Enrichment 2 Paper 1 Solution 1. Chapter: Function Topic: Representation of relation (GRAPHS) Note:

Horizontal axis or x-axis = DOMAIN = Set P Vertical axis or y-axis = CODOMAIN = Set Q Elements in Set P = OBJECT (obj.) Set Q = IMAGE (img.)

a) State the image of 2. Want: image of 2 Given: 2 is an object (hidden information) Idea: Look for object 2 and find its’ image Solution: look for object 2 from Set P, then compare it to the image in Set Q.

Answer: image of 2 is 4. ---------------------------------------Topic: Function Notation Note:

Function Notation represents the relationship between Domain and Codomain in an equation format.

b) Using the function notation, write a relation between set P and set Q. Want: relation between set P and set Q in function notation (equation format) Given: f (1) = 2 f ( 2) = 4 f (3) = 6

Idea: Find the similar relation between object and image for the above functions. Solution: f (1) = 2 = 2(1) f ( 2) = 4 = 2( 2) f (3) = 6 = 2(3) ∴ f ( x) = y = 2( x)

Answer: f ( x) = 2 x

2. Chapter: Function Topic: Inverse function Find the value of k. Want: the value of k. Given: f : x → 2 x − 3k

Idea:

f ( x) = 2 x − 3k --- (a) x f −1 : x → + 3 k x f −1 ( x) = + 3 --- (b) k

i) Use simultaneous equations ii) Find the inverse of (a) and compare with (b)

Solution: From (a) let y = f −1 ( x)

f ( y) = x 2 y − 3k = x 2 y = x + 3k x + 3k y= {Note that y = f −1 ( x) } 2 x + 3k f −1 ( x) = --- (c) {this is the inverse function of (a)} 2

Compare (b) with (c)

x x + 3k +3= k 2 x x 3k +3= + k 2 2

1 1 3 x +3= x + k k 2 2 1 1 3 x +3= x + k k 2 2 1 1 = k 2 ∴k =2

Answer: k = 2

or

3 k =3 2 ∴k =3

2 =2 3

3. Chapter: Function Topic: Representation of relation (Arrow Diagram) Note:

Left side, shows function f maps x (obj.) to y (img.). Right side, shows function g maps y (obj.) to z (img.). Elements in x=3 y=4 z=7

a) Determine f −1 (4) Want: image of inverse function f with object 4. Given: function f with object 3 maps to image 4. f : 3 → 4 or f (3) = 4 Idea: Assume function f is a road from your house to school. The object (3) is your starting point (house). The image (4) is your finishing point (school). To find the inverse function f −1 , assume you use the same road to return home from school, but in reverse. Now your object or starting point is your school and your image or finishing point is your house. f (object ) = image f (house) = school

3

4

f

−1

(object ) = image

f

−1

( school ) = house

Solution: if f (3) = 4 then f −1 (4) = 3 Answer: f −1 (4) = 3 ---------------------------------------Topic: Composite Function Note:

Composite Function is when 2 or more functions are combine into one. If f ( x) = y and g ( y ) = t , we can combine these functions as below:

g ( y) = t g[ y ] = t g[ f ( x)] = t ∴ gf ( x) = t

{since y = f (x) }

b) Determine gf (3) . Want: image of function gf with object 3. Given: f (3) = 4 , g (4) = 7 Idea: use gf ( x) = t Solution: gf ( x) = t gf (3) = 7 {from the diagram x = 3 and t = 7 } Answer: gf (3) = 7 _____________________________________________________________________ 4. Chapter: Composite Function & Inverse function Topic: Composite Function & Inverse function Find hf −1 ( x) . Want: the image of hf −1 ( x) . Given: h : x → 4x − 1 h( x) = 4 x − 1 ---(a) f : x → 2x + 3 f ( x) = 2 x + 3 ---(b)

Idea: First, find f −1 ( x) . Then, find hf −1 ( x) . Solution: From (b) let y = f −1 ( x) f ( y) = x 2y + 3 = x 2y = x − 3 x−3 y= {Note that y = f −1 ( x) } 2 x−3 f −1 ( x) = --- (c) {this is the inverse function of (b)} 2

From hf −1 ( x) = h[ f −1 ( x)] = 4[ f −1 ( x)] − 1 x−3 −1 2 = 2( x − 3) − 1 = 2x − 6 − 1 = 2x − 7

= 4

Answer: 2 x − 7

or

= h[ f −1 ( x)] x−3 2 x−3 4 −1 2 2( x − 3) − 1 2x − 6 − 1 2x − 7

= h = = = =

5. Chapter: Quadratic Equations Topic: Solving quadratic equation using quadratic formula. Solve the quadratic equation Want: value/s of x correct to three (3) decimal place. Given: x 2 = 7 x − 3 , Idea: since the answer must be given in three decimal places, it means that we need to use the quadratic formula. x=

− b ± b 2 − 4ac , a, b and c are constants 2a

Solution: First arrange the quadratic equation in general form ax 2 + bx + c = 0 . x2 = 7x − 3 x2 − 7x + 3 = 0 a = 1 , b = −7 and c = 3

Next, solve for x using the quadratic formula. x=

x=

− (−7) ± (−7) 2 − 4(1)(3) 2(1) 7 + 37 7 − 37 or x = 2 2

7 + 37 7 − 37 or x = 2 2 x = 6.541 or x = 0.459 x=

Answers: x = 6.541 or x = 0.459 __________________________________________________________________ 6. Topic: Composite Function & Inverse function

Solution Enrichment 2 Add Math

Published on Oct 28, 2010

Solution to Enrichment 2 exercise Add Math Form4

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