04.CSEC ADD MATHS 2015 nsite.com

CSEC ADD MATHS 2015

1. (a)

Data: f ( x) = x2 + 5, x ? 1 and g ( x) = 4x - 3, x ? R. Required to calculate: gf (2)

Calculation:

f (2) = (2)2 + 5

=9

\ gf (2) = g (9) = 4(9) - 3

= 33

OR

gf ( x) = 4( x2 + 5) - 3

( ) \ gf (2) = 4 (2)2 + 5 - 3

= 33

(Note: The given domain for f ( x) and g ( x) are of no consequence in the

question).

(b) Data: h ( x) = 3x + 5 , x ? R, x ? 2.

x-2

Required to calculate: h-1 ( x)

Calculation: Let y = 3x + 5 x-2

xy - 2 y = 3x + 5

xy - 3x = 5 + 2 y

x ( y - 3) = 2 y + 5

x = 2y+5 y-3

Replace y by x to obtain:

h-1 ( x) = 2x + 5 , x ? 3

x-3

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(c) Data: ( x - 2) is a factor of k ( x) = 2x3 - 5x2 + x + 2. Required to factorise: k ( x) completely.

Solution:

If ( x - 2) is a factor of k ( x) = 2x3 - 5x2 + x + 2, then k ( x) is divisible by ( x - 2), according to the remainder and factor theorem

2x2 - x -1 x - 2 2x3 - 5x2 + x + 2

- 2x3 - 4x2 - x2 + x + 2

- - x2 + 2x - x+2

- - x+2 0

We factorise 2x2 - x -1 = (2x +1)( x -1) Hence, k ( x) = ( x - 2)(2x +1)( x -1)

(d) (i)

Data: 16x + 2 = 1 4

Required to calculate: x

Calculation:

16x + 2 = 1 4

( )24

x+2

=

1

(2)2

24 x+8 = 2-2 Equating indices since the bases are equal, we obtain 4x + 8 = -2

\4x = -10

x = -2 1 2

(ii) Data: log3 ( x + 2) + log3 ( x -1) = log3 (6x - 8)

Required to calculate: x Calculation:

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log3 ( x + 2) + log3 ( x -1) = log3 (6x - 8) (Product law)

\log3 {( x + 2) ( x -1)} = log3 (6x - 8)

Remove log3, we obtain

( x + 2) ( x -1) = 6x - 8

x2 + x - 2 = 6x -8 x2 - 5x + 6 = 0

( x - 2) ( x - 3) = 0

Hence, x = 2 or 3

When x = 2, there are no terms of the equation that result in log3 (-ve) or log3 (0). When x = 3 there are no terms of the equation that result in log3 (-ve) or log3 (0).

Hence, x = 2 or 3.

2. (a)

Data: f ( x) = 3x2 - 9x + 4

(i) Required to express: f ( x) in the form a ( x + b)2 + c, where a, b and

c?R. Solution:

a(x + b)2 + c = a(x + b)(x + b) + c

( ) = a x2 + 2bx + b2 + c

= ax2 + 2abx + ab2 + c

Hence,

3x2 - 9x + 4 = ax2 + 2abx + ab2 + c Equating coefficients:

= 3

2 = -9

2(3) ? = -9

6 = -9 1

= -1 2

. + = 4 3.

3 2- 3 + = 4 2 9

3 ? + = 4 4

27 4 + = 4

3 = 4 - 6 4

3 = -2 4

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\ f ( x) = 3x2 - 9x + 4 can be written as

a = 3? R, b = -1 1 ? R and c = -2 3 ? R.

2

4

3???

x

-

1

1 2

2

? ? ?

-

2

3, 4

where

OR

( ) 3x2 - 9x + 4 = 3 x2 - 3x + 4

One half the coefficient of x is 1 (-3) = -1 1

2

2

\3(x2 - 3x) + 4

=

3???

x

-

1

1 2

2

? ? ?

+

*

?

3???

x

-

1

1 2

? ??

? ??

x

-

1

1 2

? ??

=

3???

x2

-

3x

+

9 4

? ??

= 3x2 - 9x + 6 3 4

63-23 =4 44

\* = -2 3 4

If

3x2

-

9x

+

4

=

3???

x

-

1

1 2

2

? ??

-

2

3 4

and which is of the form

a (x + b)2 + c,

where a = 3? R, b = -1 1 ? R and c = -2 3 ? R.

2

4

(ii) Required to state: The minimum point of f ( x).

Solution:

f

(x)

=

3x2

-

9x

+

4

?

3???

x

-11 2

2

? ? ?

-

2

3 4

? ??

x

-11 2

2

? ??

?

0

"x

\

3

? ??

x

-

1

1 2

2

? ? ?

?0

"x

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Hence, if the graph of f ( x) is drawn, the minimum value of f ( x) is

0 - 2 3 = -2 3. The minimum value occurs when 44

3??? x

-

1

1 2

2

? ? ?

=

0

that

is

when x = 1 1 . Therefore, the minimum point on the curve of f ( x) is

2

???1

1 2

,

-

2

3 4

???.

(b) Data: The equation 3x2 - 6x - 4 = 0 has roots a and b .

Required to calculate: The value of 1 + 1 . ab

Calculation:

Recall: If ax2 + bx + c = 0 ?a

x2 + b x + c = 0 aa

If the roots are a and b , then ( x -a )( x - b ) = 0.

\x2 - (a + b ) x + ab = 0

Equating coefficients, we get:

a + b = - b and ab = c

a

a

So, in the equation 3x2 - 6x - 4 = 0

a + b = - (-6)

3

=2

ab = - 4 3

1+1 ab

b +a = a +b ab ab

=

2 -4

3

= -1 1 2

So, 1 + 1 = -1 1 . ab 2

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