MAT1510 EXAM PACK 2023

MAT1510
EXAM PACK
2023

QUESTIONS
AND
ANSWERS

, MAT1510

PRECALCULUS​​MATHS​​1

Jan-Feb​​2017​​Solutions

QUESTION​​1
Given​​f (x) = |1−3x|
1
​​and​​g (x) = log 1 ( 3x−2
1
) − log 3 x
3




1.1 Df ​​and​​Dg
Considering​​f (x), ​​the​​function​​exist​​when
​​1 − 3x ≠0 ⇒x ≠ 31 ​​[​a ​function
​ ​is​ ​undefined
​ ​​
when ​divided
​ ​​ ​zero,
by ​ ​so
​ ​the
​ denominator
​​ ​in​
f(x) ​cannot
​ ​be
​ ​zero
​ ].​

Considering g (x) ,​​we​​can​​re-write​​the​​function​​as,
g (x) = log 1 (3x − 2)−1 − log 3 x =− log 1 (3x − 2)− log 3 x .​​This​​function​​valid​​if
3 3

3x − 2 > 0 ⇒x > 2
3 ​​and​​​if​​​x > 0 ​​combining​​the​​two,​​we​​have​​x > 2
3


Df (Domain​​of​​the​​function​​f(x))​​x : x ∈R, x ≠ 1
3
Dg (Domain​​of​​the​​function​​g(x))​​x : x ∈R, x > 2
3


1.2 f (x) > 2

1
|1−3x| >2

​​If​​1 − 3x ​​is​​greater​​than​​zero,​​then​​|( 1 − 3x )|= 1 − 3x ,​​hence
1
1−3x >2

1 > 2 (1 − 3x)

1 > 2 − 6x

6x > 1

1
x> 6



If​​1 − 3x ​​is​​less​​than​​zero,​​then​​|( 1 − 3x )|= − (1 − 3x ),​​hence
1
−(1−3x) >2

1
3x−1 >2

1 > 2 (3x − 1)

1 > 6x − 2

, 6x < 3

1
x< 2


Solution​​set:​​x∈R, 1
6 < x < 21 , x ≠ 1
3



1.3 g (x) > 0
1
log 1 ( 3x−2 ) − log 3 x > 0
3



Considering​​the​​hint,​​we​​apply​​the​​change​​of​​base​​formula
log a
log b a = logcc b

In​​general​​the,​​the​​logarithm​​of​​a​​number​​to​​the​​base​​of​​a​​fraction​​is​​always​​difficult​​to​​deal
with,​​so​​we​​choose​​to​​change​​base​​31 ​​​to​​base​​3

1
log 3 ( 3x−2 )
log 3 13
− log 3 x > 0

−log 3 (3x−2 )
−log 3 3 − log 3 x > 0

log 3 (3x − 2 ) − log 3 x > 0

log 3 ( 3x−2
x )>0

3x−2
x > 30

3x−2
x >1

3x − 2 > x

2x > 2

x>1

​​Solution​​set:​​x∈R, x > 1

QUESTION​​2


Suppose​​we​​​have​​​two​​numbers​​and​​​whose​​​difference​​i​s​​8​and​ ​.

2.1 Putting​​the​​above​​statement​​in​​a​​mathematical​​expression​​we​​have,​​y − x = 8
Therefore​​y = x + 8 ,​​and​​f (x) = x + 8


2.2 S (x, y ) = x 2 + y2 ,​​and​​from​​the​​above​​expression​​we​​have
S (x) = x2 + (x + 8)2

, S (x) = x 2 + x2 + 16x + 64

S (x) = 2x2 + 16x + 64

2.3​​Re-write​​the​​function​​in​​the​​form​​S (x) = a(x − p)2 + q ​​in​​other​​words​​completing​​the
square.​​In​​this​​form​​the​​min​​(or​​max)​​value​​of​​the​​function​​[turning​​point],​​occurs​​when
independent​​variable,​​= p ​​.​​So​​we​​have
S (x) = 2x2 + 16x + 64

S (x) = 2[x2 + 8x + 32]

S (x) = 2[(x + 4)2 − 16 + 32]

S (x) = 2[(x + 4)2 + 16]

S (x) = 2(x + 4)2 + 32



S (x) = 2(x −− 4) 2 + 32

So​​the
x =− 4

y =− 4 + 8 = 4

QUESTION​​3
3.1​​​Considering​​the​​given​​information​​we​​are​​given​​two​​point​​on​​the​​parabola​​i.e.​​(0,5)​​and
the​​tuning​​point​​(-2,9).​​Using​​the​​turning​​point​​we​​have
f (x) = a (x −− 2) 2 + 9

f (x) = a (x + 2)2 + 9

Then​​using​​the​​other​​point
5 = a (0 + 2)2 + 9

5 = 4a + 9

a =− 1

f (x) =− (x + 2)2 + 9

Where​​a =− 1, h =− 2, ​​and​​k = 9


3.2​​R​​and​​S​​are​​roots​​(x-intercepts).​​They​​occur​​when​​f (x) = 0 ,​​therefore
− (x + 2)2 + 9 = 0