SEF 1134
SEMESTER 1, 2O10/2011
QUESTION 1 [4 MARKS]
Find the limit:
(𝑥−3)2 (3−2𝑥 2 )3
𝑥→−∞ (𝑥−3)4 (2𝑥 2 −1)4
lim
[4M ]
QUESTION 2 [6 MARKS]
ax  b  2
1
x
x 0
Find numbers a and b such that lim
[6M]
QUESTION 3 [13 MARKS]
(a) If y  ln(sin 3 2 x) , prove that
2

1  1  x 
(b) Given that cos
2

1 x 
find
3
d2y
2
 dy 
    36  0
dx 2  dx 
[6M]
 ln(tan y )  a , where a is a constant ,
dy
in terms of x and y . Give your answer in the simplest form
dx
[7M]
QUESTION 4 [7 MARKS]
Given that y  ae 2 x sin( x  b) , where a and b are constants.
Find
d2y
dx 2
in terms of y and
DEPARTMENT OF MATHEMATICS
dy
.
dx
[7M]
Page 1
SEF 1134
SEMESTER 1, 2O10/2011
QUESTION 5 [12 MARKS]
𝟐𝒙𝟐
Let 𝑓(𝒙) = (𝟑−𝒙)(𝟑+𝒙)
a)
b)
c)
d)
e)
Find x and y-intercepts,
[1M]
Find all the asymptotes, if any,
[2M]
Identify the critical point(s) and intervals on which 𝑓(𝑥) is increasing and decreasing, [2M]
Identify the inflection point (s) and intervals on which 𝑓(𝑥) is concave up and down, [5M]
Use all the information obtained, sketch graph of 𝑓(𝑥).
[2M]
QUESTION 6 [13MARKS]
(a) Find the critical numbers of the function 𝑓(𝑥) = 6√𝑥 − 𝑥 √𝑥
[5M]
(b) A metal storage with tank with volume V is to be constructed in the shape of a right circular
cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?
[8M]
QUESTION 7 [12 MARKS]
Evaluate the following integrals:
𝟑 𝒙𝟐 +𝟏
(a) ∫𝟏
𝜋
[3M]
√𝒙𝟑 +𝟑𝒙
1
2
(b) ∫02 | − sin 𝑥| 𝑑𝑥
(c) ∫
1
√2𝜃+1 𝑒 √2𝜃+1
[5M]
𝑑𝜃
[4M]
QUESTION 8 [8 MARKS]
Use definition
𝐴𝑟𝑒𝑎 = lim ∑∞
𝑘=1 𝑓( 𝑥𝑘 ) ∆𝑥 with 𝑥𝑘 as the right endpoint of each n subinterval
𝑛→∞
to find the area between the graph of 𝑓(𝑥) = 2𝑥 2 and the interval [1,2].
DEPARTMENT OF MATHEMATICS
Page 2
SEF 1134
SEMESTER 1, 2O10/2011
QUESTION 9 [12 MARKS]
(a) Evaluate the definite integral
𝜋
∫04 𝑠𝑖𝑛3 𝑥 𝑠𝑖𝑛2𝑥 𝑑𝑥
[5M]
(b) Evaluate the integral
∫ 𝑡𝑎𝑛−1 √𝑥 𝑑𝑥
[7M]
QUESTION 10 [13MARKS]
(a) Find the indefinite integral
∫
2𝑥 2 +𝑥+2
𝑥(𝑥−1)2
𝑑𝑥
[8M]
(b) Find the indefinite integral
∫ √36 − 𝑥 2 𝑑𝑥
DEPARTMENT OF MATHEMATICS
[5M]
Page 3
SEF 1134
SEMESTER 1, 2O10/2011
LIST OF FORMULA
TRIGONOMETRIC IDENTITIES
sin 𝑎 sin 𝑏 =
1
[cos(𝑎 − 𝑏) − cos(𝑎 + 𝑏)]
2
cos 𝑎 cos 𝑏 =
1
[cos(𝑎 − 𝑏) + cos(𝑎 + 𝑏)]
2
sin 𝑎 cos 𝑏 =
1
[sin(𝑎 − 𝑏) + sin(𝑎 + 𝑏)]
2
1
1
sin 𝑎 + sin 𝑏 = 2 𝑠𝑖𝑛 (𝑎 + 𝑏) cos (𝑎 − 𝑏)
2
2
1
1
sin 𝑎 − sin 𝑏 = 2 𝑠𝑖𝑛 (𝑎 − 𝑏) cos (𝑎 + 𝑏)
2
2
1
1
cos 𝑎 + cos 𝑏 = 2 𝑐𝑜𝑠 (𝑎 + 𝑏) cos (𝑎 − 𝑏)
2
2
1
1
cos 𝑎 − cos 𝑏 = −2 𝑠𝑖𝑛 (𝑎 + 𝑏) sin (𝑎 − 𝑏)
2
2
sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏
sin(𝑎 − 𝑏) = sin 𝑎 cos 𝑏 − cos 𝑎 sin 𝑏
cos (𝑎 + 𝑏) = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏
cos (𝑎 − 𝑏) = cos 𝑎 cos 𝑏 + sin 𝑎 sin 𝑏
tan(𝑎 + 𝑏) =
tan(𝑎) + tan(𝑏)
1 − tan(𝑎) tan(𝑏)
tan( 𝑎 − 𝑏) =
tan(𝑎) − tan(𝑏)
1 + tan(𝑎) tan(𝑏)
𝑠𝑖𝑛2 𝑎 + 𝑐𝑜𝑠 2 𝑎 = 1
𝑡𝑎𝑛2 𝑎 + 1 = 𝑠𝑒𝑐 2 𝑎
1 + 𝑐𝑜𝑡 2 𝑎 = 𝑐𝑠𝑐 2 𝑎
cos (2𝑎) = 𝑐𝑜𝑠 2 𝑎 − 𝑠𝑖𝑛2 𝑎
cos(2𝑎) = 2𝑐𝑜𝑠 2 𝑎 − 1
cos(2𝑎) = 1 − 2 𝑠𝑖𝑛2 𝑎
DEPARTMENT OF MATHEMATICS
sin(2𝑎) = 2 sin 𝑎 cos 𝑎
Page 4
SEF 1134
SEMESTER 1, 2O10/2011
TRIGONOMETRIC FUNCTIONS
DIFFERENTIATION FORMULA
d
x  1
dx
d  x r 1 
r
2.

  x r  1
dx  r  1
1.
 
d u
du
3.
b  b u ln b 
dx
dx
d u
du
4.
e  eu 
dx
dx
d
ln u   1  du
5.
dx
u dx
d
log b u   1  du
6.
dx
u ln b dx
d
sin x  cos x
7.
dx
d
 cos x  sin x
8.
dx
d
tan x  sec 2 x
9.
dx
d
 cot x  csc 2 x
10.
dx
d
sec x  sec x tan x
11.
dx
d
 csc x  csc x cot x
12.
dx
 
DEPARTMENT OF MATHEMATICS
INTEGRATION FORMULA
 dx  x  C
x r 1
 x dx  r  1  C r  1
r
 cos xdx  sin x  C
 sin xdx   cos x  C
 sec xdx  tan x  C
 csc xdx   cot x  C
 sec x tan xdx  sec x  C
 csc x cot xdx   csc x  C
 csc xdx  ln csc x  cot x  C
 sec xdx  ln sec x  tan x  C
2
2
Page 5
SEF 1134
SEMESTER 1, 2O10/2011
Reduction Formula
1
n 1
xdx   sin n 1 x cos x 
sin n  2 xdx

n
n
1
n 1
n
n 1
n2
 cos xdx  n cos x sin x  n  cos xdx
tan n 1 x
n
tan
xdx

  tan n  2 xdx

n 1
sec n  2 x tan x n  2
n
n2
 sec xdx  n  1  n  1  sec xdx
 sin
n
INVERSE TRIGONOMETRIC FUNCTIONS FORMULAS
DIFFERENTIATION
d
1
du
[sin 1 u ] 
dx
1  u 2 dx
d
1
du
[cos 1 u ]  
dx
1  u 2 dx
d
1 du
[tan 1 u ] 
dx
1  u 2 dx
d
1 du
[cot 1 u ]  
dx
1  u 2 dx
d
1
du
[sec 1 u ] 
dx
u u 2  1 dx
d
1
du
[csc 1 u ]  
dx
u u 2  1 dx
INTEGRATION
a

u
2
1
1
u
du  tan 1  C
2
a
a
u
1
a u
2
2
du  sin 1
1
u a
2
2
du 
u
C
a
1
u
sec 1  C
a
a
DEPARTMENT OF MATHEMATICS
Page 6
SEF 1134
SEMESTER 1, 2O10/2011
SOLUTIONS
Question 1
(𝑥 − 3)2 (3 − 2𝑥 2 )3
𝑥→−∞ (𝑥 − 3)4 (2𝑥 2 − 1)4
lim
𝑥 2 (−2𝑥 2 )2
𝑥→−∞ (𝑥)4 (2𝑥 2 )4
M1A1
= lim
−8𝑥 8
𝑥→−∞ 16𝑥 8
lim
=
−8
16
=
−1
2
M1A1
Question 2
( ax  b  2) ( ax  b  2)
1
x
x 0
( ax  b  2)
lim
M1
(ax  b)  4
1
x
x 0
( ax  b  2)
A1
for limit to exists, b  4  0
M1
lim
b4
A1
ax
1
x  0 x ( ax  4  2)
lim
a
42
1
a=4
DEPARTMENT OF MATHEMATICS
M1
A1
Page 7
SEF 1134
SEMESTER 1, 2O10/2011
Question 3
(a) y  ln(sin 3 2 x)
dy (3 sin 2 2 x)( 2 cos 2 x) 6 cos 2 x


 6 cot 2 x
dx
sin 2 x
sin 3 2 x
d2y
dx
3
2
M1A1
 12 csc 2 2 x
A1
2
d2y
 dy 
    36 = 3(  12 csc 2 2 x)  (6 cot 2 x) 2  36
dx 2  dx 
2

1  1  x 
(b) cos
2

=  36(cot 2 2 x  1)  36 cot 2 2 x  36
M1
=  36 cot 2 2 x  36  36 cot 2 2 x  36 =0
A1
 ln(tan y )  a
1 x 

M1
(2 x)(1  x 2 )  (1  x 2 )( 2 x)
2 2
1 x

(1  
1 x2 


1  x 2 2
 sec 2 y  dy

+ 
=0
 tan y  dx


M1M1M1A1
  4x 
 dy
1

  

0

2
2


 cos y sin y  dx
(1  x 2 ) 2  (1  x 2 ) 2  (1  x ) 
 (1  x 2 )
2
1 x2

 dy
1
  0
 cos y sin y  dx
+ 
dy 2 sec y csc y

dx
1 x2
DEPARTMENT OF MATHEMATICS
M1
A1
A1
Page 8
SEF 1134
SEMESTER 1, 2O10/2011
Question 4
y  ae 2 x sin( x  b)
dy
 2ae  2 x sin( x  b)  ae  2 x cos( x  b)
dx
M1M1A1
dy
=  2 y  ae 2 x cos( x  b)
dx
d2y
dx 2
d2y
dx 2
d2y
dx 2
 2
dy
 2ae  2 x cos( x  b)  ae  2 x sin( x  b)
dx
= 2
= 4
dy
 dy

 2  2 y   y
dx
 dx

dy
 5y
dx
DEPARTMENT OF MATHEMATICS
M1A1
M1
A1
Page 9
SEF 1134
SEMESTER 1, 2O10/2011
Question 5
a) X-int: 𝑦 = 0 ⇒ 2𝑥 2 = 0 ⇒ 𝑥 = 0
y-int: 𝑥 = 0 ⇒ 𝑦 = 0
A1
b) VA: 𝑥 = 3, 𝑥 = −3
HA: lim 𝑓(𝑥) = −2 ⇒ 𝑦 = −2 𝑖𝑠 𝑉𝐴
M1A1
𝑥→∞
36𝑥
c) 𝑓́ (𝑥) = (𝑥2 −9)2
𝑓́ (𝑥) = 0 ⇒ 36𝑥 ⇒ 𝑥 = 0
M1
𝐼𝑓 𝑥 < 0, 𝑓́ (𝑥) < 0 ⇒ 𝑓 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝐼𝑓 𝑥 < 0, 𝑓́ (𝑥) < 0 ⇒ 𝑓 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
The graph has one critical point (0,0).
d) 𝑓 ′′ (𝑥) =
−108𝑥 2 −324
(𝑥 2 −9)3
M1
−108𝑥 2 − 324 ≠ 0 𝑓 ′′ (𝑥) ≠ 0 ⇒ 𝑛𝑜 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
For 𝑥 < −3 𝑓 ′′ (𝑥) < 0 ⇒ the graph concave down
For −3 < 𝑥 < 3, 𝑓 ′′ (𝑥) > 0 ⇒ the graph concave up
For 𝑥 > 3, 𝑓 ′′ (𝑥) < 0 ⇒ the graph concave down
e) Graph
DEPARTMENT OF MATHEMATICS
A1
M1 A1
M1 A1
A2
Page 10
SEF 1134
SEMESTER 1, 2O10/2011
Question 6
(a)
𝑓́ (𝑥) = 𝑥 3
4𝑥
𝑓́ (𝑥) =
2𝑥
2√𝑥 2 −3
+ 3𝑥 2 √𝑥 2 − 3 =
𝑥4
√𝑥 2 −3
+ 3𝑥 2 √𝑥 2 − 3
M1
4 −9𝑥 2
√𝑥 2 −3
2 (4𝑥 2
𝑓́ (𝑥) = 0, 𝑥
𝑥 = 0 𝑜𝑟 𝑥 =
A1
− 9) = 0
M1
3
∓2
A1
𝑥 2 − 3 = 0,
𝑥 = ∓√3
Critical points are: 𝑥 = ∓√3
A1
(b) 10 = (2𝑤)(𝑤). ℎ = 2𝑤 2 ℎ
ℎ=
M1
5
𝑤2
A1
The cost of material:
𝐶 = 10(2𝑤 2 ) + 6[2(2𝑤ℎ) + 2ℎ𝑤] = 20𝑤 2 + 36𝑤ℎ
5
)
𝑤2
𝐶(𝑤) = 20𝑤 2 + 36𝑤 (
= 20𝑤 2 +
M1
180
𝑤
A1
3
180
40𝑤 −180
𝐶́ (𝑤) = 40𝑤 − 𝑤 2 =
=0
𝑤2
40𝑤 3 − 180 = 0
3
𝑤=√
M1
9
2
A1
3
9
2
There is an absolute minimum for 𝑤 = √
3
3 9
Since 𝐶́ (𝑤) < 0 for 0 < 𝑤 < √ and 𝐶́ (𝑤) > 0 for 𝑤 >
2
9
2
√
3
9
3
9
𝐶 (√ ) = 20(√ )2 +
2
2
180
3 9
√
2
= $163.54
DEPARTMENT OF MATHEMATICS
M1A1
Page 11
SEF 1134
SEMESTER 1, 2O10/2011
Question 7
(a)
S1
M1
A1
(b)
𝜋
2
𝜋
𝜋
61
2
1
1
∫ | − sin 𝑥| 𝑑𝑥 = ∫ − sin 𝑥 𝑑𝑥 + ∫ 𝑠𝑖𝑛𝑥 − 𝑑𝑥 𝑺𝟏 𝑴𝟏
𝜋
2
2
2
0
0
6
𝜋
𝜋
6
1
1 2
= [ 𝑥 + 𝑐𝑜𝑠𝑥] + [𝑐𝑜𝑠𝑥 − 𝑥] 𝑨𝟏
2
2 𝜋6
0
𝜋
𝜋
𝜋
𝜋
𝜋
= [( + 𝑐𝑜𝑠 6 ) − 1] + [(− ) − (𝑐𝑜𝑠 6 − )]
12
4
12
𝑴𝟏
𝜋
= −1 − 24 𝑨𝟏
(c)
∫
1
√2𝜃 + 1 𝑒 √2𝜃+1
𝑑𝜃
𝐿𝑒𝑡 𝑢 = √2𝜃 + 1 , 𝑑𝑢 =
∴∫
1
√2𝜃 + 1 𝑒 √2𝜃+1
2
2√2𝜃 + 1
𝑑𝜃 =
1
√2𝜃 + 1
𝑑𝜃 𝑺𝟏 𝑴𝟏
𝑑𝜃
−
= ∫ 𝑒 −𝑢 𝑑𝑢 = − 𝑒 √2𝜃+1 + 𝑐 𝑨𝟏 𝑴𝟏
DEPARTMENT OF MATHEMATICS
Page 12
SEF 1134
SEMESTER 1, 2O10/2011
Question 8
∆𝒙 =
2−1 1
= ,
𝑛
𝑛
𝑥𝑘 = 1 + 𝑘
1
𝑛
𝑺𝟏
∞
𝐴𝑟𝑒𝑎 = lim ∑ 𝑓( 𝑥𝑘 ) ∆𝑥
𝑛→∞
𝑘=1
𝑛
∞
1 1
𝑘 21
= lim ∑ 𝑓(1 + 𝑘 )
= lim ∑ 2 (1 + )
𝑛→∞
𝑛→∞
𝑛 𝑛
𝑛 𝑛
𝑘=1
𝑘=1
= lim ∑𝑛𝑘=1 2 (1 +
𝑛→∞
= lim
2
𝑛→∞ 𝑛
𝑴𝟏 𝑨𝟏
2𝑘
𝑛
𝑘2
+ 𝑛2 )
2
1
𝑛
𝑴𝟏
1
2
[∑𝑛𝑘=1 1 + 𝑛 ∑𝑛𝑘=1 𝑘 + 𝑛2 ∑∞
𝑘=1 𝑘 ] 𝑴𝟏
2
2 𝑛(𝑛 + 1) 1 𝑛(𝑛 + 1)(2𝑛 + 1)
[𝑛 +
+ 2
] 𝑴𝟏
𝑛→∞ 𝑛
𝑛
2
𝑛
6
= lim
= lim [2 +
𝑛→∞
4𝑛(𝑛 + 1) 2𝑛(𝑛 + 1)(2𝑛 + 1)
+
] 𝑺𝟏
2𝑛2
6𝑛3
2
=2 + 2 + 3 =
14
3
𝑨𝟏
Question 9
𝜋
(a) ∫04 𝑠𝑖𝑛3 𝑥 𝑠𝑖𝑛2𝑥 𝑑𝑥
𝜋
= ∫04 𝑠𝑖𝑛2 𝑥 2𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥
M1
𝜋
4
= 2 ∫0 𝑠𝑖𝑛3 𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥
A1
𝜋
1
Let u = sinx
x= 4
u=
du = cosx dx
x=0
u=0
= 2 ∫ 𝑢3 𝑑𝑢
1
=2
𝑢 √2
]
4 0
=
√2
M1
A1
1
8
DEPARTMENT OF MATHEMATICS
A1
Page 13
SEF 1134
SEMESTER 1, 2O10/2011
(b) ∫ 𝑡𝑎𝑛−1 √𝑥 𝑑𝑥
Let u= 𝑡𝑎𝑛−1 √𝑥
du= 2
1
√𝑥(1+𝑥)
dv = dx
v=x
𝑑𝑥
∫ 𝑡𝑎𝑛−1 √𝑥 𝑑𝑥 = 𝑥 𝑡𝑎𝑛−1 −
Let
t = √𝑥
M1
1
2
√𝑥
A1
∫ 1+𝑥 𝑑𝑥
M1
𝑡2 = 𝑥
2𝑡𝑑𝑡 = 𝑑𝑥
𝑡2
𝑡
1
∫ 1+𝑡 2 2𝑡 𝑑𝑡 = 2 ∫ 1+𝑡 2 𝑑𝑡 = 2 [∫ 𝑑𝑡 − ∫ 1+𝑡 2 𝑑𝑡]
M1(long division)
= 2[𝑡 − 𝑡𝑎𝑛−1 𝑡] = 2[√𝑥 − 𝑡𝑎𝑛−1 √𝑥]
M1
A1
Combining :
= 𝑥𝑡𝑎𝑛−1 √𝑥 − √𝑥 − 𝑡𝑎𝑛−1 √𝑥 + 𝑐
A1
Question 10
(a) ∫
2𝑥 2 +𝑥+2
𝑥(𝑥−1)2
𝐴
𝑑𝑥 =
𝑥
𝐵
+ 𝑥−1 +
𝐶
S1
(𝑥−1)2
2𝑥 2 + 𝑥 + 2 = 𝐴(𝑥 − 1)2 + 𝐵𝑥(𝑥 − 1) + 𝐶𝑥
A=2 , C=5, B=0
2𝑥 2 +𝑥+2
𝑥(𝑥−1)2
=
2
M1
M1 A1(2 correct)
2
5
A1
+ (𝑥−1)2
𝑥
5
= ∫ 𝑥 𝑑𝑥 + ∫ (𝑥−1)2 𝑑𝑥
1
= 2𝑙𝑛𝑥 + 5 ∫ (𝑥−1)2 𝑑𝑥
M1
Let u = x-1
du = dx
1
1
∫ 𝑢2 𝑑𝑢 = − 𝑢 = −
∫
2𝑥 2 +𝑥+2
𝑥(𝑥−1)2
5
𝑑𝑥 = 2 ln 𝑥 − 𝑥−1 + 𝑐
DEPARTMENT OF MATHEMATICS
1
𝑥−1
A1
A1
Page 14
SEF 1134
(b)
SEMESTER 1, 2O10/2011
∫ √36 − 𝑥 2 𝑑𝑥
𝑙𝑒𝑡 𝑥 = 6𝑠𝑖𝑛𝜃
𝑑𝑥 = 6𝑐𝑜𝑠𝜃 𝑑𝜃
M1
∫ 6√1 − 𝑠𝑖𝑛2 𝜃 6𝑐𝑜𝑠𝜃 𝑑𝜃
A1
1
1
= 36 ∫ 𝑐𝑜𝑠 2 𝜃𝑑𝜃 = 18 [2 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 + 2 𝜃] + 𝑐
= 18 [
=
√36−𝑥 2 𝑥
6
𝑥√36−𝑥 2
2
𝑥
. 6 + 𝑠𝑖𝑛−1 (6)] + 𝑐
𝑥
+ 18 𝑠𝑖𝑛−1 (6) + 𝑐
DEPARTMENT OF MATHEMATICS
- Reduction formula
M1
- Use triangle
A1
A1
Page 15