Skip to content

Course home

1.2 Algebra and Functions

1.2 Algebra and Functions

EasyMediumHard
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384
Question 318

The vertical displacement hhh, in millimetres, of a high-precision mechanical component is modeled by the function

h(t)=t2(t+a) h(t) = t^2(t + a) h(t)=t2(t+a)

where ttt is the time in seconds and aaa is a positive constant.

a.

Sketch the curve with equation y=t2(t+a)y = t^2(t + a)y=t2(t+a).

[3]
b.

A second model for the displacement, H(t)H(t)H(t), includes a damping offset and is given by

H(t)=t2(t+a)+54 H(t) = t^2(t + a) + 54 H(t)=t2(t+a)+54

(i) Given that t+6t + 6t+6 is a factor of the polynomial H(t)H(t)H(t), use the factor theorem to show that a=4.5a = 4.5a=4.5.

[2]
bii.

State the single transformation which maps the curve with equation y=t2(t+4.5)y = t^2(t + 4.5)y=t2(t+4.5) onto the curve with equation y=t2(t+4.5)+54y = t^2(t + 4.5) + 54y=t2(t+4.5)+54.

[1]
biii.

The expression t2(t+4.5)+54t^2(t + 4.5) + 54t2(t+4.5)+54 can be written in the form (t+6)(t2+bt+c)(t + 6)(t^2 + bt + c)(t+6)(t2+bt+c). Without finding the values of bbb and ccc, use your knowledge of the transformation in part (b)(ii) and the sketch in part (a) to explain why

b2<4c b^2 < 4c b2<4c
[3]
Markscheme

1.2 Algebra and Functions Questions

  1. A Level
  2. /Maths
  3. /1.2 Algebra and Functions

566 exam-style questions on OCR A Level Maths 1.2 Algebra and Functions, covering 1.2.1 Indices, 1.2.2 Surds, 1.2.3 Simultaneous equations, 1.2.4 Quadratic functions, 1.2.5 Completing the square, 1.2.6 Solving quadratic equations, 1.2.7 Inequalities, 1.2.8 Expressing solutions of inequalities, 1.2.9 Representing inequalities graphically, 1.2.10 Manipulating polynomials, 1.2.11 Simplifying rational expressions, 1.2.12 The modulus function (A-level only), 1.2.13 Graphs of functions, 1.2.14 Sketching curves from equations, 1.2.15 Sketching reciprocal curves, 1.2.16 Interpreting solutions graphically, 1.2.17 Intersection points of graphs, 1.2.18 Proportional relationships, 1.2.19 Graph of the modulus of a linear function (A-level only), 1.2.20 Solving modulus equations graphically (A-level only), 1.2.21 Definition of a function (A-level only), 1.2.22 Inverse and composite functions (A-level only), 1.2.23 Simple graph transformations, 1.2.24 Combinations of graph transformations (A-level only), 1.2.25 Partial fractions (A-level only), and 1.2.26 Models in context. Each one has a worked solution and a mark scheme showing where the marks go.

Question bank