RELATIONS 1

Given that x є A where A= {-3≤x≤3} produce a pictorial representation of the relation, R:x onto (2x-11)

Solution

x	-3	-2	-1	0	1	2	3	4
2x-11	-17	-15	-13	-11	-9	-7	-5	-3

The diagram is shown below:

TRY THIS……….

Given that x є A, draw a pictorial representation  of

 x onto (x-7)  [2≤x≤10]

FUNCTIONS 1

Find a linear function f(x) with gradient -10 which is such that f(5)=12.

        Solution.

m=-10, points = [5,12] and [x, f(x)]

m = y₂-y₁/x₂-x₁

-10 = f(x) – 12/x-5

f(x)-12=-10(x-5) [after cross multiply]

f(x)-12=-10x-50

f(x)=-10x+50+12

f(x)=-10x+62

Hence a  linear function is f(x) = -10x+62.

TRY THIS……….

Give out a linear function f(x) with

Gradient 5 and f(2)=14

LOGARITHMS 1

Evaluate Log₂(2048 x 32).

Solution

= Log₂(2048 x 32)

= Log₂2048 +  Log₂32          (applying the product rule)

= Log₂2¹¹ +  Log₂2⁵             (2048= 2¹¹ and 32=2⁵ )

= 11Log₂2 +  5Log₂2       ( remember  Log_aa^c = cLog_aa )

= (11 x 1) +  (5 x 1)          ( remember  Log_aa = 1 )

= 11 + 5

= 16

hence Log₂(2048 x 32) = 16

TRY THIS………………

Evaluate Log₂(8 x 256). 

FACTORIZING 1

Factorize 625x² - 81y²

Solution

we use difference of two squares a² – b² = (a - b)(a + b)

625x²- 81y²  = 25²x² - 9²y²

                   = (25x)² - (3y)²

                   = (25x - 3y)( 25x + 3y)

Hence 625x² - 81y² = (25x - 3y)(25x + 3y)

TRY THIS………………………….

factorize 49m²- 36n²

MIDPOINT 1

Find the midpoint of a line from (19, 6) to (5, 10)

Solution

x1=19, x2=5, y1=6, y2=10.

Mid point = (x1 + x2, y1 + y2)

                         2             2

Mid point = (19 + 5,  6 + 10)

                        2          2

Mid point = (24, 16)

                     2    2

Hence midpoint = (12, 6)

TRY THIS……………………..

Find the midpoint of a line from (14, -8) to (-6, 22)