Date | May 2016 | Marks available | 8 | Reference code | 16M.1.hl.TZ0.5 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
Consider the curve C given by \(y = {x^3}\).
The tangent at a point P on \(C\) meets the curve again at Q. The tangent at Q meets the curve again at R. Denote the \(x\)-coordinates of \({\text{P, Q}}\) and R, by \({x_1},{\text{ }}{x_2}\) and \({x_3}\) respectively where \({x_1} \ne 0\). Show that, \({x_1},{\text{ }}{x_2},{\text{ }}{x_3}\) form the first three elements of a divergent geometric sequence.
Markscheme
attempt to find the equation of the tangent at P M1
\(y - x_1^3 = 3x_1^2(x - {x_1})\) A1
the tangent meets \(C\) when
\({x^3} - x_1^3 = 3x_1^2(x - {x_1})\) M1
attempt to solve the cubic M1
the \(x\)-coordinate of Q satisfies
\({x^2} + x{x_1} - 2x_1^2 = 0\) A1
hence \({x_2} = - 2{x_1}\) A1
hence \({x_3} = 4{x_1}\) A1
hence \({x_1},{\text{ }}{x_2},{\text{ }}{x_3}\) form the first three terms of a geometric sequence with common ratio \( - 2\) so the sequence is divergent R1AG
Note: Final R1 is not dependent on final 3 A1s providing they form a geometric sequence.
Total [8 marks]
Examiners report
This question caused a problem for many candidates and only a small number of fully correct answers were seen. Most candidates were able to find a generalised equation of a tangent, but were then unable to see what could be replaced in order to find a quadratic equation that could be solved.