Applied Mathematics

other 17 Pages

Maharashtra State Board of Technical Education

Diploma Engineering

Contributed by

Pavan xerox

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 1 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Important Instructions to Examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for
subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures
drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and
there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on
candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
1.

a)
Ans

b)
Ans
Solve any FIVE of the following:
     
 
 
 
   
2
If 3 5 7 show that 1 3 1
1 15
15
3 1 15
1 3 1
f x x x f f
f
f
f
ff
    



  

--------------------------------------------------------------------------------------------------------------------
   
 
     
 
 
   
   
 
3
3
3
2
Define odd and even function with suitable examples.
If then the function is an odd function
..
If then the function is an even function
. . 1
f x f x
e g f x x x
f x x x
f x x x
f x f x
f x f x
e g f x x
f
  

    
    
   



   
 
   
2
2
1
1
xx
f x x
f x f x
   
   
  

--------------------------------------------------------------------------------------------------------------------
10
02
1
½
½

02

½
½

½
½

22210
Page 1
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 2 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
1.

c)
Ans

d)
Ans

e)
Ans

f)

Ans

1
1
Find if
1
log 0
2
1
log
2
x a a
x a a
xa
xa
dy
y a x a x
dx
y a x a x
dy
a a ax
dx
x
dy
a a ax
dx
x


   
   
    
   

----------------------------------------------------------------------------------------------------------------------------
 
2
2
2
2
1
1
Evaluate
4
1

4
1

2
1
tan
22
dx
x
dx
x
dx
x
x
c













----------------------------------------------------------------------------------------------------------------------------
 
 
Evaluate .
.
.1

x
x
xx
xx
xx
xx
x e dx
x e dx
d
x e dx e dx x dx
dx
xe e dx
xe e dx
xe e c






  


  



----------------------------------------------------------------------------------------------------------------------------
   
1 2 1 2
12
12
22
12
12
12
If 4 5 and 3 7 find
4 5 3 7
72
72
49 4
53
z i z i z z
z z i i
z z i
zz
zz
zz
    
    
   
   
   
  

----------------------------------------------------------------------------------------------------------------------------------------
02

2

02

½

1½

02

½
½
½
½

02

1

1
22210
Page 2
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 3 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
1.

2.

g)

Ans

a)

Ans

b)
Ans

2
3
2
1
3
3
1
33
Find the area bounded by the curve 3 , the lines 1 , 3 and axis
Area
3
3
3
31
3
33
26
b
a
y x x x x
A y dx
x dx
x
   














--------------------------------------------------------------------------------------------------------------------------
Solve any THREE of the following:
 
33
33
22
22
22
2
2
2
2
Find if 3
3
3 3 3 .1
3 3 3 3
3 3 3 3
33
33
dy
x y axy
dx
x y axy
dy dy
x y a x y
dx dx
dy dy
x y ax ay
dx dx
dy
y ax ay x
dx
dy ay x
dx y ax
dy ay x
dx y ax



   


   
   







--------------------------------------------------------------------------------------------------------------------------
22
11
and 1
11
and
F
1
11
an
ind if
d

xy
tt
xy
tt
dx dy
tt
dt dt
dy
dy
dt
dx
dx
dt
dy
dx
  
  
   


02

1

1

12
04

2

½
½

1

04

1+1

22210
Page 3
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 4 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q. N.
Answers
Marking
Scheme
2.

b)

c)

Ans

d)
Ans
2
2
1
1
1
dy
t
dx
t
dy
dx


  

--------------------------------------------------------------------------------------------------------------------------
 
2
2
2
2
A bullet is fired into a mud bank and penetrates 120 3600 m. in ' ' sec. after impact.
Calculate maximum depth of penetration.
Let 120 3600
120 7200
7200
Consider 0
120 7200 0
t t t
s t t
ds
t
dt
ds
dt
ds
dt
t


  
  

  

2
2
2
120 7200
1
60
1
at
60
7200 0
The maximum depth of penetration is,
11
120 3600
60 60
1 meter
t
t
t
ds
dt
s
s



   

   

   
   


----------------------------------------------------------------------------------------------------------------
 
3
3
2
2
2
Find radius of curvature to the curve at 2,8

3
6
yx
yx
dy
x
dx
dy
x
dx





1

1

04

1
1

1

½

½

04

1

1
22210
Page 4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 5 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
2.

3.

d)

a)
Ans

 
 
 
 
2
2
2
3
2
2
2
2
3
2
2
at 2,8
3 2 12
6 2 12
1
Radius of curvature is
1 12
12
145.50
dy
dx
dy
dx
dy
dx
dy
dx




















-------------------------------------------------------------------------------------------------------------------------------------
Solve any THREE of the following:
 
 
 
 
 
 
22
22
Find the equation of the tangent to the curve 4 9 40 at 3,2
4 9 =40
8 18 0
8
18
at 3,2
83
18 2
2
3
2
slope of tangent ,
3
Equation of tangent at 3,2 is
2
2 3
3
3 6 2
xy
xy
dy
xy
dx
dy x
dx y
dy
dx
dy
dx
m
yx
yx












  
     6
2 3 12 0 xy   

-------------------------------------------------------------------------------------------------------------------------------------

½

½

½
½

12
04

1
½

1

1

½

22210
Page 5
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 6 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
3.

b)

Ans

c)
Ans

 
1
3
1
3
1
1
3
1
1
1
2
2
1
Find if sec
43
1
sec
43
Put cos cos
1
sec
4cos 3cos
1
sec
cos3
sec sec3
3
3cos
1
3
1
3
1
dy
y
dx
xx
y
xx
xx
y
y
y
y
yx
dy
dx
x
dy
dx
x






















  






















-----------------------------------------------------------------------------------------------------------------
 
 
 
 
2
2
2
log
If prove that
log 1
taking log on both sides,
log log
log log
log
log
diff.w.r.t.
1
log 1
log
log 1
log
xy
xy
xy
y
dy
ye
dx y
ye
ye
x y y e
x y y
y
x
y
y
yy
dx
y
dy
y
dx y
dy
y












04

1

½
½
½
½
1

04

1
½

2

22210
Page 6
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 7 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
3.

c)

d)

Ans

 
2
log
log 1
y
dy
dx y



OR
 
 

taking log on both sides,
log log
log log
log ------
diff.w.r.t.
1
log 1
log 1
log

1

xy
xy
ye
ye
x y y e
x y y i
x
dy dy
xy
y dx dx
x dy
y
y dx
y dy
dx
x
y




  

  







 
 
 
 
 
2
2

log
----- .
1
1
log
log

log 1
log
log 1
n
y dy
eq i
dx
y
y
dy
y dx
y
dy
dx y










----------------------------------------------------------------------------------------------------------------
 


 


22
22
1
Evaluate
sin
1

sin
Put
x
x
x
x
x
xe
dx
e
x
x
xe
dx
e
x
x
e
t
x






½

1
½

1

½

½

½

04

½

22210
Page 7
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 8 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
3.

4.

d)

a)

Ans

 
 
2
2
2
2
1
1
1

sin
cos
cot
cot
xx
x
x
xe e
dx dt
x
xe
dx dt
x
dt
t
ec t dt
tc
e
c
x






  

  





--------------------------------------------------------------------------------------------------------------------------------------
Solve any THREE of the following:
   
22
22
2
22
2
2
2
2
2
2 2 2
2
1
Evaluate
4cos 9sin
4cos 9sin
cos
4cos 9sin
cos
sec
4 9tan
Put tan
sec
49
1
or
9
23
2
3
3
tan
1
2

23
dx
xx
dx
xx
dx
x
xx
x
xdx
x
xt
x dx dt
dt
t
dt dt
t
t
t
c



























1
1
or tan
2
2
9
3
3
t
c











½

½
1
1
½

12
04

½

½

½

½
[[
½

1

22210
Page 8
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 9 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
4.

a)

b)

Ans

c)

Ans

1
1 3tan
tan
62
x
c






-------------------------------------------------------------------------------------------------------------------------------------
  
  
  
  
   
  
  
log
Evaluate:
2 log 3 log
log

2 log 3 log
Put log
1

23
consider
2 3 2 3
32
Put 2
2
Put 3
3
23
2 3 2 3
23

2 3 2 3
2log 2
x
dx
x x x
x
dx
x x x
xt
dx dt
x
t
dt
tt
t A B
t t t t
t A t B t
t
A
t
B
t
t t t t
t
dt dt
t t t t






   
    





  
   


  

   






   
   
3log 3
2log 2 log 3log 3 log
t t c
x x c
   
     

 
 
   
 
5
2
5
2
5
2
5
2
Evaluate
7
-------------------- 1
7
25

7 2 5 2 5
7
-------------------- 2
7
x
dx
xx
x
I dx
xx
x
I dx
xx
x
I dx
xx





     








½

04

½

½

½

½
½

1
½

04

1

½
22210
Page 9
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 10 of 17

WINTER– 18 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:
Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
4.

c)

d)
Ans

e)

 
55
22
5
2
5
2
5
2
add (1) and (2)
7
+
77
7
2
7
2 1
2
2 5 2
23
3
2
xx
I I dx dx
x x x x
xx
I dx
xx
I dx
Ix
I
I
I


   





  






-------------------------------------------------------------------------------------------------------------------------
 
1
1
11
22
1
2
22
1
2
22
1
2
22
1
22
2
1
2
Evaluate .tan
tan .
tan tan
1
tan
2 2 1
1
tan
2 2 1
1 1 1
tan
2 2 1
1 1 1
tan
2 2 1 1
11
tan 1
2 2 1
x x dx
x xdx
d
x xdx xdx x dx
dx
xx
x dx
x
xx
x dx
x
xx
x dx
x
xx
x dx
xx
x
x
x
























  



  



  




 
2
11
1
tan tan
22
dx
x
x x x c




   


-------------------------------------------------------------------------------------------------------------------------------------
  
Evaluate
12
x
dx
xx


½

1

1

04

½

1

½

½

½

1

04

22210
Page 10

Applied Mathematics

Applied Mathematics

Download this file to view remaining 7 pages