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Date May 2017 Marks available 4 Reference code 17M.2.sl.TZ2.8
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 8 Adapted from N/A

Question

Let \(f(x) =  - 0.5{x^4} + 3{x^2} + 2x\). The following diagram shows part of the graph of \(f\).

M17/5/MATME/SP2/ENG/TZ2/08

 

There are \(x\)-intercepts at \(x = 0\) and at \(x = p\). There is a maximum at A where \(x = a\), and a point of inflexion at B where \(x = b\).

Find the value of \(p\).

[2]
a.

Write down the coordinates of A.

[2]
b.i.

Write down the rate of change of \(f\) at A.

[1]
b.ii.

Find the coordinates of B.

[4]
c.i.

Find the the rate of change of \(f\) at B.

[3]
c.ii.

Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis, the line \(x = b\) and the line \(x = a\). The region \(R\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

[3]
d.

Markscheme

evidence of valid approach     (M1)

eg\(\,\,\,\,\,\)\(f(x) = 0,{\text{ }}y = 0\)

2.73205

\(p = 2.73\)     A1     N2

[2 marks]

a.

1.87938, 8.11721

\((1.88,{\text{ }}8.12)\)     A2     N2

[2 marks]

b.i.

rate of change is 0 (do not accept decimals)     A1     N1

[1 marks]

b.ii.

METHOD 1 (using GDC)

valid approach     M1

eg\(\,\,\,\,\,\)\(f’’ = 0\), max/min on \(f’,{\text{ }}x =  - 1\)

sketch of either \(f’\) or \(f’’\), with max/min or root (respectively)     (A1)

\(x = 1\)     A1     N1

Substituting their \(x\) value into \(f\)     (M1)

eg\(\,\,\,\,\,\)\(f(1)\)

\(y = 4.5\)     A1     N1

METHOD 2 (analytical)

\(f’’ =  - 6{x^2} + 6\)     A1

setting \(f’’ = 0\)     (M1)

\(x = 1\)     A1     N1

substituting their \(x\) value into \(f\)     (M1)

eg\(\,\,\,\,\,\)\(f(1)\)

\(y = 4.5\)     A1     N1

[4 marks]

c.i.

recognizing rate of change is \(f’\)     (M1)

eg\(\,\,\,\,\,\)\(y’,{\text{ }}f’(1)\)

rate of change is 6     A1     N2

[3 marks]

c.ii.

attempt to substitute either limits or the function into formula     (M1)

involving \({f^2}\) (accept absence of \(\pi \) and/or \({\text{d}}x\))

eg\(\,\,\,\,\,\)\(\pi \int {{{( - 0.5{x^4} + 3{x^2} + 2x)}^2}{\text{d}}x,{\text{ }}\int_1^{1.88} {{f^2}} } \)

128.890

\({\text{volume}} = 129\)     A2     N3

[3 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.

Syllabus sections

Topic 6 - Calculus » 6.3 » Local maximum and minimum points.
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