Name : __________________________________

Form : ________________________ R2011_1

1(a) Diagram 1 shows the relation between two sets of numbers

P = { – 2 , – 1 , 0 , 1 , 2 } Q = { – 1 , 0 , 1 , 2, 3, 4 }

Diagram 1 Based on the above information, the relation between set P and Q is defined by the set of ordered pairs: [ { – 2, – 1 } , { – 1 , 1 } ,( 1, 2 ) , ( 2, 3 ) ] State ( i ) the object of 1 (ii) the image of 2 (iii) the type of relation between set P and Q [ (i) – 1 (ii) 3 (iii) one – to – one

1(b) Diagram 2 shows the relation between two sets of numbers

P = { d , e , f } Q = { 2 , 4 , 6 , 8, 10 }

Diagram 2 Based on the above information, the relation between set P and Q is defined by the set of ordered pairs [ { d, 4 } , { e , 4 } ,( f, 10 ) ] State (i) the domain (ii) the range (iii) the type of relation between set P and Q [ (i) {d,e,f } (ii) {4,10} (iii) many to – one

1(c) Diagram 3 shows the relation between two sets of numbers

(i) Express the relation ordered pairs. (ii) the object of 1 (iii) the type of relation [ (i) (1,1],(2,2),(2,3), (3,1) (ii){1, 3} (iii) many – to – many )

*1(d) Diagram 4 shows the graph of f(x) = | 5 – x |

y

0 5 x Diagram 4State,

i) Domainii) Rangeiii) Type of relation

[ (i) domain – 2 < x < 6 (ii) 0 < f(x) < 7, (iii) many to one ]

Halus_1 1

1 2 3

Set Q

Set PDiagram 34

3

2

1

(–2,7)

( 6,1)

2(a) Given that f : x = 5x + 1 , find ( i ) f –1 (x) (ii) f –1 (11 ) (iii) f –1 (16) = p (iv) f –1( 2w+1) = 4 [(i) ( x – 1 )/ 5 (ii) 2 (iii) p =3 (iv) w =10 ]

2(b) Given that f : x = 4x – 7 find ( i ) f –1 (x) (ii) f –1 (1 ) (iii) f –1 (5) = p *(iv) f –1( 3w+3) = w [(i) ( x +7)/4 (ii) 2 (iii) p=3 (iv) w=10 ]

2(c) Given that f : x = 6 – 5x , find ( i ) f –1 (x) (ii) f –1 (– 4 ) (iii) f –1 (– 9) = p (iv) f –1( 6 – 2 w) = 4 [(i) ( x – 6 )/ –5= 6 – x / 5 (ii) 2 (iii) p=3 (iv) w=10 ]

2(d) Given that f : x = 4 – 3x find ( i ) f –1 (x) (ii) f –1 (– 2 ) (iii) f –1 (–5) = p (iv) f –1( 14 – 4 w) = w [(i) ( x – 4) / – 3 = ( 4 – x )/3 (ii) 2 (iii) p=3 (iv) w=10 ]

Halus_1 2

3(a). The function f is defined by f : x 3x – 2 and g:x 5x + 1. Given the composite function fg:x px + q, find the values of p and q. [ p = 15 , q = 1 ]

3(b). The function f is defined by f : x 7x + 3 and g:x 6x – 5. Given the composite function gf :x rx + s, find the values of r and s. [ r = 42 , s = 13 ]

3(c).The function f is defined by f : x mx + n and g:x 5x + 1. Given the composite function fg:x 15x – 2, find the values of m and n. [ m = 3 , n = – 5 ]

3(d). The function f is defined by f : x 7x + 3 and g:x mx + n . Given the composite function fg :x 10 – 14 x, find the values of m and n [ m = – 2 , n = 1 ]

3(e) The function f is defined by f : x mx + 3 and g:x 3x + n . Given the composite function fg :x 12x – 5, find the values of m and n. [ m = 4 , n = – 2 ]

3(f) The function f is defined by f : x mx – 1 and g:x 4x + 3 . Given the composite function fg :x 24x +n, find the values of m and n. [ m = 6 , n = 17 ]

Halus_1 3

3*(g). The function f is defined by f: x x2 – 4 and the function g is defined by g: x m x + n where m > 0 , n > 0 Given the composite function fg : x 4 x2 + 12 x + 5 , find the values of m and n. [ m=2 , n = 3 ]

3(h) The function f is defined by f: x 3x2 + 2 and the function g is defined by g: x m x + n where m > 0 , n > 0 Given the composite function fg : x 12 x2 + 36 x + 29 , find the values of m and n. [ m=2 , n = 3 ]

3(i). The function f is defined by f: x mx2 +n and the function g is defined by g: x 3 x + 2 where m > 0 , n > 0 Given the composite function fg : x 18 x2 + 24 x + 11 , find the values of m and n. [ m=2 , n = 3 ]

3(j)The function f is defined by f: x mx2 + n and the function g is defined by g: x 5 x – 3 where m > 0 , n > 0 Given the composite function fg : x 50 x2 – 60 x + 21 , find the values of m and n. [ m=2 , n = 3 ]

4(a). Given that 2 is a root to x2 + 2x + p = 0. Find the value of p and the other root.

Halus_1 4

For examiner’s

use only

[ p = – 8 , the other root = – 4 ]

4(b) Given that 3 is a root to x2 + px + 12 = 0. Find the value of p and the other root. [ p = – 7 , the other roots = = 4 ]

4(c).Given that p is a root to x2 + x – 5p = 0. Find the value of p where p 0 and the other root. [ p = 4 , the other root = = – 5 ]

*4(d). Given that 2+p is a root to x2 + 2x – 8 = 0. Find the value of p and the root of the equation [ p = – 6 or 0 , roots = – 4 , 2 ]

4(f). Given that 3 – p is a root to x2 – 7x +10 = 0. Find the value of p and the root of the equation. [ p = – 2 or 1 , roots = 2 , 5 ]

4(g). Given that 2p is a root to x2 – 4x –12 = 0. Find the value of p and the root of the equation. [ p = – 1 or 3 , roots = – 2 , 6 ]

5(a) Diagram show the graph of f(x) = ( x + 2 )2 + 3Halus_1 5

Find the values of p and q. [ p = – 2 , q = 3 ]

5(b) Diagram show the graph of f(x) = ( x – 2 )2 –5 Find the values of p and q. [ p = 2 , q = – 5 ]

5(c) Diagram show the graph of f(x) = ( x –3 )2 + q Find the values of p and q. [ p = 3 , q = 8 ]

5(d) Diagram show the graph of f(x) = ( x +p )2 – 7 Find the values of p and q. [ p = 3 , q = –7 ]

5(e) Diagram show the graph of f(x) = – 4( x + 1 )2 + 10 Find the values of p and q. [ p = – 1 , q = 10 ]

5(f) Diagram show the graph of f(x) = 3 ( x – 2 )2 –5 Find the values of p and q. [ p = 2 , q = – 5 ]

5(g) Diagram show the graph of f(x) = 2( x –3 )2 + q Find the values of p and q. [ p = 3 , q = 8 ]

5(h) Diagram show the graph of f(x) = ( x +p )2 + 7 Find the values of p and q. [ p = 3 , q = 7 ]

6(a) Find the range of values of x for which ( x – 3 ) ( x + 2 ) < 0 [ – 2 < x < 3 ]

Halus_1 6

0

( p , q )

x

y

0

( p , q )

x

y

0

( p , 8 )

x

y

0

(–3 , q )

x

y

0

( p , q )

x

y

0

( p , q )

x

y

0

( p , 8 )x

y

0

(–3 , q )

x

y

6(b) Find the range of values of x for which ( x – 3 ) ( x + 2 ) < 6 [ – 3 < x < 4 ]

6(c) Find the range of values of x for which ( x – 3 ) ( x + 2 ) < 4x [ – 1 < x < 6 ]

6(d) Find the range of values of x for which ( x – 3 ) ( x + 2 ) < 3x – 9 [ 1 < x < 3 ]

6(e) Find the range of values of x for which ( x – 3 ) ( x + 2 ) < ( x – 3 ) [ – 1 < x < 3 ]

6(f) Find the range of values of x for which 2x2 + 9x + 9 < 0 [ –3 < x < –3/2 ]

6(g) Find the range of values of x for which 2x2 + 9x + 9 < 5 [ – 4 < x < – ½ ]

6(h) Find the range of values of x for which 2x2 + 9x + 9 < – 2x [ – 9/2 < x < – 1 ]

6(i) Find the range of values of x for which 2x2 + 9x + 9 < 15 – 2x [ – 6 < x < ½ ]

6(j) Find the range of values of x for which 2x2 + 9x + 9 < 9 – x2

[ – 3 < x < 0 ]

*7(a) Solve the equation 16 x + 2 – 4 x – 4 = 0 [ x = – 8 ]

Halus_1 7

7(b) Solve the equation 125 x + 2 – 5 x – 4 = 0 [ x = – 5 ]

7(c) Solve the equation 9 x + 2 – 27 x – 4 = 0 [ x = 16 ]

7(d) Solve the equation 3 (9 x + 2 )– 27 x – 4 = 0 [ x = 17 ]

*7(e) Solve the equation 2 x + 2 – 2 x +1 = 8 [ x = 2 ]

7(f) Solve the equation 3 x + 2 – 3 x +1 = 54 [ x = 2 ]

7(g) Solve the equation 5 x + 2 – 5 x +1 = 500 [ x = 2 ]

7(h) Solve the equation 2 x + 2 – 2 x – 1 = 28 [ x = 3 ]

8. Given that log5 2 = h , and log 5 3 = k. Express the following in terms of h and k. (a) log 5 1.5 [ k – h ]

Halus_1 8

(b) log 5 13.5 [ 3k – h ]

(c) log 5 0.3 [ k – h – 1 ]

*(d) log 5 0.5 [ log 5 ½ = log 5 2 -1 = – h ]

(e) log 5 0.48 [ 2h +k – 2 ]

(f) log 5 12 [ 2h + k ]

(g) log 5 450 [ 2+h +2k ]

(h) log 10 12 [ (2h +k)/(h+1) ]

(i) log 15 18 [ (h + 2k /(k+1)]

(j) log 25 54 [ (h+3k]/2 ]

(k) log 100 24 [ (3h+k)(2+2h) ]

(l) log 30 60 [ (2h +k +1)/(h+k+1)]

9.(a)Solve log3 ( 2x + 1) – log 3 ( x – 1 ) = 1 [ x = 4 ]

Halus_1 9

(b) Solve log2 ( 2x + 1) – log 2 ( x – 1 ) = 2 [ x = 5/2 ]

(c) Solve log3 ( 2x + 1) – log 3 ( x – 1 ) = 3 [ x = 28/ 25 ]

(d) Solve log5 ( 2x – 3) – log 5 ( x – 1 ) = 0 [ x = 2 ]

9.(e) Solve log3 ( 2x – 3) – log 3 ( x – 1 ) = –1 [ x = 8/5 ]

*(f) Solve log3 ( 2x + 1) + log 3 ( 3x – 2 ) = 1 [ (6x + 5 ) ( x – 1) = 0 ; x = 1 discard x = – 5/6 ]

(g) Solve log5 ( 2x + 1) + log 5 ( 3x – 2 ) = 1 [ (6x – 7 ) ( x + 1) = 0 ; x = 7/6 discard x = –1 ]

Name : __________________________________

Form : ________________________ R2011_2

10 (a) Given that point P ( h,k ) divide the line segment A (– 3 , 5 ) and B ( 7,20 ) in the ratio AP : PB = 2 : 3 , find the value of h and of k. [ h = 1, k = 11 ]

Halus_1 10

10 (b) Given that point P ( h,k ) divide the line segment A (3 , – 10 ) and B ( 28, 5 ) in the ratio AP : PB = 1 : 4 , find the value of h and of k. [ h = 8, k = – 7 ]

10 (c) Given that point P ( 10,5 ) divide the line segment A (– 5 , – 4 ) and B ( h, k ) in the ratio AP : PB = 3 : 2 , find the value of h and of k. [ h = 20, k = 11 ]

10 (d) Given that point P ( 14, –6 ) divide the line segment A ( h, k ) and B ( 29, 4 ) in the ratio AP : PB = 2 : 5 , find the value of h and of k. [ h = 8, k = – 10 ]

10 (e) Given that point P ( – 2 ,12 ) divide the line segment A (– 4 , k ) and B ( h, 24 ) in the ratio AP : PB = 1 : 2 , find the value of h and of k. [ h = 2 , k = 6 ]

10 (f) Given that point P (–7 , –6 ) divide the line segment A (h , –10 ) and B ( 2, k ) in the ratio AP : PB = 1 : 3 , find the value of h and of k. [ h = – 10, k = 6 ]

10 (g) Given that point P ( h, 14 ) divide the line segment A (– 8 , 6 ) and B ( 22, 26 ) in the ratio AP : PB = 2 : k , find the value of h and of k. [ h = 4, k = 3 ]

*10 (h) Given that point P ( 18, – 2 ) divide the line segment A (15 , – 8 ) and B ( 20, 2 ) in the ratio AP : PB = h : k , find the ratio of h : k. [ h : K =3 : 2 ]

Halus_1 11

11(a)The vertices of a triangle are A (1,6) , B(– 2 ,5) and C ( 9,2) . find the area of triangle ABC [ 10 ]

11(b) The vertices of a quadrilateral are A (4,7) , B(– 2 ,3) and C ( 6 , – 3 ) and D(10,3) Find the area of the quadrilateral.

[ 60 ]

11(c) The vertices of a pentagon are A (9,5) , B(3 ,7) and C ( – 3, 4) , D(3 ,– 4 ) and E (11,2). Find the area of the pentagon [ 84 ]

11(d)The vertices of a triangle are A (3,6) , B(– 2 ,3) and C ( 8,p) . Given that the area of the triangle is 20 unit 2 , find the values of p. [ p = 1 , 17 ]

11(e) The vertices of a triangle are A (4,7) , B(– 2 ,3) and C ( p,1) . Given that the area of the triangle is 20 unit 2 , find the values of p. [ p = 5 , – 15 ]

12.(a) P moves such that its distance is always 4

Halus_1 12

units from Q ( 4 , –1 ). Find the equation of locus P. [ x2 + y2 – 8x + 2y + 1 = 0 ]

12.(b) P moves such that its distance is always 5 units from Q ( 3 , –2 ). Find the equation of locus P. [ x2 + y2 – 6x + 4y – 12 = 0 ]

*12.(c) Q ( –1,5 ). and R( 7, –1 ) is the diameter to the moving points P .Find the equation of locus P. [ centre C ( 3, 2 ) ;r = 5, x2 + y2 – 6x –4y – 12 = 0 ]

12.(d) Q ( –6,0 ). and R( 4, 6 ) is the diameter to the moving points P .Find the equation of locus P. [ centre C (–1 , 3 ) ;r = 34, x2 + y2 + 2x –6y – 24 = 0 ]

12.(e) P moves such that its distance from Q ( 4 , –5 ) is always twice the distance from R ( 6,8) Find the equation of locus P. [ PQ2 = 4 PR 2 ; 3x2+ 3y2 – 40x – 74y + 359 = 0 ]

12.(f) P moves such that its distance from Q ( 4 , –5 ) and R ( 6,8) is always in the ratio of PQ:PR= 2:3 Find the equation of locus P. [ 9PQ2 = 4PR 2 ; 5x2+ 5y2 – 24 x +154y – 31 = 0 ]

Halus_1 13

13(a) Find the mean, variance and the standard deviation of the following data.

(i) 1 , 7 , 7, 25 ( x1, x2, x3 , x4 )

Hence or otherwise, deduce the means, standard deviation and variance of

*(ii) 0, 6, 6, 24 (x1– 1,x2 –1 ,x3 –1 ,x4 –1)

(iii) 6,12, 12,20 ( x1+5 ,x2+5 , x3 +5 ,x4 +5)

(iv) 2 , 14,14,50 ( 2x1, 2x2, 2 x3 , 2 x4)

(v) 3 , 21, 21,75 ( 3x1 ,3x2 ,3 x3 , 3 x4)

(vi) 7,31,31, 103 (4x1+3,4x2+3, 4x3+3, 4x4+3)

mean StandardDeviation ( )

Variance(2)

(i) x1 10 9 81(ii) x1– 1 10 – 1 = 9 9 81(iii) x1+5 10+5 = 15 9 81(iv) 2x1 2(10) = 20 2(9)=18 182=324(v) 3x1 3(10) = 30 3(9)=27 272=729(vi) 4x1+3 4(10)+3 = 43 4(9)=36 362=1296

13(b) Find the range , first quartile and the third quartile for the following sets of data.

( i ) 1, 2 , 3, 4, 6 ( x1, x2, x3 , x4 , x5 )

(ii) 11,12,13,14, 16 ( x1+10, ……. , x5 +10)

(iii) 2, 4, 6, 8 , 12 ( 2x1, ………. , 2x5 )

(iv) 7, 11, 15, 19, 27 ( 4x1+3………. , 4x5 +3)

Range First Quartile ( Q1 )

Third Quartile( Q3 )

(i) x1 5 1.5 5(ii) x1+10 5 1.5+10=11.5 5+10=15(iii) 2x1 2(5) = 10 2(1.5)=3 2(5)=10(iv) 4x1+3 4(5)=20 4(1.5)+3= 9 4(5)+3= 23

13(c) Find the range , first quartile and the third quartile for the following sets of data.

( i ) 1, 2 , 5, 6, 7, 8 ( x1, x2, x3 , x4 , x5 , x6)

(ii) 11,12,15,16, 17,18 ( x1+10, ……. , x6 +10)

(iii) 3, 6, 15, 18 , 21,24 ( 3x1, ………. , 3x6 )

(iv) 7, 11, 23, 27, 31,35 ( 4x1+3………. , 4x6 +3)

Range First Quartile ( Q1 )

Third Quartile( Q3 )

(i) x1 7 2 7(ii) x1+10 7 2+10=12 7+10=17(iii) 3x1 3(7) = 21 3(2)= 6 3(7)= 21(iv) 4x1+3 4(7) =28 4(2)+3= 11 4(7)+3= 31

Halus_1 14

Standard deviation = = 2 = variance

13(d) The mean of the 4 numbers is 7 and the sum of the squares of the numbers is 238 Find the variance and standard deviation. [ Var = 10.5, Std dev= 3.240 ]

13(e) The variance of the 4 numbers is 7.5 and the sum of the squares of the number is 226 Find the mean and standard deviation. [ Mean = 7 , Std dev= 7.5 = 2.739 ]

13(f) The mean of the 4 numbers is 6.5 and standard deviation 2.5,find the sum of the squares of the number and the variance. [ sum of squares = 194, Var= 6.25 ]

13(g) Given that the value of and of a set of 6 numbers is 25 and 125 respectively. If number 4 is taken out from the set of data, find the new value of standard deviation. [ std deviation = 2.040 ]

13(h) Given that the value of and of a set of 6 numbers is 30 and 160 respectively. If number 8 is added to the set of data, find the new value of standard deviation. [ std deviation = 2.531=1.591 ]

13.(i) It is given that the sum of 6 numbers is m. The sum of the squares of number is 360 and the variance is 5p2. Express m in terms of p m = 180( 12 – p2 )

Halus_1 15

14(a) Diagram below shows two sectors OBC and OAD with common centre O.

The length of the arc BC is 12.8 cm and OA : AB = 3 : 1, find

(i) the length of OB,

(ii) the perimeter of the shaded region.

(ii) the area of the shaded region. [ (i) OB = 16 cm (ii) 30.4 cm (iii) 44.8 cm2 ]

14 (b) The diagram shows a semicircle ABCO of centre O and radius r.

The length of the arc AB is 8.52 cm and the

angle of COB is 2.29 radians. Calculate,(i) the value of r, (ii) the area of the shaded region.

[Use ] [ (i) r = 10 (ii) 114.5 cm2 ]

14 (c) The diagram shows a semicircle ABCO of centre O and radius r.`

The length of the arc BC is 7 cm and the angle of AOB is 0.72 radians. Calculate,

(a) the value of r, (b) the area of the shaded region.

[Use ] [(i) r = 25 cm (ii) 87.5 = 275 cm2 ]

14(c) The diagram shows a semicircle ABCO of centre O and radius r.`

The length of the arc AB is 9.6 cm and the length of minor AB is 9.6 radians. Calculate,

(a) the value of r, (b) the area of the shaded region.

[Use ] [(i) r = 12 cm (ii) 86.4 = 271.47 cm2 ]

Halus_1 16

O

D

A

C

B

0.8 rad

A O C

B

0.72

A O C

B

2.29 rad

A O

C

B

0.8

15(a) Given y = 5x ( x – 4 ), find (i) dy/dx, (ii) the value of x when x is a minimum (iii) the minimum value of y [ (i) 10 x – 20 (ii) x = 2 (iii) – 20 ]

15(b) Given y = 4x ( 5 – x ) find (i) dy/dx, (ii) the value of x when x is a maximum (iii) the maximum value of y [ (i) 20 – 8 x (ii) x = 2.5 (iii) 25 ]

15(c) Given y = ( 7 + 2x )( 4 – x ), find (i) dy/dx, (ii) the value of x when x is a maximum (iii) the maximum value of y [ (i) 1 – 4 x (ii) x = ¼ (iii) 28 1/8 ]

15(d) Given y = 2x3 – 5x2– 4x – 6 , find (i) dy/dx, (ii) find the turning points (maximum and minimum point) of the curve [ (i) 6x2 – 10 x – 4 (ii) ( – 1/3 , – 5 8/27 ) maximum ( 2 , – 18 ) minimum point

15(e) Given y = – x3 + x2 + 8 x + 7 , find (i) dy/dx, (ii) find the turning point (maximum and minimum point ) of the curve [ (i) – 3x2 + 2x + 8 (ii) ( – 4/3 , 13/27 or 0.48 ) minimum ( 2 , 19 ) maxmum point ]

Halus_1 17

16(a) The sides of a square increase at the rate of 2 cms –1. Find the rate of the area of the square changes when the side is 5 cm, [ dA/dx x dx/dt = 10 x 2 = 20 cm2 s –1 ]

16(b) The sides of a square decrease at the rate of 3 cms –1. Find the rate of the perimeter of the

square changes when the side is 6 cm, [ dP/dx x dx/dt = 4 x – 3 = – 12 cm s –1 ]

16(c) The radius of a circle increase at the rate of 2 cms –1. Find the rate of the area of the circle changes when the radius is 5 cm, [ dA/dr x dr/dt = 10 x 2 = 20 cm2 s –1 ]

16(d) The radius of a circle decrease at the rate of 3 ms –1. Find the rate of the perimeter of the circle changes when the radius is 6 cm, [ dP/dr x dr/dt = 2 x – 3 = – 6 cm s –1 ]

17(a) The sides of a square increase from 5 cm to 5.01 cm. Find the approximate increase in area of the square. [ δA = dA/dxxδx = 10 x 001 = 0.1 ]

17(b) The sides of a square decrease from 6 cm to 5.98 cm. Find the small changes in perimeter of the square. [ δP = dP/dxxδx = 4 x –002 = – 0.08cm ]

17(c) The radius of a circle increase from 5 cm to 5.02 cm. Find the small increment in the area of the ircle [ δA = dA/drxδr = 10 x 002 = 0.2 ]

17(b) The radius of a circle decrease from 6.0 cm to 5.97 cm. Find the small cahnge of perimeter of the circle [ δP = dP/drxδr = 2 x –003 = – 0.06 ]

Halus_1 18

16(e) The perimeter of a square increase at the rate of 3 ms –1. Find the rate of change of the side when the area of the square is 100 cm2, [ dP/dt =dP/dr x dr/dt ; 3 = 4 x dr/dt , dr/dt = 3/4 cm s –1 ]

16(f) The area of a square decrease at the rate of 4 cm2s –1. Find the rate of the change of the side of a square when the area of the square is 100 cm2 [ dA/dt =dA/dr x dr/dt ; –4 = 20 x dr/dt , dr/dt = –1/5 cm s –1 ]

16(g) The perimeter of a circle increase at the rate of 5 cms –1. Find the rate of change of the radius when the aea of the circle is 144 cm2, [ dP/dt =dP/dr x dr/dt ; 5 = 2 x dr/dt , dr/dt = 5/2 cm s –1 ]

16(h) The area of a circle decrease at the rate of 4 cm2s –1. Find the rate of the side of a square when the area of the square is 64 cm2 [ dA/dt =dA/dr x dr/dt ; –4 = 16 x dr/dt , dr/dt = –1/4 cm s –

1 ]

17(e) Given that y = 3x2 , find the new value of y when x changes from 5 to 5.01 [ y + δy = 3(52) + 6(5)(0.01) = 75.3 ]

17(f) Given that y = 5x3 , find the new value of y when x changes from 5 to 5.01 [ y + δy = 625 + 15(52)(0.01) = 628.75 ]

17(f) Given that y = 3x2 + 5x3 , find the new value of y when x changes from 5 to 5.01 [ y + δy =700 + (405)(0.01) = 704.05

Halus_1 19

Halus_1 20

Name : __________________________________

Form : ________________________ R2011_3

18(a)The first three terms of an an arithmetic progression are 6 , p – 3 , 14. Find (i) the value of p (ii) the tenth term (iii) sum of first ten terms [ (i) p = 13; 6, 10, 14; a = 6 , d = 4 (ii) T 10 = 42 (III) Sn = 10 = 240 ]

18(b)The first three terms of an an arithmetic progression are p – 5 , 14, 2p – 6 Find (i) the value of p (ii) the twelve term (iii) sum of first twelve terms [ (i) p = 13; 8,14,20, a = 8 , d = 6 (ii) T 12 = 74 (III) Sn = 12 = 492 ]

18(c)The first three terms of an an arithmetic progression are p – 3 , p + 5 , 2p Find (i) the value of p (ii) the fifteen term (iii) sum of first fiftheen terms [ (i) p = 13; 10,18,26 ; a = 10, d = 8 (ii) T15 = 122 (III) Sn = 15 = 990 ]

18(d) Three consecutive terms of an arithmetic progression are 2 – 3p, 6 and 4p. Find the first term and common difference of the progression.

[ p = 10 a = – 28 , d = 34 ]

18(e)Three consecutive terms of an arithmetic progression are 2 – 4p, p – 4 and 5p. Find the first terms and the common difference of the progression. [ p = 10 a = – 38 , d = 44 ]

18(f)Three consecutive terms of an arithmetic progression are p– 3, 2p – 3 and 2p + 7. Find the common difference of the progression.

[ p = 10 a = 7 , d = 10 ]

Halus_1 21

19(a)In a geometric progression, the first term is 64 and the fourth term 8. Calculate (i) the common ratio (ii) the sum to infinity [ (i) r = ½ ( ii) S =128 ]

19(b)In a geometric progression, the first term is 243 and the third term 27. Calculate (i) the common ratio (ii) the sum to infinity [ (i) r = 1/3 ( ii) S =364 ½ ]

19(c)In a geometric progression, the first term is 625 and the fifth term is 1. Calculate (i) the common ratio (ii) the sum to infinity [ (i) r = 1/5 ( ii) S =781 ¼ ]

19(d) In a geometric progression, the second term is 14 and the fourth term 56. Calculate (i) the common ratio (ii) the sum of first four terms [ (i) r = 2 ( ii) S n=4 = 105 ]

19(e) In a geometric progression, the third term is 18 and the sixth term 486. Calculate (i) the common ratio (ii) the sum of first six terms [ (i) r = 3 ( ii) S n=6 = 728 ]

19(f) In a geometric progression, the fourth term is 1000 and the sixth term 25000. Calculate (i) the common ratio (ii) the sum of first six terms [ (i) r = 5 ( ii) S n=6 = 31248 ]

Halus_1 22

20 (a)

(i) Write the equation of the graph [ m = 2 , c = 4 ; xy = 2 x + 4 ]

(ii) Find the value of m and of n given that the variables x and y are related by (a) [ m = 2 , n = 4 ]

(b) [ m = 4 , n = 2 ]

(c) mxy = n x + 1 [ m = ¼ , n = ½ ]

20 (a)

(iii) Write the equation of the graph [ m = – ½ , c = 4 ; y2 = – ½ x + 4 ]

(iv) Find the value of m and of n given that the variables x and y are related by (a) [ m = – ½ , n = 4 ]

(b) [ m = 4 , n = – ½ ]

(c) my2 = n x + 1 [ m = ¼ , n = – 1/8 ]

21(a) Given that O(0,0), A ( 3, 1 ) , B ( 6, – 3)

Halus_1 23

4

( 2, 8)

x

xy

4

( 4, 2)

x

and C ( – 2 , 3 ) Express the following in terms of unit vector i and j

(i) AB(ii) The unit vector in the direction of AB(iii) BC(iv) The unit vector in the direction of BC(v) CA(vi) The unit vector in the direction of CA

[ (i) 3i – 4 j (ii) (3i – 4j)/5 (iii) – 8 i + 6 j (iv) ( – 4 i + 3j)/5 (v) 5 i – 2 j ( vi ) ( 5 i – 2 j )/ 29

21(b) Given and

(i) Find 2 r – s

(ii) | 2 r – s | [ 13 ]

21(c)

Express the following vectors in terms of vector i and j

(i) PR(ii) unit vector in the direction of PR(iii) SR(iv) unit vector in the direction of SR

[ (i) – 5i – 12 j (ii) (– 5i – 124j)/13 (iii) 8 i – 9 j (iv) (8 i – 9 j )/ 145

21(d) Given and

and | v – w | = 5 find the values of p . [ p = – 2, 6 ]

Halus_1 24

0 x

y P(6,7)

Q(8,2)

R(1,– 5 )

S(–7, 4)

22(a) Diagram shows triangle ABC.

AD = 2DB and AC = x and AB = y Find in terms of x and y (i) CB (ii) CD [ (i) – x + y (ii) – x + (2y) /3 ]

22(b) Diagram shows triangle ABC.

M is the midpoint of AC,3AD = 2DB and AC = 2x and AB = 5y Find in terms of x and y (i) CB (ii) CD (iii) BM [ (i) –2 x +5 y (ii) – 2x + 2y (iii) – 5y + x ]

22(c)Diagram show triangle ABC

AD = 1/3 AB ; BE : EC = 3 : 1, AD = 2rAC = 4S; express the following vector in terms of r and s

(i) AB (ii) BC (iii) DC (iv) AE (v) DE

[ (i) 6r(ii) – 6r + 4s (iii) – 2r + 4s (iv)( 1(6r)+3(4s))/4=3/2(r+2s) (v) 3s – ½ r

22(d)Diagram show triangle ABC

CE=ED=DB, AB = x , AC = y ; express the following vector in terms of x and y(i) CB (ii) AE (iii) AD [ (i) – y + x (ii) ( 1 x + 2y)/ 3 (iii) (2x+1y)/3= (2x+y)/3

Halus_1 25

A B

C

D

A B

C

D

M

A B

C

D

E

A B

C

D

E

23.(a) Given that , find the value of

(i) (ii)

(iii) (iv)

(v) (vi)

and the value of m if

(vii)

(viii)

(ix)

[ (i) 10 (ii) – 15 (iii) 21 (iv) – 55 (v) 253 (vi) 21/5 (vii) m = ½ (viii) – 5 (ix) – 3 ]

23(b). Given that , find the value of

k if (i)

(ii)

(iii)

[ (i) ½ (ii) ½ (iii) 3/8 ]

23(c) Find the value of n and k for each of the following

(i)

(ii)

(iii)

[ (i) k = – ¾ n = – 1 (ii) k= 1/4 , n = 3 (iii) k = 3/16, n = 4 ]

24(a)Express function f(x)= x2 – 4x + 5 inHalus_1 26

( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(II) By means of calculus, determine the maximum or minimum value of f(x) = x2 – 4x + 5 and state the value of x then it happens.

(III) Sketch the graph of f(x)= x2 – 4x + 5

[ (i) ( x – 2 )2 + 1, min because a = 1 > 0, f(x) min = 1, x= 2 (ii) dy/dx = 2x – 4, d2y/dx2 = 2 > 0 , min, dy/dx = 2x – 4 = 0 , x = 2 , f(2) = (2)2 – 4(2) + 5 = 1(iii) y–int = 5, min at ( 2,1), parabolic curve]

24(b)Express function f(x)= x2 + 7 x – 8 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(ii) By means of calculus, determine the maximum or minimum value of f(x) = x2 + 7 x – 8 and state the value of x then it happens.

(iii) Sketch the graph of f(x)= x2 + 7 x – 8

[ (i) ( x + 7/2 )2 – 49/4 – 8 = ( x + 7/2 )2 – 81/4 min because a = 1 > 0, f(x) min = – 81/4 x= – 7/2 (ii) dy/dx = 2x +7, d2y/dx2 = 2 > 0 , min, dy/dx = 2x +7 = 0 ,

x = – 7/2 , f(– 7/2) = (– 7/2)2 – 4(– 7/2) – 8 = – 13.25(iii) y–int = – 1 , min at (– 7/2, – 13.25), parabolic curve]

24(c)Express function f(x)= 12 + 4x – x2 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(II) By means of calculus, determine the maximum or minimum value of f(x) = 12 + 4x – x2 and state the value of x then it happens.

(III) Sketch the graph of f(x)= 12 + 4x – x2

[ (i) – ( x – 2 )2 + 16, max because a = – 1 < 0, f(x) max = 16 when x= 2 (ii) dy/dx = 4 – 2x, d2y/dx2 = –2 < 0 , max, dy/dx = 4 – 2x = 0 , x = 2 , f(2) = 12 + 4((2) – (2)2 = 16(iii) y–int = 12, max at ( 2,16), parabolic curve]

24(d)Express function f(x)= 10 – 3x – x2 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(ii) By means of calculus, determine the maximum or minimum value of f(x) = 10 – 3x – x2 and state the value of x then it happens.

(iii) Sketch the graph of f(x)= 10 – 3x – x2

[ (i) – ( x + 3/2 )2 + 12 ¼ , max because a = – 1 < 0, f(x) max = 12 ¼ when x= 1 (ii) dy/dx = – 3 – 2x, d2y/dx2 = –2 < 0 , max, dy/dx – 3 – 2x = 0 , x = – 3/2 , f( – 3/2 ) = 10 – 3(–3/2) – (– 3/2 )2 = 12 ¼ (iii) y–int = 10, max at ( 7/4,12 ¼ ), parabolic curve]

Halus_1 27

24(e)Express function f(x)= 2x2 – 8x + 5 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(ii) By means of calculus, determine the maximum or minimum value of f(x) = 2x2 – 8x + 5 and state the value of x then it happens.

(iii) Sketch the graph of f(x)= 2x2 – 8x + 5

[ (i) 2( x – 2 )2 –3, min because a = 2 > 0, f(x) min = –3 when x= 2 (ii) dy/dx = 4x – 8, d2y/dx2 = 4 > 0 , min, dy/dx = 4x – 8 = 0 , x = 2 , f(2) = 2(2)2 – 8(2) + 5 = –3(iii) y–int = 5, min at ( 2, –3), parabolic curve]

24(f)Express function f(x)=2 x2 + 14 x – 1 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(ii) By means of calculus, determine the maximum or minimum value of f(x) = 2x2 + 14 x – 1 and state the value of x then it happens.

(iii) Sketch the graph of f(x)= 2x2 + 14 x – 1

[ (i) 2( x +7/2 )2 – 25 ½ , min because a = 2 > 0, f(x) min = – 25 ½ when x= – 7/2 (ii) dy/dx = 4x +14, d2y/dx2 = 4 > 0 , min, dy/dx = 4x +14 = 0 , x = – 7/2 , f(– 7/2) = 2(– 7/2)2 – 14(– 7/2) – 1 = – 25 ½ (iii) y–int = – 1 , min at (– 7/2, – 25 ½ ), parabolic curve]

24(g)Express function f(x)= 3 + 8x – 2x2 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(ii) By means of calculus, determine the maximum or minimum value of f(x) = 3 + 8x – 2x2 and state the value of x then it happens.

(iii) Sketch the graph of f(x)= 3 + 8x – 2x2

[ (i) – 2( x – 2 )2 + 11, max because a = – 2 < 0, f(x) max = 11 when x= 2 (ii) dy/dx = 8 – 4x, d2y/dx2 = –4 < 0 , max, dy/dx = 4 – 2x = 0 , x = 2 , f(2) = 3 + 8((2) –2 (2)2 = 11(iii) y–int =3, max at ( 2,11), parabolic curve]

24(h)Express function f(x)= 9 + 14x – 2x2 in ( i ) a( x + b) 2 + c, hence deduce its maximum or minimum value and state the value of x when it occurs.

(ii) By means of calculus, determine the maximum or minimum value of f(x) = 9 + 14x – 2x2 and state the value of x then it happens.

(iii) Sketch the graph of f(x)= 9 + 14x – 2x2

[ (i) – 2( x – 7/2 )2 + 33 ½ , max because a = – 2 < 0, f(x) max = 33 ½ when x= 7/2 (ii) dy/dx = 14 – 4x, d2y/dx2 = –4 < 0 , max, dy/dx = 14 – 4x = 0 , x = 7/2 , f(7/2) = 9 + 14((7/2) – 2(7/2)2 = 33 ½ (iii) y–int = 9, max at ( 7/2,33 ½ ), parabolic curve]

Halus_1 28

25(a) Find the equation of the curve for its gradient function is 2x + 3 and passes through (i) ( 1, 8 ) [ y = x2 + 3x + 4 ]

(ii) ( 2, 8 ) [ y = x2 + 3x – 2 ]

(iii ) (–2, 8 ) [ y = x2 + 3x + 10]

25(b) Find the equation of the curve for its gradient function is 6x(x + 1) – 3 and passes through (i)( –1, 9 ) [ y = 2x3 + 3x2 – 3x + 5 ]

(ii)( 1, 9 ) [ y = 2x3 + 3x2 – 3x + 7 ]

(iii)( 2, 3 ) [ y = 2x3 + 3x2 – 3x – 19 ]

26(a) Find the equation of the tangent for curve

y = 3x(2x – 5) – 4 and passes through (i) ( 1, – 13 ) [ y = – 3x – 10]

(ii) ( 2, – 10 ) [ y = 9x – 28 ]

(iii) (–2, 50 ) [ y = – 39x –28]

26(b) Find the equation of the tangent for function f(x) =6x(x + 1) – 3 and passes through (i)( –1, –3 ) [ y = – 6 x – 9 ]

(ii)( 1, 9 ) [ y = 18 x – 9 ]

(iii)( 2, 33 ) [ y = 30 x – 27]

Halus_1 29

27(a) Solve cos 2 A + cos A + 1 = 0 for . Cos A ( 2cos A + 1 ) = 0 ; A= 90º 270o ,120º, 240o ]

(b) Solve 2 cos 2 A + 3 cos A + 1 = 0 for

. ( 4cosA – 1 ) ( cos A + 1 ) = 0 ; A= 75º 31’ ,284º29’ 180o

(c) Solve 3 cos 2 A + 5 cos A +2 = 0 for

. ( 6cosA – 1 ) ( cos A + 1 ) = 0 ; A= 80º 24’ ,279º36’ 180o

27(d) Solve cos 2 A + sin A + 2 = 0 for . ( 3 – 2sin A ) ( 1 + sin A ) = 0 ; A= 270º ]

(e) Solve 2 cos 2 A + 4 sin A + 1 = 0 for

. ( 3 – 2sin A ) ( 1 + 2sin x ) = 0 ; A= 210º ,330o ]

(f) Solve 3 cos 2 A + sin A – 2 = 0 for . ( 1+ 3 sinA ) ( 1 – 2 sin A ) = 0 ; A= 199º 28’ ,340º32’

30o , 150o

Halus_1 30

28(a) Given that sin A = p and A is an acute angle, express the following in terms of p. (i) tan A ( ii ) cos A (iii) sin 2A ( iv ) cos 2A [ (i) p / ( 1 – p2 ) ( ii) ( 1 – p2 ) (iii) 2p( ( 1 – p2 ) (iv) 1 – 2p2 )

`

(b) Given that cos B = m and B is an acute

angle, express the following in terms of m. (I) tan B ( ii ) sin B (iii) sin 2B (iv) cos 2B [ (i) ( 1 – m2 ) /m( ii) ( 1 – m2 ) (iii) 2m( ( 1 – m2 ) ) (iv) 2m2 – 1

(c) Given that tan C = n and C is an acute

angle, express the following in terms of n. ( i ) cos C ( ii ) sin C (iii) sin 2C (iv) cos 2C (v) tan 2C [ (i) 1/ ( 1 + n2 ) ( ii) n / ( 1 + n2 ) (iii) 2n/( 1 + n2 ) (iv) 2/(1+n2 )– 1 =( 1 – n2]/( 1+n2) (v) 2n/(1-n2)

(d) Given that tan C = w and C is in third quadrant, express the following in terms of n. (I) cos C ( ii ) sin C (iii) sin 2C (iv) cos 2C [ (i) – 1/ ( 1 + w2 ) ( ii) – w / ( 1 + w2 ) (iii) – 2w( 1 + w2 ) (iv) 2/(1+w2 )– 1 =( 1 – w2]/( 1+w2)

Halus_1 31