Relationships and Calculus Assessment Standard 1



|Questions |Assessment Standard |Description |

|1,3 |RC1.1 |Factorise and solve a cubic a≠1 |

|2 |RC1.1 |Given nature use discriminant to find coefficient |

|4,6 |EF1.2 |Trig identities |

|5 |EF1.2 |Application of addition and double angle formulae |

|7,8 |RC1.2 |Solve equations in degrees or radians using trig formulae/identities |

|9,10a |A1.2 |Determine and use the equation of a circle |

|10b |A1.2 |Use properties of tangency |

|11 |RC1.4 |Integrate algebraic functions expressed in powers of x |

|12 |A1.4 |Find the area between curve and x axis |

|13 |A1.4 |Find the area between line and curve or two curves |

Relationships and Calculus Assessment Standard 1.1

1 This rectangular box shown in the diagram has

a volume of 12 m3. Its dimensions are in metres.

(a) Show that x3 + 3x2 – 4x – 12 = 0.

(b) Hence determine algebraically the dimensions

of the box.

2 The graph of the function f(x) = px2 – x + 3 does not touch

or cross the x-axis. What are the possible values for p?

3 Solve the cubic equation f(x) = 0 given:

• f(x) has a turning point at (3,0)

• (x + 1) is a factor of f(x)

Expressions and Functions Assessment Standard 1.2

4 Show that (3 – 2sinx)(3 + 2sinx) = 5 + 4cos2x

5 If sinx = ¾ then

(a) Find the exact values of cos x

(b) Show that the exact value of sin 2x is [pic]

6 (a) Show that [pic] (based on a suitable domain) (2)

(b) Hence state the minimum value of [pic][pic]

justifying your answer

#2.2

Relationships and Calculus Assessment Standard 1.2

7 Solve [pic] for [pic]

8 Solve [pic]

Applications Assessment Standard 1.2

9 A circle has radius 7 units and centre (5,-2).

Write down the equation of the circle.

10 a) A circle has equation x2 + y2 + 4x - 10y + 20 = 0

Write down the coordinates of its centre (2)

and the length of its radius.

b) Find the equation of the tangent to this circle at (1,5) (2)

Relationships and Calculus Assessment Standard 1.4

11 Find [pic] [pic]

Applications Assessment Standard 1.4

12 The curve with equation

Y = x3 – 7x2 + 7x + 15 is shown in the diagram

with Q at (3,0).

Calculate the shaded area

13 The line with equation y = 5 - x and the curve

with equation y = 5 – x2 meet at the points where [pic] and [pic].

Calculate the area enclosed by them.

|1(a) |•1 Sets up equation |•1 x(x – 1)(x + 4) = 12 |

| |•2 Multiplies out and = zero |•2 [pic] |

|(b) |#2.1 Uses correct strategy |#2.1 Evidence of complete factorisation, ie 3 linear brackets |

| |•3 Strategy to start solving, eg direct |•3 f (a) = b or synthetic division |

| |substitution or synthetic division | |

| | |[pic] |

| | | |

| | |To be awarded •3 a must be a factor of 12. |

| |•4 Obtains correct linear factor (or root) |•4 |

| | |[pic] |

| | |Remainder = 0, (x – 2) is a factor or x = 2 is a root |

| |•5 Factorises |•5 [pic] |

| |•6 Solves |•6 x = 2, x = -2, and x = -3 |

| |•7 Shows clearly only one solution |•7 Shows clearly x = -2, and x = -3 have been disregarded |

| |#2.2 Communicates clearly and relates to the |#2.2 shows that x = 2 is the only solution, therefore, the box |

| |context |measures |

| | |2 metres by 1 metre by 6 metres |

|Notes:•4 is only available for obtaining (x – 2) as a factor or x = 2 as a root. |

|#2.1 for evidence of an attempt at solving [pic]can only be given if •4 has been awarded. |

|2 |#2.1 Knows to use the discriminant and |#2.1 b2 - 4ac < 0 |

| |substitutes correctly |(-1)2- 4 x p x 3 < 0 |

| |•1 completes substitution process and starts to|•1 1 – 12p < 0 |

| |solve the inequality | |

| |•2 solves the inequality |•2 1/12 < 0 |

|3 |#2.2 Communicates solution from the context |#2.2 evidence of correctly communicating the context by providing the |

| | |correct solution |

| | |x = 3 and x = -1 |

|4 |•1 expands LHS |•1 9 -4sin2x |

| |•2 substitutes for sin2x |•2 9 - 4(1-cos2x) |

| |•3 simplifies and completes |•3 9 – 4 + 4cos2x = 5 + 4 cos2x = RHS |

| |#2.1 evidence of selecting an appropriate trig |#2.1 substitute for “cosine squared” |

| |identity | |

| |Alternatively use RHS: |•1 5 + 4( 1 – sin2x) |

| |•1 substitutes for cos2x |•2 9 – 4sin2x |

| |•2 simplifies |•3 (3 - 2sinx)(3 + 2sinx) =LHS |

| |•3 factorises and completes | |

|NB: Do not award •2 if candidate omits brackets, ie. 9 – 4 - 4 cos2x |

|5a |#2.1 Correct strategy |#2.1 Attempts to find the adjacent and then use appropriate formula |

| |•1 Finds unknown side |•1 √5 |

|5b |•1 Starts to find [pic] |•1 [pic] and substitutes for [pic] or [pic] |

| |•2 Finds [pic] |•2 2 x 3/4 x √5/4 = 6√5/16 =3√5/8 |

|Notes: |

|Do not award •2 without adequate working |

|Award 2 points of process to candidates who find the adjacent and subsequently evaluate [pic]correctly. |

|#2.1 can be awarded where the candidate finds the adjacent and attempts to use the formula for[pic]. |

|6(a) |Proof |

| |•1 Strategy |•1 |eg take LHS and substitutes cosx/sinx for 1/[pic] |

| |•2 Simplify |•2 |[pic] |

|6(b) |–3, with valid explanation. |

| |#2.2 Communicates |#2.2 |[pic] |

| | | |and the minimum value of [pic] is –1 |

| | | |[pic] minimum value of |

| | | |[pic] |

| | | | |

| | | | |

| | | | |

|7 |31·8º and 128·2º | |

| |#2.1 Express in standard form [pic] |#2.1 |[pic] |

| |•1 Solve equation for one value of [pic] |•1 |41·8º |

| |•2 Process second solution |•2 |138·2º |

| |•3 Process solutions for x |•3 |31·8º and 128·2º |

|Note: Accept final solutions in radians as long as the 10( has also been converted. |

|8 |[pic] | |

| |•1 Substitutes for cos2x |•1 |[pic] |

| |•2 Simplifies and equates to 0 |•2 |[pic] |

| |•3 Factorises |•3 |[pic] |

| |•4 Solves for [pic] |•4 |[pic] |

| |•5 Solves for [pic] |•5 |[pic] |

|Notes: |

|Candidates must clearly show that they are disregarding [pic]. |

|Do not penalise solutions outwith [pic]. |

|9 |( x – 5 )2 + ( y + 2 )2 = 49 |

| |•1 Interpret centre |•1 |( x – (5) )2 + ( y – (- 2 )2 |

| |•2 Interpret radius and complete equation |•2 |= 49 |

|Note: In •2 is not awarded for 72, this must be simplified to 49 |

|10a) |Centre (-2,5) radius 3 units |

| |•1 State centre of circle |•1 (-2,5) |

| |•2 Know how to and find radius of circle |•2 r = 3 |

|10b) |X = 1 |

| |•1 Find gradient of radius |•1 |m = 0 |

| |•2 knows tangent is vertical line and |•2 |X = 1 |

| |states equation | | |

|11 |•1 Starts to express in integrable form |•1 [pic] or [pic] |

| |•2 Completes |•2 [pic] |

| |•3 Integrates one term correctly |•3 [pic] |

| |•4 Integrates other term correctly |•4 [pic] |

| |•5 constant of integration |•5 … + c ** |

|Note: •4 is only available as a result of integrating a negative power. |

| |

| |

| |

| |

|12 |•1 know to integrate with correct limits |•1 3 ∫0 |

| |•2 integrates |•2 ¼ x4 -7/3 x3 + 7/2 x2 + 15x |

| |•3 substitutes in the limits |•3 ¼ x34 -7/3 x33 + 7/2 x32 + 15x3 -(0) |

| |•4 process limits |•4 33 ¾ |

|Note: •3 and 4 are only available to those who integrate |

|13 |1/6 units2 |

| |•1 Know to integrate and state limits |•1 |[pic] |

| |•2 Use ‘upper – lower’ |•2 |[pic] stated or implied by •3 |

| |•3 Interpret upper – lower |•3 |x – x2 |

| |•4Integrate |•4 |½ x2 -1/3 x3 |

| |•5 Substitute limits |•5 |½ 12 -1/3 13 – (0) |

| |•6 Process limits |•6 |1/6 or equivalent |

|Notes: Do not penalise candidates who work with 5 – x2 – 5 + x throughout in an unsimplified form. |

|Candidates who use 5 - x2 - 5 – x leading to - x2 - x lose •3 but all other marks are still available. |

|Differentiation loses •4, •5 and •6. |

|•2 is lost for subtracting the wrong way round and subsequently •6 may be lost for statements such as: |

|‘-1/6 units2’ or ‘-1/6 = 1/6’ or ‘-1/6 so ignore negative’ |

|•6 may be gained for statements such as ‘-1/6 so area is 1/6’ |

|[pic]is correct and so all six marks are still available |

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(2)

#2.1

(5)

#2.2

#2.1

(2)

#2.2

#2.1

(3)

#2.1

(1)

(2)

#2.1

(3)

(5)

(2)

(4)

(4)

(6)

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