MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

FUNCTION
Concept and Definition
1.

Determine the domain and range for each of the given function,
(a) f  x   x 1
Answer:
Since it is undefined for –ve values, it must be that x - 1  0 ie x  1
1
 Domain  , 

(b) f  x  



1 x 4
2

Answer: f (x) is undefined if x 2  4  0 , ie  x  2 x  2   0 x = 2 or x = -2
The domain is all Real Numbers except 2 and -2.
 domain  ,  2 atau 2,   .
2
(c ) f  x   x  3

Answer:
All the real numbers ie  ,   

(d) f  x   x  5
Answer:

f  x is defined for all real numbers
0 , x-50 x5 Domain 5, 

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MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

x 1
(e) f ( x)  x 2  6 x  5
Answer:
f (x ) is undefined when,

x 2  x  5x  0
( x  5)( x  1)  0 x  5

or x  1
 Domain for f (x ) are all real numbers except –5 and –1

 3
(f) f ( x)  
2
4  x

jika jika x 1 x 1

Find the value of f ( 4 ) .
Answer:
f ( 4)  4  ( 4) 2 = 4  16  20.

(g) f  x   2 x  3
Answer:
–2x + 3 > 0
–2x > –3

2x < 3 x < 3/2 = 1.5
Then the domain is "all x < 3/2".

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MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

2.

Given g ( x )  3 x 2  x  5 , (i) find the domain and (ii) value of the function at g(z), g ( r 2 ) and g ( x  h )

Answer:
Domain: All real numbers. g(z) = 3 z 2  z  5 g( r 2 ) = 3( r 2 ) 2  r 2  5 = 3r 4  r 2  5 g(x + h) = 3 ( x  h) 2  ( x  h)  5
= 3( x 2  2hx  h 2 )  x  h  5
= 3 x 2  6 hx  3h 2  x  h  5

3. If f ( x )  x 2 , find

f ( x  h)  f ( x ) h Answer:

( x  h)2  x 2 f ( x  h)  f ( x )
=
h h x 2  2hx  h 2  x 2
=
h
=

2hx  h2 h =

h( 2 x  h)
= 2x + h h 4. Determine whether the given equation is a function
(i)
(ii)

2y + 3x = 6 y2 + 3x = 6

Answer:
2y + 3x = 6 is a function, because you can solve for y:

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MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

2y + 3x = 6
2y = –3x + 6 y = (–3/2)x + 3 y2 + 3x = 6 is not a function

Linear and Quadratic Functions
1. If a straight line passes through the point (1, a) and (4, -2) and is parallel to the line passing through the point (2, 8) and (-7, a + 4), what is the value of a?
Answer:
Since the two lines are parallel, m1 = m2



2a a4

3
9

-9 (-2 – a) = 3 (a – 4)
18 + 9a = 3a -12
6a = -30 a = -5
2. Given the equation y = 4x - 3, what is the corresponding change in y if
(a)

x increased by 1 unit?

(b) x decreases by 2 units?
Answer:(a)

If x increases 1 unit, y decreases 4 units.

(b)

If x decreases 2 units, y decreases 4 (-2) = -8 units.

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MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

3.

Given 2x + 3y = 4,

(a) Is the gradient of the equation positive or negative?
(b) When the value of x increases, does y increase or decrease?
(c) If x decreased by 2 units, what is the corresponding change in y?
Answer:
2x + 3y = 4
Deriving the slope-intercept form of the equation,
3y = 4 – 2x y= 4 2
 x
3 3

2
, hence the gradient is negative.
3

(a)

m= 

(b)

Since the gradient is negative, y decreases if x increases.

(c)

If x decreases by 2 units, y will increse by 

2
 2  4 unit.
3
3

4. (a) Determine the equation of the line that passes through the point (-2, 2) and parallel to the line 2x - 4y - 8 = 0.
(b) Find the equation of the line that passes through the point (2, 4) and perpendicular to the line 3x + 4y - 22 = 0.
Answer:
2x – 4y – 8 = 0
4y = 2x – 8 y= 1 x – 2.
2

Since the lines are parallel, m =

 y–2=

1
.
2

1
[x – (-2)]
2

y=

1 x+3 2
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MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

(a)

3x + 4y – 22 = 0
4y = -3x + 22 y=  3 22
+
4
4

Since the lines are perpendicular, m =

 the equation of the line is , y – 4 =

y=

4
.
3

4
(x – 2)
3

4
4
x+
3
3

5. Given the quadratic equation y = –x2 + 6x +7. Determines
(i)
(ii)
(iii)

if the parabola opens up or opens down. the coordinates of the vertex the x intercepts

Answer:
(i)

the parabola opens down because a < 0.

(ii)

the coordinates of the vertex
=

=

=

(

)

=

=3
(

)( )
(

)

= 16

 the coordinates are (3, 16)
(iii)
6.

the x intercepts

Solve the simultaneous equations
5x−3y = 26 (1)
4x+ 2y = 34 (2) .

Solution
If y is eliminated, multiply equation (1) by 2 and equation (2) by 3. This gives
10x−6y = 52 (3)
12x+ 6y = 102 (4)

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MANUAL TUGASAN PELAJAR :MATEMATIK UNTUK EKONOMI DAN PENGURUSAN

Therefore,
10x − 6y = 52

(3)

12x + 6y = 102

(4)

22x + 0y = 154
So x = 154/ 22 = 7.
By substituting,

5(7)−3y = 26
35−3y = 26
−3y = 26−35
−3y = −9 y= 3

Hence, x = 7, y = 3.

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