| Date | May 2009 | Marks available | 2 | Reference code | 09M.2.sl.TZ2.5 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
A function is defined by \(f(x) = \frac{5}{{{x^2}}} + 3x + c,{\text{ }}x \ne 0,{\text{ }}c \in \mathbb{Z}\).
Write down an expression for \(f ′(x)\).
Consider the graph of f. The graph of f passes through the point P(1, 4).
Find the value of c.
There is a local minimum at the point Q.
Find the coordinates of Q.
There is a local minimum at the point Q.
Find the set of values of x for which the function is decreasing.
Let T be the tangent to the graph of f at P.
Show that the gradient of T is –7.
Let T be the tangent to the graph of f at P.
Find the equation of T.
T intersects the graph again at R. Use your graphic display calculator to find the coordinates of R.
Markscheme
\(f'(x) = \frac{{ - 10}}{{{x^3}}} + 3\) (A1)(A1)(A1)(A1)
Note: Award (A1) for −10, (A1) for x3 (or x−3), (A1) for 3, (A1) for no other constant term.
[4 marks]
4 = 5 + 3 + c (M1)
Note: Award (M1) for substitution in f (x).
c = −4 (A1)(G2)
[2 marks]
\(f '(x) = 0\) (M1)
\(0 = \frac{{ - 10}}{{{x^3}}} + 3\) (A1)(ft)
(1.49, 2.72) (accept x = 1.49 y = 2.72) (A1)(ft)(A1)(ft)(G3)
Notes: If answer is given as (1.5, 2.7) award (A0)(AP)(A1).
Award at most (M1)(A1)(A1)(A0) if parentheses not included. (ft) from their (a).
If no working shown award (G2)(G0) if parentheses are not included.
OR
Award (M2) for sketch, (A1)(ft)(A1)(ft) for correct coordinates. (ft) from their (b). (M2)(A1)(ft)(A1)(ft)
Note: Award at most (M2)(A1)(ft)(A0) if parentheses not included.
[4 marks]
0 < x < 1.49 OR 0 < x ≤ 1.49 (A1)(A1)(ft)(A1)
Notes: Award (A1) for 0, (A1)(ft) for 1.49 and (A1) for correct inequality signs.
(ft) from their x value in (c) (i).
[3 marks]
For P(1, 4) \(f '(1) = - 10 + 3\) (M1)(A1)
\(= -7\) (AG)
Note: Award (M1) for substituting \(x = 1\) into their \(f '(x)\). (A1) for \(-10 + 3\).
\(-7\) must be seen for (A1) to be awarded.
[2 marks]
\(4 = -7 \times 1 + c\) \(11 = c\) (A1)
\(y = -7 x + 11\) (A1)
[2 marks]
Point of intersection is R(−0.5, 14.5) (A1)(ft)(A1)(ft)(G2)(ft)
Notes: Award (A1) for the x coordinate, (A1) for the y coordinate.
Allow (ft) from candidate’s (d)(ii) equation and their (b) even with no working seen.
Award (A1)(ft)(A0) if brackets not included and not previously penalised.
[2 marks]
Examiners report
A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
Most candidates could score 4 marks.
A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
A good number of candidates correctly substituted into the original function.
A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
Very few managed to answer this part algebraically. Those candidates who were aware that they could read the values from their GDC gained some easy marks.
A large number of candidates were not able to answer this question well because of the firstterm of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
Proved to be difficult for many candidates as increasing/decreasing intervals were not well understood by many.
A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
As this part was linked to part (a), candidates who could not find the correct derivative could not show that the gradient of T was −7. Many candidates did not realize that they had to substitute into the first derivative. For those who did, finding the equation of T was a simple task.
A large number of candidates were not able to answer this question well because of the firstterm of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
As this part was linked to part (a), candidates who could not find the correct derivative could not show that the gradient of T was −7. Many candidates did not realize that they had to substitute into the first derivative. For those who did, finding the equation of T was a simple task.
A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.
For those who had part (d) (ii) correct, this allowed them to score easy marks. For the others, it proved a difficult task because their equation in (d) would not be a tangent.
