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Date May 2018 Marks available 2 Reference code 18M.2.sl.TZ1.4
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 4 Adapted from N/A

Question

Consider the function \(f\left( x \right) = \frac{{48}}{x} + k{x^2} - 58\), where x > 0 and k is a constant.

The graph of the function passes through the point with coordinates (4 , 2).

P is the minimum point of the graph of f (x).

Find the value of k.

[2]
a.

Using your value of k , find f ′(x).

[3]
b.

Use your answer to part (b) to show that the minimum value of f(x) is −22 .

[3]
c.

Write down the two values of x which satisfy f (x) = 0.

[2]
d.

Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.

[4]
e.

Markscheme

\(\frac{{48}}{4} + k \times {4^2} - 58 = 2\)    (M1)
Note: Award (M1) for correct substitution of x = 4 and y = 2 into the function.

k = 3     (A1) (G2)

[2 marks]

a.

\(\frac{{ - 48}}{{{x^2}}} + 6x\)     (A1)(A1)(A1)(ft) (G3)

Note: Award (A1) for −48 , (A1) for x−2, (A1)(ft) for their 6x. Follow through from part (a). Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

b.

\(\frac{{ - 48}}{{{x^2}}} + 6x = 0\)     (M1)

Note: Award (M1) for equating their part (b) to zero.

x = 2     (A1)(ft)

Note: Follow through from part (b). Award (M1)(A1) for \(\frac{{ - 48}}{{{{\left( 2 \right)}^2}}} + 6\left( 2 \right) = 0\) seen.

Award (M0)(A0) for x = 2 seen either from a graphical method or without working.

\(\frac{{48}}{2} + 3 \times {2^2} - 58\,\,\,\left( { =  - 22} \right)\)   (M1)

Note: Award (M1) for substituting their 2 into their function, but only if the final answer is −22. Substitution of the known result invalidates the process; award (M0)(A0)(M0).

−22     (AG)

[3 marks]

c.

0.861  (0.860548…), 3.90  (3.90307…)     (A1)(ft)(A1)(ft) (G2)

Note: Follow through from part (a) but only if the answer is positive. Award at most (A1)(ft)(A0) if answers are given as coordinate pairs or if extra values are seen. The function f (x) only has two x-intercepts within the domain. Do not accept a negative x-intercept.

[2 marks]

d.

(A1)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.

[4 marks]

e.

Examiners report

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a.
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b.
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c.
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d.
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e.

Syllabus sections

Topic 7 - Introduction to differential calculus » 7.5 » Values of x where the gradient of a curve is zero.
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