a repeated linear factor (ax+b)^2 gets two terms, one over (ax+b) and one over (ax+b)^2; an irreducible quadratic factor gets a linear numerator over it.
- Find the unknown constants either by substituting values of x that make one bracket zero (fastest for distinct linear factors) or by comparing coefficients of each power of x (needed for repeated or quadratic factors, or as a check).
- For composite and inverse functions with restricted domains, work out the range of the inner function first (for a composite, to check it lies in the outer function's domain) or the range of the original function (for an inverse, since that becomes the inverse's domain).
- For modulus equations and inequalities, identify the critical values where each modulus expression is zero, reason about each region separately (or square both sides if every term is a modulus or is known to be non-negative), and verify every candidate solution in the ORIGINAL statement, since squaring or splitting into cases can introduce extra roots that do not actually satisfy it.
- For a sequence of graph transformations, apply them to the equation one at a time in the order given; a horizontal stretch or translation is applied to x itself (so it can look 'reversed'), while a vertical one is applied to the whole function value.
- Where a question asks you to show that two expressions are identical, manipulate one side only where possible, or compare coefficients of every power of x rather than checking only one or two numerical instances.
Worked example
Express (5x^2 - 2x - 1)/((x+1)(x-1)^2) in partial fractions.
- Since the denominator has a distinct linear factor and a repeated linear factor, write (5x^2 - 2x - 1)/((x+1)(x-1)^2) = A/(x+1) + B/(x-1) + C/(x-1)^2.
- Multiply both sides by (x+1)(x-1)^2 to clear denominators: 5x^2 - 2x - 1 = A(x-1)^2 + B(x+1)(x-1) + C(x+1).
- Substitute x = 1 (this makes the A and B terms vanish): 5 - 2 - 1 = 2 = C(1+1), so C = 1.
- Substitute x = -1 (this makes the B and C terms vanish): 5 + 2 - 1 = 6 = A(-1-1)^2 = 4A, so A = 3/2.
- Compare the coefficients of x^2 on both sides (the left-hand side has 5): A + B = 5, so B = 5 - 3/2 = 7/2.
- Final answer: (5x^2 - 2x - 1)/((x+1)(x-1)^2) = 3/(2(x+1)) + 7/(2(x-1)) + 1/(x-1)^2.
Practice questions
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Q1State the general partial-fraction form used when a denominator contains a repeated linear factor (px+q)^2, and explain why a single constant over (px+q)^2 alone is not general enough.Show answer
Answer: A/(px+q) + B/(px+q)^2. Combining these two terms over a common denominator gives (A(px+q)+B)/(px+q)^2, a general linear numerator; a single term B/(px+q)^2 on its own can only ever produce a constant numerator, which is not general enough to represent every proper fraction with this denominator.
Q2Express (3x^2 + 2x - 5)/(x^2 - 4) in the form A + B/(x-2) + C/(x+2), stating the values of A, B and C.Show answer
Answer: A = 3, B = 11/4, C = -3/4. (Divide first since the numerator and denominator both have degree 2: 3x^2+2x-5 = 3(x^2-4) + 2x+7, then decompose (2x+7)/((x-2)(x+2)).)
Q3The function f is defined by f(x) = x^2 + 6x + 5 for x >= -3, x is real. Find f^-1(x), stating its domain.Show answer
Answer: f^-1(x) = sqrt(x+4) - 3, domain x >= -4. (Complete the square: f(x) = (x+3)^2 - 4, which has range [-4, infinity) for x >= -3; this range becomes the domain of f^-1.)
Q4Solve the inequality |2x - 3| < |x + 4|.Show answer
Answer: -1/3 < x < 7. (Square both sides: (2x-3)^2 < (x+4)^2, which simplifies to 3x^2 - 20x - 7 < 0, i.e. (3x+1)(x-7) < 0.)
Q5Functions f and g are defined by f(x) = sqrt(x-2) for x >= 2, and g(x) = x^2 + 1 for all real x. Find the set of values of x for which the composite function fg(x) is defined, and find fg(3).Show answer
Answer: fg(x) is defined for x <= -1 or x >= 1 (since g(x) must be at least 2, i.e. x^2 + 1 >= 2). fg(3) = sqrt(9-1) = 2sqrt(2).
Q6Explain why, when solving an equation of the form |ax+b| = cx+d by squaring both sides, it is essential to check each solution in the original equation.Show answer
Answer: Squaring both sides gives (ax+b)^2 = (cx+d)^2, which is also satisfied whenever ax+b = -(cx+d), introducing the solution of the 'negative case' as well as the true one. Since |ax+b| is never negative, a solution is only valid if cx+d is also non-negative there; any root making cx+d negative solves the squared equation but must be rejected.
Q7The graph of y = f(x) has a single minimum point at (2, -5). State the coordinates of the minimum point on the graph of y = 3f(x+1) - 4.Show answer
Answer: (1, -19). (f(x+1) shifts the minimum to (1,-5); 3f(x+1) stretches the y-value to (1,-15); subtracting 4 gives (1,-19).)
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
f(x) = (7x-3)/((2x-1)(x+2)), x is real, x not equal to 1/2 or -2. (a) Express f(x) in partial fractions. (4) (b) Hence write down the equations of the three asymptotes of the curve y = f(x). (2)
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The function f is defined by f(x) = (3x+1)/(x-3), x is real, x not equal to 3. (a) Find f^-1(x), stating its domain. (4) (b) Show that ff(x) = x for all x in the domain of f, and state what this shows about the relationship between f and f^-1. (4)
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Solve the inequality (2x-1)/(x+3) >= 1, x not equal to -3, giving your answer in set notation. (5)
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Free printable worksheet
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