| Date | May 2009 | Marks available | 2 | Reference code | 09M.1.sl.TZ1.15 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 15 | Adapted from | N/A |
Question
The straight line, L, has equation \(2y - 27x - 9 = 0\).
Find the gradient of L.
Sarah wishes to draw the tangent to \(f (x) = x^4\) parallel to L.
Write down \(f ′(x)\).
Find the x coordinate of the point at which the tangent must be drawn.
Write down the value of \(f (x)\) at this point.
Markscheme
y = 13.5x + 4.5 (M1)
Note: Award (M1) for 13.5x seen.
gradient = 13.5 (A1) (C2)
[2 marks]
4x3 (A1) (C1)
[1 mark]
4x3 = 13.5 (M1)
Note: Award (M1) for equating their answers to (a) and (b).
x = 1.5 (A1)(ft)
[2 marks]
\(\frac{{81}}{{16}}\) (5.0625, 5.06) (A1)(ft) (C3)
Note: Award (A1)(ft) for substitution of their (c)(i) into x4 with working seen.
[1 mark]
Examiners report
The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.
Most were successful in this part.
The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.
This part was usually well attempted.
The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.
Only the best candidates succeeded in this part.
The structure of this question was not well understood by the majority; the links between parts not being made. Again, this question was included to discriminate at the grade 6/7 level.
Only the best candidates succeeded in this part.
