Marks available | 3 |

Reference code | 09N.2.sl.TZ0.9 |

## Question

Let \(f(x) = 5\cos \frac{\pi }{4}x\) and \(g(x) = – 0.5{x^2} + 5x – 8\) for \(0 \le x \le 9\) .

On the same diagram, sketch the graphs of *f* and *g* .

Consider the graph of \(f\) . Write down

(i) the *x*-intercept that lies between \(x = 0\) and \(x = 3\) ;

(ii) the period;

(iii) the amplitude.

Consider the graph of *g* . Write down

(i) the two *x*-intercepts;

(ii) the equation of the axis of symmetry.

Let *R* be the region enclosed by the graphs of *f* and *g* . Find the area of *R*.

## Markscheme

** A1A1A1 N3**

**Note**: Award * A1* for

*f*being of sinusoidal shape, with 2 maxima and one minimum,

*for*

**A1***g*being a parabola opening down,

*for*

**A1****two**intersection points in approximately correct position.

**[3 marks] **

(i) \((2{\text{, }}0)\) (accept \(x = 2\) ) **A1 N1**

(ii) \({\text{period}} = 8\) **A2 N2**

(iii) \({\text{amplitude}} = 5\) **A1 N1 **

**[4 marks] **

(i) \((2{\text{, }}0)\) , \((8{\text{, }}0)\) (accept \(x = 2\) , \(x = 8\) ) **A1A1 N1N1**

(ii) \(x = 5\) (must be an equation) **A1 N1**

**[3 marks] **

**METHOD 1**

intersect when \(x = 2\) and \(x = 6.79\) (may be seen as limits of integration) **A1A1**

evidence of approach **(M1)**

e.g. \(\int {g – f} \) , \(\int {f(x){\rm{d}}x – \int {g(x){\rm{d}}x}}\) , \(\int_2^{6.79} {\left( {( – 0.5{x^2} + 5x – 8) – \left( {5\cos \frac{\pi }{4}x} \right)} \right)}\)

\({\text{area}} = 27.6\) **A2 N3**

** METHOD 2**

intersect when \(x = 2\) and \(x = 6.79\) (seen anywhere) **A1A1**

evidence of approach using a sketch of *g* and *f* , or \(g – f\) . **(M1)**

e.g. area = \(A + B – C\) , \(12.7324 + 16.0938 – 1.18129 \ldots \)

\({\text{area}} = 27.6\) **A2 N3**

** [5 marks]**

## Examiners report

Graph sketches were much improved over previous sessions. Most candidates graphed the two functions correctly, but many ignored the domain restrictions.

Many candidates found parts (b) and (c) accessible, although quite a few did not know how to find the period of the cosine function.

Many candidates found parts (b) and (c) accessible, although quite a few did not know how to find the period of the cosine function.

Part (d) proved elusive to many candidates. Some used creative approaches that split the area into parts above and below the *x*-axis; while this leads to a correct result, few were able to achieve it. Many candidates were unable to use their GDCs effectively to find points of intersection and the subsequent area.

Marks available | 3 |

Reference code | 13N.2.sl.TZ0.2 |

## Question

Let \(f(x) = (x – 1)(x – 4)\).

Find the \(x\)-intercepts of the graph of \(f\).

The region enclosed by the graph of \(f\) and the \(x\)-axis is rotated \(360^\circ\) about the \(x\)-axis.

Find the volume of the solid formed.

## Markscheme

valid approach *(M1)*

*eg* \(f(x) = 0\), sketch of parabola showing two \(x\)-intercepts

\(x = 1,{\text{ }}x = 4{\text{ }}\left( {{\text{accept (1, 0), (4, 0)}}} \right)\) *A1A1 N3*

*[3 marks]*

attempt to substitute either limits or the function into formula involving \({f^2}\) *(M1)*

*eg* \(\int_1^4 {{{\left( {f(x)} \right)}^2}{\text{d}}x,{\text{ }}\pi \int {{{\left( {(x – 1)(x – 4)} \right)}^2}} } \)

\({\text{volume}} = 8.1\pi {\text{ (exact), 25.4}}\) *A2 N3*

*[3 marks] *

## Examiners report

Marks available | 3 |

Reference code | 14M.2.sl.TZ2.2 |

## Question

Let \(f(x) = 5 – {x^2}\). Part of the graph of \(f\)is shown in the following diagram.

The graph crosses the \(x\)-axis at the points \(\rm{A}\) and \(\rm{B}\).

Find the \(x\)-coordinate of \({\text{A}}\) and of \({\text{B}}\).

The region enclosed by the graph of \(f\) and the \(x\)-axis is revolved \(360^\circ \) about the \(x\)-axis.

Find the volume of the solid formed.

## Markscheme

recognizing \(f(x) = 0\) *(M1)*

*eg* \(f = 0,{\text{ }}{x^2} = 5\)

\(x = \pm 2.23606\)

\(x = \pm \sqrt 5 {\text{ (exact), }}x = \pm 2.24\) *A1A1 N3*

*[3 marks]*

attempt to substitute either limits or the function into formula

involving \({f^2}\) *(M1)*

*eg* \(\pi \int {{{\left( {5 – {x^2}} \right)}^2}{\text{d}}x,{\text{ }}\pi \int_{ – 2.24}^{2.24} {\left( {{x^4} – 10{x^2} + 25} \right)} ,{\text{ }}2\pi \int_0^{\sqrt 5 } {{f^2}} } \)

\(187.328\)

volume \(= 187\) *A2 N3*

*[3 marks]*