Date | November 2010 | Marks available | 4 | Reference code | 10N.2.sl.TZ0.5 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Sketch | Question number | 5 | Adapted from | N/A |
Question
Consider the function f (x) = x3 – 3x– 24x + 30.
Write down f (0).
Find \(f'(x)\).
Find the gradient of the graph of f (x) at the point where x = 1.
(i) Use f '(x) to find the x-coordinate of M and of N.
(ii) Hence or otherwise write down the coordinates of M and of N.
Sketch the graph of f (x) for \( - 5 \leqslant x \leqslant 7\) and \( - 60 \leqslant y \leqslant 60\). Mark clearly M and N on your graph.
Lines L1 and L2 are parallel, and they are tangents to the graph of f (x) at points A and B respectively. L1 has equation y = 21x + 111.
(i) Find the x-coordinate of A and of B.
(ii) Find the y-coordinate of B.
Markscheme
30 (A1)
[1 mark]
f '(x) = 3x2 – 6x – 24 (A1)(A1)(A1)
Note: Award (A1) for each term. Award at most (A1)(A1) if extra terms present.
[3 marks]
f '(1) = –27 (M1)(A1)(ft)(G2)
Note: Award (M1) for substituting x = 1 into their derivative.
[2 marks]
(i) f '(x) = 0
3x2 – 6x – 24 = 0 (M1)
x = 4; x = –2 (A1)(ft)(A1)(ft)
Notes: Award (M1) for either f '(x) = 0 or 3x2 – 6x – 24 = 0 seen. Follow through from their derivative. Do not award the two answer marks if derivative not used.
(ii) M(–2, 58) accept x = –2, y = 58 (A1)(ft)
N(4, – 50) accept x = 4, y = –50 (A1)(ft)
Note: Follow through from their answer to part (d) (i).
[5 marks]
(A1) for window
(A1) for a smooth curve with the correct shape
(A1) for axes intercepts in approximately the correct positions
(A1) for M and N marked on diagram and in approximately
correct position (A4)
Note: If window is not indicated award at most (A0)(A1)(A0)(A1)(ft).
[4 marks]
(i) 3x2 – 6x – 24 = 21 (M1)
3x2 – 6x – 45 = 0 (M1)
x = 5; x = –3 (A1)(ft)(A1)(ft)(G3)
Note: Follow through from their derivative.
OR
Award (A1) for L1 drawn tangent to the graph of f on their sketch in approximately the correct position (x = –3), (A1) for a second tangent parallel to their L1, (A1) for x = –3, (A1) for x = 5 . (A1)(ft)(A1)(ft)(A1)(A1)
Note: If only x = –3 is shown without working award (G2). If both answers are shown irrespective of workingaward (G3).
(ii) f (5) = –40 (M1)(A1)(ft)(G2)
Notes: Award (M1) for attempting to find the image of their x = 5. Award (A1) only for (5, –40). Follow through from their x-coordinate of B only if it has been clearly identified in (f) (i).
[6 marks]
Examiners report
The value of f (0) and the derivative function, f '(x) were well done in parts (a) and (b). In part (c) many candidates found f (1) instead of f '(1) . In part (d) many students did not use their f (x) to find the x-coordinates of M and N and instead used their GDC. The sketch was generally well done although some students forgot to label M and N or did not use the specified window. The last part of the question was a clear discriminator. Examiners were pleased to see how this challenging question was solved using alternative methods.
The value of f (0) and the derivative function, f '(x) were well done in parts (a) and (b). In part (c) many candidates found f (1) instead of f '(1) . In part (d) many students did not use their f (x) to find the x-coordinates of M and N and instead used their GDC. The sketch was generally well done although some students forgot to label M and N or did not use the specified window. The last part of the question was a clear discriminator. Examiners were pleased to see how this challenging question was solved using alternative methods.
The value of f (0) and the derivative function, f '(x) were well done in parts (a) and (b). In part (c) many candidates found f (1) instead of f '(1) . In part (d) many students did not use their f (x) to find the x-coordinates of M and N and instead used their GDC. The sketch was generally well done although some students forgot to label M and N or did not use the specified window. The last part of the question was a clear discriminator. Examiners were pleased to see how this challenging question was solved using alternative methods.
The value of f (0) and the derivative function, f '(x) were well done in parts (a) and (b). In part (c) many candidates found f (1) instead of f '(1) . In part (d) many students did not use their f (x) to find the x-coordinates of M and N and instead used their GDC. The sketch was generally well done although some students forgot to label M and N or did not use the specified window. The last part of the question was a clear discriminator. Examiners were pleased to see how this challenging question was solved using alternative methods.
The value of f (0) and the derivative function, f '(x) were well done in parts (a) and (b). In part (c) many candidates found f (1) instead of f '(1) . In part (d) many students did not use their f (x) to find the x-coordinates of M and N and instead used their GDC. The sketch was generally well done although some students forgot to label M and N or did not use the specified window. The last part of the question was a clear discriminator. Examiners were pleased to see how this challenging question was solved using alternative methods.
The value of f (0) and the derivative function, f '(x) were well done in parts (a) and (b). In part (c) many candidates found f (1) instead of f '(1) . In part (d) many students did not use their f (x) to find the x-coordinates of M and N and instead used their GDC. The sketch was generally well done although some students forgot to label M and N or did not use the specified window. The last part of the question was a clear discriminator. Examiners were pleased to see how this challenging question was solved using alternative methods.