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Higher Mathematics
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Differentiation
Paper 1 Section A
Each correct answer in this section is worth two marks.
1. Differentiate 2 3√x with respect to x .
A. 6√
x
B. 32
3√x4
C. − 43 3√x2
D. 23 3√x2
Key Outcome Grade Facility Disc. Calculator Content SourceD 1.3 C 0.83 0.38 NC C2, C3 HSN 091
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[END OF PAPER 1 SECTION A]
Paper 1 Section B
2.[SQA] Find the coordinates of the point on the curve y = 2x2 − 7x + 10 where the tangentto the curve makes an angle of 45◦ with the positive direction of the x -axis. 4
Part Marks Level Calc. Content Answer U1 OC34 C NC G2, C4 (2, 4) 2002 P1 Q4
•1 sp: know to diff., and differentiate•2 pd: process gradient from angle•3 ss: equate equivalent expressions•4 pd: solve and complete
•1 dydx = 4x − 7
•2 mtang = tan 45◦ = 1•3 4x − 7 = 1•4 (2, 4)
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3.[SQA]
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4.[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
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y
(−a, 0)
(0, b)
(c, d)
(e, 0)
y = f (x)
Part Marks Level Calc. Content Answer U1 OC33 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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5.[SQA] If y = x2 − x , show that dydx = 1 +
2yx . 3
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6.[SQA] Given f (x) = 3x2(2x − 1) , find f ′(−1) . 3
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7.[SQA] Find dydx where y =
4x2 + x
√x· 4
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8.[SQA] Find f ′(4) where f (x) =x − 1√x · 5
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9.[SQA] Given that y = 2x2 + x , find dydx and hence show that x
(
1 +dydx
)
= 2y . 3
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10.[SQA] Differentiate 2√
x(x + 2) with respect to x . 4
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11.[SQA] If f (x) = kx3 + 5x − 1 and f ′(1) = 14, find the value of k . 3
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12.[SQA]
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13.[SQA] A function f is defined by the formula f (x) = (x − 1)2(x + 2) where x ∈ R .
(a) Find the coordinates of the points where the curve with equation y = f (x)crosses the x - and y-axes. 3
(b) Find the stationary points of this curve y = f (x) and determine their nature. 7
(c) Sketch the curve y = f (x) . 2
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14.[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
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y
(12, 0)
P
Part Marks Level Calc. Content Answer U1 OC35 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 dP
dx = . . . − 4x3
•2 dPdx = 36x2 − 4x3
•3 dPdx = 0
•4 x = 0 and x = 9•5 nature table about x = 0 and x = 9
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15.[SQA]
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16.[SQA]
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17.[SQA] Find the x -coordinate of each of the points on the curve y = 2x3 − 3x2 − 12x + 20at which the tangent is parallel to the x -axis. 4
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18.[SQA]
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19.[SQA] Calculate, to the nearest degree, the angle between the x -axis and the tangent tothe curve with equation y = x3 − 4x − 5 at the point where x = 2. 4
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20.[SQA]
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21.[SQA] The point P(−1, 7) lies on the curve with equation y = 5x2 + 2. Find the equationof the tangent to the curve at P. 4
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22.[SQA] Find the equation of the tangent to the curve y = 4x3 − 2 at the point wherex = −1. 4
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23.[SQA] Find the equation of the tangent to the curve y = 3x3+ 2 at the point where x = 1. 4
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24.[SQA] Find the equation of the tangent to the curve with equation y = 5x3 − 6x2 at thepoint where x = 1. 4
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25.[SQA] A ball is thrown vertically upwards. The height h metres of the ball t seconds afterit is thrown, is given by the formula h = 20t − 5t2 .
(a) Find the speed of the ball when it is thrown (i.e. the rate of change of heightwith respect to time of the ball when it is thrown). 3
(b) Find the speed of the ball after 2 seconds.Explain your answer in terms of the movement of the ball. 2
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26.[SQA] For what values of x is the function f (x) = 13 x3 − 2x2 − 5x − 4 increasing? 5
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27.[SQA]
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28.[SQA] The point P(−2, b) lies on the graph of the function f (x) = 3x3 − x2 − 7x + 4.
(a) Find the value of b . 1
(b) Prove that this function is increasing at P. 3
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29.[SQA] A sketch of the graph of y = f (x) where f (x) = x3 − 6x2 + 9x is shown below.
The graph has a maximum at A and a minimum at B(3, 0) .
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y A y = f (x)
B(3, 0)
(a) Find the coordinates of the turning point at A. 4
(b) Hence sketch the graph of y = g(x) where g(x) = f (x + 2) + 4.Indicate the coordinates of the turning points. There is no need to calculatethe coordinates of the points of intersection with the axes. 2
(c) Write down the range of values of k for which g(x) = k has 3 real roots. 1
Part Marks Level Calc. Content Answer U1 OC3(a) 4 C NC C8 A(1, 4) 2000 P1 Q2(b) 2 C NC A3 sketch (translate 4 up, 2
left)(c) 1 A/B NC A2 4 < k < 8
•1 ss: know to differentiate•2 pd: differentiate correctly•3 ss: know gradient = 0•4 pd: process
•5 ic: interpret transformation•6 ic: interpret transformation
•7 ic: interpret sketch
•1 dydx = . . .
•2 dydx = 3x2 − 12x + 9
•3 3x2 − 12x + 9 = 0•4 A = (1, 4)
translate f (x) 4 units up, 2 units left
•5 sketch with coord. of A′(−1, 8)•6 sketch with coord. of B′(1, 4)
•7 4 < k < 8 (accept 4 ≤ k ≤ 8)
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30.[SQA] A curve has equation y = x4 − 4x3+ 3.
(a) Find algebraically the coordinates of the stationary points. 6
(b) Determine the nature of the stationary points. 2
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31.[SQA] A curve has equation y = 2x3+ 3x2
+ 4x − 5.
Prove that this curve has no stationary points. 5
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32.[SQA]
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33.[SQA]
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34.[SQA] Find the values of x for which the function f (x) = 2x3 − 3x2 − 36x is increasing. 4
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35.[SQA]
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36.[SQA] A function f is defined by the formula f (x) = 4x2(x − 3) where x ∈ R .
(a) Write down the coordinates of the points where the curve with equationy = f (x) meets the x - and y-axes. 2
(b) Find the stationary points of y = f (x) and determine the nature of each. 6
(c) Sketch the curve y = f (x) . 2
(d) Find the area completely enclosed by the curve y = f (x) and the x -axis. 4
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37.[SQA] The diagram shows a sketch of the graphs of y = 5x2 − 15x − 8 andy = x3 − 12x + 1.
The two curves intersect at A and touch at B, i.e. at B the curves have a commontangent.
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O x
y
A
y = x3 − 12x + 1
B
y = 5x2 − 15x − 8
(a) (i) Find the x -coordinates of the point of the curves where the gradients areequal. 4
(ii) By considering the corresponding y-coordinates, or otherwise,distinguish geometrically between the two cases found in part (i). 1
(b) The point A is (−1, 12) and B is (3,−8) .Find the area enclosed between the two curves. 5
Part Marks Level Calc. Content Answer U2 OC2(ai) 4 C NC C4 x = 1
3 and x = 3 2000 P1 Q4(aii) 1 C NC 0.1 parallel and coincident(b) 5 C NC C17 21 1
3
•1 ss: know to diff. and equate•2 pd: differentiate•3 pd: form equation•4 ic: interpret solution
•5 ic: interpret diagram
•6 ss: know how to find area betweencurves
•7 ic: interpret limits•8 pd: form integral•9 pd: process integration•10 pd: process limits
•1 find derivatives and equate•2 3x2 − 12 and 10x − 15•3 3x2 − 10x + 3 = 0•4 x = 3, x = 1
3
•5 tangents at x = 13 are parallel, at
x = 3 coincident
•6 ∫
(cubic − parabola)or
∫
(cubic) −∫
(parabola)
•7 ∫ 3−1 · · · dx
•8 ∫
(x3 − 5x2 + 3x + 9)dx or equiv.•9 [ 1
4 x4 − 53 x3 + 3
2 x2 + 9x]3−1 or equiv.
•10 21 13
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38.[SQA]
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39.[SQA]
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40.[SQA] Differentiate sin 2x +2√x with respect to x . 4
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41.[SQA] If f (x) = cos2 x − 23x2 , find f ′(x) . 4
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42.[SQA] Differentiate 4√
x + 3 cos 2x with respect to x . 4
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[END OF PAPER 1 SECTION B]
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Paper 21.[SQA]
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2.[SQA]
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3.[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√
32
(
x2 +16x
)
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
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x
y
Part Marks Level Calc. Content Answer U1 OC36 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
√3
2 (2x − 16x−2) or 3√
3x − 24√
3x−2
•3 A′(x) = 0•4 − 16
x2 or − 24√
3x2
•5 x = 2•6 x 2− 2 2+
A′(x) −ve 0 +veso x = 2 is min.
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4.[SQA] The shaded rectangle on this maprepresents the planned extension to thevillage hall. It is hoped to provide thelargest possible area for the extension.
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Village hall
Manse Lane
The
Venn
el
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) .
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O x
y
lb
(a, 0) (8, 0)
(0, 6)
(a) (i) Show that l = 54 a .
(ii) Express b in terms of a and hence deduce that the area, A m2 , of theextension is given by A = 3
4 a(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 0.1 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. nature
table
•1 proof of l = 54 a
•2 b = 35 (8 − a)
•3 complete proof leading to A = . . .
•4 dAda = . . . = 0
•5 6 − 32 a
•6 a = 4•7 e.g. nature table, comp. the square
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5.[SQA]
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6.[SQA]
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8.[SQA]
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9.[SQA]
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10.[SQA]
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16.[SQA] A curve has equation y = x − 16√x , x > 0.
Find the equation of the tangent at the point where x = 4. 6
Part Marks Level Calc. Content Answer U1 OC36 C CN C4, C5 y = 2x − 12 2001 P2 Q2
•1 ic: find corresponding y-coord.•2 ss: express in standard form•3 ss: start to differentiate•4 pd: diff. fractional negative power•5 ss: find gradient of tangent•6 ic: write down equ. of tangent
•1 (4,−4) stated or implied by •6
•2 −16x− 12
•3 dydx = 1 . . .
•4 . . . + 8x− 32
•5 mx=4 = 2•6 y − (−4) = 2(x − 4)
17.[SQA] A ball is thrown vertically upwards.
After t seconds its height is h metres, where h = 1·2 + 19·6t − 4·9t2 .
(a) Find the speed of the ball after 1 second. 3
(b) For how many seconds is the ball travelling upwards? 2
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19.[SQA]
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20.[SQA]
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21.[SQA] The diagram shows a sketch of thegraph of y = x3 − 3x2 + 2x .(a) Find the equation of the
tangent 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
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yy = 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)
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22.[SQA] The diagram shows part of the graph of thecurve with equation y = 2x3 − 7x2 + 4x + 4.(a) Find the x -coordinate of the maximum
turning point. 5
(b) Factorise 2x3 − 7x2 + 4x + 4. 3
(c) State the coordinates of the point A andhence find the values of x for which2x3 − 7x2 + 4x + 4 < 0. 2
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y
A(2, 0)
y = f (x)
Part Marks Level Calc. Content Answer U2 OC1(a) 5 C NC C8 x = 1
3 2002 P2 Q3(b) 3 C NC A21 (x − 2)(2x + 1)(x − 2)
(c) 2 C NC A6 A(− 12 , 0), x < − 1
2
•1 ss: know to differentiate•2 pd: differentiate•3 ss: know to set derivative to zero•4 pd: start solving process of equation•5 pd: complete solving process
•6 ss: strategy for cubic, e.g. synth.division
•7 ic: extract quadratic factor•8 pd: complete the cubic factorisation
•9 ic: interpret the factors•10 ic: interpret the diagram
•1 f ′(x) = . . .•2 6x2 − 14x + 4•3 6x2 − 14x + 4 = 0•4 (3x − 1)(x − 2)•5 x = 1
3
•6· · · 2 −7 4 4
· · · · · · · · ·
· · · · · · · · · 0•7 2x2 − 3x − 2•8 (x − 2)(2x + 1)(x − 2)
•9 A(− 12 , 0)
•10 x < − 12
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23.[SQA]
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24.[SQA]
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27.[SQA] Find the equation of the tangent to the curve y = 2 sin(x − π
6 ) at the point wherex = π
3 . 4
Part Marks Level Calc. Content Answer U3 OC24 C CN C5, C20 y =
√3x + 1 − π√
3 2002 P2 Q6
•1 pd: find derivative•2 ss: know derivative at x = . . .
represents grad.•3 pd: find corresponding y-coordinate•4 ic: state equation of tangent
•1 dydx = 2 cos(x − π
6 )
•2 m =√
3•3 yx= π
3= 1
•4 y − 1 =√
3(x − π
3 )
[END OF PAPER 2]
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