Memorandum: MAT1613 May2016.

1. (a)The y intercept is where x = 0 :

1 − 2x2
f (x) =
x2 − 1
then
1
f (0) = = −1
−1
(b) For the horizontal asymptote we divide by the hghest power of x :
1 − 2x2 1/x2 − 2
f (x) = =
x2 − 1 1 − 1/x2

hence

1/x2 − 2 −2
lim f (x) = lim = = −2
x→±∞ x→±∞ 1 − 1/x2 1
and so the horizontal asymptote is y = −2.
Fot the vertical asymptote: The function is not defined for x = ±1.We check whether
or not there are vertical asymptotes at x = ±1 by calculating the following limits:

1 − 2x2 1 − 2x2
lim+ = −∞; lim− =∞
x→1 x2 − 1 x→1 x2 − 1

1 − 2x2 1 − 2x2
lim + = ∞; lim − = −∞
x→−1 x2 − 1 x→−1 x2 − 1
Hence the verical asymptotes are x = 1 and x = −1.
(c)
0 (x2 − 1)(−4x) − (1 − 2x2 )2x −4x3 + 4x − 2x + 4x3 2x
f (x) = = =
(x2 − 1)2 (x2 − 1)2 (x2 − 1)2
and
2(x2 − 1)2 − 2x2(x2 − 1)2x
f 00 (x) =
(x2 − 1)4
2(x − 1)(x2 − 1 − 4x2 )
2
=
(x2 − 1)4
2(−3x2 − 1) −2(3x2 + 1)
= =
(x2 − 1)3 (x2 − 1)3
−2(3x2 + 1)
=
(x2 − 1)2 (x − 1)(x + 1)

1
Since (x2 − 1)2 ≥ 0 and (3x2 + 1) > 0 and f 00 (x) is undefined at x = ±1 we consider y = −2
, the undefined values and y = (x − 1) and y = (x + 1) in the sign pattern.
Sign pattern for f 00 (x) :

y =x−1 − − +
y =x+1 − + +
y = −2 − − −
f 00 (x) − −1 + 1 −

i) f is concave up on (−1, 1) and concave down on (∞, −1) ∪ (1, ∞)

−2(3x2 +1)
ii) Since f 00 (x) = 0 where (x2 −1) 2
2 (x−1)(x+1) = 0 .which is never possible since −2(3x + 1) < 0

for all while x, there is no reflection point.

2.
Find the exact value of
1
sin[tan−1 (−1) + 3 cos−1 (− )]
2
Let tan−1 (−1) = θ then tan θ = −1 and so θ = − π4
Let cos−1 (− 21 ) = φ then cos φ = − 12 and so φ = 2π 3
Then
π 2π π −1
sin[− + 3( )] = sin[− ] = √
4 3 4 2
3. √ 1
f (x) = 2 x = 2x 2 f (1) = 2
1
f 0 (x) = x− 2 f 0 (1) = 1
3
f 00 (x) = − 12 x− 2 f 00 (1) = − 12
5 000
f 000 (x) = 34 x− 2 f (1) = 43
7
f iv (x) = − 15 8
x− 2 f iv (1) = − 15 8

(x − 1) f 0 (1) (x − 1)2 f 00 (1) (x − 1)3 f 000 (1) (x − 1)4 f 00v (1)

P4,1 f (x) = f (1) + + + +
1! 2! 3! 4!
1 3 15
= 2 + (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4
2.2! 4.3! 8.4!
1 1 5
= 2 + (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4
4 8 64

4.a) lim+ x2 ln x
x→0

ln x −∞
lim+ x2 ln x = lim+ 2
which is of the form
x→0 x→0 1/x ∞
1/x
= lim+
x→0 −2x−3
x2
= lim+
x→0 −2
= 0

2
b)

1 x (x − 1) − x2 ln x 0
lim+ ( − ) = lim+ 2
which is of the form
x→1 x ln x x − 1 x→1 (x − x) ln x 0
2
1 − [x /x + 2x ln x] 0
= lim 2
which is of the form
x→1+ (x − x)/x + ln x(2x − 1) 0
1 − x − 2x ln x 0
= lim which is of the form
x→1+ x − 1 + (2x − 1) ln x 0
−1 − [2x/x + 2 ln x]
= lim
x→1+ 1 + (2x − 1)/x + 2 ln x
−3 − 2 ln x
= lim
x→1+ 1 + 2 − 1/x + 2 ln x
−3
=
2

c) lim (1 + tan x)1/x We have a form 1∞ .

x→0

ln lim (1 + tan x)1/x = lim ln(1 + tan x)1/x

x→0 x→0
ln(1 + tan x) 0
= lim which is of the form
x→0 x 0
1 2
(1+ tan x)
sec x
= lim
x→0 1
1
= sec2 0
1 + tan 0
= 1

Hencelim (1 + tan x)1/x = e1 = e

x→0
5.
Rotation around the y q− axis so all notaion in terms of y.
y−1
Outer radius : R(y) = 2
y−1
Inner radius: r(y) = 2
q
Intersection points: ⇔ y−12
= y−1
2
so 2(y −1) = (y −1)2 i.e. y 2 −4y +3 = (y −3)(y −1) = 0
then y = 1 or y = 3.

3
 3
r
y−1 2 y−1 2
Z
V = π ( ) −( ) ]dy
2 2
Z1 3
= π [(y − 1)/2 − (y 2 − 2y + 1)/4]dx
Z1 3
π
= [−y 2 + 4y − 3]dy
4 1
π −1 3
= [ y + 2y 2 − 3y] 31
4 3
π π 4 π
= [−9 + 18 − 9 + 1/3 − 2 + 3] = . =
4 4 3 3

6. Z √
sec2 (3+ln x)
a) x
dx ,x > 0
√
Let u = 3 + ln x , du
dx
= 1
2x
i.e. 2du = dx
x

√
sec2 (3+ ln x)
Z Z
dx = 2 sec2 udu
x
= 2 tan u + c
√
= 2 tan(3+ ln x) + c

Z
2
b) ln x4 dx, x > 1
2
u = ln x4 dv = 1
du
dx
= x42 . 2x
4
= 2
x
v=x

x2 x2
Z Z
ln dx = x ln − 2 1dx
4 4
x2
= x ln − 2x + c
4
Z 1
dx
c) 3
−∞ (3−2x)
We first look at the indefinite integral:
Let u = 3 − 2x then dudx
= −2
So we have
Z Z
dx 1
= u−3 du
(3 − 2x)3 −2
1 −2
= u +c
4
1
= +c
4(3 − 2x)2

4
 Z 1
dx 1
= lim |1
−∞ (3 − 2x)3 t→−∞ 4(3 − 2x)2 t

1 1
= − lim
4 t→−∞ 4(3 − 2t)2
1
=
4

7.
x−1
Z
dx
(x + 1)(x2 + x + 1)
We see that for y = x2 + x + 1, b2 − 4ac = −3 < 0, so

x−1 A Bx + C
= +
(x + 1)(x2 + x + 1) x + 1 x2 + x + 1
A(x2 + x + 1) + (Bx + C)(x + 1)
=
(x + 1)(x2 + x + 1)

Then

x − 1 = A(x2 + x + 1) + (Bx + C)(x + 1) = (A + B)x2 + (A + B + C)x + (A + C)

Then
A + B = 0, A + B + C = 1 and A + C = −1
We have

So C = 1, A = −2
and B = 2

x−1 −2
Z Z Z
2x + 1
dx = dx + dx
(x + 1)(x2 + x + 1) x+1 x2 + x + 1
Z Z
1 1
= −2 du + dv where u = x + 1, du/dx = 1, v = x2 + x + 1, dv/dx =
u v
= −2 ln |u| + ln |v| + c
= −2 ln |x + 1| + ln x2 + x + 1 + c
(x2 + x + 1)
= ln +c
(x + 1)2

8.a) g(x) = x2 + 2x + 5 = (x + 1)2 + 4

5
b) (2)
Use (a) and trigonometric substitution to determine the integral.
Z Z
dx dx
√ . dx. = p
x2 + 2x + 5 (x + 1)2 + 4
Z
1 dx
= q
2 ( x+1 )2 + 1
2

Let
x+1 dx
= tan θ i.e. x = 2 tan θ − 1 then = 2 sec2 θ
2 dθ
Then
2 sec2 θdθ
Z Z
1 dx 1
q = √
2 ( x+1 )2 + 1 2 tan2 θ + 1
2
Z
= sec θdθ
= ln |sec θ + tan θ| + c
p
(x + 1)2 + 4 x + 1
= ln + +c
2 2
√
x2 + 2x + 5 + x + 1
= ln +c
2

6
9.
Z Z 2
dx 1+z 2
= 2z 1−z 2
dz
0 1 + sin x + cos x 1 + 1+z 2 + 1+z 2
Z 2
1+z 2
= 1+z 2 +2z+1−z 2
dz
1+z 2
Z
2
= dz
2(z + 1)
Z
1
= dz
(z + 1)
= ln |z + 1| + c
x
= ln tan + 1 + c
2

Z π
2 dx π
= ln tan + 1 − ln |tan 0 + 1| = ln 2 − ln 1 = ln 2
0 1 + sin x + cos x 4