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Higher Mathematics
Exam Revision 4(non-calculator)
1.[SQA]
Part Marks Level Calc. Content Answer U1 OC1
(a) 3 C CN G5, G3 1994 P2 Q2
(b) 2 C CN G1
(c) 2 C CN G1
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Higher Mathematics
2.[SQA] P, Q and R have coordinates (1,−2) , (6, 3) and (9, 14) respectively and are threevertices of a kite PQRS.
(a) Find the equations of the diagonals of this kite and the coordinates of the pointwhere they intersect. 7
(b) Find the coordinates of the fourth vertex S. 2
Part Marks Level Calc. Content Answer U1 OC1
(a) 7 C CN G8, G3, G5 1990 P2 Q2
(b) 2 C CN G8
3.[SQA] The graph of a function f intersects thex -axis at (−a, 0) and (e, 0) as shown.
There is a point of inflexion at (0, b) and amaximum turning point at (c, d) .
Sketch the graph of the derived function f ′ . 3
O x
y
(−a, 0)
(0, b)
(c, d)
(e, 0)
y = f (x)
Part Marks Level Calc. Content Answer U1 OC3
3 C CN A3, C11 sketch 2002 P1 Q6
•1 ic: interpret stationary points•2 ic: interpret main body of f•3 ic: interpret tails of f
•1 roots at 0 and c (accept a statement tothis effect)
•2 min. at LH root, max. between roots•3 both ‘tails’ correct
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Higher Mathematics
4.[SQA] A company spends x thousandpounds a year on advertisingand this results in a profit of Pthousand pounds. A mathematicalmodel , illustrated in the diagram,suggests that P and x are related byP = 12x3 − x4 for 0 ≤ x ≤ 12.Find the value of x which gives themaximum profit. 5
O x(12, 0)
P
Part Marks Level Calc. Content Answer U1 OC3
5 C NC C11 x = 9 2001 P1 Q6
•1 ss: start diff. process•2 pd: process•3 ss: set derivative to zero•4 pd: process•5 ic: interpret solutions
•1 dPdx = 36x2 . . . or dPdx = . . .− 4x3•2 dPdx = 36x2 − 4x3•3 dPdx = 0•4 x = 0 and x = 9•5 nature table about x = 0 and x = 9
5.[SQA] Find the equation of the tangent to the curve with equation y = 5x3 − 6x2 at thepoint where x = 1. 4
Part Marks Level Calc. Content Answer U1 OC3
4 C NC C4, G3 1992 P1 Q1
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Higher Mathematics
6.[SQA] A sequence is defined by the recurrence relation un+1 = 0·3un + 5 with first termu1 .
(a) Explain why this sequence has a limit as n tends to infinity. 1
(b) Find the exact value of this limit. 2
Part Marks Level Calc. Content Answer U1 OC4
(a) 1 C CN A12 1996 P1 Q11
(b) 2 C CN A13
7.[SQA]
Part Marks Level Calc. Content Answer U2 OC1
(a) 4 C NC A4 1996 P2 Q4
(b) 7 C NC A17, A18, A6
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Higher Mathematics
8.[SQA] The diagram shows a sketch of thegraph of y = x3 − 3x2+ 2x .(a) Find the equation of thetangent to this curve at thepoint where x = 1. 5
(b) The tangent at the point (2, 0)has equation y = 2x− 4. Findthe coordinates of the pointwhere this tangent meets thecurve again. 5
Ox
y
y = x3 − 3x2+ 2x
Part Marks Level Calc. Content Answer U2 OC1
(a) 5 C CN C5 x+ y = 1 2000 P2 Q1
(b) 5 C CN A23, A22, A21 (−1,−6)
•1 ss: know to differentiate•2 pd: differentiate correctly•3 ss: know that gradient = f ′(1)•4 ss: know that y-coord = f (1)•5 ic: state equ. of line
•6 ss: equate equations•7 pd: arrange in standard form•8 ss: know how to solve cubic•9 pd: process•10 ic: interpret
•1 y′ = . . .•2 3x2 − 6x+ 2•3 y′(1) = −1•4 y(1) = 0•5 y− 0 = −1(x− 1)
•6 2x− 4 = x3 − 3x2 + 2x•7 x3 − 3x2 + 4 = 0
•8· · · 1 −3 0 4
· · · · · · · · ·· · · · · · · · · · · ·
•9 identify x = −1 from working•10 (−1,−6)
9. (a)[SQA] Given that x+ 2 is a factor of 2x3 + x2 + kx+ 2, find the value of k . 3
(b) Hence solve the equation 2x3 + x2 + kx+ 2 = 0 when k takes this value. 2
Part Marks Level Calc. Content Answer U2 OC1
(a) 3 C CN A21 k = −5 2001 P2 Q1
(b) 2 C CN A22 x = −2, 12 , 1
•1 ss: use synth division orf (evaluation)
•2 pd: process•3 pd: process
•4 ss: find a quadratic factor•5 pd: process
•1 f (−2) = 2(−2)3 + · · ·•2 2(−2)3 + (−2)2 − 2k+ 2•3 k = −5
•4 2x2 − 3x + 1 or 2x2 + 3x − 2 orx2 + x− 2
•5 (2x− 1)(x− 1) or (2x− 1)(x+ 2) or(x+ 2)(x− 1)and x = −2, 12 , 1
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Higher Mathematics
10. (a) (i) Show that (x− 1) is a factor of f (x) = 2x3 + x2 − 8x+ 5.
(ii) Hence factorise f (x) fully. 5
(b) Solve 2x3 + x2− 8x+ 5 = 0. 1
(c) The line with equation y = 2x − 3 is a tangent to the curve with equationy = 2x3 + x2 − 6x+ 2 at the point G.
Find the coordinates of G. 5
(d) This tangent meets the curve again at the point H.
Write down the coordinates of H. 1
Part Marks Level Calc. Content Answer U2 OC1
(a) 5 C CN A21 (x− 1)(x− 1)(2x+ 5) 2010 P1 Q22
(b) 1 C CN A22 x = 1,− 52(c) 5 C CN A23 (1,−1)(d) 1 C CN A23 (− 52 ,−8)
•1 ss: know to use x = 1•2 ic: complete evaluation•3 ic: state conclusion•4 pd: find quadratic factor•5 pd: factorise completely
•6 ic: state solutions
•7 ss: set ycurve = yline•8 ic: express in standard form•9 ss: compare with (a) or factorise•10 ic: identify xG•11 pd: evaluate yG
•12 pd: state solution
•1 evaluating at x = 1...•2 2+ 1− 8+ 5 = 0•3 (x− 1) is a factor•4 (x− 1)(2x2 + 3x− 5)•5 (x− 1)(x− 1)(2x+ 5)
•6 x = 1 and x = − 52
•7 2x3 + x2 − 6x+ 2 = 2x− 3•8 2x3 + x2 − 8x+ 5 = 0•9 (x− 1)(x− 1)(2x+ 5) = 0•10 x = 1•11 y = −1
•12 (− 52 ,−8)
11.[SQA] One root of the equation 2x3 − 3x2 + px+ 30 = 0 is −3.Find the value of p and the other roots. 4
Part Marks Level Calc. Content Answer U2 OC1
4 C NC A21 1993 P1 Q7
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Higher Mathematics
12.[SQA] Functions f and g are defined on the set of real numbers by f (x) = x − 1 andg(x) = x2 .
(a) Find formulae for
(i) f(
g(x))
(ii) g(
f (x))
. 4
(b) The function h is defined by h(x) = f(
g(x))
+ g(
f (x))
.
Show that h(x) = 2x2 − 2x and sketch the graph of h . 3
(c) Find the area enclosed between this graph and the x -axis. 4
Part Marks Level Calc. Content Answer U2 OC2
(a) 4 C NC A4 1999 P2 Q6
(b) 3 C NC A4
(c) 4 C NC C16
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Higher Mathematics
13.[SQA]
Part Marks Level Calc. Content Answer U2 OC2
(a) 2 C NC A23 1993 P2 Q2
(b) 4 C NC C16
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Higher Mathematics
14.[SQA] A firm asked for a logo to bedesigned involving the letters Aand U. Their initial sketch isshown in the hexagon.
A mathematical representationof the final logo is shown in thecoordinate diagram.
The curve has equationy = (x + 1)(x − 1)(x − 3) andthe straight line has equationy = 5x − 5. The point (1, 0) isthe centre of half-turn symmetry.
Calculate the total shaded area. 7
x
y
(−2,−15)
(4, 15)
−2 −1 0 1 2 3 4
Part Marks Level Calc. Content Answer U2 OC2
7 C CN C17 4012 units2 2001 P2 Q8
•1 ss: express in standard form•2 ss: split area and integrate•3 ss: subtract functions•4 pd: process•5 pd: process•6 pd: process•7 ic: use symmetry or otherwise fortotal area
•1 y = x3 − 3x2 − x+ 3
•2∫ 41 (. . .)dx or
∫ 1−2(. . .)dx
•3∫ [
(5x− 5) − (x3 − 3x2 − x+ 3)]
dxor
∫ [
(x3 − 3x2 − x+ 3) − (5x− 5)]
dx•4
∫
(−x3 + 3x2 + 6x− 8)dx•5
[
− 14x4 + x3 + 3x2 − 8x]
•6 2014 or −2014 depending on chosenintegrals
•7 4012
15.[SQA] A curve for whichdy
dx= 6x2 − 2x passes through the point (−1, 2) .
Express y in terms of x . 3
Part Marks Level Calc. Content Answer U2 OC2
3 C NC C18 1998 P1 Q10
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Higher Mathematics
16.[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
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Higher Mathematics
17.[SQA] The diagram shows the graph of acosine function from 0 to π .
(a) State the equation of the graph. 1
(b) The line with equation
y = −√3 intersects this
graph at point A and B.
Find the coordinates of B. 3
O x
y
A B y = −√3
π
2π
2
−2
Part Marks Level Calc. Content Answer U2 OC3
(a) 1 C NC T4 y = 2 cos 2x 2002 P1 Q8
(b) 3 C NC T7 B( 7π12 ,−√3)
•1 ic: interpret graph
•2 ss: equate equal parts•3 pd: solve linear trig equation inradians
•4 ic: interpret result
•1 2 cos 2x
•1 2 cos 2x = −√3
•2 2x = 5π6 ,7π6
•3 x = 7π12
18.[SQA]
Part Marks Level Calc. Content Answer U2 OC3
3 C CN T8 1996 P1 Q15
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Higher Mathematics
19.[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 + 610 · 513
•4 126130
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Higher Mathematics
20.[SQA]
Part Marks Level Calc. Content Answer U2 OC4
(a) 4 C CN G5, G3 1991 P2 Q2
(b) 6 C CN G10, G1
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Higher Mathematics
21.[SQA] Find the equation of the tangent at the point (3, 1) on the circlex2 + y2 − 4x+ 6y− 4 = 0. 5
Part Marks Level Calc. Content Answer U2 OC4
5 C CN G11 1991 P1 Q8
22.[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
23.[SQA]
Part Marks Level Calc. Content Answer U3 OC1
(a) 3 C CN G10 1998 P2 Q6
(b) 3 C CN G9, G25
(c) 3 A/B CN CGD
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Higher Mathematics
24.[SQA] VABCD is a pyramid with a rectangular base ABCD.
Relative to some appropriate axes,
−→VA represents −7i − 13 j − 11k−→AB represents 6i + 6 j − 6k−→AD represents 8i − 4 j + 4k .
K divides BC in the ratio 1 : 3.
Find−→VK in component form. 3
A B
CD
V
K1
3
Part Marks Level Calc. Content Answer U3 OC1
3 C CN G25, G21, G20
1−8−16
2000 P1 Q7
•1 ss: recognise crucial aspect•2 ic: interpret ratio•3 pd: process components
•1 −→VK =
−→VA +
−→AB +
−→BK or−→
VK =−→VB+
−→BK
•2 −→BK = 1
4
−→BC or 14
−→AD or
2−11
or
−1−7−17
•3 −→VK =
1−8−16
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Higher Mathematics
25.[SQA]
Part Marks Level Calc. Content Answer U3 OC1
1 C CN G26 1997 P1 Q13
3 A/B CN G29, G27
26.[SQA] Given that f (x) = (5x− 4) 12 , evaluate f ′(4) . 3
Part Marks Level Calc. Content Answer U3 OC2
1 C CN C21 58 2000 P2 Q8
2 A/B CN C21
•1 pd: differentiate power•2 pd: differentiate 2nd function•3 pd: evaluate f ′(x)
•1 12(5x− 4)−12
•2 ×5•3 f ′(4) = 5
8
27. (a)[SQA] Find the derivative of the function f (x) = (8− x3) 12 , x < 2. 2
(b) Hence write down∫
x2
(8− x3) 12dx . 1
Part Marks Level Calc. Content Answer U3 OC2
(a) 2 A/B CN C21 − 32x2(8− x3)−12 2002 P1 Q10
(b) 1 A/B CN C24 − 23(8− x3)12 + c
•1 pd: process differentiation•2 pd: use the chain rule
•3 ic: interpret answer from (a)
•1 12(8− x3)−12
•2 . . .×−3x2
•3 − 23 f (x) or − 23(8− x3)12
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Higher Mathematics
28.[SQA] Given f (x) = (sin x+ 1)2 , find the exact value of f ′(π
6 ) . 3
Part Marks Level Calc. Content Answer U3 OC2
3 A/B NC C21, C20, T2 1998 P1 Q16
29.[SQA] Find∫
(
6x2 − x+ cos x)
dx . 4
Part Marks Level Calc. Content Answer U3 OC2
4 C NC C23 1995 P1 Q3
30.[SQA] The curve y = f (x) passes through the point ( π
12 , 1) and f′(x) = cos 2x .
Find f (x) . 3
Part Marks Level Calc. Content Answer U3 OC2
3 A/B NC C23 1997 P1 Q15
31.[SQA] Find x if 4 logx 6− 2 logx 4 = 1. 3
Part Marks Level Calc. Content Answer U3 OC3
3 C NC A32, A28, A31 x = 81 2001 P1 Q8
•1 pd: use log-to-index rule•2 pd: use log-to-division rule•3 ic: interpret base for logx a = 1 andsimplify
•1 logx 64 − logx 42•2 logx 6
4
42
•3 all processing leading to x = 81
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Higher Mathematics
32. (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
33.[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◦)
[END OF QUESTIONS]
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