MAT1613 EXAM PACK 2023

, Memorandum Nov2017 MAT1613 paper

a) f (x) = ln(2x + 1) − 3x , x > 0

2
fJ(x) = −3
2x + 1
2 − 3(2x + 1)
=
2x + 1
−1 − 6x
=
2x + 1
y = —6x − 1 + + + −-
-
y = 2x + 1 - + + +
Sign pattern: x=0 +
f J (x) — + + −
-
-1/2 -1/6

i) Since we are only looking at values > 0 , f decreases on (0, ∞)
ii) f J (x) = 0 when −2x+1
1−6x = 0 i.e.when x = − 1 but - 1 < 0 so no extreme value.
6 6
b)

−6( 2x + 1) + 2(6x + 1)
fJJ(x) =
(2x + 1)2
−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 2 = θ then cos θ = −√32 .but cos−1 is negative in the 2nd quadrant so the
angle with the x-axis in the 2nd quadrant is π6 and so the angle θ = 5π/6

see the diagram:

Now
π 1 π 5π 2+5
tan + cos−1 − = tan( + )) = tan π
3 2 3 6 6
1 7π
= tan = √
6 3




1

,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 (1) (x − 1)3 f JJJ (1) (x − 1)4 fiv (1)
P3,1f (x) = f (1) + + + +
1! 2! 3! 4!

24 192
= 1 − 4 (x − 1) + (x − 1)2 − (x − 1)3
2! 2
3! 3
= 1 − 4 (x − 1) + 12 (x − 1) − 32 (x − 1)
4.a)
lim e2x − ex − x
x→0 x2
We write
e2x − ex − x 0
limx→0 which is of the form
2e 2x − exx2 − 1 0
= limx→0 which is of the form 0
2x 0
= limx→0 4e − e
2x x

3 2
=
2

b)
(cos 4x) + 1
lim
x→π4 (− sin 2x) + 1


(cos 4x) + 1 0
lim is of the form
x→π4 (− sin 2x) + 1 0

−4 sin 4x 0
= lim is of the form
x→π4 −2 cos 2x 0

−4.4 cos 4x cos 4x cos π
= lim = − 4 lim
sin 2x = −4 =4
x→π4 2.2 sin 2x x→π4 sin π2




c)
1
lim (cos 2x)x2
x→0

, 2