Further Mathematics · AQA 7367 · Topic D

Further algebra and functions

Roots of polynomials, summing series, Maclaurin series, limits, inequalities, the modulus function, rational functions and conic sections.

  • 16 sub-topics
  • 10 in Year 12, 4 in Year 13
  • 257 practice questions
  • 16 animated lessons

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. D1Roots and coefficients of polynomials up to quarticsY12
  2. D2Polynomials with transformed rootsY12
  3. D3Standard results for sums of integers, squares and cubesY12
  4. D4Method of differences, including with partial fractionsY12
  5. D5Find the Maclaurin series of a function, including the general termY13
  6. D6Standard Maclaurin series and their ranges of validityY12
  7. D7Evaluate limits using Maclaurin series or l'Hôpital's ruleY13
  8. D8Solve cubic and quartic polynomial inequalitiesY12
  9. D9Solve inequalities of the form (ax + b)/(cx + d) < (ex + f)/(gx + h)Y12
  10. D10Modulus functions and related inequalitiesY13
  11. D11Sketch y = 1/f(x) given the graph of y = f(x)Y13
  12. D12Graphs of rational functions (ax + b)/(cx + d)Y12
  13. D13Rational functions with a quadratic numerator or denominatorY13
  14. D14Ranges and stationary points of rational functions without calculusY12
  15. D15Sketch the parabola, ellipse, hyperbola and rectangular hyperbolaY12
  16. D16Transformations of curvesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 257 questions in this topic, free, no account needed.

Further algebra and functions

Easy

Roots of a quadratic · Year 12 · 3 marks

The roots of x2−5x+7=0x^2 - 5x + 7 = 0 are α\alpha and β\beta.
  1. (a)
    Write down α+β\alpha + \beta and αβ\alpha\beta.
    (1)
  2. (b)
    Find α2+β2\alpha^2 + \beta^2.
    (2)
Show worked solutionHide worked solution

Part (a)1 mark

  1. Sum =−ba= -\tfrac{b}{a}, product =ca= \tfrac{c}{a}.α+β\alpha + \beta=5, αβ{}= 5,\ \alpha\beta=7{}= 7

Part (b)2 marks

  1. Squaring the sum gives the squares plus 2αβ2\alpha\beta.α2+β2\alpha^2 + \beta^2=(α+β)2{}= (\alpha + \beta)^2−2αβ{}- 2\alpha\beta
  2. =25−14=11= 25 - 14 = 11

Total3

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
How did it go?

Report a problem with this question

Further algebra and functions

Medium

Roots of a cubic and their squares · Year 12 · 5 marks

The roots of 2x3−3x2+4x−5=02x^3 - 3x^2 + 4x - 5 = 0 are α\alpha, β\beta and γ\gamma.
  1. (a)
    Write down α+β+γ\alpha + \beta + \gamma, αβ\alpha\beta+βγ{}+ \beta\gamma+γα{}+ \gamma\alpha and αβγ\alpha\beta\gamma.
    (2)
  2. (b)
    Find α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 and explain what your answer tells you about the roots.
    (3)
Show worked solutionHide worked solution

Part (a)2 marks

  1. For ax3+bx2+cx+dax^3 + bx^2 + cx + d: ∑α=−ba\sum\alpha = -\tfrac{b}{a}, ∑αβ=ca\sum\alpha\beta = \tfrac{c}{a}, αβγ=−da\alpha\beta\gamma = -\tfrac{d}{a}.∑α\sum\alpha=32,{}= \tfrac{3}{2},∑αβ\quad \sum\alpha\beta=2,{}= 2,αβγ\quad \alpha\beta\gamma=52{}= \tfrac{5}{2}

Part (b)3 marks

  1. Squaring the sum of the roots gives the squares plus twice every pair.α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2=(∑α)2{}= (\sum\alpha)^2−2∑αβ{}- 2\sum\alpha\beta
  2. Substitute the values from part (a).=94−4=−74= \tfrac{9}{4} - 4 = -\tfrac{7}{4}
  3. Two of the roots must be a complex conjugate pair.The squares of real numbers cannot add to a negative, so not all the roots are real.

Total5

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
How did it go?

Report a problem with this question

Further algebra and functions

Hard

Roots of a quartic · Year 12 · 6 marks

The roots of x4+2x3−x+3=0x^4 + 2x^3 - x + 3 = 0 are α\alpha, β\beta, γ\gamma and δ\delta.
  1. (a)
    Find 1α\dfrac{1}{\alpha}+1β{}+ \dfrac{1}{\beta}+1γ{}+ \dfrac{1}{\gamma}+1δ{}+ \dfrac{1}{\delta}.
    (3)
  2. (b)
    Find α2\alpha^2+β2{}+ \beta^2+γ2{}+ \gamma^2+δ2{}+ \delta^2.
    (3)
Show worked solutionHide worked solution

Part (a)3 marks

  1. For ax4+bx3+cx2+dx+eax^4 + bx^3 + cx^2 + dx + e: ∑αβγ\sum\alpha\beta\gamma=−da{}= -\tfrac{d}{a} and αβγδ\alpha\beta\gamma\delta=ea{}= \tfrac{e}{a}.∑αβγ\sum\alpha\beta\gamma=1,{}= 1,αβγδ\quad \alpha\beta\gamma\delta=3{}= 3
  2. Put the four fractions over the common denominator αβγδ\alpha\beta\gamma\delta.∑1α\sum\dfrac{1}{\alpha}=∑αβγαβγδ{}= \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}
  3. Substitute the values.=13= \tfrac{1}{3}

Part (b)3 marks

  1. ∑α=−ba\sum\alpha = -\tfrac{b}{a} and ∑αβ=ca\sum\alpha\beta = \tfrac{c}{a}; there is no x2x^2 term, so c=0c = 0.∑α\sum\alpha=−2,{}= -2,∑αβ\quad \sum\alpha\beta=0{}= 0
  2. The same identity as for a cubic: it works for any number of roots.∑α2\sum\alpha^2=(∑α)2{}= (\sum\alpha)^2−2∑αβ{}- 2\sum\alpha\beta
  3. Substitute the values.=4−0=4= 4 - 0 = 4

Total6

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
How did it go?

Report a problem with this question