University of South Africa MAT1613 Exam
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Memorandum Nov2017 MAT1613
paper a) f (x) = ln(2x + 1) − 3x , x > 0
2 −3
fJ(x) =
2x + 1
2 − 3(2x + 1)
=
2x + 1
−1 − 6x
=
2x + 1
y = —6x − 1 + + + −-
-
y = 2x + 1 - + + +
Sign x=0
f (x)
J — + + −
pattern: +
-1/2 -1/6
-
i) Since we are only looking at values > 0 , f decreases on (0, ∞)
ii) f J (x) = 0 when −1−6x 6 1 but
2x+1 = 0 i.e.when x = − 6 - 1 < 0 so no extreme value.
b)
−6( 2x + 1) + 2(6x + 1)
fJJ(x) =
(2x + 1)2
University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
,University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
−4
=
(2x + 1)2
Since we have that the denominator is always posiitve on the intervalx >− 0
and∞ y = 4 always negative the derivative is always negative and thus the
function is concave down on (0, ).
2.
√ !!
π 3
tan + cos−1 − .
3 2
Let cos−1 −√3 = θ then cos θ = −√3 .but cos−1 is negative in the 2nd
quadrant so the
2 2
angle with the x-axis in the 2nd 6 and so the angle θ = 5π/6
π
quadrant is
see the diagram:
No
w π 1 π 5π 2+5
tan + cos−1 − = tan( + )) = tan π
3 2 3 6 6
University of South Africa MAT1613 Exam
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,University of South Africa MAT1613 Exam
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7π 1
= tan = √
6 3
1
University of South Africa MAT1613 Exam
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, University of South Africa MAT1613 Exam
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3.
f (x) = (2x − 1)−2 f (1) = 1
f J (x) = −2(2x − 1)−3 2 = −4(2x − 1)−3 f J (1) = −4
fJJ (x) = 12(2x − 1)−4 2 = 24(2x − 1)−4 fJJ (1) = 24
f JJJ (x) = −96(2x − 1)−52 = −192(2x − 1)−5 f JJJ (1) = −192
(x − 1) f J (1) (x 1)2 fJJ (x 1)3 f JJJ (x − 1)4 fiv
P3,1f (x) = f + (1) (1) (1)
− −
(1) + 1! + +
2! 3!
4!
24 192
= 1 − 4 (x − 1) + 2! (x − 1)2 − 3! (x − 1)3
2 3
= 1 − 4 (x − 1) + 12 (x − 1) − 32 (x − 1)
4.a)
e2x − ex − x
We write
University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
questions and answers best solved latest
update 2022
Memorandum Nov2017 MAT1613
paper a) f (x) = ln(2x + 1) − 3x , x > 0
2 −3
fJ(x) =
2x + 1
2 − 3(2x + 1)
=
2x + 1
−1 − 6x
=
2x + 1
y = —6x − 1 + + + −-
-
y = 2x + 1 - + + +
Sign x=0
f (x)
J — + + −
pattern: +
-1/2 -1/6
-
i) Since we are only looking at values > 0 , f decreases on (0, ∞)
ii) f J (x) = 0 when −1−6x 6 1 but
2x+1 = 0 i.e.when x = − 6 - 1 < 0 so no extreme value.
b)
−6( 2x + 1) + 2(6x + 1)
fJJ(x) =
(2x + 1)2
University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
,University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
−4
=
(2x + 1)2
Since we have that the denominator is always posiitve on the intervalx >− 0
and∞ y = 4 always negative the derivative is always negative and thus the
function is concave down on (0, ).
2.
√ !!
π 3
tan + cos−1 − .
3 2
Let cos−1 −√3 = θ then cos θ = −√3 .but cos−1 is negative in the 2nd
quadrant so the
2 2
angle with the x-axis in the 2nd 6 and so the angle θ = 5π/6
π
quadrant is
see the diagram:
No
w π 1 π 5π 2+5
tan + cos−1 − = tan( + )) = tan π
3 2 3 6 6
University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
,University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
7π 1
= tan = √
6 3
1
University of South Africa MAT1613 Exam
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update 2022
, University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
3.
f (x) = (2x − 1)−2 f (1) = 1
f J (x) = −2(2x − 1)−3 2 = −4(2x − 1)−3 f J (1) = −4
fJJ (x) = 12(2x − 1)−4 2 = 24(2x − 1)−4 fJJ (1) = 24
f JJJ (x) = −96(2x − 1)−52 = −192(2x − 1)−5 f JJJ (1) = −192
(x − 1) f J (1) (x 1)2 fJJ (x 1)3 f JJJ (x − 1)4 fiv
P3,1f (x) = f + (1) (1) (1)
− −
(1) + 1! + +
2! 3!
4!
24 192
= 1 − 4 (x − 1) + 2! (x − 1)2 − 3! (x − 1)3
2 3
= 1 − 4 (x − 1) + 12 (x − 1) − 32 (x − 1)
4.a)
e2x − ex − x
We write
University of South Africa MAT1613 Exam
questions and answers best solved latest
update 2022
