Applied Mathematics

other 18 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 18

SUMMER– 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:
 
 
3
1
3
3
3
1
If 64 log , find
3
11
64 log
33
1
4 log 3
3
1
41
3
1
3
3
x
f x x f
f
f
f
f




   

   
   

  



  







--------------------------------------------------------------------------------------------------------------------
       
   
 
3
3
3
If sin , show that 3 3 4
34
3sin 4sin
sin3
3
f x x f x f x f x
f x f x
xx
x
fx
  





OR

 
3fx

10
02

½
½

½

½

02
½
1
½

22210
Page 1
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 2 of 18

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

b)

c)
Ans

d)
Ans

e)
Ans

f)
   
3
3
sin3
3sin 4sin
34
x
xx
f x f x




-------------------------------------------------------------------------------------------------------------------
1
1
1
2
1
2
Find if sin
sin
1
sin
1
1
sin
1
x
x
xx
x
dy
y e x
dx
y e x
dy
e x e
dx
x
dy
ex
dx
x






  


  




-----------------------------------------------------------------------------------------------------------------------------
 
 
 
 
2
2
2
32
4 3 2
Evaluate: 1
1
2 1
2
2
4 3 2
x x dx
x x dx
x x x dx
x x x dx
x x x
c


  
  
   





-----------------------------------------------------------------------------------------------------------------------------
 
2
2
2
Evaluate: sin 2
sin 2
1
2sin 2
2
1
1 cos4
2
1 sin 4
24
x dx
x dx
x dx
x dx
x
xc



  







-----------------------------------------------------------------------------------------------------------------------------
2
Find the area bounded by the curve , -axis and ordinates 0 to 3y x x x x  

½
1
½

02

1+1

02

½
½

1

02

½

½

1

02
22210
Page 2
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 3 of 18

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

2.

f)
Ans

g)

Ans

a)

Ans

3
2
0
3
3
0
3
Area
3
3
0
3
9
b
a
A y dx
x dx
x














-------------------------------------------------------------------------------------------------------------------------
2
22
1
Express in form , where 1 and , are real number
1
1

1
11
11
12
1
1 2 1
11
2
2
0

i
z a ib i a b
i
i
z
i
ii
z
ii
ii
z
i
i
z
i
z
z i i

   





  









    

--------------------------------------------------------------------------------------------------------------------------
Attempt any THREE of the following:
 
2 2 3
2 2 3
22
22
If 13 2 1 , find at 1, 2
13 2 1
26 2 2 3 0
26 2 4 3 0
dy
x x y y
dx
x x y y
dy dy
x x y x y
dx dx
dy dy
x x xy y
dx dx
   
  

    


    

½

½
½

½

02

½
½

½

½

12
04

1

22210
Page 3
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 4 of 18

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

a)

b)
Ans

c)

Ans
 
 
22
22
22
2 3 26 4
2 3 26 4
26 4
23
at 1, 2
18 9
14 7
dy dy
x y x xy
dx dx
dy
x y x xy
dx
dy x xy
dx x y
dy
dx
    
    







----------------------------------------------------------------------------------------------------------------
   
   
   
 
If sin , 1 cos , find at =
2
sin 1 cos
1 cos 0 sin sin
sin
1 cos
sin
1 cos
at =
dy
x a y a
dx
x a y a
dx dy
a a a
dd
dy
dy a
d
dx
dx a
d
dy
dx

   
  
  








   
   
    




2
sin
2
1
1 cos
2
dy
dx



  


-------------------------------------------------------------------------------------------------------------------------------------
4000
The rate of working of an engine is given by the expression 10 , where is the
speed of the engine.Find the speed at which the rate of working is the least.
4000
The rate of working is, 10
VV
V
WV
V
d



2
4000
10
W
dV V


1

1

1

04

1+1

1

1

04

½

22210
Page 4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 5 of 18

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

c)

d)

Ans
 
2
23
2
2
2
2
3
2
8000
Consider 0
4000
10 0
4000
10
400
20, 20
at 20
8000
10
20
The speed is 20 at which the rate of working is least
dW
dV V
dW
dV
V
V
V
V
V
dW
dV
V


  


  

   


---------------------------------------------------------------------------------------------------------------
2
2
A telegraph wire hangs in the form of a curve 2sin sin2 . Find the radius of curvature
of the wire at the point
2
2sin sin 2
2cos 2cos2
2sin 4sin 2
at
2
2cos 2cos
2
y x x
x
y x x
dy
xx
dx
dy
xx
dx
x
dy
dx






  
   


  


 
2
2
3
2
2
3
2
2
2
2
22
2
2sin 4sin 2 2
22
1
12
Radius of curvature
2
dy
dx
dy
dx
dy
dx






   
     
   
   









   
  


½

½

½

1

1

04

½

½

½

½

1

22210
Page 5
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 6 of 18

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

3.

d)

a)
Ans

b)

Ans

Radius of curvature 5.59 i.e. 5.59  

----------------------------------------------------------------------------------------------------------------
Solve any THREE of the following:
 
2
2
11
Find the equation of the tangent to the curve 9 12 7 which is parellel to the axis
9 12 7
18 12
tangent is parellel to the axis
0
18 12 0
2
3
3
2
, ,3
3
slope of tan
y x x x
y x x
dy
x
dx
x
dy
dx
x
x
y
xy
   
  
  


  






 gent , 0
2
Equation of tangent at ,3 is
3
2
3 0
3
30
m
yx
y





  


  

------------------------------------------------------------------------------------------------------------------
   
 
2
sin
Find if log
1 cos
sin
log
1 cos
1 sin
sin
1 cos
1 cos
1 cos cos sin 0 sin
1 cos
sin
1 cos
dy x
y
dx x
x
y
x
dy d x
x
dx dx x
x
x x x x
dy x
dx x
x

















  







1

12
04

½

½
½
½

½

1
½

04

1

1

22210
Page 6
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 7 of 18

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

b)

22
2
2
1 cos cos sin
sin 1 cos
1 cos 1
sin 1 cos
1
cos
sin
sin
log
1 cos
2sin cos
22
log
2cos
2
log tan
2
11
sec
22
tan
2
co
dy x x x
dx x x
dy x
dx x x
dy
ecx
dx x
OR
x
y
x
xx
y
x
x
y
dy x
x
dx
dy
dx












  















  

  
  

   
 
 
 
2
22
s
1
2
2sin cos 2sin cos
2 2 2 2
1
cos
sin
sin
log
1 cos
log sin log 1 cos
11
cos sin
sin 1 cos
cos 1 cos
sin
sin 1 cos
cos cos sin
sin 1 cos
x
x x x x
dy
ecx
dx x
OR
x
y
x
y x x
dy
xx
dx x x
xx
dy x
dx x x
dy x x x
dx x x

  





   
   


  





1

1

1

1

2

1
1

½

1

22210
Page 7
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 8 of 18

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

b)

c)

Ans

d)
Ans

 
cos 1 1
cos
sin 1 cos sin
dy x
ecx
dx x x x

   


-----------------------------------------------------------------------------------------------------------------
 
 
2
log
If ,then prove that
1 log
log log
log log
y x y
y x y
y x y
dy x
xe
dx
x
xe
xe
y x x y e









 
     
 
 
 
 
 
2
2
2
2
log
log
log 1
log 1
log 1 log 1
log 1
1
log 1 1 0
log 1
log 1 1
log 1
log
log 1
y x x y
y x y x
y x x
x
y
x
dd
x x x x
dy
dx dx
dx
x
xx
dy
x
dx
x
dy x
dx
x
dy x
dx
x
  
  
  


  



   










--------------------------------------------------------------------------------------------------------------------------
 
2
2
1
1
cos
Evaluate:
1 sin
Putsin
cos
1
tan
tan sin
x
dx
x
xt
xdx dt
dt
t
tc
xc












½

04

½
½

½

½

1

1

04
1
1

1
1
22210
Page 8
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2013 Certified)
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Page 9 of 18

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

4.

a)

Ans

b)

Ans
Solve any THREE of the following:
   
  
  
  
   
  
  
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
   
     

--------------------------------------------------------------------------------------------------------------------------
22
2
2
Evaluate:
3 2sin

3 2sin
22
Put tan , , sin
2 1 1
2
1
2
32
1
dx
x
dx
x
x dt t
t dx x
tt
dt
t
t
t


  











12

04

½

½

½

½
½

1
½

04

1
22210
Page 9
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 10 of 18

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

b)

c)

Ans
 
 
2
2
2
2
2
2
2
2
2
1
1
2
3 1 2 2
2
3 3 4
2
3 4 3
2
4
3
1
3
1 4 4
..
2 3 9
2
4 4 4
3
1
3 9 9
2
3
25
39
2
3
25
33
2
21
3
tan
3
55
33
2
tan
2
23
tan
55
3
dt
tt
dt
tt
dt
tt
dt
tt
TT
dt
tt
dt
t
dt
t
t
c
x











  



   





































c








------------------------------------------------------------------------------------------------------------------------------------
1
2
1
2
sin
Evaluate:
1
sin

1
xx
dx
x
xx
dx
x







½

½

½

½

½

½

04

22210
Page 10

Applied Mathematics

Applied Mathematics

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