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– 19 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

c)

Ans
Attempt 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
    



  

--------------------------------------------------------------------------------------------------------------------
 
 
         
 
   
4 2 2
4 2 2
42
2
4 2 2
State whether the function 3 5 3cos 2sin is even or odd
3 5 3cos 2sin
3 5 3cos 2sin
3 5 3cos 2sin
function is an even function
f x x x x x
f x x x x x
f x x x x x
f x x x x x
f x f x
    
    
          
      
  


--------------------------------------------------------------------------------------------------------------------
1
1
1
2
Find if .sin
.sin
1
. sin .
1
x
x
xx
dy
y e x
dx
y e x
dy
e x e
dx
x





  


10
02
1

1

02

½
½
½
½

02

1+1

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

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

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

c)

d)
Ans

e)
Ans

f)
Ans

1
2
1
sin
1
x
dy
ex
dx
x


  




----------------------------------------------------------------------------------------------------------------------------


 
 
2log
2log
2 log
2 log .1
1
2 log . .
2 log 1
2 log
2 log 1
Evaluate
x dx
x dx
x dx
x dx
x x x dx
x
x x dx
x x x
c
xx
c
e
e
e
e
e
e
e
e






















----------------------------------------------------------------------------------------------------------------------------
 
2
2
2
Evaluate sin
sin
1
2sin
2
1
1 cos2
2
1 sin 2
22
x dx
x dx
x dx
x dx
x
xc



  







----------------------------------------------------------------------------------------------------------------------------
2
3
2
0
3
3
0
Find the area under the curve from 0 to 3 with axis.
Area
3
b
a
y x x x x
A y dx
x dx
x
   









02

½

½
½
½

02

1

1

02

½

½

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

SUMMER– 19 EXAMINATION
Subject Name: Applied Mathematics Model Answer Subject Code:

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

2.

f)

g)
Ans

a)
Ans

3
3
0
3
9






--------------------------------------------------------------------------------------------------------------------------
   
 
 
22
22
1 1 1
Express 1 in Polar form.
1
1, 1
1 1 2
1
2 tan 2 tan 2 tan 1
1
7
2
44
Polar form is cos sin
77
1 2 cos sin
44
zi
zi
xy
r z x y
y
x
z r i
ii
   




  


   
       
    
     
   
   
  
  

   
   
   

   


--------------------------------------------------------------------------------------------------------------------------
Attempt any THREE of the following:
 
22
22
Find if 4
4
2 2 4 .1
2 2 4 4
2 4 4 2
42
24
2
2
dy
x y xy
dx
x y xy
dy dy
x y x y
dx dx
dy dy
x y x y
dx dx
dy
y x y x
dx
dy y x
dx y x
dy y x
dx y x



   


   
   







--------------------------------------------------------------------------------------------------------------------------
½

½

02

½

½

1

12

04

2

1

1

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

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

Q.
No.
Sub
Q. N.
Answers
Marking
Scheme
2.

b)

Ans

c)

Ans

   
   
   
 
If sin , 1 cos , find at =
2
sin 1 cos
1 cos 0 sin sin
sin
1 cos
sin
1 cos
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

   
  
  






   
   
    




at =
2
sin
2
1
1 cos
2
dy
dx





  


--------------------------------------------------------------------------------------------------------------------------
2
2
2
2
11
Find the radius of curvature of the curve 1 at ,
44
1
11
0
22
11
22
1
2
2 2 2
xy
xy
dy
dx
xy
y
dy
dx
x
dy
xy
dx
yx
dy
dx x
y y y
x
y x x x
dy
dx x x





  
  





  
   


  
   
   
   
    

11
at ,
44




04

1+1

1

1

04

½

½

½

½

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

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
2.

d)

d)
Ans

 
2
2
3
2
2
3
2
2
2
2
1
4
1
1
4
1
1
4
2
1
2
4
4
1
4
1
11
Radius of curvature
4
0.707
dy
dx
dy
dx
dy
dx
dy
dx
    







   









   
  


-------------------------------------------------------------------------------------------------------------------------------------
 
     
32
32
2
2
2
2
2
2
32
max
Find the maximum and minimum value of 9 24
Let 9 24
3 18 24
6 18
Consider 0
3 18 24 0
2 or 4
at 2 6 2 18 6 0
is maximum at 2
2 9 2 24 2 20
at
x x y
y x x x
dy
xx
dx
dy
x
dx
dy
dx
xx
xx
dy
x
dx
yx
y

  
   
  

  
  
      

   
 
     
2
2
32
min
4 6 4 18 6 0
is minimum at 4
4 9 4 24 4 = 16
dy
x
dx
yx
y
     

  

Note: If students attempted to solve the question give appropriate marks.

½

½

½

½

04

½
½

1

½

½

½

½
22210
Page 5
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Page 6 of 18

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
3.

a)
Ans

Attempt any THREE of the following:
 
 
 
 
 
22
22
Find equation of tangent and normal to the curve 2 3 18 at 3,1
2 3 18
4 .1 6 0
4 6 0
64
4
6
at 3,1
1 4 3
6 1 3
11
3
slope of t
x xy y
x xy y
dy dy
x x y y
dx dx
dy dy
x x y y
dx dx
dy
y x y x
dx
dy y x
dx y x
dy
dx
dy
dx
  
  

    


    
   









 
 
 
 
11
angent ,
3
Equation of tangent at 3,1 is
11
1 3
3
3 3 11 33
11 3 36 0
13
slope of normal , '
11
Equation of normal at 3,1 is
3
1 3
11
11 11 3 9
3 11 2 0
m
yx
yx
xy
m
m
yx
yx
xy



  
    
   

  
  
   
   

-----------------------------------------------------------------------------------------------------------------
12

04

½

½

½

1
½

1

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

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
3.

b)

Ans

c)
Ans

 
 
 
 
Find if sin
Let
Taking log on both sides,
log log
log log
11
log .1
1
1 log
1 log
Let sin
taking log on both sides,
log log sin
log logsin
11
sin
x
x
x
x
x
x
dy
y x x
dx
ux
ux
u x x
du
xx
u dx x
du
x
u dx
du
ux
dx
vx
vx
v x x
dv
x
v dx




  
  
  




 
     
.cos logsin .1
1
cot logsin
cot logsin
1 log sin cot logsin
x
x
xx
x
dv
x x x
v dx
dv
v x x x
dx
dy du dv
dx dx dx
dy
x x x x x x
dx

  
  
  
    

----------------------------------------------------------------------------------------------------------------
 
3sec 4tan
3sec 4tan
3sec 4tan 2
If find
3sec tan 4sec
xx
xx
xx
dy
ye
dx
ye
dy
e x x x
dx





   

------------------------------------------------------------------------------------------------------------------------------------
04

½

½

½

½

½

½

1

04

4

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

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
3.

d)

Ans

  
  
  
  
   
2
2
2
sec
Evaluate
1 tan 3 tan
sec

1 tan 3 tan
Put tan
sec
1

13
1
+
1 3 1 3
1 3 1
x
dx
xx
x
dx
xx
xt
xdx dt
dt
tt
AB
t t t t
A t B t







   
    



  
  

1
Put 1 ,
2
1
Put 3 ,
2
11
1
22
1 3 1 3
1
13
tA
tB
t t t t
d
tt
   

  
  
   


   
   
11
22

13
11
log 1 log 3
22
11
log 1 tan log 3 tan
22

t dt
tt
t t c
x x c






    
    


OR
  
  
2
2
sec

1 tan 3 tan
Put tan
sec
1

13
x
dx
xx
xt
xdx dt
dt
tt







04

1

½

½

½

1

½

1

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

SUMMER– 19 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

43
4
Third Term = 4
4
1

4 4 4 3
1

21
1 2 1
log
2 2 1
11
log
23
1 tan 1
log
2 tan 3
dt
tt
dt
tt
dt
t
t
c
t
t
c
t
x
c
x




   















-------------------------------------------------------------------------------------------------------------------------------------
Attempt any THREE of the following:
 
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




   


½

1

1

½

12
04

1

½

½

½

½

1

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

SUMMER– 19 EXAMINATION

Subject Name: Applied Mathematics Model Answer Subject Code:

Q.
No.
Sub
Q.N.
Answers
Marking
Scheme
4.

b)

Ans

c)

Ans

   
 
 
2
22
2
2
2
22
22
2
2
2
Evaluate
4 5cos
4 5cos
21
Put tan , , cos
2 1 1
2
1

1
45
1
2
4 1 5 1
2
4 4 5 5
2
9
2
3
13
2 log
2 3 3
3 tan
1
2
log
3
3 tan
2
dx
x
dx
x
x dt t
t dx x
tt
dt
t
t
t
dt
tt
dt
tt
dt
t
dt
t
t
c
t
x
c
x



  










  

  


















-------------------------------------------------------------------------------------------------------------------------------------
   
   
   
        
2
2
2
2
25
Evaluate
1 2 3
25
1 2 3
25
Consider
1 2 3 1 2 3
2 5 2 3 1 3 1 2
Put 1
7 12
x
dx
x x x
x
dx
x x x
x A B C
x x x x x x
x A x x B x x C x x
x
A

  

  

  
     
          





04

1

½

½
½

1

½

04

½

22210
Page 10

Applied Mathematics

Applied Mathematics

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