MAT1510 PRECALCULUS MATHS 2022

MAT1510
PRECALCULUS MATHS

2022

, MAT1510

PRECALCULUS​​MATHS​​1
Jan-Feb
​​2022​​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

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