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Higher Mathematics GCC Applied Calculus 1. [SQA] hsn .uk.net Page 1 Questions marked ‘[SQA]’ c SQA All others c Higher Still Notes

GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

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Page 1: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

GCC Applied Calculus

1.[SQA]

hsn.uk.net Page 1

Questions marked ‘[SQA]’ c© SQAAll others c© Higher Still Notes

Page 2: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

2.[SQA]

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Page 3: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

3.[SQA]

4.[SQA]

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Questions marked ‘[SQA]’ c© SQAAll others c© Higher Still Notes

Page 4: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

5.[SQA] A goldsmith has built up a solid which consists of a triangularprism of fixed volume with a regular tetrahedron at each end.

The surface area, A , of the solid is given by

A(x) =3√3

2

(

x2 +16

x

)

where x is the length of each edge of the tetrahedron.

Find the value of x which the goldsmith should use tominimise the amount of gold plating required to cover thesolid. 6

x

6.[SQA] The shaded rectangle on this maprepresents the planned extension to thevillage hall. It is hoped to provide thelargest possible area for the extension.

Village hall

Manse Lane

TheVennel

8 m

6 m

The coordinate diagram represents theright angled triangle of ground behindthe hall. The extension has length lmetres and breadth b metres, as shown.One corner of the extension is at the point(a, 0) .

O x

y

lb

(a, 0) (8, 0)

(0, 6)

(a) (i) Show that l = 54a .

(ii) Express b in terms of a and hence deduce that the area, A m2 , of theextension is given by A = 3

4a(8− a) . 3

(b) Find the value of a which produces the largest area of the extension. 4

hsn.uk.net Page 4

Questions marked ‘[SQA]’ c© SQAAll others c© Higher Still Notes

Page 5: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

7. The parabolas with equations y = 10− x2 and y = 25(10− x2) are shown in the

diagram below.

= 10 – 2

R

ST

Q

P

22

5(10 )= −

O

x

x

x y

y

y

A rectangle PQRS is placed between the two parabolas as shown, so that:

• Q and R lie on the upper parabola.

• RQ and SP are parallel to the x-axis.

• T , the turning point of the lower parabola, lies on SP.

(a) (i) If TP = x units, find an expression for the length of PQ.

(ii) Hence show that the area, A , of rectangle PQRS is given by

A(x) = 12x− 2x3· 3

(b) Find the maximum area of this rectangle. 6

8.[SQA]

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Page 6: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

9.[SQA]

10.[SQA]

11.[SQA]

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Page 7: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

12.[SQA] The graphs of y = f (x) and y = g(x) areshown in the diagram.

f (x) = −4 cos(2x) + 3 and g(x) is of theform g(x) = m cos(nx) .

(a) Write down the values of m and n . 1

(b) Find, correct to one decimal place,the coordinates of the points ofintersection of the two graphs in theinterval 0 ≤ x ≤ π . 5

(c) Calculate the shaded area. 6

π

= f( )

= g( )

7

3

0

–1

–3x

x

x

y

y

y

[END OF QUESTIONS]

hsn.uk.net Page 7Questions marked ‘[SQA]’ c© SQA

All others c© Higher Still Notes

Page 8: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

GCC Applied Calculus

1.[SQA]

Part Marks Level Calc. Content Answer U1 OC3

(a) 1 C CN CGD 1996 P2 Q11

(a) 3 A/B CN CGD

(b) 2 C CN C11

(b) 3 A/B CN C11

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Page 9: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

2.[SQA]

Part Marks Level Calc. Content Answer U1 OC3

(a) 1 C CN CGD 1997 P2 Q10

(a) 3 A/B CN CGD

(b) 3 C CN C11

(b) 3 A/B CN C11

hsn.uk.net Page 2

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Page 10: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

3.[SQA]

Part Marks Level Calc. Content Answer U1 OC3

(a) 3 A/B CR CGD 1998 P2 Q10

(b) 3 C CR C11

(b) 3 A/B CR C11

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Page 11: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

4.[SQA]

Part Marks Level Calc. Content Answer U3 OC2

(a) 1 C NC A6 1993 P2 Q11

(a) 2 A/B NC A6

(b) 1 C NC C11, C21

(b) 6 A/B NC C11, C21

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Page 12: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

5.[SQA] A goldsmith has built up a solid which consists of a triangularprism of fixed volume with a regular tetrahedron at each end.

The surface area, A , of the solid is given by

A(x) =3√3

2

(

x2 +16

x

)

where x is the length of each edge of the tetrahedron.

Find the value of x which the goldsmith should use tominimise the amount of gold plating required to cover thesolid. 6

x

Part Marks Level Calc. Content Answer U1 OC3

6 A/B CN C11 x = 2 2000 P2 Q6

•1 ss: know to differentiate•2 pd: process•3 ss: know to set f ′(x) = 0•4 pd: deal with x−2•5 pd: process•6 ic: check for minimum

•1 A′(x) = . . .

•2 3√32 (2x− 16x−2) or 3

√3x− 24

√3x−2

•3 A′(x) = 0

•4 − 16x2or − 24

√3

x2

•5 x = 2•6 x 2− 2 2+

A′(x) −ve 0 +veso x = 2 is min.

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Page 13: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

6.[SQA] The shaded rectangle on this maprepresents the planned extension to thevillage hall. It is hoped to provide thelargest possible area for the extension.

Village hall

Manse Lane

TheVennel

8 m

6 m

The coordinate diagram represents theright angled triangle of ground behindthe hall. The extension has length lmetres and breadth b metres, as shown.One corner of the extension is at the point(a, 0) .

O x

y

lb

(a, 0) (8, 0)

(0, 6)

(a) (i) Show that l = 54a .

(ii) Express b in terms of a and hence deduce that the area, A m2 , of theextension is given by A = 3

4a(8− a) . 3

(b) Find the value of a which produces the largest area of the extension. 4

Part Marks Level Calc. Content Answer U1 OC3

(a) 3 A/B CN CGD proof 2002 P2 Q10

(b) 4 A/B CN C11 a = 4

•1 ss: select strategy and carrythrough

•2 ss: select strategy and carrythrough

•3 ic: complete proof

•4 ss: know to set derivative to zero•5 pd: differentiate•6 pd: solve equation•7 ic: justify maximum, e.g. naturetable

•1 proof of l = 54a

•2 b = 35(8− a)

•3 complete proof leading to A = . . .

•4 dAda = . . . = 0

•5 6− 32a

•6 a = 4•7 e.g. nature table, comp. the square

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Page 14: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

7. The parabolas with equations y = 10− x2 and y = 25(10− x2) are shown in the

diagram below.

= 10 – 2

R

ST

Q

P

22

5(10 )= −

O

x

x

x y

y

y

A rectangle PQRS is placed between the two parabolas as shown, so that:

• Q and R lie on the upper parabola.

• RQ and SP are parallel to the x-axis.

• T , the turning point of the lower parabola, lies on SP.

(a) (i) If TP = x units, find an expression for the length of PQ.

(ii) Hence show that the area, A , of rectangle PQRS is given by

A(x) = 12x− 2x3· 3

(b) Find the maximum area of this rectangle. 6

Part Marks Level Calc. Content Answer U1 OC3

(ai) 2 B CN C11 6− x2 2010 P2 Q5

(aii) 1 B CN C11 2x× (6− x2) = A(x)

(b) 6 C CN C11 max is 8√2

•1 ss: know to and find OT•2 ic: obtain an expression for PQ•3 ic: complete area evaluation

•4 ss: know to and start to differentiate•5 pd: complete differentiation•6 ic: set derivative to zero•7 pd: obtain•8 ss: justify nature of stationary point•9 ic: interpret result and evaluatearea

•1 4•2 10− x2 − 4•3 2x(6− x2) = 12x− 2x3

•4 A′(x) = 12 · · ·•5 12− 6x2•6 12− 6x2 = 0•7

√2

•8 x · · ·√2 · · ·

A′(x) + 0 −•9 Max and 8

√2

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Page 15: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

8.[SQA]

Part Marks Level Calc. Content Answer U1 OC3

(a) 2 C NC A6 1992 P2 Q5

(a) 2 A/B NC A6

(b) 4 C NC C11

(b) 2 A/B NC C11

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Page 16: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

9.[SQA]

Part Marks Level Calc. Content Answer U1 OC3

(a) 4 C NC CGD 1994 P2 Q7

(b) 3 C NC C11

(b) 5 A/B NC C11

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Page 17: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

10.[SQA]

Part Marks Level Calc. Content Answer U2 OC2

(a) 2 C NC A6 1999 P2 Q10

(b) 7 A/B NC C17

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Page 18: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

11.[SQA]

Part Marks Level Calc. Content Answer U2 OC2

3 C CN C17, CGD 1994 P2 Q10

6 A/B CN C17, CGD

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Page 19: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

12.[SQA] The graphs of y = f (x) and y = g(x) areshown in the diagram.

f (x) = −4 cos(2x) + 3 and g(x) is of theform g(x) = m cos(nx) .

(a) Write down the values of m and n . 1

(b) Find, correct to one decimal place,the coordinates of the points ofintersection of the two graphs in theinterval 0 ≤ x ≤ π . 5

(c) Calculate the shaded area. 6

π

= f( )

= g( )

7

3

0

–1

–3x

x

x

y

y

y

Part Marks Level Calc. Content Answer U3 OC2

(a) 1 C CN T4 m = 3 and n = 2 2009 P2 Q5

(b) 5 C CR T6 (0·6, 1·3), (2·6, 1·3)(c) 6 B CR C17, C23 12·4

•1 ic: interprets graph

•2 ss: knows how to find intersection•3 pd: starts to solve•4 pd: finds x-coord in 1st quadrant•5 pd: finds x-coord in 2nd quadrant•6 pd: finds y-coordinates

•7 ss: knows how to find area•8 ic: states limits•9 pd: integrate•10 pd: integrate•11 ic: substitute limits•12 pd: evaluate area

•1 m = 3 and n = 2

•2 3 cos 2x = −4 cos 2x+ 3•3 cos 2x = 3

7•4 x = 0·6•5 x = 2·6•6 y = 1·3, 1·3

•7∫ (

−4 cos 2x+ 3− 3 cos 2x)

dx

•8∫ 2·60·6 · · ·

•9 “−7 sin 2x”•10 3x− 7

2 sin 2x

•11 (3× 2·6− 72 sin 5·2)− (3× 0·6− 7

2 sin 1·2)•12 12·4

[END OF QUESTIONS]

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Page 20: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

GCC Higher Circles

1.[SQA]

2.[SQA] Find the equation of the circle which has P(−2,−1) and Q(4, 5) as the end pointsof a diameter. 3

3.[SQA] The line y = −1 is a tangent to a circle which passes through (0, 0) and (6, 0) .

Find the equation of this circle. 6

4.[SQA] Find the equation of the tangent at the point (3, 4) on the circlex2 + y2 + 2x− 4y− 15 = 0. 4

5.[SQA]

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Page 21: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

6.[SQA]

7.[SQA]

8.[SQA]

(a) Show that the point P(5, 10) lies on circle C1 with equation(x+ 1)2 + (y− 2)2 = 100. 1

(b) PQ is a diameter of this circle asshown in the diagram. Find theequation of the tangent at Q. 5P(5, 10)

Q

O x

y

(c) Two circles, C2 and C3 , touch circle C1 at Q.

The radius of each of these circles is twice the radius of circle C1 .

Find the equations of circles C2 and C3 . 4

9.[SQA] Explain why the equation x2 + y2 + 2x+ 3y+ 5 = 0 does not represent a circle. 2

10.[SQA] For what range of values of k does the equation x2 + y2 + 4kx − 2ky− k − 2 = 0represent a circle? 5

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Page 22: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

11.[SQA] Circle P has equation x2 + y2 − 8x − 10y+ 9 = 0. Circle Q has centre (−2,−1)and radius 2

√2.

(a) (i) Show that the radius of circle P is 4√2.

(ii) Hence show that circles P and Q touch. 4

(b) Find the equation of the tangent to the circle Q at the point (−4, 1) . 3

(c) The tangent in (b) intersects circle P in two points. Find the x -coordinates of

the points of intersection, expressing you answers in the form a± b√3. 3

12.[SQA]

13.[SQA]

[END OF QUESTIONS]

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Page 23: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

GCC Higher Circles

1.[SQA]

Part Marks Level Calc. Content Answer U2 OC4

2 C CN G10 1999 P1 Q4

2.[SQA] Find the equation of the circle which has P(−2,−1) and Q(4, 5) as the end pointsof a diameter. 3

Part Marks Level Calc. Content Answer U2 OC4

3 C CN G10 1995 P1 Q9

3.[SQA] The line y = −1 is a tangent to a circle which passes through (0, 0) and (6, 0) .

Find the equation of this circle. 6

Part Marks Level Calc. Content Answer U2 OC4

1 C CN G10 1996 P1 Q20

5 A/B CN G9, G15

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Page 24: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

4.[SQA] Find the equation of the tangent at the point (3, 4) on the circlex2 + y2 + 2x− 4y− 15 = 0. 4

Part Marks Level Calc. Content Answer U2 OC4

4 C CN G2, G5, G9 1996 P1 Q4

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Page 25: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

5.[SQA]

Part Marks Level Calc. Content Answer U3 OC1

8 C CN G9, G10, G25 1995 P2 Q8

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Page 26: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

6.[SQA]

Part Marks Level Calc. Content Answer U3 OC1

5 C CN G9, G25 1997 P1 Q12

7.[SQA]

Part Marks Level Calc. Content Answer U2 OC4

(a) 3 C CN G9 1993 P1 Q5

(b) 1 C CN G10

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Page 27: GCCAppliedCalculus · • Q and R lie on the upper parabola. • RQ and SP are parallel to the x-axis. • T , the turning point of the lower parabola, lies on SP. ... 6 A/B CN C11

Higher Mathematics

8.[SQA]

(a) Show that the point P(5, 10) lies on circle C1 with equation(x+ 1)2 + (y− 2)2 = 100. 1

(b) PQ is a diameter of this circle asshown in the diagram. Find theequation of the tangent at Q. 5P(5, 10)

Q

O x

y

(c) Two circles, C2 and C3 , touch circle C1 at Q.

The radius of each of these circles is twice the radius of circle C1 .

Find the equations of circles C2 and C3 . 4

Part Marks Level Calc. Content Answer U2 OC4

(a) 1 C CN A6 proof 2009 P2 Q4

(b) 5 C CN G11 3x+ 4y+ 45 = 0

(c) 4 A NC G15 (x− 5)2 + (y− 10)2 = 400,(x+ 19)2+(y+ 22)2 = 400

•1 pd: substitute

•2 ic: find centre•3 ss: use mid-point result for Q•4 ss: know to, and find gradient ofradius

•5 ic: find gradient of tangent•6 ic: state equation of tangent

•7 ic: state radius•8 ss: know how to find centre•9 ic: state equation of one circle•10 ic: state equation of the other circle

•1 (5+ 1)2 + (10− 2)2 = 100

•2 centre = (−1, 2)•3 Q = (−7,−6)•4 mrad = 8

6

•5 mtgt = − 34•6 y− (−6) = − 34(x− (−7))

•7 radius = 20•8 centre = (5, 10)•9 (x− 5)2 + (y− 10)2 = 400•10 (x+ 19)2 + (y+ 22)2 = 400

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Higher Mathematics

9.[SQA] Explain why the equation x2 + y2 + 2x+ 3y+ 5 = 0 does not represent a circle. 2

Part Marks Level Calc. Content Answer U2 OC4

2 C CN G9 1993 P1 Q18

10.[SQA] For what range of values of k does the equation x2 + y2 + 4kx − 2ky− k − 2 = 0represent a circle? 5

Part Marks Level Calc. Content Answer U2 OC4

5 A NC G9, A17 for all k 2000 P1 Q6

•1 ss: know to examine radius•2 pd: process•3 pd: process•4 ic: interpret quadratic inequation•5 ic: interpret quadratic inequation

•1 g = 2k, f = −k, c = −k− 2stated or implied by •2

•2 r2 = 5k2 + k+ 2•3 (real r ⇒) 5k2 + k+ 2 > 0 (accept ≥)•4 use discr. or complete sq. or diff.•5 true for all k

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Higher Mathematics

11.[SQA] Circle P has equation x2 + y2 − 8x − 10y+ 9 = 0. Circle Q has centre (−2,−1)and radius 2

√2.

(a) (i) Show that the radius of circle P is 4√2.

(ii) Hence show that circles P and Q touch. 4

(b) Find the equation of the tangent to the circle Q at the point (−4, 1) . 3

(c) The tangent in (b) intersects circle P in two points. Find the x -coordinates of

the points of intersection, expressing you answers in the form a± b√3. 3

Part Marks Level Calc. Content Answer U2 OC4

(a) 2 C CN G9 proof 2001 P1 Q11

(a) 2 A/B CN G14

(b) 3 C CN G11 y = x+ 5

(c) 3 C CN G12 x = 2± 2√3

•1 ic: interpret centre of circle (P)•2 ss: find radius of circle (P)•3 ss: find sum of radii•4 pd: compare with distance betweencentres

•5 ss: find gradient of radius•6 ss: use m1m2 = −1•7 ic: state equation of tangent

•8 ss: substitute linear into circle•9 pd: express in standard form•10 pd: solve (quadratic) equation

•1 CP = (4, 5)•2 rP =

√16+ 25− 9 =

√32 = 4

√2

•3 rP + rQ = 4√2+ 2

√2 = 6

√2

•4 CPCQ =√62 + 62 = 6

√2 and “so

touch”

•5 mr = −1•6 mtgt = +1•7 y− 1 = 1(x+ 4)

•8 x2+(x+ 5)2− 8x− 10(x+ 5)+ 9 = 0•9 2x2 − 8x− 16 = 0•10 x = 2± 2

√3

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Higher Mathematics

12.[SQA]

Part Marks Level Calc. Content Answer U2 OC4

(a) 4 C CN G10 1990 P2 Q8

(b) 4 C CN G10

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Higher Mathematics

13.[SQA]

Part Marks Level Calc. Content Answer U2 OC4

5 C CN G9, G1, G15 1992 P1 Q16

[END OF QUESTIONS]

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Higher Mathematics

GCC Trig Equations and Expressions

1.[SQA] Solve the equation 3 cos 2x◦ + cos x◦ = −1 in the interval 0 ≤ x ≤ 360. 5

2. (a)[SQA] Solve the equation sin 2x◦ − cos x◦ = 0 in the interval 0 ≤ x ≤ 180. 4

(b) The diagram shows parts of twotrigonometric graphs, y = sin 2x◦

and y = cos x◦ .

Use your solutions in (a) to writedown the coordinates of the point P. 1

O x

yy = sin 2x◦

y = cos x◦P

90180

3.[SQA]

(a) Using the fact that 7π12 = π

3 + π

4 , find the exact value of sin(

7π12

)

. 3

(b) Show that sin(A+ B) + sin(A− B) = 2 sin A cos B . 2

(c) (i) Express π

12 in terms ofπ

3 andπ

4 .

(ii) Hence or otherwise find the exact value of sin(

7π12

)

+ sin(

π

12

)

. 4

4.[SQA] Functions f (x) = sin x , g(x) = cos x and h(x) = x+ π

4 are defined on a suitableset of real numbers.

(a) Find expressions for:

(i) f (h(x)) ;

(ii) g(h(x)) . 2

(b) (i) Show that f (h(x)) = 1√2sin x+ 1√

2cos x .

(ii) Find a similar expression for g(h(x)) and hence solve the equationf (h(x)) − g(h(x)) = 1 for 0 ≤ x ≤ 2π . 5

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Higher Mathematics

5.[SQA] On the coordinate diagram shown, A is thepoint (6, 8) and B is the point (12,−5) .Angle AOC = p and angle COB = q .

Find the exact value of sin(p+ q) . 4

O x

yA(6, 8)

C

B(12,−5)

pq

6.[SQA]

7.[SQA]

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Higher Mathematics

8.[SQA]

9.[SQA]

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Higher Mathematics

10.[SQA]

[END OF QUESTIONS]

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Higher Mathematics

GCC Trig Equations and Expressions

1.[SQA] Solve the equation 3 cos 2x◦ + cos x◦ = −1 in the interval 0 ≤ x ≤ 360. 5

Part Marks Level Calc. Content Answer U2 OC3

5 A/B CR T10 60, 131·8, 228·2, 300 2000 P2 Q5

•1 ss: know to usecos 2x = 2 cos2 x− 1

•2 pd: process•3 ss: know to/and factorise quadratic•4 pd: process•5 pd: process

•1 3(2 cos2 x◦ − 1)•2 6 cos2 x◦ + cos x◦ − 2 = 0•3 (2 cos x◦ − 1)(3 cos x◦ + 2)•4 cos x◦ = 1

2 , x = 60, 30

•5 cos x◦ = − 23 , x = 132, 228

2. (a)[SQA] Solve the equation sin 2x◦ − cos x◦ = 0 in the interval 0 ≤ x ≤ 180. 4

(b) The diagram shows parts of twotrigonometric graphs, y = sin 2x◦

and y = cos x◦ .

Use your solutions in (a) to writedown the coordinates of the point P. 1

O x

yy = sin 2x◦

y = cos x◦P

90180

Part Marks Level Calc. Content Answer U2 OC3

(a) 4 C NC T10 30, 90, 150 2001 P1 Q5

(b) 1 C NC T3 (150,−√32 )

•1 ss: use double angle formula•2 pd: factorise•3 pd: process•4 pd: process

•5 ic: interpret graph

•1 2 sin x◦ cos x◦•2 cos x◦(2 sin x◦ − 1)•3 cos x◦ = 0, sin x◦ = 1

2•4 90, 30, 150

or

•3 sin x◦ = 12 and x = 30, 150

•4 cos x◦ = 0 and x = 90

•5(

150,−√32

)

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Higher Mathematics

3.[SQA]

(a) Using the fact that 7π12 = π

3 + π

4 , find the exact value of sin(

7π12

)

. 3

(b) Show that sin(A+ B) + sin(A− B) = 2 sin A cos B . 2

(c) (i) Express π

12 in terms ofπ

3 andπ

4 .

(ii) Hence or otherwise find the exact value of sin(

7π12

)

+ sin(

π

12

)

. 4

Part Marks Level Calc. Content Answer U2 OC3

(a) 3 C NC T8, T3

√3+ 1

2√2

2009 P1 Q24

(b) 2 C CN T8 proof

(c) 3 B NC T11 π

12 = π

3 − π

4

(c) 1 C NC T11√62 or

32

•1 ss: expand compound angle•2 ic: substitute exact values•3 pd: process to a single fraction

•4 ic: start proof•5 ic: complete proof

•6 ss: identify steps•7 ic: start process (identify ‘A’ & ‘B’)•8 ic: substitute•9 pd: process

•1 sin π

3 cosπ

4 + cos π

3 sinπ

4

•2√32 × 1√

2+ 12 × 1√

2

•3√3+ 1

2√2or equivalent

•4 sin A cos B+ cos A sin B+ · · ·•5 · · · + sin A cos B − cos A sin B andcomplete

•6 π

12 = π

3 − π

4

•7 2×√32 × 1√

2

•8√62 or

32

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Higher Mathematics

4.[SQA] Functions f (x) = sin x , g(x) = cos x and h(x) = x+ π

4 are defined on a suitableset of real numbers.

(a) Find expressions for:

(i) f (h(x)) ;

(ii) g(h(x)) . 2

(b) (i) Show that f (h(x)) = 1√2sin x+ 1√

2cos x .

(ii) Find a similar expression for g(h(x)) and hence solve the equationf (h(x)) − g(h(x)) = 1 for 0 ≤ x ≤ 2π . 5

Part Marks Level Calc. Content Answer U2 OC3

(a) 2 C NC A4 (i) sin(x + π

4 ), (ii)cos(x+ π

4 )2001 P1 Q7

(b) 5 C NC T8, T7 (i) proof, (ii) x = π

4 ,3π4

•1 ic: interpret composite functions•2 ic: interpret composite functions

•3 ss: expand sin(x+ π

4 )•4 ic: interpret•5 ic: substitute•6 pd: start solving process•7 pd: process

•1 sin(x+ π

4 )•2 cos(x+ π

4 )

•3 sin x cos π

4 + cos x sin π

4 and

complete

•4 g(h(x)) = 1√2cos x− 1√

2sin x

•5 ( 1√2sin x+ 1√

2cos x)− ( 1√

2cos x− 1√

2sin x)

•6 2√2sin x

•7 x = π

4 ,3π4 accept only radians

5.[SQA] On the coordinate diagram shown, A is thepoint (6, 8) and B is the point (12,−5) .Angle AOC = p and angle COB = q .

Find the exact value of sin(p+ q) . 4

O x

yA(6, 8)

C

B(12,−5)

pq

Part Marks Level Calc. Content Answer U2 OC3

4 C NC T9 6365 2000 P1 Q1

•1 ss: know to use trig expansion•2 pd: process missing sides•3 ic: interpret data•4 pd: process

•1 sin p cos q+ cos p sin q•2 10 and 13•3 8

10 ·1213 + 6

10 ·513

•4 126130

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Higher Mathematics

6.[SQA]

Part Marks Level Calc. Content Answer U2 OC3

(a-c) 3 C NC CGD 1999 P2 Q8

(d) 4 A/B NC T8, G2

(e) 2 A/B NC G2, G5

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Higher Mathematics

7.[SQA]

Part Marks Level Calc. Content Answer U2 OC3

1 C NC T9 1996 P1 Q18

4 A/B NC T9

8.[SQA]

Part Marks Level Calc. Content Answer U2 OC3

1 C NC T10 1991 P1 Q20

3 A/B NC T10

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Higher Mathematics

9.[SQA]

Part Marks Level Calc. Content Answer U2 OC3

(a) 4 C CR T10 1992 P2 Q7

(b) 1 C CR A7

(c) 2 C CR A6

(d) 2 C CR T2

(d) 1 A/B CR T2

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Higher Mathematics

10.[SQA]

Part Marks Level Calc. Content Answer U2 OC3

(a) 2 C CN CGD 1991 P2 Q3

(a) 1 A/B CN CGD

(b) 5 C CN T10, T11

[END OF QUESTIONS]

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Higher Mathematics

GCCWaveFunction

1.[SQA] Express 2 sin x◦ − 5 cos x◦ in the form k sin(x− α)◦ , 0 ≤ α < 360 and k > 0. 4

2.[SQA] Express 8 cos x◦− 6 sin x◦ in the form k cos(x◦ + a◦) where k > 0 and 0 < a < 360. 4

3.[SQA] Find the maximum value of cos x− sin x and the value of x for which it occurs inthe interval 0 ≤ x ≤ 2π . 6

4.[SQA]

5. (a) 12 cos x◦ − 5 sin x◦ can be expressed in the form k cos(x + a)◦ , where k > 0and 0 ≤ a < 360.

Calculate the values of k and a . 4

(b) (i) Hence state the maximum and minimum values of12 cos x◦ − 5 sin x◦ .(ii) Determine the values of x , in the interval 0 ≤ x < 360, at which thesemaximum and minimum values occur. 3

6. (a)[SQA] Write sin(x) − cos(x) in the form k sin(x − a) stating the values of k and awhere k > 0 and 0 ≤ a ≤ 2π 4

(b) Sketch the graph of y = sin(x) − cos(x) for 0 ≤ x ≤ 2π , showing clearly thegraph’s maximum and minimum values and where it cuts the x -axis and they-axis. 3

7.[SQA]

8.[SQA] Solve the equation 2 sin x◦ − 3 cos x◦ = 2·5 in the interval 0 ≤ x < 360. 8

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Higher Mathematics

9.[SQA]

10. (a) The expression 3 sin x− 5 cos x can be written in the form R sin(x+ a) whereR > 0 and 0 ≤ a < 2π .

Calculate the values of R and a . 4

(b) Hence find the value of t , where 0 ≤ t ≤ 2, for which

t∫

0

(3 cos x+ 5 sin x) dx = 3.

7

11.[SQA] The displacement, d units, of a wave after t seconds, is given by the formula

d = cos 20t◦ +√3 sin 20t◦ .

(a) Express d in the form k cos(20t◦ − α◦) , where k > 0 and 0 ≤ α ≤ 360. 4

(b) Sketch the graph of d for 0 ≤ t ≤ 18. 4

(c) Find, correct to one decimal place, the values of t , 0 ≤ t ≤ 18, for which thedisplacement is 1·5 units. 3

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Higher Mathematics

12.[SQA]

13.[SQA]

[END OF QUESTIONS]

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Higher Mathematics

GCCWaveFunction

1.[SQA] Express 2 sin x◦ − 5 cos x◦ in the form k sin(x− α)◦ , 0 ≤ α < 360 and k > 0. 4

Part Marks Level Calc. Content Answer U3 OC4

4 C CR T13 k =√29, α = 68·2 1997 P1 Q11

2.[SQA] Express 8 cos x◦− 6 sin x◦ in the form k cos(x◦ + a◦) where k > 0 and 0 < a < 360. 4

Part Marks Level Calc. Content Answer U3 OC4

4 C CR T13 10 cos(x◦ + 36·9◦) 2001 P2 Q5

•1 ss: expand k cos(x◦ + a◦)•2 ic: compare coefficients•3 pd: process•4 pd: process

•1 k cos x cos a − k sin x sin a statedexplicitly

•2 k cos a = 8 and k sin a = 6 statedexplicitly

•3 k = 10•4 a = 36·9

3.[SQA] Find the maximum value of cos x− sin x and the value of x for which it occurs inthe interval 0 ≤ x ≤ 2π . 6

Part Marks Level Calc. Content Answer U3 OC4

6 A/B CN T14 max value√2 when

x = 7π4

2000 P1 Q10

•1 ss: use e.g. k cos(x+ a)•2 ic: expand chosen rule•3 pd: compare coefficients•4 pd: process•5 pd: process•6 ic: interpret trig expression

•1 e.g. use k cos(x+ a)•2 k cos x cos a− k sin x sin a•3 k cos a = 1 and k sin a = 1•4 k =

√2

•5 tan a = 1, a = π

4 (45◦ is bad form)

•6 max. value =√2 when x = 7π

4 (donot accept 45◦)

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Higher Mathematics

4.[SQA]

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CR T13 1992 P1 Q7

(b) 1 C CR T14

(b) 1 A/B CR T14

5. (a) 12 cos x◦ − 5 sin x◦ can be expressed in the form k cos(x + a)◦ , where k > 0and 0 ≤ a < 360.

Calculate the values of k and a . 4

(b) (i) Hence state the maximum and minimum values of12 cos x◦ − 5 sin x◦ .(ii) Determine the values of x , in the interval 0 ≤ x < 360, at which thesemaximum and minimum values occur. 3

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CN T13 k = 13, a = 22·6 2010 P2 Q2

(bi) 1 C CN T14 max 13, min −13(bii) 2 C CN T14 max at 337·4, min at 157·4

•1 ss: use addition formula•2 ic: compare coefficients•3 pd: process k•4 pd: process a

•5 ss: state maximum and minimum•6 ic: find x corresponding to max.value

•7 pd: find x corresponding to min.value

•1 k cos x◦ cos a◦ − k sin x◦ sin a◦•2 k cos a◦ = 12 and k sin a◦ = 5•3 13 (do not accept

√169)

•4 22·6 (accept any answer whichrounds to 23)

•5 13, −13•6 maximum at 337·4 and no others•7 minimum at 157·4 and no others

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Higher Mathematics

6. (a)[SQA] Write sin(x) − cos(x) in the form k sin(x − a) stating the values of k and awhere k > 0 and 0 ≤ a ≤ 2π 4

(b) Sketch the graph of y = sin(x) − cos(x) for 0 ≤ x ≤ 2π , showing clearly thegraph’s maximum and minimum values and where it cuts the x -axis and they-axis. 3

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C NC T13√2 sin(x− π

4 ) 2002 P1 Q9

(b) 3 C NC T15, T14 sketch

•1 ss: know to expand, and expand•2 ic: compare coefficients•3 pd: write down the value of k•4 pd: process a

•5 ic: sketch a sine curve•6 ic: int/com max. and min. values•7 pd: process intercepts

•1 k sin x cos a − k cos x sin a statedexplicitly

•2 k cos a = 1 and k sin a = 1 statedexplicitly

•3 k =√2

•4 a = π

4 accept in degrees

•5 correct shape of graph (i.e. sin) butnot passing through the origin

•6 graph lies between√2 and −

√2

•7 ( π

4 , 0), (5π4 , 0), (0,−1) accept only

answers in radians

7.[SQA]

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CR T13 1996 P2 Q7

(b) 3 C CR T16

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Higher Mathematics

8.[SQA] Solve the equation 2 sin x◦ − 3 cos x◦ = 2·5 in the interval 0 ≤ x < 360. 8

Part Marks Level Calc. Content Answer U3 OC4

8 A/B CR T16 100·2◦, 192·4◦ 1999 P2 Q9

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Higher Mathematics

9.[SQA]

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CR T13 1994 P2 Q5

(b) 4 C CR T16

(c) 2 A/B CR T16

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Higher Mathematics

10. (a) The expression 3 sin x− 5 cos x can be written in the form R sin(x+ a) whereR > 0 and 0 ≤ a < 2π .

Calculate the values of R and a . 4

(b) Hence find the value of t , where 0 ≤ t ≤ 2, for which

t∫

0

(3 cos x+ 5 sin x) dx = 3.

7

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CN T13 R =√34, a = 5·253 2011 P2 Q6

(b) 7 B CN C23, T3, T16 t = 0·6

•1 ss: use compound angle formula•2 ic: compare coefficients•3 pd: process R•4 pd: process a

•5 pd: integrate given expression•6 ic: substitute limits•7 pd: process limits•8 ss: know to use wave equation•9 ic: write in standard format•10 ss: start to solve equation•11 pd: complete and state solution

•1 R sin x cos a+ R cos x sin a•2 R cos a = 3 and R sin a = −5•3

√34 (accept 5·8)

•4 5·253 (accept 5·3)

•5 3 sin x− 5 cos x•6 (3 sin t− 5 cos t) − (3 sin 0− 5 cos 0)•7 3 sin t− 5 cos t+ 5•8

√34 sin(t+ 5·3) + 5

•9 sin(t+ 5·3) = − 2√34

•10 t+ 5·3 = 3·5, 5·9•11 t = 0·6

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Higher Mathematics

11.[SQA] The displacement, d units, of a wave after t seconds, is given by the formula

d = cos 20t◦ +√3 sin 20t◦ .

(a) Express d in the form k cos(20t◦ − α◦) , where k > 0 and 0 ≤ α ≤ 360. 4

(b) Sketch the graph of d for 0 ≤ t ≤ 18. 4

(c) Find, correct to one decimal place, the values of t , 0 ≤ t ≤ 18, for which thedisplacement is 1·5 units. 3

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CR T13 1991 P2 Q8

(b) 2 C CR T1

(b) 2 A/B CR T1

(c) 1 C CR T7

(c) 2 A/B CR T7

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Higher Mathematics

12.[SQA]

Part Marks Level Calc. Content Answer U3 OC4

(a) 1 C CR CGD 1993 P2 Q9

(a) 4 A/B CR CGD

(b) 4 C CR T13

(c) 1 C CR T16

(c) 3 A/B CR T16

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Higher Mathematics

13.[SQA]

Part Marks Level Calc. Content Answer U3 OC4

(a) 4 C CR T13 1989 P2 Q9

(b) 2 C CR T1, A3

(b) 4 A/B CR T1, A3

(c) 1 A/B CR CGD

(d) 2 A/B CR CGD

[END OF QUESTIONS]

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