| Date | May 2009 | Marks available | 3 | Reference code | 09M.2.sl.TZ1.5 |
| Level | SL only | Paper | 2 | Time zone | TZ1 |
| Command term | Differentiate | Question number | 5 | Adapted from | N/A |
Question
Consider the function \(f(x) = 3x + \frac{{12}}{{{x^2}}},{\text{ }}x \ne 0\).
Differentiate \(f (x)\) with respect to \(x\).
Calculate \(f ′(x)\) when \(x = 1\).
Use your answer to part (b) to decide whether the function, \(f\) , is increasing or decreasing at \(x = 1\). Justify your answer.
Solve the equation \(f ′(x) = 0\).
The graph of f has a local minimum at point P. Let T be the tangent to the graph of f at P.
Write down the coordinates of P.
The graph of f has a local minimum at point P. Let T be the tangent to the graph of f at P.
Write down the gradient of T.
The graph of f has a local minimum at point P. Let T be the tangent to the graph of f at P.
Write down the equation of T.
Sketch the graph of the function f, for −3 ≤ x ≤ 6 and −7 ≤ y ≤ 15. Indicate clearly the point P and any intercepts of the curve with the axes.
On your graph draw and label the tangent T.
T intersects the graph of f at a second point. Write down the x-coordinate of this point of intersection.
Markscheme
\(f' (x) = 3 - \frac{24}{x^3}\) (A1)(A1)(A1)
Note: Award (A1) for 3, (A1) for –24, (A1) for x3 (or x−3). If extra terms present award at most (A1)(A1)(A0).
[3 marks]
\(f '(1) = -21\) (M1)(A1)(ft)(G2)
Note: (ft) from their derivative only if working seen.
[2 marks]
Derivative (gradient, slope) is negative. Decreasing. (R1)(A1)(ft)
Note: Do not award (R0)(A1).
[2 marks]
\(3 - \frac{{24}}{{{x^3}}} = 0\) (M1)
\(x^3 = 8\) (A1)
\(x = 2\) (A1)(ft)(G2)
[3 marks]
(2, 9) (Accept x = 2, y = 9) (A1)(A1)(G2)
Notes: (ft) from their answer in (d).
Award (A1)(A0) if brackets not included and not previously penalized.
[2 marks]
0 (A1)
[1 mark]
y = 9 (A1)(A1)(ft)(G2)
Notes: Award (A1) for y = constant, (A1) for 9.
Award (A1)(ft) for their value of y in (e)(i).
[2 marks]
(A4)
Notes: Award (A1) for labels and some indication of scale in the stated window.
Award (A1) for correct general shape (curve must be smooth and must not cross the y-axis).
Award (A1) for x-intercept seen in roughly the correct position.
Award (A1) for minimum (P).
[4 marks]
Tangent drawn at P (line must be a tangent and horizontal). (A1)
Tangent labeled T. (A1)
Note: (ft) from their tangent equation only if tangent is drawn and answer is consistent with graph.
[2 marks]
x = −1 (G1)(ft)
[1 mark]
Examiners report
Many students did not know the term “differentiate” and did not answer part (a).
However, the derivative was seen in (b) when finding the gradient at x = 1. The negative index of the formula did cause problems for many when finding the derivative. The meaning of the derivative was not clear for a number of students.
Part (d) was handled well by some but many substituted x = 0 into f '(x).
It was clear that most candidates neither knew that the tangent at a minimum is horizontal nor that its gradient is zero.
It was clear that most candidates neither knew that the tangent at a minimum is horizontal nor that its gradient is zero.
It was clear that most candidates neither knew that the tangent at a minimum is horizontal nor that its gradient is zero.
There were good answers to the sketch though setting out axes and a scale seemed not to have had enough practise.
Those who were able to sketch the function were often able to correctly place and label the tangent and also to find the second intersection point with the graph of the function.
Those who were able to sketch the function were often able to correctly place and label the tangent and also to find the second intersection point with the graph of the function.
