Final Term Exam
Date: December 26, 2015
Course: Mathematics 1 – A
Duration: 3 hours
Benha University
Faculty of Engineering- Shoubra
Eng. Mathematics & Physics Department
Preparatory Year
(‫)تخلفات‬
 The Exam consists of one page
 Answer All Questions
Question 1
Find y` from the following:
2
(a) 𝑦 = 𝑥 4 + 2𝑥 + 3𝑥
(c) 𝑦 = cos 𝑥 2 − ln sin 𝑥
(e) 𝑦 4 = 𝑥 log(𝑥 + 𝑦)
Question 2
(a)Find the following limits:
i Lim
x→1
x−1
7
x −1
ii Lim
x→0
 No. of questions: 4
 Total Mark: 100
24
2
(b) 𝑦 = cosh 𝑥 . sec 2𝑥
(d) y = tan−1 x + tan−3 x
(f) y = t sec t, 𝑥 = t sinh t
ln(1 + 3x)
2x − 3x
iii Lim
x→0
x − sin x
x3 + x2
iv Lim
x→∞
x 8 + 2x
x + x9
(b)Write the Maclurin’s series of the function: f x = x sin x.
1
1+x
(c)Show that : tanh−1 x = 2 ln 1−x .
(d)Determine the extrema of : f x = 𝑥 3 − 6𝑥 2
g x = 𝑥3 + 3
Question 3
Integrate the following:
(a) 
(c)
x3
(4  x )
2 1/ 2
e x  e x
ex
e
x
dx
dx
(e )  x (x 2  1) dx
sin x  cos x
(b) 
dx
sin x  cos x
12
4
4
6
30
(d)  tan 4 x dx
(f)  e 3x sinh 2x dx
Question 4
(a)Find the area of the surface of revaluation generated by revolving about x7
axis the cycloid x  a(t  sin t), y  (1  cos t), 0  t  2
3
7
(b) Find the area bounded by the curves: y  x , x  2, x  5, y  0 .
(c) Find the volume generated by revolving, about x-axis, the area bounded
6
2
y

x
,
y

0,
x

10
by:
Good Luck
Dr. Mohamed Eid
Dr. Fathi Abdsallam
1
Answer
Answer of Question 1
2
(a)(i) 𝑦` = 4𝑥 3 + 2 𝑥 . ln 2 . 2𝑥 + 3
(ii) 𝑦` = cosh 𝑥 2 . sec 2𝑥 . tan 2𝑥 . 2 + sinh 𝑥 2 . 2𝑥. sec 2𝑥
(iii) 𝑦` = −2𝑥 sin 𝑥 2 −
(iv) 𝑦` =
1
1+𝑥 2
cos 𝑥
sin 𝑥
− 3 𝑡𝑎𝑛−4 𝑥. 𝑠𝑒𝑥 2 𝑥
(v) 4𝑦 3 . 𝑦` = 1. log(𝑥 + 𝑦) + 𝑥
Then 𝑦` 4𝑦 3 −
𝑥
1
.
ln 10 𝑥+𝑦
𝑥
Then 𝑦` =
(vi) 𝑦` =
𝑦
𝑥
1
.
1+𝑦`
ln 10 𝑥+𝑦
= log(𝑥 + 𝑦) +
𝑥
.
1
ln 10 𝑥+𝑦
1
log (𝑥+𝑦)+ln 10 .𝑥 +𝑦
𝑥
1
4𝑦 3 −ln 10 .𝑥 +𝑦
=
𝑡 sec 𝑡.tan 𝑡+sec 𝑡
𝑡 cosh 𝑡+sinh 𝑡
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Answer of Question 2
a i Lim
x→0
x − 1 0 1/2
1
=
=
=
x7 − 1 0
7
14
ln(1 + 3x) 0
ii Lim
= = Lim
x→0 2x − 3x
x→0
0
iii Lim
x→0
ln(1 + 3x)
ln(1 + 3x)
1
3
x
x
= Lim x
=
x
x
((2/3) − 1)
x→0 3 ((2/3) − 1)
ln(2/3)
3x
x
x
x − sin x 0
=
x3 + x2
0
Using L’Hopital’s rule, we get
Lim
x→0
x − sin x
1 − cos x 0
sin x
(0)
= Lim 2
= = Lim
=
=0
3
2
x→0 3x + 2x
x +x
0 x→0 6x + 2 0 + 2
2
1 2
x 8 + 2x ∞
x + x8 = 0 + 0 = 0
iv Lim
=
=
Lim
x→∞ x + x 9
∞ x→∞ 1 + 1 0 + 1
x8
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(b)Since sin x = x −
x3
3!
+
x5
5!
….
x4
x6
Then 𝑓 𝑥 = 𝑥 sin x = x 2 − + ….
3!
5!
--------------------------------------------------------------------------------4-Marks
(c)Proof
--------------------------------------------------------------------------------4-Marks
(d) 𝑆𝑖𝑛𝑐𝑒 𝑓` 𝑥 = 3𝑥 2 − 12𝑥 = 3𝑥 𝑥 − 4 = 0
Then x = 0 , x = 4.
Then, the point (0, 0) is maximum because 𝑓`` 0 = 6𝑥 − 12 = −12
the point (4, – 32) is minimum because 𝑓`` 4 = 6𝑥 − 12 = 12
𝑆𝑖𝑛𝑐𝑒 𝑔` 𝑥 = 3𝑥 2 = 0
Then x = 0 .
Using the first derivative test, we get
𝑔` 0− = 𝑔` −1 = 3𝑥 2 = 3 = 𝑔` 1 = 𝑔` 0+
Then, the point (0, 3) is neither maximum nor minimum.
--------------------------------------------------------------------------------4-Marks
Dr. Mohamed Eid
3
Answer of Question (3)
x3
(a) 
(4  x )
2 1/ 2
dx put x  2sin 
dx  2cos d  , cos  4  x 2
4  x 2  4  4sin 2   2cos ,
Substitute in the problem we have
x3

dx  
2 1/ 2
(4  x )

8sin 3  cos d
 4 sin 3  d  4 sin 2  .sin  d
2cos

 4  1  cos 2  sin  d  4  sin  d  4  cos 2  sin  d
1
 4cos  cos3   c
3
 4 4  x 2 
(b) 

1
4  x2
3

3/ 2
c
sin x  cos x
dx
sin x  cos x
sin x  cos x
sin x  cos x sin x  cos x
(sin x  cos x) 2
dx  

dx  
dx

sin x  cos x
sin x  cos x sin x  cos x
sin 2 x  cos 2 x
  (sin x  cos x)2dx   (sin 2 x  2sin x cos x  cos 2 xdx
1
  (1  2sin x cos x)dx   (1  sin 2x)dx  x  cos 2x  c
2
4
(c)
e x  e x
ex
e
x
dx  ln(e x  e  x )  C
(d)  tan 4 x dx   tan 2 x tan 2 xdx   tan 2 x (sec 2 x  1)dx
  (tan 2 x sec 2 x  tan 2 x)dx
  (tan 2 x sec2 x  (sec2 x  1)dx
  (tan 2 x sec 2 x  sec 2 x  1)dx
1
 tan 3 x  tan x  x  c
3
(e )  x (x 2  1) dx
put y  x 2  1 then dy  2xdx


3/ 2
1
1 2 3/ 2
1 2
2
1/ 2
x
(
x

1)
dx

y
dy

.
y

C

x

1
C


2
2 3
3
e
(f)  e3 x cosh 2 x dx   e3 x
1
 
2
2x
 e2 x
dx
2
 e5x  e x  dx   101 e5x  12 e x   c
-----------------------------------------------------------------------------30-Marks
Answer of Question (4)
(a) x  a(t  sin t ), y  a(1  cost ), 0  t  2
x  a (t  sin t ), y  a (1  cost )
2
dx
 a (1  cost ),
dt
2
dy
 a sin t
dt
t
 dx   dy 
dL       dt  a 2 (1  cos t )2  a 2 sin 2 t  2a sin dt
2
 dt   dt 
5
2
 dx 
 dy 
 S x  2  y      dt  4 a  a (1  cos t )sin t dt
2
 dt 
 dt 
a
0
2
b
2
 8 a 
2
0
2
2
sin 3 t dt  8 a 2
2
2
2t
t
 sin 2 sin 2 dt
0
 8 a 2  (1  cos 2 t )sin t dt
2
2
0
2
 8 a 2  (sin t  cos 2 t sin t ) dt
2
2
2
0
2
2 3t
t
 8 a  2cos  cos  
2 3
2

0
2
 8 a 2 (2cos   2 cos3  )  (2cos0  2 cos3 0) 
2 3
2
3


64
 8 a 2 0  (2  2 )   a 2
3 

3
5
5
5
1
1 
A   ydx   x dx   x 4   54  24 

 4 2 4 
(b)
2
2
625  16 609


square unit
4
4
x 10
(c)V   
x 0
3
10
y dx    x 4dx 
2
0
  5 10
x
 20000 cubic unit
5
0
-----------------------------------------------------------------------------20-Marks
Dr. Fathi Abdessalam
6