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2013 Higher Paper 1

Click to jump to question:

Paper 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Paper 2: 1 2 3 4 5 6 7 8 9 Question 1 Question 2 Gradient of tangent at (5, 12) Using gives Question 3 Question 4

The graph of this function has an amplitude of 4, a period of 180 , and is shifted down by 1 unit. So, the solution is A.

g(f(x)) = g(x

2

+ 1) = 3(x

2

+ 1) − 4

= 3x

2

+ 3 − 4

= 3x

2

− 1

y = x

2

− 4x + 7

dy

dx

= 2x − 4

= 2 ∙ 5 − 4 = 6

y − b = m(x − a)

y − 12 = 6(x − 5)

y − 12 = 6x − 30

y = 6x − 28

a = 2, b = 4, c = 5

b

2

− 4ac = 4

2

− 4 ∙ 2 ∙ 5

= 16 − 40

= − 24

°

(2)

Rearranging gives

Gradient of this line is

Since L is parallel it also has gradient

Using with gives

Rearranging gives

Question 6 Let

The remainder is 4 when is divided by

Question 7

5x + 3y − 6 = 0

y = − 5

3

x + 2

− 5

3

− 5

3

y − b = m(x − a)

(−2, − 1)

y − (−1) = − 5

3

(x − (−2))

y = − 5

3

x − 8

3

f(x) = x

3

+ 3x

2

− 5x − 6

f(2) = 2

3

+ 3 ∙ 2

2

− 5 ∙ 2 − 6 = 4

f(x)

(x − 2)

∫ x(3x + 2) dx

= ∫(3x

2

+ 2x) dx

= x

3

+ x

2

+ c

(3)

Question 8

Since the sequence does have a limit as n The answer is C) Only statement 2) is correct

Question 9 Question 10 Question 11

Compared to the graph of this graph is inverted and shifted ‘k’ units to the right By inspection B is the correct solution

u

n+1

= 0.1u

n

+ 8

u

1

= 0.1u

0

+ 8

11 = 0.1u

0

+ 8

3 = 0.1u

0

30 = u

0

−1 < 0.1 < 1

→ ∞

sin2x = 2sinxcosx

= 2 ∙ 1

5

∙ 2

5

= 4

5

cos(270 − a) = cos270cosa + sin270sina

= 0 ∙ cosa + ( − 1) ∙ sina

= − sina

y = − f(x − k)

(4)

Question 13 Let

So cannot be in the domain of

Question 14

where is the angle between and itself

Question 15

so is our reference angle

Using CAST tan is negative in quadrants 2 & 4 So, the solutions are and

However, these solutions are for giving

f_ + g_ =

5

4

5

f_ + g_ = 5

2

+ 4

2

+ 5

2

= 64

= 8

x

2

− 7x + 12 = 0

(x − 4)(x − 3) = 0

x = 3, x = 4

x = 3, x = 4

f(x)

a

_ . ( a

_ + b)

_ = a

_  . a

_ + a

_  . b

_

= a

_ a

_ cosθ + 5

θ

_

a

= 3 ∙ 3 ∙ cos0 + 5

= 9 + 5

= 14

tan π

4

= 1 π

4

π − π

4

= 3π

4

π + π

4

= 5π

4

x

2

x = 3π

2

,  x = 5π

2

(5)

But is out of our range so

Question 16

Question 17

Rewriting this gives

This shows that the roots are at So, by inspection,

So

Substituting the point (1, 3) gives

Question 18

2

x = 3π

2

∫ (1 − 6x)

− 1 2 

dx

= (1 − 6x)

1 2 1 2

∙ ( − 6)

+ c

= − 1

3 (

1 − 6x)

1 2

+ c

y = kx(x + a)

2

y = k(x − 0)(x + a)(x + a)

x = 0, x = − a

a = 2

y = kx(x + 2)

2

3 = k(1 + 2)

2

3 = 9k

k = 1

3

y = sin(x

2

− 3)

dy

dx

= cos(x

2

− 3) ∙ 2x

= 2xcos(x

2

− 3) 

(6)

Multiplying through by -1 gives

Consider

These are the roots of the quadratic. By considering the graph we see It is negative (i.e. for

The solutions are all such that

Question 20

The equation of the line is since gradient and y-intercept

Rewriting as an exponential gives

Question 21

Now complete the square on the inside of the bracket

1 − 2x − 3x

2

> 0

3x

2

+ 2x − 1 < 0

(3x − 1)(x + 1) < 0

(3x − 1)(x + 1) = 0

x = − 1, x = 1

3

y < 0)

−1 < x < 1

3

x

−1 < x < 1

3

log

3

y = 2x

= 2

= 0

log

3

y = 2x

y = 3

2x

2x

2

+ 12x + 1

= 2(x

2

+ 6x) + 1

= 2[(x + 3)

2

− 9] + 1

= 2(x + 3)

2

− 18 + 1

= 2(x + 3)

2

− 17

(7)

Question 22

a) Centre , Radius

b) Let C be the centre of

Therefore since perpendicular gradients

Using with (3, 2) gives

c) Radius of

Since the centre of we have

Expanding brackets and collecting terms gives

d) Substituting into gives

Expanding and simplifying gives

Using the discriminant with gives

Since the discriminant equals 0 there is only one solution Therefore, there is only one point at which the line Intersects with . In other words, it is a tangent

x

2

+ y

2

+ 2x + 4y − 27 = 0

= ( − 1,  − 2)

= 1

2

+ 2

2

+ 27 = 32

C

1

m

cp

= 1

m

tan

= − 1

y − b = m(x − a)

y − 2 = − 1(x − 2)

y = − x + 5

c

2

=

32

2

= 8

c

2

= (10,  − 1)

(x − 10)

2

+ (y + 1)

2

= ( 8)

2

(x − 10)

2

+ (y + 1)

2

= 8

x

2

+ y

2

− 20x + 2y + 93 = 0

y = − x + 5

c

2

x

2

+ (5 − x)

2

− 20(5 − x) + 2y + 93 = 0

x

2

− 16x + 64 = 0

a = 1, b = − 16, c = 64

b

2

− 4ac = (−16)

2

− 4 ∙ 1 ∙ 64 = 0

y = − x + 5

c

2

(8)

a) Let

By inspection we have and

Giving by exact values

Since we want to be such that both and are positive, by CAST, must be in quadrant 1. So, is correct

Thus

b)

from a)

has a maximum value of 10

So has a maximum value of 14

3sinx − cosx = ksin(x − a) = ksinxcosa − kcosxsina

= kcosasinx − ksinacosx

3 = kcosa

1 = ksina

k = ( 3)

2

+ 1

2

= 2

ksina

kcosa

= 1

3

= tana

a = 30°

a

sina

cosa

30°

3sinx − cosx = 2sin(x − 30)°

4 + 5cosx − 5 3sinx = 4 − 5( 3sinx − cosx)

= 4 − 5 ∙ 2sin(x − 30)

= 4 − 10sin(x − 30)

= − 10sin(x − 30) + 4

−10sin(x − 30)

(9)

Question 24

a) i)

This shows that & are parallel but since T is a common point, It follows that are collinear.

ii) The ratio is 2 : 3

b) Since C lies on the x-axis it has a y co-ordinate of 0 and a z co-ordinate of 0. Let

Then

since they are perpendicular So,

AT = 10

10

4

      →

TB = 15

15

6

AT = 2 55

2

      →

TB = 3 55

2

1

2

AT = 1

3

TB

3→

AT = 2 →

TB

AT →

TB

A,  T  B

C = (c,  0, 0)

TC = C − T = c − 3

0

0

TB ∙ →

TC = 0

TB ∙ →

TC = 15(c − 3) + (15 ∙ 0) + (6 ∙ 0) = 0

15c − 45 = 0

c = 3

c = (3, 0, 0)

(10)

Question 1

Substituting and gives

Substituting and gives

Now we have simultaneous equations to solve for (1)

(2) (2) – (1) gives

Substituting into (2) gives

So,

Question 2 a)

since are perpendicular

Using with we have

u

1

= 4, u

2

= 7, u

3

= 16

u

n+1

= mu

n

+ c

u

1

u

2

u

2

= mu

1

+ c

7 = 4m + c

u

2

u

3

u

3

= mu

2

+ c

16 = 7m + c

m  c

7 = 4m + c

16 = 7m + c

9 = 3m

m = 3

m = 3

16 = 7 ∙ 3 + c

c = − 5

m = 3, c = − 5

m

PQ

= 6 − 2

5 − 7

= − 2

m

QR

= 1

2

PQ  QR

y − b = m(x − a)

Q = (5, 6)

(11)

b)

Equating QR with PT gives

Multiply through by 6 to simplify

Substituting into gives

c) So, So,

y − 6 = 1

2

(x − 5)

y = 1

2

x + 7

2

x + 3y = 13

y = − 1

3

x + 13

3

1

2

x + 7

2

= − 1

3

x + 13

3

3x + 21 = − 2x + 26

x = 1

x = 1

y = 1

2

x + 7

2

y = 1

2

+ 7

2

= 4

T = (1, 4)

QT = (

−4

2 )

R = (1 − 4, 4 − 2) = ( − 3, 2)

QP = (

    2

−4 )

S = (−3 + 2, 2 − 4) = ( − 1,  − 2)

(12)

a) Let

Using synthetic division gives

Note that does not factorise

b) Let

Stationary Points occur where giving

from a)

The solution to this equation is since has no solutions (

Thus has only one stationary point

Question 4

Equation the line and curves equations gives

f(x) = x

3

+ 3x

2

+ x − 5

1 1 3 1 -5 1 4 5 1 4 5 0

f(x) = (x − 1)(x

2

+ 4x + 5)

x

2

+ 4x + 5

g(x) = x

4

+ 4x

3

+ 2x

2

− 20x + 3

g′(x) = 4x

3

+ 12x

2

+ 4x − 20

g′(x) = 0

4x

3

+ 12x

2

+ 4x − 20 = 0

x

3

+ 3x

2

+ x − 5 = 0

(x − 1)(x

2

+ 4x + 5) = 0

x = 1

x

2

+ 4x + 5

b

2

− 4ac = − 4 < 0)

g(x)

2x + 3 = x

3

+ 3x

2

+ 2x + 3

x

3

+ 3x

2

= 0

x

2

(x + 3) = 0

(13)

When So Area Question 5

x = 0, x = − 3

x = − 3, y = 2 ∙ (−3) + 3 = − 3

B = ( − 3,  − 3)

= ∫

−30

(x

3

+ 3x

2

+ 2x + 3) − (2x + 3) dx

= ∫

−30

(x

3

+ 3x

2

)dx

=

[

x

4

4

+ x

3

]

0 −3

= (0) − (

81

4

− 27) =

27

4

units

2

log

5

(3 − 2x) + log

5

(2 + x) = 1

log

5

[(3 − 2x)(2 + x)] = 1

5

1

= (3 − 2x)(2 + x)

5 = 6 − x − 2x

2

2x

2

+ x − 1 = 0

(2x − 1)(x + 1) = 0

x = − 1, x = 1

2

(14)

(

There are no other solutions in

Question 7

a) by inspection of the diagram Area

Substituting into gives

b) i)

a 0

5sin3x dx = 10

3

[−

5

3

cos3x]

a 0

= 10

3

− 5

3

cos3a) − ( − 5

3

cos0) = 10

3

− 5

3

cos3a + 5

3

= 10

3

− 5

3

cos3a = 5

3

cos3a = − 1

3a = π,  a = π

3

0 ≤ a < π

L = 3x + 4y

= 24 = 2xy

y = 24

2x

= 12

x

y = 12

x

L

L = 3x + 4 ∙ 12

x

= 3x + 48

x

L(x) = 3x + 48

x

= 3x + 48x

−1

L′(x) = 3 − 48x

−2

(15)

Minimum values occur where

Since is a length it cannot be negative, so

gives a minimum value of

ii) Minimum length occurs where Minimum cost

= 3 − 48

x

2

L′(x) = 0

3 − 48

x

2

= 0

3x

2

− 48 = 0

x

2

− 16 = 0

(x + 4)(x − 4) = 0

x

x = 4

x = 4 

L(x)

x = 4

L(4) = 3 ∙ 4 + 48

4

= 24m

= 24 ∙ £8.25 = £198

(16)

Separating into two equations gives Question 9 a)

Since the concentration has halved

to 2 significant figures

sin2x = 2 cos

2

x

2sinxcosx = 2 cos

2

x

2sinxcosx−2 cos

2

x = 0

2cosx(sinx − cosx) = 0

2cosx = 0

sinx − cosx = 0

cosx = 0

sinx = cosx

x = π

2

,   3π

2

cosx

sinx

= 1

tanx = 1

x = π

4

,   5π

4

p

t

= p

0

e

−kt

p

t

=

p

2

0

p

0

2

= p

0

e

−25k

0.5 = e

−25k

log

e

0.5 = log

e

e

−25k

log

e

0.5 = − 25k

k =

log

e

0.5

−25

k = 0.028

(17)

b)

Let and Substitute to give Rounding gives So, is 11% of

Therefore, the concentration has decreased by 89%

p

t

= p

0

e

−kt

t = 80

k = 0.028 .

p

t

= p

0

e

−0.028 x 80

p

t

= p

0

e

−2.24

p

t

= 0.1065 p

0

p

t

= 0.11 p

0

p

t

p

0

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