07. CSEC ADD MATHS 2018

[Pages:34]CSEC ADD MATHS 2018

SECTION I Answer BOTH questions. ALL working must be clearly shown.

1. (a) (i)

Given that f ( x) = x2 - 4 for x ? 0 , find the inverse function, stating its

domain. SOLUTION:

Data: f ( x) = x2 - 4 for x ? 0 Required to Find: f -1 ( x), stating its domain.

Solution:

Let y = f ( x)

\ y = x2 - 4 y + 4 = x2

x2 = y + 4

x= y+4 Replace y by x, we obtain:

= + 4

f -1 ( x) = x + 4

$ -ve

So, f -1 ( x) = x + 4 for x ? -4 .

(ii) On the grid provided below, sketch f -1 ( x).

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SOLUTION:

Data: Graph showing f ( x) = x2 - 4 Required To Draw: The graph of f -1 ( x)

Solution:

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(iii) State the relationship between f ( x) and f -1 ( x).

SOLUTION:

Required to state: The relationship between f ( x) and f -1 ( x)

Solution:

( ) ( ) f x ?R?efle?ction?in ?y = x?? f -1 x

The domain of f is the co-domain of f -1. The co-domain of f is the domain of f -1.

If (a, b) is a point in f ( x) then (b, a) will be the corresponding point on f -1 ( x).

(b) Derive the polynomial, P( x) , of degree 3 which has roots equal to 1, 2 and -4 .

SOLUTION:

Required to derive: Polynomial, P( x) , of degree 3 with roots equal to 1, 2 and

-4 . Solution:

If 1, 2 and -4 are roots of the polynomial P( x) , then according to the

Remainder and Factor Theorem, ( x -1), ( x - 2) and ( x - (-4)) will be three

factors of P( x) .

\ P ( x) = ( x -1)( x - 2)( x + 4)

= (x2 - 3x + 2)(x + 4)

= x3 - 3x2 + 2x + 4x2 -12x + 8 = x3 + x2 -10x + 8

Hence, P( x) = x3 + x2 -10x + 8 is the required polynomial of degree 3.

(c) An equation relating V and t given by V = kat where k and a are constants.

(i) Use logarithms to derive an equation of the form y = mx + c that can be used to find the values of k and a.

SOLUTION:

Data: V and t are related by the equation V = kat , where k and a are

constants. Required to express: The equation in the form of a straight line

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Solution:

V = kat

Taking lg:

lgV = (lg kat )

lgV = lg k + lg at lgV = lg k + t lg a

This is of the form y = mx + c , where y = lgV (a variable), m = lg a (a constant), x = t (a variable) and c = lg k (a constant).

(ii) If a graph of y versus x from the equation in Part (c) (i) is plotted, a straight line is obtained. State an expression for the gradient of the graph. SOLUTION: Data: The graph of y versus x from the equation in Part (c) (i) is a straight line. Required to state: An expression for the gradient of the graph Solution:

lgV = (lg a)t + lg k

The above diagram gives an indication of what the sketch may look like. When y vs x is drawn, a straight line of gradient m is obtained and which cuts the vertical axis at c.

So, when the equivalent form of lgV vs t is drawn, the straight line obtained will have a gradient of lg a . (The intercept on the vertical axis will be lg k .)

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2. (a) (i)

Given that g ( x) = -x2 + x - 3, express g ( x) in the form a ( x + h)2 + k ,

where a, h and k are constants.

SOLUTION:

Data: g ( x) = -x2 + x - 3 Required to express: g ( x) in the form a ( x + h)2 + k , where a, h and k

are constants. Solution:

g (x) = -x2 + x -3

g (x) = -(x2 - x)-3

Half the coefficient of x is 1 (-1) = - 1

2

2

So,

g

(

x

)

=

-

? ??

x

-

1 2

2

? ? ?

+ *,

where

* is

to

be determined

Consider

-

? ??

x

-

1

2

?

2 ??

=

-

? ??

x

-

1 2

? ??

? ??

x

-

1? 2 ??

=

-

? ??

x2

-

x

+

1 4

? ??

= -x2 + x - 1 + 4

3 -2

4

- x2 + x - 3

So, * = -2 3 4

Hence,

g

(

x

)

=

-

? ??

x

-

1 2

2

? ? ?

-

2

3 4

is of the form

a( x + h)2 + k ,

where

a = -1, h = - 1 and k = -2 3 .

2

4

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Alternative Method:

a(x + h)2 + k = a(x + h)(x + h) + k

( ) = a x2 + 2hx + h2 + k

= ax2 + 2ahx + ah2 + k So ax2 + 2ahx + ah2 + k = -x2 + x - 3 Equating coefficients:

a = -1

2(-1) = 1 1

=- 2

(

-1)

? ??

-

1 2

2

? ??

+

k

=

-3

k = -2 3 4

So,

g

(

x

)

=

-

? ??

x

-

1 2

2

? ??

-23 4

and

which

is of

the required form

where

a = -1, h = - 1 and k = -2 3 .

2

4

(ii) On the grid provided below, sketch the graph of g ( x), showing the

maximum point and the y-intercept.

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SOLUTION:

Required to sketch: The graph of g ( x), showing the maximum point

and the y-intercept.

Solution:

g

(

x

)

=

-

? ??

x

-

1 2

2

? ??

-

2

3 4

? ??

x

-

1 2

2

? ??

?

0

"x

So

g ( x)

has

a

maximum

value

of

-(0)- 2 3

=

3 -2

44

at

-

? ??

x

-

1 2

2

? ? ?

=0

i.e. x = 1 . So, on the graph of g ( x), the coordinates of the maximum

2

point

are

? ??

1 2

,

-

2

3 4

? ??

.

When x = 0 , g (0) = -(0)2 + (0) - 3 = -3

\ g ( x) cuts the vertical axis at (0, - 3).

Let g ( x) = 0

\-x2 + x - 3 = 0 is of the form ax2 + bx + c = 0, where a = -1, b = 1, c = -3

b2 = (1)2 =1 and 4ac = 4(-1)(-3) =12

Notice b2 < 4ac, hence, g ( x) has no real solutions and so does not cut

the horizontal axis.

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(b) In a geometric progression, the 3rd term is 25 and the sum of the 1st and 2nd terms is 150. Determine the sum of the first four terms, given that r > 0 .

SOLUTION: Data: Geometric series with the 3rd term = 25 and the sum of the first and second terms = 150 . The common ratio, r > 0 . Required to calculate: The sum of the first four terms

Calculation: Let Tn = nth term for the geometric progression

Tn = ar n - 1, where a = 1st term Hence, T3 = ar2 = 25 (data)

T1 = a and T2 = ar So a + ar = 150 (data)

Let

ar2 = 25

...

a + ar = 150

...

From :

25 a = r2

Substitute into : 25 25 r2 + r2 ? r = 150

?r 2

25 + 25r = 150r2

1+ r = 6r2

6r2 - r -1 = 0

(3r +1)(2r -1) = 0

So

r = - 1 or 1

32

r > 0 (data), so r = 1 , Substitute r = 1 in :

2

2

a

? ??

1 2

2

? ??

=

25

a = 100

a (1- rn )

Sn = 1- r , r ................
................

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