Exam (elaborations)
MAT1510 EXAM PACK 2023
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- University Of South Africa
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MAT1510 EXAM PACK2023
LATEST QUESTIONS
AND UPDATED
ANSWERS
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, MAT1510
PRECALCULUSMATHS1
Jan-Feb2017Solutions
QUESTION1
Givenf (x) = |1−3x|
1
andg (x) = log 1 ( 3x−2
1
) − log 3 x
3
1.1 Df andDg
Consideringf (x), thefunctionexistwhen
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) ,wecanre-writethefunctionas,
g (x) = log 1 (3x − 2)−1 − log 3 x =− log 1 (3x − 2)− log 3 x .Thisfunctionvalidif
3 3
3x − 2 > 0 ⇒x > 2
3 andifx > 0 combiningthetwo,wehavex > 2
3
Df (Domainofthefunctionf(x))x : x ∈R, x ≠ 1
3
Dg (Domainofthefunctiong(x))x : x ∈R, x > 2
3
1.2 f (x) > 2
1
|1−3x| >2
If1 − 3x isgreaterthanzero,then|( 1 − 3x )|= 1 − 3x ,hence
1
1−3x >2
1 > 2 (1 − 3x)
1 > 2 − 6x
6x > 1
1
x> 6
If1 − 3x islessthanzero,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
Solutionset:x∈R, 1
6 < x < 21 , x ≠ 1
3
1.3 g (x) > 0
1
log 1 ( 3x−2 ) − log 3 x > 0
3
Consideringthehint,weapplythechangeofbaseformula
log a
log b a = logcc b
Ingeneralthe,thelogarithmofanumbertothebaseofafractionisalwaysdifficulttodeal
with,sowechoosetochangebase31 tobase3
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
Solutionset:x∈R, x > 1
QUESTION2
Supposewehavetwonumbersandwhosedifferenceis8and .
2.1 Puttingtheabovestatementinamathematicalexpressionwehave,y − x = 8
Thereforey = x + 8 ,andf (x) = x + 8
2.2 S (x, y ) = x 2 + y2 ,andfromtheaboveexpressionwehave
S (x) = x2 + (x + 8)2
, S (x) = x 2 + x2 + 16x + 64
S (x) = 2x2 + 16x + 64
2.3Re-writethefunctionintheformS (x) = a(x − p)2 + q inotherwordscompletingthe
square.Inthisformthemin(ormax)valueofthefunction[turningpoint],occurswhen
independentvariable,= p .Sowehave
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
Sothe
x =− 4
y =− 4 + 8 = 4
QUESTION3
3.1Consideringthegiveninformationwearegiventwopointontheparabolai.e.(0,5)and
thetuningpoint(-2,9).Usingtheturningpointwehave
f (x) = a (x −− 2) 2 + 9
f (x) = a (x + 2)2 + 9
Thenusingtheotherpoint
5 = a (0 + 2)2 + 9
5 = 4a + 9
a =− 1
f (x) =− (x + 2)2 + 9
Wherea =− 1, h =− 2, andk = 9
3.2RandSareroots(x-intercepts).Theyoccurwhenf (x) = 0 ,therefore
− (x + 2)2 + 9 = 0
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