349252033 Cape Pure Maths Unit 1 Solutions 2011 2016 PDF

CAPE PURE MATHEMATICS UNIT 1 CAPE UNIT 1 2016 SOLUTIONS Question 1

a.

i.

  2   . 3 3  23  3  3 0 54930 363 1 1 10 1 1  21 1  1 1  1 1 10 2110 13 250, 25 25 25 13,12   2   2512 3 2 74   3 2  2512 2  6 Given

If

----------

Also

is a factor then

(1)

then

------

(2)

Subtracting (1) from (2) we have

Substitute

ii.

into (2) we have

is a factor therefore using long division we have

7 25 7 21 412 412 0

2 74     32 74    3  421  0,3,  , 4 When

1

3 3  0.

SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016

b.

1,6 15 1 . 6 15   . 1 6+ 16 166  1  5  65  5  561 1   1, anand   1.  →  ˅ ∧ ˅ ˅ →  ∧  When

which is divisible by 5 therefore the statement is true for

Assume the statement is true for

When

Therefore

where

we have

which is divisible by 5.

Therefore the statement is true for Since the statement is true for

c.

p T T F F

ii.

q T F T F

T F T T

It is true for all natural numbers n.

T T T F

 →  ˅ ˅ →  ∧  and

T F F F

are logically equivalent because both have the same

truth values in their output column.

Question 2

a.

T F T T

log10 log   4 log10 10  4 10  2 10  16  10160   2  8  0 2,8 2

b.

  +− , ≠1.  ≠ , ≠,,≠1

Given the function

If the function is one-to-one then,

  31 ≠   31   3  1 ≠   3  1 33≠33 4≠4  ≠    , where  ≠ 1 −  + −   3   3 ( −)   1    13  31 1  333 31   44   ,  ,where ,≠1,   − <>    2  5  4  6  0 areare , ,a ndnd .   52  230        0  52 ,       2, and   3 ,

Therefore a and b are distinct and hence a maps to f(a f(a), and b maps to f(b f(b) For any

and

Therefore the function is a one-to-one and onto because for

c.

i.

Given the roots of the equation

3


c.

ii.

An equation whose roots are

 ,  and    

has

1  1  1         2     52 19  2 23 3 3  9 1  1  1       2  52 22 1  3 3   4 1 1    13  19  3   199   14   19  0 36 76 940

4

Question 3

a.

i.

Prove

 sec   −

1sin LHS  1 sin sin 1sin  1sin  sin  1sin1    cos1   sec  . ii.

Given

   − 

sec    ,cos     cos± √     6 , 56 , 76 , 116  sincos sinsincossincos cos1 and sin1 tan1,  ,and  √ 2 sincos √ 2sin 2 sin   is √ 2       this gives an acute angle

Therefore

b.

i.

Therefore Hence,

ii.

The max value of

This occurs at

Therefore the smallest non-negative value of

5

 is 


c.

Prove

tantantantantantan tan       1tantantantantantan tantan tantan tantan        1tantan tantan tan  1tan1tantan tantan 1tantan  tan 1tantan tantan 1tantantantan  1tantantan 1tantan tantantantantantan  1tantantantantantan Proven

Question 4

a.

i.

sin,sin    1sin cos  cos√1  sin tan cos  √1   Given

6

b.

tan2 and sin 2tan  , and tan     1tan √1  2    √1   1     √1 2   1√1  1  2   1 √1 1     21  √112  2√1   12 1 2   32  and    15 ||  √ 194 1 94 √ 1414 ||  √ 4125 4 125 √ 3030 cos |∙|||||  √ 2310 14× 14 × √ 3030 0.439 , ,  is 2 √4    √ 3   √ 3 anand      √ 3,3,   √ 3

ii.

Given

i.

Given

ii.

c.

At any time the point

from the origin and a from the xthe x-axis. axis. Therefore its

distance from the y the y-axis -axis is given by Hence,

.

7

using Pythagoras Theorem.


d.

230     9 23   ( (23 23)  9   4 1299 5 120 512  0 0,  0,3     ,2  3  0,3 and   ,  ------

(1)

-------

From (1),

(2)

-----

(3)

Substituting (3) into (2) we have

When When

Therefore the points of intersection are

Question 5

a.

∫  1/    1,     .   1/ /  /     4  3  34 /    34   1/  

Given

, Using the substitution substitution

8

1

we have

b.

c.

V   ∫−    1,     1/     1/     −  1/    35   1/ 10  35 cubic units     >0 ∫    ∫     −    −  −  (−−− −)   −   −      −   −  1    12 ln−  1 10   12 ln−  1  ln  1   12 ln  1ln| ln|  1|   12 ln|  1| lnln| lnln|  1|   12 1 1  12 Given

Dividing both numerator and denominator by

9

we have


d.

i.

Bacteria grow exponentially at a rate of 2% per hour where



is the

number of bacteria present t hours later is given by the differential equation

 0.02   0.02  ln0.02   .+   . 0,1000, 1000,1000. 2000, 20001000 . 2  . ln20.02,   0.ln022  34.6666 hrshrs   2  5 12   6 101 3, 3  63  103 183 3, 3,3  23  53  3 12112 3 112833 83137

Separating variables and integrating both sides we have

When

ii.

therefore

When the bacteria population is double

Question 6

a.

Given

The gradient of the tangent at the point where

is is given by

When

Therefore the equation of the tangent at a t the point where

10

is given by

b.

i.

ii.

 23 ≤0   { >0 l→im   0  20 33 lim   0  →   0, 0, →lim  lim →  3,    Given

For

to be continuous at

Therefore

iii.

Given For

and

.

+− 0 lim→ + 

≤0

lim→ 0    3 0  3 lim→ 33  lim→    >0   20    3  0  20  3   0     0 lim→    2 33    lim→    2  lim→  lim→ 22    0 ′0 0  2 2 For

If the

is differentiable at

then

Therefore

11


c.

  √  therefore   ℎ  √   ℎ  ℎ  ℎ  ℎ  √ √ √  lim→ √ ℎ × ℎ √ ℎ  ℎ √   ℎ   lim→ ℎ(√ ℎ  ℎ √ )) ℎ lim→ ℎ(√ ℎ  ℎ √ )) 1 lim→ (√ ℎ  ℎ √ ))  2√ 12 Given

12

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS FOR 2015 EXAM

1.

a.

i.

The inverse

ii.

iii.

~→~ ~→~ ~ ~  →  ~→~ and the contrapositive

p

q

T

T

F

F

T

T

T

F

F

T

F

F

F

T

T

F

T

T

F

F

T

T

T

T

 →  ~→~         5 5  5  5  5 0 1252550 25120………. 1   1 1  1  1  1 24 1124 24 ………. 2 24144, 6 624,30 and

are logically equivalent because both final columns

are the exactly same.

b.

Given i.

If

is a factor then

When divided by

the remainder is 24 therefore

Subtract (2) from (1) we have

13


ii.

    6 30

 6   5   6 30   5      5 630 630 0

c.

  6 30   5 6 6    5  3  2  55  5  5 ⋯5 .. and 4  5+  5 ...  1,..4 1,..4 1 4×520, and R.H.S5 520 1 , 4  45  5  5  5 ⋯5  5+  5 1,  4 4  1  5+  5  45  5  5  5 ⋯5  5+  45+ 4( 4()  45+  5+ 54 545+ 5×5+  5  5+  5 1.   1,  and   1,

Given When

Therefore result is true for Assume result is true for

When

.

therefore

we have

R.H.S L.H.S

R.H.S = L.H.S therefore result is is true for Since the result is true for

it is true for all positive integer n.

14

2.

a.

i.

A function is one-to-one if each element in the domain maps to one and only one image in the co-domain and each element in the range is the

::  → 

image of only one element in the domain. Therefore given that

:→  °  :→   °      and

are one-to-one functions,

 ° 

is a one-to-one one-to-one function

because the co-domain of f is used as the domain for g for g and this makes a one-to-one function.

b

ii.

A function is onto if each element in the co-domain is mapped unto at

::  → 

least one element in the domain. Therefore given that

:→  °  :→ and

are onto functions,

 ° 

is a onto function because the

co-domain of f is used as the domain for g, for g, and this makes

b.

i.

an onto function.

3      0

3  94  94  0 39  49  4  0 mult multipliplying ying both both sidesdes by 9 39  29  2  0 factori factorising the equation 39  2 0, 9   23 , not possible 9  2  0 9  2   log2 log9 0.315 15


ii.

c.

Given i. ii.

3.

a.

i.

|56| 56| 5 56>0 565 411    56<0 565 565 16    3005 0,3001301 3301 9033005 5 603   log603 hoursrs log5  3.9898 hou cos3cos2 cos2cossin2sin  2cos   1 cos2sincossin 2cos cos2sin cos 2cos cos2 cos21cos  cos 2cos cos2cos2cos  4cos 3cos when

we have

When

we have

When

When

we have

16

ii.

b.

i.

ii.

cos6cos20 cos64cos 23cos2 4cos 23cos2cos20 4cos 24cos20 4cos2 cos 21  0 4cos20 2  2 , 32 , 52 , 72   4 , 34 , 54 , 74 cos 210 cos 21 cos2±1 2  0,  22, 3 44 0, 2 , , 32 , 2 2 2 3sin24cos2 sin2 sin2cossincos2 cos3 and sin4 tan  , and  √ 3  4 tan− 0.927 and 5 3sin24cos25sin 3sin24cos25sin20.927 27   5 1 1  7     7  5  12   5 1 1 1   7   7  5 12 Maximum value of occurs when

Minimum value of occurs when

17


4.

a.

i.

:   √ 10cos3; 1 0cos3;  √ 10sin2 1 0sin2 :   4co 4cos s  3;   4si4sin n  2 Given

From

 : cos +√  ; sin −√ 

cos sin   1   3    2  1 √ 1010 √ 1010   3    2  (√ 10) 10) From

ii.

 : cos − ; sin −

 4 3   4 2  1   3    2  4   3    2  (√ 10) 10)   2  (√ 10) 10)    3   3    2  4   2  4    3 (√ 10) 10)    3  4    3 10 10   69 16 69 10 6916 69 101666 612    12

From (1)

……….

(1)

………..

(2)

…….. (3)

…………..

Equating (2) and (4) we have

18

(4)

     1   2  4  23  4416 494  4 14  0 4 1610  16±√16   8  16 16±48 √ 1515   16±4√   3.94, 0.0635   ,3.94 and   , 0.0635  0,30, 3 , 5,25, 2     3  4  5    2    694 694 1025 44 44    694 401004  1616 3  3 40101070     403   103   1073  0  25 107  400 20 5    3   9  3  9  3  0  104  20 5    3   3  9  √   ,  

Substituting

into (4) we have

Points of intersection are

b.

If the point

moves so that its distance from a fixed point

times the distance from the fixed point

This is the equation of a circle with centre c entre

19

is two

then;

and radius

.


5.

a.

Given

  si n     4 ifif ≠0,≠0 0 0, l→im sin lim→ 4   4 l→im sin sin   4 l→im sin lim→ sin  4 l→im sin  1 4  sin sin2 2  lim→   ℎℎ  lim→ sin2  ℎℎ sin2 22ℎ2 22ℎ2 2cos sin  2 2 lim→ ℎ 42ℎ   2cos sin ℎ 2 lim→ ℎ lim→ 2cos 2cos2ℎ 2ℎ ×lim→ sinℎℎ 2cos2 If f If f is continuous at

then

Multiplying numerator and denominator by a we have

b.

Given

Differentiating from first principles we have

20

c. Given ven   √ 12         and     − ×′  1 −       1   2 2  1     √ 2   2 (√ 1   )  2  2 1    √  1  √  1     2  2 1    1  √ 1      2   1  2/ 1   √ 1   i.

Using the quotient and power rule:

Multiplying both sides by x by x we we have

ii.

   1  2√ 1    1 1 √ 12     1    1  2/  21  −/   2  32 1  −/2   1 6/   1  6√ 1     1  32 √ 1  

From (i)

21


6.

Given i.

Since  √ 12     1 3   1 3  0   3 7, 7,     9 and 3   3 37, 9 33 37 9 379 416 4 49 5 4,5 AB is the line

AC is the line

and BC is the line

. The lines AB intersects intersects AC at the point A therefore the coordinates

of A is found by solving these equations simultaneously. ……. (1)

……. (2)

Substituting (1) into into (2) we have

From (2)

Therefore the coordinates of A is

AB intersects BC at the point B therefore solving these equations gives the coordinates of the point B.

37 33 337 37 3 9213 824 3

…… (1) …… (2)

Substituting (1) into into (2) we have

22

From (1)

  33 72

Therefore the coordinates of B is

3,2

AC intersects BC at the point C therefore solving these equations gives the coordinates of the point C.

9 33 9 39   3 2733 424 6 963 ……

(1)

…… (2)

From (1)

…. (3)


6,3   3 37   9 9        3     4  6 1  6 3    2 7 3 9 2  4  3  2 3 3 4  7.5   36 36  28 36  13.5  4 28 13 36 4 0,6   3 83 i.  3 83    4 3 0,66 6     4 36 Therefore the coordinates of C is

ii.

The area bounded by these three lines is given by

sq. units

b.

Given that

at the point

The curve passes through the point

The equation of the curve is therefore 23

therefore


ii.

At the stationary point

  0

therefore

3 830 31 31  3  0    ,and 3    ,   4   3   6     6.5 3, 3, 3 3  43 3  33 3  6 12  ,6.5 and 3,12  68    ,  68>0  ,6.5 3,3  63 8<0 3,122  Max 3,12 12   4 36 When

When

Therefore the stationary points are

When

Therefore

is a minimum point

When

Therefore

iii.

is a maximum point.

3

2 1 0 1 1 3 6

Min13 ,6.5

24



CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS FOR 2014 EXAM Question 1

a. p T T T T F F F F b.

i.

q T T F F T T F F Given

r T F T F T F T F

 →   →   →   ∧   →   T T F F T T T T

T T F T T T F T

T T F F T T F T

⊕       5516 ⊕       5516 ⊕⊕ ⊕  2⊕2    2   52516  8 4 10516      5414    1 1  0 1  1  1  51 4140 154140 5100 2 2,     2 2 54 542 2  14     2 56 1  therefore

ii.

a.

is commutative in R

We have

If

is a factor then

Therefore

,

b.

When

is a factor of

therefore using long division we have.

25


 6

  1   2 56      5    66 66 0   2 56   12 6    1  2  3   1,   2 and   3 1  3  5  ⋯  21   4  1 1,  1  1,   41  1  1 1  1  3  5  ⋯  21   4  1 1  + 4  1  1 1  3  5  ⋯  21  2  1  1 1  3  5  ⋯  21 21   4  1   4  1  2  1  1   21 2121 21  21 21  21 21  21 21  21 21  21 21 3 321 Therefore factors are

c.

When

L.H.S

and R.H.S

L.H.S = R.H.S therefore result is true for Assume result is true for

, therefore

When

R.H.S

L.H.S =

L.H.S

26

  L.H.S 21 2 53 3 53  21 23  1 3 23   3 1 21 2123 23   3 1 4 83 83   3 1 4  2}  3   3 1 4  1 43 43   3 1 4  1  1

1 1,,   1

L.H.S = R.H.S therefore result is true for

Since result is true for

and and


Question 2

a.

Given i.

  2 1,    −−   2    1  22  1  1  24  4  1  1  8  8 21  8  8  3   2  1  −  −  2     1  2 −   1 11  −   −   a.

b.

ii.

Since

27


b.

    35       33   3

Given that

We know that Adding

to both sides we have

    33 3 5    8  8     2    log+  log 3log 2 loglog 3log 2 log2log

[Log both sides]

c.

i.

   20

  2 10 [   ,  21   1 1,    1 , ,  0 log  1 log31  2

Multiplying both sides by

Let

ii.

  1 2log 2 log 31   1 log 2 log 31 31 1 4 1431 1124 113    113

28

 

d.

−  √ + +  √ − −  √ + + √ − + √ − − √ + + √ − − √ +

(√ 3  1)  (√ 3  1)  (√ 2  1)  (√ 2  1) (√ 3  1)(√ )(√ 3  1) (√ 2  1)(√ )(√ 2  1) 4  2√ 342 33 42 3 3  2 232 2 32 2 √ √ √  1 21 82  61 4610

Question 3

a.

i.

ii.

⁄ ⁄− −   + ⁄+ ⁄

cossinsincos  sincossincos / sinsin sinsin sinsin  sincossincos × sinsin cossinsincos sincossincos  cossinsincos  sisinn   − 1, 0≤≤2, + sin , sin−    ,cos  √  for 0 ≤  ≤  cotcot sincossincos  cotcot cossinsincos sincossincos cossinsincos  1 When

−√   1 √ +

Multiplying both numerator & denominator by 2

cos √ 3sin 3sin  1 3 sincos √ 3sincos cos √ 3sin 3 sin √ 3sincos 3 sincos 29


b.

i.

0  2√ 3sin 3sin sin0   0, , 2 for 0 ≤  ≤ 2.  3sin24cos2 sin2 sin2 sin2cossincos2   √ 3  4  5 3sin2 n 2  4cos2 4cos2   sin2co n 2cos   sinc n  cos2 cos3, sin4 sin 4 tan  cos   3 tan− 0.9 27  sin2 5sin20.927  Given

to be written in the form

We have

Comparing we have

rad

ii.

a.

is at a minimum when

20.927 27  32 2  32 0.927   34  0.9227

  1.89 rarad

b.

The maximum value of

1 1 1   7 7  5 2

 − −

is when

And the minimum value of

1 1 1   7 7  5 5 12 30

 −

 5 so

is when

  5

Question 4

a.

Given the equations of i.

  10 50 240 2 2, 523 and

  10 50 and

are

and

intersects at the centre of the circle therefore

Solving the equations simultaneously we have ------ (1) ------ (2)

Adding (1) and (2)

When

from (2)

5

Therefore the coordinate of the centre of the circle is (2, 3)

ii.

Let A (1, 2) and B (a, b) be the coordinates of the endpoints of the diameter of the circle and the coordinates of the it’s centre it’s centre (2, 3) is the midpoint of the line AB. AB. Therefore in calculating the midpoint we have

 2 1 2, 3  2 2 3 4

Therefore B has coordinates (3, 4)

iii.

√ 2   2    3  (√ 2)2)  44 692    46110

The point p point p moves moves in a circular path with centre (2, 3) and radius The equation of the path of p of p is given by

31

.


b.

Given

  + ,   −  and

  1 1  1    1 1 1   1     1    1  1 1     1 1  × 1   , subs substittitututinging   1 1  we have   1  , substituting   1   we have 1       1  1        11    1      21

c.

i.

3,2,1 3,2,1,1,,5 2,1,4 ⃗  ⃗  ⃗ 3 1 4 ⃗   21    5   2 4  ⃗ 4 2     4 ⃗  ⃗  ⃗

Given

and

32

⃗   15   421   193  ⃗ 3 3 1     9 ⃗  ⃗  ⃗ ⃗   421   231   315  ⃗ 35 ii.

Given PQ is the hypotenuse therefore RQ and RP are perpendicular to each other.

 ∙   0 3 3  1     9 9 ∙ 35 35 33 33  1  95  0 333450 345 ,

  45 3  15

Question 5

a.

Given i.

<3   {2,   , ≥3 lim   3  9 → lim   3 3  2 →  3 lim   →lim  → 932 62, .

If

is continuous at

, then

Therefore

  13

33


2  i .    4 Given that

lim 2lim→  we have →

2 2 1 0 2 1 14  0 04

2  3 5  24 6  5  12

125 7 b. Let   √ 1 ,   ℎ  √ 1 ℎ

Using differentiation from first principle, we have

 lim→   ℎℎ   lim→ √ 1 ℎℎ  √ 1 √ (√  √ ℎ)(√ ℎ) lim→ ℎ lim→ ℎ(√ √ √ ℎ)(√ ℎ) lim→ ℎ(√ √ √ ℎ)(√ ℎ) × √ √   √ √   ℎℎ       ℎ lim  → ℎ(√   ℎ)(√ )(√   √   ℎ) ℎ lim  → ℎ(√   ℎ)(√ )(√   √   ℎ)

34

 lim→  1 (√   ℎ)(√ )(√   √   ℎ)  (√ )(√ 1)(√   √ )   21√    12 −/ ii.

Given

  √ ++

   ,   − ,  √ 1     1   12 1  −/  2√ 11 

Using the quotient rule when We have

1   1   1    ( ) √   2 √ 1   (√ 1  ) 1    √    1 1 2√ 1   ( 1   )(2 )( 2 1   )   √ √   2√ 11   (√ 1  )(2)(2√ 1  )   2√ 1  1     21    2 1       2  2 1  

c.

Given

cos, θ sin,

sin  cos

  / / cos cot  sin

35


Question 6

a.

i.

a.

Given

  3 41 

Integrating both sides we have

3 41 + +    3 2  1  4 1  1 

    2  1, 4 4 4  1  21  1   44 0     2     0  3 410 31 31  1  0    ,or 1    ,   2    When

b.

At the stationary points

therefore

When

4   271  29  13  169  27 27

Therefore coordinate of the stationary point is When

1,1  21 10

 ,  

Therefore coordinate of the stationary point is (1, 0 )

36

 64

  ,   6 42  < 0 ,      1,   61 42  > 0 1,01, 0    0,   0 0,0,0 0, 0,   2 0  210 210   1  0 0,0,1 0,00, 0 and 1,0 1, 0 When

Therefore

is a maximum

When

Therefore

ii.

is a minimum

y-intercept when

x-intercept when

When

x-intercepts x-intercepts

 max13 , 274 

427 0

min1,01, 0 23 1

13

37




b.

i.

2√ 1    ∫ 2√ Using the substitution method with

  √ 1  

   1  −/ ×2     √+   √ 1  ,    ,  

We have

3, √  √ 1  3  √ 1010 0,  √ 101 1 01 √      2    √   2 √ 10  2  3  110

When When

 21 2 10) 1 0 ( ) √  3  3   23  1010 120.42

ii.

Volume of revolution about the x the x-axis -axis is given by Volume

  ∫(2√ (2√ 1  ) 

   4  1        4   4  4 4 2

  3  5 0  42   4 2    3  5   0   323  1285   54415  cubic units 38

 ∫  

therefore from b. (i)

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS FOR 2013 EXAM

Question 1

a.

i. p T T F F

q T F T F

p T T F F

Q T F T F

ii.

b.

Given

→ T F T T

 ∧  ~ ∧  T F F F

F T T T

⊕   25 2⊕2   2252 2⊕4 410 2⊕ 98 2⊕0  980   1  8  0 1,8

39


c.

When

1, 5 38

which is divisible by 2

1  5  3 1,we have 5+ 355  3  12 5  3 5+  3   1, ,an,and   1     9 16    1 1 1  0 1 1  91  1 160 60, 6 Therefore statement is true for

Assume statement is true when

Therefore

is divisible by 2

When

is assumed to be divisible by 2 and 12 is divisible by 2

Therefore

is divisible by 2

Since the statement is true for

d.


Given i.

If

is a factor factor then

ii.

therefore

 1016

  1   9 616

   10  6 10 10 1616 1616 0

iii.

  9 616   12 1016    1  2  8   9 616 616   1  2  8  0 1,2,8

Therefore

40

Question 2

a.

Given

   , ≥1            

By completing the square we have

The function has a parabolic shape with axis of symmetry The domain given is

≥1,

  .

and this section of the graph is a one to one function and

due to it passing the horizontal line test.



  12

      

1 0

12

    ≠           0           0   1 1  0   , or   1  1 ≥1     1   ≥1

Alternatively, if we assume that

Then So

where

is not a one to one function when

.

,or 

Therefore For

which eliminates

,

Therefore the function is one to one for

41

.

is only true true if



.


b.

i.

Given a.

 32,and    32  and    3  2, therefore    3 2 Let

interchanging

−   3 2

    and     lnln ln2    ln  −   ln    3  2  3  2 3  2  and  3  2 Let

interchanging

we have

Natural log of both sides gives.

Therefore

b.

ii.

interchanging

  −

we have

Natural log of both sides.

ln  ln 3 2

2ln 3 2 , ther therefeforeore    12 ln  3 2 [ ]−  12 ln  3 2

  12 ln(−) − −  12 ln  3 2

Therefore

[[]]−  − − 42

we have

c.

i.

3 41≤5 3 44≤0 32  2 ≤ 0 2,  ≤2 2≤≤ 23  ≥ 23 32      2    32  2    2≤≤ 

Critical points

Therefore

from the table table is negative or from the graph

is the part of the graph that is below b elow the x the x-axis. -axis.

ii.

|  2| 35   2  35 35  449 3025 8 26210 8 1214210 4 423 23  723 23  0 4723  0 Squaring both sides we have

   74 ,not possible  32 only answer

Alternatively, for

For

2

  2 ≥ 0

we have

  2 < 0,

235

23,  32

  2 35 235

we we have

47,  74 not possible 43

23

2≤≤ 


Question 3

a.

i.

L.H.S

 + 

substituting

 tan  =

sin 2 2tan We have 1tan   1  cos  sin  cos

Multiplying denominator and numerator by

cos 

2tan 2sincos    cos sin 1    1tan  cos sin  2tan 1tan  2sincos

ii.

2sincossin2 sin2tan0 Given

2tan 1tan  tan0

2tantan 2tantan 1tan   0 2tantantan   0 tantan   0 tan 1tan   0 tan1tan1tan  0 tan0,1,1 tan−0 0,,2  tan−1   0, 4 , 34 , 54 , 74 Acute angle for

44

b.

i.

 3cos4sin cos cos   coscossinsin cos3, sin4 Given

sin 4 tan cos 3 tan− 430.927

ii.

   3  4  5 3cos4sin5cos 3cos4sin5cos0.927 27  5cos0.927  1≤cos≤1 1≤cos≤1  +  a.

Therefore maximum value of

b.

Minimum value of

is 5

is when

is maximum

1  8 1 5  131 Therefore 8

iii.

a.

Given that A, B and C are the angles of a triangle where their sum is π.

              sinsin sinsin      sincos   sin sin    sincos sin   cos sin0, cos1 sin     0sin   1  sin   sin    sin sinsin   sinsin   sinsin sinsin    sinsin sinsin    sinsinsinsin   sin    sin    Taking sine of the angles

Therefore

b.

from (iii).a.

therefore

and

so

45


Question 4

a.

i.

Given

   6440

By completing the square we have

 6    4  6 4   6 2     2    4 2     2  40

 69  9   44 440   3    2 93                , ,  3,23, 2 √ 9  3 3,23, 2 22 6,26, 2 63  0 6,26, 2 2 2 6,26, 2   6    ,24 The equation of a circle is given by

Where

is the centre and r its radius.

Therefore the circle has centre

ii.

a.

and radius

The gradient of the line between the centre

circumference

is given by

,

Therefore equation of the normal to the circle at

b.

From 24,   2 4 Substituting

at

, which is a vertical line parallel pa rallel to the

y-axis. y-axis.

Given

is given by

The tangent is perpendicular to the normal therefore the line is perpendicular to

b.

and a point on the

  +      into

 4   4 We have  2   2  816   4   4  2  81624   4  1024   4 

4 1024

46

c.

i.

ii.

3,1,2, 1,2,4 and and 1 1,, 1,2 1, 2 ⃗   ⃗  ⃗ 3 1   12   42  2  63  ⃗  236 ⃗  ⃗  ⃗ 1 1   42   21  2  12  ⃗ 22 168  ∙0 ∙0 0 2 ∙16  ∙   3 8 6  0×2  16×3  8×6 048480 ⃗  0 2 ∙16  ∙   1 8 2  0×2 0×2  16×1 16×1  8×2 8×2 016160 ⃗ Given

Given Then

, if

is perpendicular to the plane through A, B, and C

and

Therefore r is perpendicular to

Therefore r is perpendicular to

47


iii.

The vector equation of a plane is given by

∙∙

 168 32 ∙∙   ∙ 168 168  32 32 ∙ 168 168 1681616 1680 20 where r is any vector

on the plane, n is a vector normal to

the plane and a is the position vector for a point on the plane. Using

and

we have

Question 5

a.

Given i.

ii.

<2   {2, , >2  l→im  lim   →  2  4 lim  lim → →   2 224 lim  lim →  → lim   4 →  2 .

is not continuous at

because

48

2

is undefined.

 23  b. LetLet     2

−       ,   

Using the quotient rule

   23  22     2   3  2 × 2   6 6  2   2          2 × 22 23 × 6  2  23       2  222  6    2  12 18   2  2 446     2   144   4 10   2 13cos, 2sin  03sin3sin  2cos Let

c.

Given

  / /   2cos 3sin   23 cot

d.

i.

    3 4 4  3  430

------ (1)

------

(2)

Substitute

4

into (1)

49


  1  3  0   1,or,or 3 1,4 1,41  4 3,43  12 1,41, 4 and 3, 3,12 12 When

ii.

Area of the shaded region is given by A

    3     4    3  3      2 1   3 3 1 3 

  23   2   3 33 33  13 3

  16 16  18 18  3    sq. units

Question 6

a.

i.

∫ 1    1, 1,  1, 

Let

1   1 1            4  3    1       1        1     4  3  

  121 11  13 13  50

ii.

Given

 2cos,  4sin53cos

      2c 2 cos s  4si4sin5 n 5  3co 3cos  5cos4sin5     5sin 45 cos5          2c 2 cos s   4s 4 sin5 n 5  3cos 3cos  2sin 45 cos5 cos5 3sin           5sin n   45 cos5 A, B, and C are merely constants of the integrals therefore

b.

i.

Length of rectangle is x, is x, width width of rectangle is 2r 2 r and and length of semi-circle is

   

Perimeter of track is given by

ii.

22600 2   6002   6002 2

Area of track is given by

  22      2 2   

 6002  6002   2 2  2     2  2      6002  12004    2     2  2      2 1  12008 22  2 26002 60022 2   2 1  12008 22  6002 6002 51




When  0, 2 1  12008 22  6002 6002  0

12008 22  6002 6002  0 2  12008 12008  2 26002 6002  0 24001612008120040 24001640 1642400 4 44   2400 4  42400 

  + ≈ 84 memetretress

   2 1  822  2 2  2 1  24   8   4  8 > 2   Therefore  < 0

  + sin2cos   cossin  2 sin    cossin2sin sincos  cos cossin coscossin ′′sin

Therefore

c.

i.

Let

give the maximum area.

52

ii.

sin2cos When 0, 1, 12 3 ,6, 3 6sin2cos3 623 1    sin2cos    3 We have

When

we have

The specific solution is

53

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS TO 2012 EXAM Question 1

a.

Given

i.

  2   10 1  1  0 1  21  1  1 100 2100 8  1 6 1 1  21 1  1 1  1 1 106 2106 6 214, 7 7 76, 1   2  7 10 is a factor of

therefore

------- (1)

When

is divided by

it gives a remainder of

-------- (2)

Adding equations (1) and (2) we have therefore

Substituting

into (2) give

therefore

ii.

2 910

  1 2  7 10 2  2 9   9  9 1010 1010 0

54

therefore

2  7 10   122 910 25    1  225    1,   2, and 25 (√    )) 16 √ 240 240 (√    )) 2    2 16  16 √ 240 240 16 16 2   √ 240 240 240 √ 4×602√ 4 ×602√ 6060 √ 240√ 2 2√  2√ 6060 60 16  60 16  60  16600   610 10  0 6,10 6, 16610 10,16106 6,10 and 10,6 Therefore the factors of

b.

Given

Therefore

are

--------

(1)

------- (2)

Therefore

--------

(3)

------

(4)

--------

(5)


When When

55

c.

|37| 37| ≤ 5 37 37 ≤ 25 9 4249≤25 9 4224≤0 33 148 ≤ 0 332  4 ≤ 0    , 4


Given

Squaring both sides we have

|37| 37| ≤ 5 23

4

Critical points

 ≤ 23 23 ≤≤4  ≥ 4 32      4    32  4     ≤≤4  37 ≥ 0 37≤5 3≤12, ≤4 37 37 ≤ 0 37 37 ≤ 5 37≤5 3≤2  ≥   ≤≤4  |37| 5≤0 |37| ≥ 0

Therefore

Alternatively, when

we have

When

we have

Therefore ii.

Modulus always give the value of o f the function as positive and

5 is also greater than zero, therefore if we add the two together a number nu mber greater than zero will be the result.

Therefore

|37|  5

cannot be less than zero for any real value of x of x and this

function will not intersect the x the x-axis -axis resulting in us having no real solution. 56

Question 2

a.

Given i.

ii.

 →   3 (()      3      3   3    6 93    6  6 (()3   6  6    3  3   6 6 693   6 6 66   7 6 6 76 76 1  76 1 1  71 1 61760 2  76 2  72 681460 3  76 3  73 6272160   7 6  1  2  3 0,1,2,3    4 310            4 310        0           2        2    2    is a factor of

is a factor of

is a factor of

Therefore

b.

i.

Given

are the roots the equation

Dividing both sides by 4 we have

Therefore

ii.

57


iii.

  Sum of roots Given

and

 

are the roots of a quadratic equation we have

2  2  2 2      2    

1  18 2  16  1   1  4 16

2

4 Product of roots 2 × 2    14 4

4×1664  2640 log log log log log   log  ×  ×  ×  ×   log  log 10−  1 Therefore the quadratic equation is

c.

i.

ii.

  1 2 98 99 ∑log   log  log  ⋯log  log          1 2 3 99 100 = log 12 × 23×…×9899 10099  log 1001  log 10−

 2

58

Question 3

a.

i.

ii.

cos    coscossinsin cos22cos   1 cos3cos cos3cos2 2 cos2cossin2sin  2cos   1 cos cos 2sincos 2sincos sin sin22sincos sin22sincos   2cos   1 cos2sin cos cos cos 2cos 2cos   1 2sin  2coscos sin      sin6sin2 sin6sin2 sinsin2cos+ sin−   sin6sin2    2cos+sin−     cos4sin2 cos42cos 21 21  2cos 21 sin2 sin6sin20 2cos 21 21 sin20 2cos 210 Given

and

L.H.S

Using the factor formulae

iii.

cos 2  12 cos2 ± √ 12 0≤≤ 2 2  4 , 34   8 , 38

sin20 20, 0, 0,

0≤≤ 

2 2 , 8 , 38 59


b.

Given

cot cos0  cos 2 sin  cos0 2cos sin cos0 2cos   1cos  cos0 2cos coscos   0 cos 2cos cos0 cos cos cos 2cos1 2cos1  0 cos0 cos 2cos10 Using the quadratic formula

−±√ −   −±√ 

    1    2± 2± 2 2  4 1 1 cos 2 cos 2 ±2√ 8, √ 8  √ 2×42√ 2 ×42√ 2 cos 2 ±22√ 2  1 ± √ 2

Question 4

a.

i.

cos1 √ 2,2 , cos≠1 √ 2 cos0,1 √ 2 3sec,and 3tan sec 1tan  Given

sec 3 , tan 3 3 13 9  1  9

 9

60

ii.

 9   √ 10 10  10 109   1090   1  9  0 1,9 1,   √ 1010 (1,√ (1, √ 10) 10) 9, √  √ 903√ 9 03√ 1010 (9,3√ 10) 10) 34 6  34  6 22 ∙ 34 34 ∙ 6 6 32427 ∙ || × || cos ------ (1) ------ (2)

--------- (3)

squaring (2)

Substitute (3) into (1)

When

point

When

b.

i. ii.

iii.

iv.

point

and

cos || ×∙ ||

||   3  4  √ 255 2 55 ||   1  6  √ 3737

cos 5 ×27√ 3737

cos0.888 cos− 0.88827.4 61


Question 5

a.

i.

Let

    +−



is discontinuous when

Therefore

ii.

2,2

 40

+  lim − →−

  8    2 24

 24       2  l→−im   224   2  24   22      24 2 2 l→−im   2  222  4  412  3

ii.

l→im +

22 4 lim sin2 → 2 2 l→im sin2 2 lim   2  0  2 → l→im sin2 2 1   4 2 lim sin2  2 → 62

b.

Given i.

  1, >1,   {4, <1. lim   1 12 → lim   4  1 → lim  →lim   →lim  → 42 2 1  2  1    a.

b.

When

exist

Therefore

ii.

c.

Given

For to be continuous at

 2 2 2−  2 2

1,1 1   , 2   When

therefore

------ (1)

When

therefore

354  2 22  22 354 4 4

3516 3417, 2 2 12, 3 2,3 ----- (2)

Add eq. (1) and (2) we have

From (1) when

we have

Therefore

63

.


Question 6

a.

i.

Given

√4  7

  12 4  7−/ × 8   4 √4  7  44  7   4

√4  7    4

ii.

    √  −

using the quotient rule

−       ,   

4,   4    44  7   √44  7   √4  74 4√4 4  7 (√4  7) 4(√4  7)(√4 444 )(√4  7)  4 √44 77    7 16   4 4  4  7√4  7  2816  16  4  7√4  7  4  28 7√4  7 64

  44  7    428  7     428  7    4   √4  7  16  4  7       428  7  416 7   28 16  4  7   7  4 4  4  7  4 Therefore

b.

i.

         4

  3  6  Integrating both sides we have

3  6      3   1, 0 0  1  31   013 4     3  4 When

65


ii.

  3  6  At the stationary points

  0 

therefore

3 60 3  2  0 0,2 0, 4 0,40, 4 2,2  32  4 81240 2,02, 0 0,40, 4 2,02, 0 When When

Therefore the stationary points are

iii.

 66  When

0,  6 0,40, 4

Therefore When

is a maximum

2,

Therefore

2,02, 0

  62 66  is a minimum minimum

66

and

iv.

0   3 40 0,2    2    2      3  4 44, 1   3  4    2  1 2,and1 1,0 and 2,02, 0 The curve meets the x-axis when

therefore

The minimum point has

Therefore

is a factor

Equating the constants we have

Therefore the curve meets the x-axis at

v.

 4

 0,4

2

1,0 2 1

67

0

2

2,02, 0

 2,0 

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS TO 2011 EXAM

Question 1

a.

(√ 75 75  √ 12) 12)  (√ 75 75  √ 12) 12) [(√ [(√ 75 75  √ 12)(√ 1 2)(√ 75 75  √ 12)][(√ 12)][(√ 75 75  √ 12)(√ 1 2)(√ 75 75  √ 12)] 12)] (2√ (2√ 75)(2√ 75)(2√ 12) 12) (2√ (2√ 25×3)(2√ 25×3)(2√ 4×3) 4 ×3) 2 × 5√ 3×2×2√ 3 ×2×2√ 3 40×3120 27/ × 9/ × 81/         3  × 3  × 3   3 × 3 × 3 3++ 3  9       i.

ii.

b.

x ) ( x

(0, 4)

Q

0

1

68

2

x

i.

0,0,  4 0,1,2 1 140 5 2  2  2  2 40 4212 26 1 1 1 15 4      44    44   1  2   4  1 122 ×  42 2 2 20 2 2 When

therefore is the y the y-intercept -intercept

From the graph the y the y-intercept -intercept is where the curve cuts the y the y-axis. -axis. Therefore

ii.

From the graph when

therefore

------ (1)

----- (2)

---- (3)

dividing (2) by 2

(1) Subtract (2)

From (1) when

we have

So

iii.

Equating the constants we have

Therefore the third factor is

The x The x coordinate coordinate of the point Q is

69

.


c.

i.

Given

 log log √   log log  log   12 log  log         14  4  40   4  0 0,4 0,log 0   2  1 4,log   4   2  16 1,16   || 12<0 , ≥0 ||  , <0  12<0, for ≥0   3  4 < 0 3,4 Let

therefore

Squaring both sides we have

When

When

Therefore

ii.

Given

Because of We have

Critical points

70

 12 3

4

4

 12<0

 12



, has inequality for values of between

3

3<<4

 12<0, for <0   3  4 < 0 3,4  12<0

We have

Critical points

, therefore for

 4<<3

, has inequality

for values of ,

Taking the union of both sets we have inequality for values of

Question 2

a.

i.

 and   240      0,       24,        2    224 24    48    33,  4833  810   9  9  0   9, or 9

Given

are the roots of

a.

b.

ii.

Given

71

, 4<<4


b.

Given i.

ii.

iii.

c.

23 23  2 2  3 0  6 0, 3  2 20  3  2  6  3 12315 2, 23  3  23  3 6  3  215  3 9 30333 3 23  3  23  3 63 63  2 23 3  3 3  23  3 323 3 3 3  3 2  21    1  221 2 221 21  21 1 and

When

we have

When

we have

When

we have

An even number can be express as

where is an integer.

A odd number can be express as

where is an integer.

For two consecutive numbers one must be even and the other odd, therefore The product of two consecutive integers can be Where

.

.

Two times any number makes it even. Therefore the product of two consecutive consecu tive integers is an even integer.

72

d.

Given to prove that

1 11  5  6

When

  5

is divisible by 6

we have

which is divisible by 6

1 

Therefore the statement is true for Assume statement is true when

, therefore

     5 1   1  1  5   1 215 215   1 26   2 6 26   3 86  53 36   5  3 3  1  6   5     1 1 1, and 1, is divisible by 6

When

We assumed

we have

is divisible by 6,

is an even integer which when

multiplied by 3 is divisible by 6 and 6 is divisible by 6.

Therefore when

the statement is true.

Since the statement is true for

it is true true for all all positive integer integer n.

Question 3

a.

      and       with || 13 and ||  10 ||       13 and and ||       10    ∙             ∙                          13  10 169100  69 i.

73


ii.

2    11 anand so    2  11    211 and    211    311     11     3     11 and          11     3  11 113  and       11 11       ∙     3 3  11 11  11 11 3 ×   69  3 44 1213  69 3(  )44 520 ||   ((  )  10     100  3100 100 44 520 44 352   8    100  √1008  ±6   8  6 or 8  6 211   286 11512   286 86 11512 10    2150    220 0,10, 1 10   0 and   1 Therefore

b.

i.

Given the line L has equation

and the circle C has equation

The general equation of a circle is given by

where f and g are the coordinates of the centre of the circle.

From the equation of the circle the coordinate of the centre is Therefore from the line equation

74

when

We have

0110

therefore it is shown that the line L passes p asses through the

centre of the circle.

ii.

L intersects C at P and Q therefore we solve simultaneously the equations of L and C.

   2150 10 1     1  2  1 150    2122150 2 160   8   ±√ 8±2√ 8 ±2√ 2 1±2√ 2 (2√ (2√ 2,12√ 2 ,12√ 2)2) and (2√ (2√ 2,12√ 2 ,12√ 2)2)    c  cos and     sisinn  -------- (1)

-------- (2)

From (2)

------- (3)

Substitute (3) into (1) we have

The coordinates of P and Q are

iii.

Given the parametric equations

cos − sin −   − −   cos sin   1         1 We have

and

so

                   1 1150     1 164   0, ,  1, and   4

The equation of C written in this form is given by

Therefore

75

.


iv.

 ,,           16  0    1    16   1    16 1. Let the circle

The circle

has centre

and radius 4 therefore



has equation

touches the line L at the centre of C (0, 1) and has the same radius

Therefore

Also the line through the centres of the circle is perpendicular to the line L therefore it has gradient

   01  1

1   1     1      2150 1    1    16 21    16 1      8 1±√ 1±√ 8±2√ 8 ±2√ 2 12√ 12√ 2, 2 , 12√ 12√ 2   1     1  (1 (1  2√ 2)2 2 )2√ 2   1  (1 (1  2√ 2)2 2 )2√ 2 (2√ 2,12 2 ,12√ 2)2) and (2√ 2,12 2 ,12√ 2 )   (2√ (2√ 2)2) (12√ (12√ 2)2)  16 (2√ 2)2) (12√ 2)2)  16   (2√ (2√ 2)2) (12√ (12√ 2)2)  16 (2√ (2√ 2)2) (12√ (12√ 2)2)  16 Substituting

into

we have

Therefore the centres are

The possible equations are

and

76

Question 4

a.

i.

8cos 10cos 30 cos  8 1030 8 6430 243  43  0 21 2143 43  0    or  cos    , so cos± √  cos− √    0≤≤     or     cos    , so cos± √  cos− √       or       4 , 34 , 6 or 56   90    6 co cos , so  6co 6cos  8 sin, so 8sin 6cos8sin Given Let

then

The acute angle is

therefore is in the first and second quadrants .

The angles are

for the second quadrant.

The acute angle is The angles are

b.

i.

From the diagram angle

triangle

for the second quadrant.

therefore triangle

and as a result angle

77

 .

is equal to angle

is similar to


ii.

|| |  7 6cos8sin7 sin sin   6cos8sin sin sin   sincossincos 8cos  6sin Given

then

Using

We have

Therefore

sin 6 tan cos 8 tan 68  34

tan− 36.87 or 0.644   √ 6  8  10 6cos8sin10sin 6cos8sin10sin0.644 44 10sin0.644  7 sin0.644 44   0.644sin−   0.6 4444.4 2 or 0.775 0.7 750. 644   0.131 131 radrad or 7.7.55 6cos8sin10sin 6cos8sin10sin0.644 44 sin0.644 || 15 rad

rad

iii.

Therefore the maximum value of BC is 10 because maximum value of 1. So

c.

i.

is NOT possible

−     

sin tan  cos

78

has a

ii.

a.

−     

sin2 tan2  cos2

b.

−     

sin3 tan3  cos3

iii.

From the above identities it can be seen that

1cos2 sin2 tan

1cos2tansin2 1tansin2cos2   ∑tansin2cos2  ∑ 1   = = Therefore

Question 5

a.

++  lim −− →−

l→−im   22  33 1 l→−im   33  23   23 5

b.

Given

i. ii. iii.

 1 if ≥2   {1 if <2 2  2 15 lim   →lim  1  2 15 → lim  lim → →1  21 79


iv.

c.

Given

2 lim   →lim  → 521 42 > 2    32 1,21, 2   7  3 23  1,  7, 7323 432 1,2 232 3 622 10 3 310 > 13 10 13 32   7  7 2    1 7141 715 1 1 1,0 715 0,15 15114 if f is continuous at

then

and at

We have When

------

---------

When

-----

(1)

(2)

(3)

--------

(4)

------

(5)

Multiple (4) by 2

Subtract (5) from (2)

Therefore the equation is

ii.

The gradient of the tangent at T is

Therefore the gradient of the normal is

The equation of the normal is given by

iii.

The line

cuts the x the x-axis -axis at

, therefore coordinates of M is

the normal has equation

when

therefore coordinates for N

,

M and N is on the x the x-axis -axis therefore length of

80

Question 6

a.

i.

    12,  12   3  12   0, 3  120  3 40 2,2 2,2 2,22  12 12 16 2,22  12  16 2,16 16 and 2,166 0,  12   0    0          12 0   12 12  0 0,±√ 12±2 1 2±2√ 3

Given

We have

At the stationary points

When When

Therefore the stationary points have coordinates

ii.

At the origin

this is the gradient of the tangent.

The gradient of the normal is therefore

origin is given by iii.

The curve

, and the equation of the normal at the

,

cuts the x the x-axis -axis when

.

therefore

, and

The area between the curve and the positive x positive x-axis -axis is given by

√        12 12 √     12    6  4 √ 01212 12)  (√ 12)

  6(√ 12) 12)  4  612 12  1444  36 sq.sq.unitsts 81


b. i.

ii.

  Using the resul result        We havesin      sin      sin   sin therefore sin      sin    a. sin      sin     sinsin sinsin    πsin sin      sin sin       . sin sin sin      2sin sin   

 cos cos0 cos  cos  cos cos0 [(1)  1]

 2sin 2   sin  

82

349252033 Cape Pure Maths Unit 1 Solutions 2011 2016 PDF

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