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.

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]

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.