# 2014 Summer -17301-Maths III-Model Answer Sheet

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Winter 2014 Examination
Subject & Code: Applied Maths (17301) Model Answer Page No: 1/28
Que.
No.
Sub.
Marks

Important Instructions to the Examiners:
1) The Answers should be examined by key words and not as
word-to-word as given in the model answer scheme.
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 the candidate and those in the 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 the
6) In case of some questions credit may be given by judgment
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.
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 2/28
Que.
No.
Sub.
Marks
1)
a)
Ans.
b)
Ans.
Attempt any TEN of the following:
Find the gradient of the tangent of the curve 3
y x = at x = 4.
( )
3 3/ 2
1/ 2
1/ 2
3
2
3
4 3
2
y x x
dy
x
dx
x
dy
dx
= =
∴ =
∴ =
= =
OR
( )
( )
3
3
3
2
3
2
3
1
2
1
3
2
1
3 4 3
2 4
y x
dy d
x
dx dx x
x
x
x
dy
dx
=
∴ = ⋅
= ⋅
∴ =
= ⋅ =
-------------------------------------------------------------------------------------
Find the radius of the curvature of the curve 2
y ax = 4 at the
point (a a , 2 ).
( )
( )
( )
2
2
2 2
2
2 2
2
2
3/ 2 5/ 2
2 2
3 3
2 2
4
2 4
4 2
2
2
, 2 ,
2 2 1 1 1
2 2 2
1
2 1 1
2 2 2
1 1 1
y ax
dy
y a
dx
dy a a
dx y y
d y a dy
dx y dx
at a a
dy a d y a and
dx a dx a a
d y
dx a
a a
dy
dx
κ
=
∴ =
∴ = =
∴ = − ⋅

= = = − ⋅ = −

∴ = = = − = −

      +
+            
1
1
1
1
½
½
½
2
2
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Subject & Code: Applied Maths (17301) Page No: 3/28
Que.
No.
Sub.
Marks
1)
c)
Ans.
( )
5/ 2
2
2
3/ 2 5/ 2
2
2
3
2
3
2
1
2
1
1 1
2 2 2 1
2
a
OR
dy
dx
a a
d y
a
dx
ρ
κ
ρ
∴ = = −
      +
      +      
= = = − ⋅ = −

OR
( )
2
2 2 2
2 2
2 2
2
2
2
5/ 2
5/ 2
4
2 4
2 2 0 2 2 0
1
, 2 ,
4 1 1 1 1
2 2 2 2
1
2
2
y ax
dy
y a
dx
d y dy dy d y dy
y or y
dx dx dx dx dx
d y dy
dx y dx
at a a
dy a d y and
dx a dx a a
a
a
κ
ρ
=
∴ =
  ∴ + ⋅ = + =    
  ∴ = −    

= = = − ⋅ = −

∴ = −
∴ = −
-------------------------------------------------------------------------------------
Evaluate ( )2
tan cot x x dx + ∫
( )
( )
( )
( )
( )
2
2 2
2 2
2 2
2 2
tan cot
tan 2 tan cot cot
tan 2 cot
sec 1 2 cos 1
sec cos
tan cot
x x dx
x x x x dx
x x dx
x ec x dx
x ec x dx
x x c
+
= + + ⋅
= + + ⋅
= − + + − ⋅
= + ⋅
= − +

Note: In the solution of any INDEFINITE integration
problems, if the constant c is not added, ½ mark may
be deducted.
½
OR
½
½
½
½
½
½
½
1
2
2
2
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Subject & Code: Applied Maths (17301) Page No: 4/28
Que.
No.
Sub.
Marks
1) d)
Ans.
e)
Ans.
f)
Ans.
Evaluate ( ) 2 1
sec log x dx
x

( )
( )
2 2
log
1
sec log sec 1
tan
tan log
Put x t
x dx tdt
x dx dt
x
t c
x c
=
=
∴ =
= +
= +
∫ ∫
OR
( )
( )
2 2
log
1
1
sec log sec
tan
tan log
Put x t
dx dt
x
x dx tdt
x
t c
x c
=
∴ =
∴ =
= +
= +
∫ ∫
-------------------------------------------------------------------------------------
Evaluate x
xe dx ∫
.
( ) x x x
x x
x x
d
xe dx x e dx e dx x dx
dx
xe e dx
xe e c
= − ⋅    
= −
= − +
∫ ∫ ∫ ∫

-------------------------------------------------------------------------------------
Evaluate 2
1
3 2
dx
x x + + ∫
2 2
2 2
2 2 2
9 9 3 1 3 2 3 2
4 4 2 2
1 1
3 2 3 1
2 2
3 1
1 2 2 log
1 3 1
2
2 2 2
1
log
2
x x x x x
dx dx
x x
x
x
c
x
x
c
x
   
+ + = + + − + = + −        
=
+ +         + −    
  + −  
= +       + +      
  +
= +     +
∫ ∫
½ + ½
½
½
½
½
½
½
½
½
½+½
½+½
½
½
2
2
2
2
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Subject & Code: Applied Maths (17301) Page No: 5/28
Que.
No.
Sub.
Marks
1)
g)
Ans.
OR
2
2
2 2
1 1
3 2 9 9 3 2
4 4
1
3 1
2 2
3 1
1 2 2 log
1 3 1
2
2 2 2
1
log
2
dx dx
x x
x x
dx
x
x
c
x
x
c
x
=
+ + + + − +
=
        + −    
 
+ −  
= +       + +      
  +
= +     +
∫ ∫

OR
( )( )
( )( )
( )
( )( )
( ) ( )
2
1 1
3 2 1 2
1
1 2 1 2
1
1 *
1 1 1
1 2 1 2
1 1
1 2
log 1 log 2
I dx dx
x x x x
A B
x x x x
A
B
x x x x
I dx
x x
x x c
= =
+ + + +
= +
+ + + +
∴ =
= − − − − − − − − − − −

= +
+ + + +
  − ∴ = +     + +
= + − + +
∫ ∫

Note (*): There are various methods to find the values of A and
B to partially factorize the given expression including
direct method. Students may apply any one of the
methods. Take in count all such methods.
-------------------------------------------------------------------------------------
Evaluate
2
1 3 2
dx
x −

( )
( ) ( )
2
2
1
1
log 3 2
3 2 3
log 6 2 log 3 2
3 3
log 4 log1
3 3
log 4
3
dx x
x
  −
=  
−  
− −
= −
= −
=

½
½
½
½
½
½
1
1
½
½
2
2
2
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Subject & Code: Applied Maths (17301) Page No: 6/28
Que.
No.
Sub.
Marks
1)
h)
Ans.
i)
Ans.
Find the area above the x-axis bounded by y x = sin and the
ordinates
6
x
π
= and
3
x
π
= .
[ ]
/3
/ 6
/3
/6
sin
cos
cos cos (*)
3 6
1 3 0.366
2
A xdx
x
or
π
π
π
π
π π
=
= −
   
= − − − − − −        
− +
=

OR
[ ]
/ 6
/3
/6
/3
sin
cos
cos cos (*)
6 3
3 1 0.366
2
1 3 0.366
2
A xdx
x
or
Area A or
π
π
π
π
π π
=
= −
   
= − − − − − −        
− +
= −
− +
∴ =

Note: Due to the use of advance non-programmable scientific
calculators, writing directly the value of the step (*) as
0.366 or -0.366 is permissible. No marks to be deducted
for calculating directly the value.
-------------------------------------------------------------------------------------
Find the order and degree of the equation
2
²
2 3 1 0
²
d y dy
y
dx dx
   
+ − − =          
Order = 2
2
2 2
1 ² 1 2
3 ²
1 ² 1 2
9 ²
dy d y
y
dx dx
dy d y
y
dx dx
   
− = −        
    ∴ − = −        
Degree = 2
½
½
1
½
½
½
½
1
1
2
2
2
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Subject & Code: Applied Maths (17301) Page No: 7/28
Que.
No.
Sub.
Marks
1)
j)
Ans.
k)
Ans.
Verify that x x y Ae Be−
= + is a solution of
²
0
²
d y
y
dx
− =
( )
( )
2
2
2
2
2
2
1
1
0
x x
x x x x
x x
x x
y Ae Be
dy Ae Be Ae Be
dx
d y Ae Be
dx
Ae Be
d y
y
dx
d y
y
dx

− −

= +
∴ = + ⋅ − = −
∴ = − ⋅ −
= +
∴ =
∴ − =
-------------------------------------------------------------------------------------
A bag contains 7 white balls, 5 black balls, and 4 red balls. If
two balls are drawn at random from the bag, find the
probability that both the balls are white.
16
2
7
2
Total 7 5 4 16
120
21
21 7 0.175
120 40
n C
m C
m
p or
n
= + + =
= =
∴ = =
∴ = = =
OR
7
2
16
2
Total 7 5 4 16
21 7 0.175
120 40
C
p
C
or
= + + =
∴ =
= =
Note: Due to the use of advance non-programmable scientific
calculators which is permissible in the board
examination, writing directly the values of nCr
or
n r n r C p q r

is permissible. No marks to be deducted for
calculating directly the value.
-------------------------------------------------------------------------------------
½
½
½
½
½
½
1
1
1
2
2
2
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Subject & Code: Applied Maths (17301) Page No: 8/28
Que.
No.
Sub.
Marks
1)
2)
l)
Ans.
a)
Ans.
What is the probability of getting more than 4 in a single throw
of a die?
{ }
6
5,6
2
2 1 0.333
6 3
n
A
m
m
p or
n
=
=
∴ =
∴ = = =
-------------------------------------------------------------------------------------
Attempt any four.
Find the equations of the tangent and normal to the curve
2 2 4 9 40 x y + = at the point (1, 2).
( )
( )
( )
2 2 4 9 40
4 2 9 2 0 8 18 0
18 8
8 4
18 9
1, 2 , the slope of tangent is
4 1 2
9 2 9
the equation of tangent is
2
2 1
9
9 18 2 2
2 9 20 0 2 9 20 0
1, 2 ,
x y
dy dy
x y or x y
dx dx
dy
y x
dx
dy x x
dx y y
at
dy
m
dx
y x
y x
x y or x y
at
+ =
∴ ⋅ + ⋅ = + =
∴ = −
− ∴ = = −

= = − = −

− = − −
∴ − = − +
∴ + − = − − + =

( )
the slope of normal is
9
2
the equation of tangent is
9
2 1
2
2 4 9 9
9 2 5 0 9 2 5 0
m
y x
y x
x y or x y
=

− = −
∴ − = −
∴ − − = − + + =
½
½
1
½
½
½
½
½
½
½
½
2
4
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Subject & Code: Applied Maths (17301) Page No: 9/28
Que.
No.
Sub.
Marks
2)
b)
Ans.
c)
Ans.
A beam is bent in the form of the curve y x x = − 2sin sin 2 , find
the radius of curvature of the beam at this point
2
x
π
= .
( )
2
2
2
2
2
2
2
2
2
2
3
2
3
2
2sin sin 2
2cos 2cos 2
& 2sin 4sin 2
,
2
2cos 2cos 2 2
2 2
2sin 4sin 2 2
2 2
1
1 2
5.590
2
1
y x x
dy
x x
dx
d y
x x
dx
at x
dy
dx
d y and
dx
dy
dx
d y
dx
OR
d y
dx
dy
π
π π
π π
ρ
κ
= −
∴ = −
= − +
∴ =
   
= − =        
   
= − + = −        
      +       +       ∴ = = = −

=
+
( ) 2 2
3 3
2 2
2
0.1789
1 2
1
5.590
dx
ρ
κ

= = −
      +
           
∴ = = −
-------------------------------------------------------------------------------------
A metal wire 36 cm long is bent to form a rectangle. Find its
dimensions when its area is maximum.
Let x and y be the sides of rectangle.
( ) 2
2 2 36 18
18
18 18
18 2
x y or x y
y x
But area A xy x x x x
dA
x
dx
∴ + = + =
∴ = −
= = − = −
∴ = −
1
1
½
½
1
OR
½
½
½
1
½
4
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Subject & Code: Applied Maths (17301) Page No: 10/28
Que.
No.
Sub.
Marks
2)
d)
Ans.
e)
Ans.
2
2
2
2
2
, 0
18 2 0
9
9, 2 0
9, max
18 9
d A
dx
dA For stationary values
dx
x
x
d A At x
dx
At x A has imum value
and the other side is
y x
∴ = −
=
∴ − =
∴ =
= = − <
∴ =
= − =
-------------------------------------------------------------------------------------
Evaluate 3 2
3
3 16 48
x
dx
x x x

− − + ∫
( )( ) 3 2 2
2
2 2
3 3
3 16 48 3 16
1
16
1
4
1 4 log
8 4
x x dx dx
x x x x x
dx
x
dx
x
x
c
x
− −
=
− − + − −
=

=

  −
= +     +
∫ ∫

OR
( )( )( )
( )( )
( ) ( )
3 2
3 3
3 16 48 3 4 4
1
4 4
1/ 8 1/8
4 4
1 1 log 4 log 4
8 8
x x dx dx
x x x x x x
dx
x x
dx
x x
x x c
− −
=
− − + − − +
=
− +
  −
= +     − +
= − − + +
∫ ∫

-------------------------------------------------------------------------------------
Evaluate
( )2
1
9 log
dx
x x   +
 

log
1
Put x t
dx dt
x
=
∴ =
½
½
½
½
1
1
2
1
1
2
1
4
4
4
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Subject & Code: Applied Maths (17301) Page No: 11/28
Que.
No.
Sub.
Marks
2)
f)
( )2 2
2 2
1
1
1 1
9 log 9
1
3
1
tan
3 3
1 log
tan
3 3
dx dt
x x t
dt
t
t
c
x
c

∴ =   + +
 
=
+
 
= +    
 
= +    
∫ ∫

OR
( )2
2
2 2
1
1
log
1
1
9 log
1
9
1
3
1
tan
3 3
1 log
tan
3 3
Put x t
dx
x x dx dt
x
dt
t
dt
t
t
c
x
c

=

  + ∴ =  
=
+
=
+
 
= +    
 
= +    

-------------------------------------------------------------------------------------
Evaluate ( )( )
2
sec
1 tan 3 tan
x
dx
+ + x x ∫
( )( )
( )( )
( )( )
2
2
sec tan
1 tan 3 tan sec
1
1 3
1
1 3 1 3
1
2
1
2
x Put x t
dx
x x xdx dt
dt
t t
A B
t t t t
A
B
=
+ + ∴ =
=
+ +
∴ = +
+ + + +
∴ =
= −

[Please refer the note written in the question 1 (f).]
½
½
1
1
1
½
½
1
1
1
1
1
4
4
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Subject & Code: Applied Maths (17301) Page No: 12/28
Que.
No.
Sub.
Marks
2)
3)
a)
Ans.
( )( )
( )( )
[ ] [ ]
[ ] [ ]
1 1
1 2 2
1 3 1 3
1 1
1 2 2
1 3 1 3
1 1 log 1 log 3
2 2
1 1 log 1 tan log 3 tan
2 2
t t t t
dt dx
t t t t
t t c
x x c

∴ = +
+ + + +
 

 
= +  
+ + + +  
= + − + +
= + − + +
∫ ∫
-------------------------------------------------------------------------------------
Attempt any four.
Evaluate
/ 4 2
0
x xdx sec
π

( ) ( )
( )
( )
/ 4
/ 4 2 2 2
0
0
/ 4
0
/ 4
0
sec sec sec
tan tan
tan log sec
tan log sec 0 log sec0
4 4 4
log 2 0.439
4
d
x xdx x xdx xdx x dx
dx
x x xdx
x x x
or
π
π
π
π
π π π
π
 
= −    
= −    
= −    
   
= − − −            
= −
∫ ∫ ∫ ∫

Note: In case of definite integrations, the problem may be
solved by without limits and then the limits would be
applied, as illustrated below:
( ) ( )
( )
( )
( )
2 2 2
/ 4 / 4 2
0 0
sec sec sec
tan tan
tan log sec
sec tan log sec
tan log sec 0 log sec0
4 4 4
log 2 0.439
4
d
x xdx x xdx xdx x dx
dx
x x xdx
x x x
x xdx x x x
or
π π
π π π
π
= −
= −
= −
= −    
   
= − − −            
= −
∫ ∫ ∫ ∫

½
½
1
1
1
½
½
1
1
1
½
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 13/28
Que.
No.
Sub.
Marks
3)
b)
Evaluate
2
1 3
x
dx
x x + − ∫
[ ]
2
1
2
1
2
1
2
1
2
1
Re 3
3
3
3
3
3
2
3
2 1
2
2 2 1
1
2
place x x
x
I dx x x
x x
x
I dx
x x
x x I dx
x x
I dx
I x
I
I
→ −
= ∴ − →
+ −
− ∴ =
− +
+ −
∴ =
+ −
∴ = ⋅
∴ =
∴ = −
∴ =

OR
( )
[ ]
2
1
2
1
2
1
2
1
2
1
2
1
3
3
3 3 3
3
3
3
2
3
2 1
2
2 2 1
1
2
x
I dx
x x
x
dx
x x
x
I dx
x x
x x I dx
x x
I dx
I x
I
I
=
+ −

=
− + − −
− ∴ =
− +
+ −
∴ =
+ −
∴ = ⋅
∴ =
∴ = −
∴ =

-------------------------------------------------------------------------------------
1
1
½
1
½
½
½
1
½
1
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 14/28
Que.
No.
Sub.
Marks
3)
c)
Ans.
Find the area bounded by the curve 2
y x = and the line y x = .
( )
( )
[ ]
2
2
2
2 1
1
2
0
1
2 3
0
,
0
0, 1
2 3
1 1 0 0
2 3
0.167
b
a
Given y x y x
x x
x x
x
A y y dx
x x dx
x x
= =
∴ =
∴ − =
∴ =
∴ = −
= −
 
= −    
 
= − − −    
=

OR
( )
( )
[ ]
2
2
2
2 1
1
2
0
1
3 2
0
,
0
0, 1
3 2
1 1 0 0
3 2
1
0.167
6
1
0.167
6
b
a
Given y x y x
x x
x x
x
A y y dx
x x dx
x x
or
Area A or
= =
∴ =
∴ − =
∴ =
∴ = −
= −
 
= −    
 
= − − −    
= − −
∴ =

-------------------------------------------------------------------------------------
1
½
1
½
1
1
½
1
½
½
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 15/28
Que.
No.
Sub.
Marks
3)
d)
Ans.
e)
Ans.
Solve ( ) 2
1 2 cos 2 dy
x y
dx
− = −
( )
( )
2
2
2
2
1 2 cos 2
2
1 2
cos
cos
sec
tan
tan 2
dy
x y
dx
Put x y v
dy dv
dx dx
dv
v
dx
dv dx
v
vdv dx
v x c
x y x c
− = −
− =
∴ − =
∴ =
∴ =
∴ =
∴ = +
∴ − = +
∫ ∫
-------------------------------------------------------------------------------------
Evaluate sin dy y y
dx x x
= +
( )
sin
sin
sin
cos
cos
log cos cot log
log cos cot log
dy y y
dx x x
y
Put v or y vx
x
dy dv
v x
dx dx
dv
v x v v
dx
dv
x v
dx
dx ecv dv
x
dx ecv dv
x
ec v v x c
y y
ec x c
x x
= +
= =
∴ = +
∴ + = +
∴ =
∴ =
∴ =
∴ − = +
  ∴ − = +    
∫ ∫
1
1
½
1
½
1
½
½
½
½+½
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 16/28
Que.
No.
Sub.
Marks
3)
4)
f)
Ans.
a)
Ans.
Evaluate ( ) ( )2
1 1 dy x
x y e x
dx
+ − = +
( ) ( )
( )
( )
( )
( )
2
1
log 1 1
1 1
1
1
1
1
and 1
1
1
1
1 1 1
1 1
1
1
1
1
x
x
x
pdx dx x x
x
x
x
dy
x y e x
dx
dy
y e x
dx x
P Q e x
x
IF e e e
x
y IF Q IF dx c
y e x dx
x x
y e dx
x
y e c
x

− + +
+ − = +
∴ − ⋅ = +
+
= − = +
+
∫ ∫ ∴ = = = =
+
∴ ⋅ = ⋅ ⋅ +
∴ ⋅ = ⋅ + ⋅
+ +
∴ ⋅ = ⋅
+
∴ ⋅ = +
+

-------------------------------------------------------------------------------------
Attempt any four.
Evaluate
/ 2
0
cos
cos sin
x
dx
x x
π
+

[ ]
/ 2
0
/ 2
0
/ 2
0
/ 2
0
/ 2
0
Replace / 2
cos sin cos
cos sin & cos sin
sin
sin cos
cos sin 2
cos sin
2 1
2
0
2
4
x x
x
I dx x x
x x
x x
x
I dx
x x
x x I dx
x x
I dx
I x
I
π
π
π
π
π
π
π
π
→ −
= ∴ →
+

∴ =
+
+
∴ =
+
∴ = ⋅
∴ =
= −
∴ =

1
1
1
1
1
1
½
1
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 17/28
Que.
No.
Sub.
Marks
4)
b)
Ans.
OR
[ ]
/ 2
0
/ 2
0
/ 2
0
/ 2
0
/ 2
0
/ 2
0
cos
cos sin
cos
2
cos sin
2 2
sin
sin cos
cos sin 2
cos sin
2 1
2
0
2
4
x
I dx
x x
x
dx
x x
x
I dx
x x
x x I dx
x x
I dx
I x
I
π
π
π
π
π
π
π
π π
π
π
=
+
    −
 
=
        − + −    
∴ =
+
+
∴ =
+
∴ = ⋅
∴ =
= −
∴ =

-------------------------------------------------------------------------------------
Evaluate
1
2
0
x x dx 1− ⋅ ∫
( ) ( )
( )
( )
[ ]
1
2
0
1 2
0
1
2
0
1 3 5
2 2
0
1
5 7
2 2 3
2
0
1
3 5 7
2 2 2
0
3 5 7
2 2 2
1
1 1 1
1 2
2
2
2
3 5 7
2 2
2 4 2
3 5 7
2 4 2 1 1 1 0 0 0
3 5 7
16 0.152
105
I x x dx
I x x dx
x x x dx
x x x dx
x x
x
x x x
or
= − ⋅
∴ = − − − ⋅
= − + ⋅
= − + ⋅
 
= − +  
 
 
= − +    
 
= − + − − +    
=

½
½
1
½
1
½
½
1
1
1
½
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 18/28
Que.
No.
Sub.
Marks
4)
c)
Ans.
d)
Ans.
Find by integration the area of the circle 2 2 2 x y a + = .
( ) ( )
2 2 2
2 2 2
2 2
2 2
2 2
0
2
2 2 1
0
2 2
1 1
2
2
0, 0
,
4
4
4 sin
2 2
4 0 sin 1 0 sin 0
2 2
4
2 2
b
a
a
a
x y a
y a x
y a x
At y a x
x a a
A ydx
a x dx
x a x
a x
a
a a
a
a
π
π

− −
+ =
∴ = −
∴ = −
= − =
∴ = −
∴ =
= −
   
= − +        
   
= + − +        
 
= ⋅    
=

OR
( ) ( )
2 2
2
2 2 1
2 2
1 1
2 2
2
,
2
2
2 sin
2 2
2 0 sin 1 0 sin 1
2 2
2
2 2 2 2
b
a
a
a
a
a
x a a
A ydx
a x dx
x a x
a x
a
a a
a a
a
π π
π

− −
= −
∴ =
= −
   
= − +        
   
= + − + −        
 
= ⋅ + ⋅    
=

-------------------------------------------------------------------------------------
Evaluate 2 3 3 4 ² dy x y y e x e
dx
− − = +
( )
2 3 3
2 3 3
2 3
4 ²
4 ²
4 ²
x y y
x y y
x y
dy
e x e
dx
e e x e
e x e
− −
− −

= +
= ⋅ +
= +
1
½
1
½
½
½
1
½
1
½
½
½
1
4
4
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Subject & Code: Applied Maths (17301) Page No: 19/28
Que.
No.
Sub.
Marks
4)
e)
Ans.
f)
Ans.
( )
( )
2
3
3 2
3 2
3
4 ²
4 ²
4
3 2 3
x
y
y x
y x
dy e x dx
e
e dy e x dx
e e
x c
∴ = + −
∴ = +
∴ = + +
∫ ∫
-------------------------------------------------------------------------------------
Evaluate (2 ² ² 2 sin 0 xy y dx x xy y dy + + + + = ) ( )
( ) ( )
( )
tan
2
2
2 ² ² 2 sin 0
2 ²
2 2
² 2 sin
2 2
.
2 ² sin
2 ² cos
2
² cos
y cons t terms free from x
xy y dx x xy y dy
M xy y
M
x y
y
N x xy y
N
x y
x
the equation is exact
Mdx Ndy c
xy y dx ydy c
x
y y x y c
or x y xy y c
+ + + + =
= +

∴ = +

= + +

∴ = +

+ =
+ + =
∴ ⋅ + − =
+ − =
∫ ∫
∫ ∫
-------------------------------------------------------------------------------------
Show that 2 2 y ax = is a solution of
2
2 0 dy dy
x y ax
dx dx
    − + =   .
2 2
2 2
2 3
2
2 3
2
2 2
2
2
2 2
2
2
0
y ax
dy
y ax
dx
dy ax ax
dx y y
dy dy ax ax
x y ax x y ax
dx dx y y
a x
ax ax
y
a x
ax ax
ax
=
∴ =
∴ = =
   
∴ − + = − ⋅ +        
= − +
= − +
=
1
1
1
1
½
½
1
1
1
1
1
1
4
4
4
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Subject & Code: Applied Maths (17301) Page No: 20/28
Que.
No.
Sub.
Marks
5)
a)
Ans.
b)
Ans.
Attempt any four.
A husband and wife appear in an interview for two vacancies
in the same post. The probability of husband’s selection is 1
7
and that of wife selection is 1
5
. What is probability that:
1) Both of them will be selected,
2) None of them will be selected.
( ) ( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
1 1 6 ' 1
7 7 7
1 1 4 ' 1
5 5 5
( & )
1 1
7 5
1
0.0286
35
se ( '& ')
' '
6 4
7 5
24 0.686
35
P H P H
P W P W
P Both selected P H W
P H P W
or
P None is lected p H W
P H P W
or
= = − =
= = − =
=
= ⋅
= ⋅
=
=
= ⋅
= ⋅
=
-------------------------------------------------------------------------------------
The overall percentage of failures in a certain examination is 20.
If six candidates appear in an examination, what is the
probability that at least five pass the examination?
% of failure = 20%
∴ % of passing = 80%
( ) ( ) ( )
( ) ( ) ( ) ( ) 5 1 6 0 6 6
5 6
80 0.8 1 0.8 0.2
100
6
5 5 6
0.8 0.2 0.8 0.2
0.6553
p q
n
p at least p p
C C
∴ = = = − =
=
∴ = +
= +
=
½
½
1
½
1
½
1
1
1
1
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 21/28
Que.
No.
Sub.
Marks
6)
c)
Ans.
d)
Ans.
Note: Due to the use of advance non-programmable scientific
calculators, writing directly the values of nCr
or n r n r C p q r

is permissible. No marks to be deducted for calculating
directly the value.
-------------------------------------------------------------------------------------
A skilled typist, on routine work, kept a record of mistakes per
day during 300 working days. Fit a Poisson distribution to the
set of observations.
x 0 1 2 3 4 5 6
y 143 90 42 12 9 3 1
x y xy
0 143 0
1 90 90
2 42 84
3 12 36
4 9 36
5 3 15
6 1 6
300 267
( ) 0.89
267 0.89
300
0.89
! !
r m r
mean m
e m e
p
r r
− −
∴ = =
∴ = =
-------------------------------------------------------------------------------------
Evaluate
1 sin cos
dx
+ +x x ∫
2
2 2 2
2
2
2 2
2
2 2
2
tan
2
2 2 1 , sin , cos
1 1 1
2
1
sin cos 1 2 1 1
1 1
2
1
1 2 1
1
x
Put t
dt t t dx x x
t t t
dt
dx t
x x t t
t t
dt
t
t t t
t
=
− ∴ = = =
+ + +
+ =
+ + −
+ +
+ +
+ =
+ + + −
+
∫ ∫

1
2
1
1
½
4
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Subject & Code: Applied Maths (17301) Page No: 22/28
Que.
No.
Sub.
Marks
5)
e)
Ans.
f)
Ans.
[ ]
2
2 2
1
log 1
log 1 tan
2
dt
t
dt
t
t c
x
c
=
+
=
+
= + +
 
= + +    

-------------------------------------------------------------------------------------
Evaluate ( )
1
3/ 2
0
x x dx 1− ∫
( ) ( ) ( )
( )
( )
( )
1 1
3/ 2 3/ 2
0 0
1
3/ 2
0
1
3/ 2 3/ 2
0
1
3/ 2 5/ 2
0
1
5/ 2 7/ 2
0
1 1 1 1
1
5 / 2 7 / 2
1 1 0
5 / 2 7 / 2
4
0.114
35
x x dx x x dx
x x dx
x x x dx
x x dx
x x
or
− = − − −    
= −
= − ⋅
= −
 
= −    
 
= − −    
=
∫ ∫

-------------------------------------------------------------------------------------
Solve cos sin
sin cos
dy y x y y
dx x x y x
+ +
= −
+ +
( ) ( )
cos sin
sin cos
cos sin sin cos 0
cos sin
cos cos 1
dy y x y y
dx x x y x
y x y y dx x x y x dy
M y x y y
M
x y
y
+ +
= −
+ +
∴ + + + + + =
= + +

∴ = + +

½
1
1
½
½
½
½+½
1
½
1
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 23/28
Que.
No.
Sub.
Marks
6)
6)
a)
Ans.
( )
tan
sin cos
cos cos 1
.
cos sin 0
sin sin
y cons t terms free from x
N x x y x
N
x y
x
the equation is exact
Mdx Ndy c
y x y y dx dy c
y x x y xy c
= + +

∴ = + +

+ =
+ + + =
∴ + + =
∫ ∫
∫ ∫
-------------------------------------------------------------------------------------
Attempt any four.
A coin is tossed and a die is rolled. Show that the events head
and six are independent and mutually exclusive.
Case I) consider the experiment of two events “A coin is tossed
and a die is rolled” are taken together.
∴ = S
{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),
(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.
∴n = 12.
Let A = event of occurring head,
A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
m = 6

( ) 6 1
12 2
m
P A
n
∴ = = =
Let B = event of occurring six
B = {(H, 6), (T, 6)}
m = 2

( ) 2 1
12 6
m
P B
n
∴ = = =
Now A B ∩ = {(H, 6)}
∴m = 1
The probability of happening head and six is
( ) 1
12
m
P A B
n
∴ ∩ = =
But ( ) ( ) 1 1 1
2 6 12
P A P B = × = .
∴ ∩ = P A B P A P B ( ) ( ) ( )
∴ the events are independent.
½
½
1
1
½
½
½
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 24/28
Que.
No.
Sub.
Marks
6)
b)
Ans.
But
P A B or also P A B P A P B P A B ( ∩ ≠ ∪ = + − ∩ ) 0  ( ) ( ) ( ) ( )
 
∴ the events are not mutually exclusive.
Case II) Consider the experiment of two events “A coin is
tossed and a die is rolled” are not taken together and
done exclusively.
i. the set of tossing coin is { H, T}. Consequently n =
2. Now let A = event of occurring head, then m = 1
and hence ( ) 1
2
m
P A
n
= = .
ii. the set of rolling of die is {1, 2, 3, 4, 5, 6}.
Consequently n = 6. Now let B = event of occurring
six, then m = 1 in this case and hence ( ) 1
6
m
P B
n
= =
Now here in this case A B ∩ = Φ and hence P A B ( ∩ =) 0
shows that the events are mutually exclusive but the events are
not independent as:
( ) ( ) ( ) 1 1 1
2 6 12
P A P B P A B = × = ≠ ∩ .
-------------------------------------------------------------------------------------
If A and B are two events such that ( ) ( ) 1 1
,
2 3
P A P B = = and
( ) 7
12
P A B ∩ = , find P A B ( ' ' ∩ )
( ) ( )
( )
( ) ( ) ( )
' ' '
1
1
1 1 7 1
2 3 12
3
0.75
4
P A B P A B
P A B
P A P B P A B
or
∩ = ∪
= − ∪
= − + − ∩    
 
= − + −    
=
OR
½
½
½
½
½
1
1
1
1
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 25/28
Que.
No.
Sub.
Marks
6)
c)
Ans.
OR
( ) ( ) ( ) ( )
( ) ( )
( )
1 1 7
2 3 12
1
0.25
4
' ' '
1
1
1 1 0.25
4
3
0.75
4
P A B P A P B P A B
or
P A B P A B
P A B
or
or
∪ = + − ∩
= + −
=
∩ = ∪
= − ∪
= − −
=
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In a sample of 1000 cases the mean of a certain test is 14 and
standard deviation is 2.5. Assuming the distribution to be
normal, find 1) how many students score between 12 and 15?
and 2) how many students score above 18?
( ) ( )
( ) ( )
( ) ( )
14 2.5 1000
12 14 15 14 1) 0.8 0.4
2.5 2.5
12 15 0.8 0.4
0.8 0 0 0.4
0 0.8 0 0.4
Given x N
z z
P x P z
P z P z
P z P z
= = = σ
− −
= = − = =
∴ ≤ ≤ = − ≤ ≤
= − ≤ ≤ + ≤ ≤
= ≤ ≤ + ≤ ≤
Note: To get the further solution of the problem, the students
are required the set of the values of the area under
standard normal curves. And this set of values is not
provided with this question. These values can only be
obtained from the Table of Area Under Standard Normal
Curve. This table is provided at the end of the solution for
The further solution is formed using the values obtained
from this table only.
(12 15 0.2881 0.1554 )
0.4435
. 1000 0.4435 443.5 . ., 444
P x
no of students N P i e
∴ ≤ ≤ = +
=
∴ = ⋅ = × =
1
1
½
1
½
½
½
½
½
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 26/28
Que.
No.
Sub.
Marks
6)
d)
Ans.
( ) ( )
( )
18 14 2) 1.6
2.5
18 1.6
0.5 0 1.6
0.5 0.4452
0.0548
. 1000 0.0548 54.8 . ., 55
z
P x P z
P z
no of students N P i e

= =
∴ ≤ = ≤
= − ≤ ≤
= −
=
∴ = ⋅ = × =
Note: If the students have adopted any other assumptions, due
credit may be given.
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Show that the equation of the tangent to the curve x
2
a
m m
y
b
        + =    
at the point (a, b) is 2
x y
a b
+ = .
1 1
1 1
2
1 1 0
at ( , ),
1 1 0
1 1 0
slope of tangent at ( , )
the equation of tangent is
m m
m m
m m
x y
a b
x y dy
m m
a a b b dx
a b
a b dy
m m
a a b b dx
dy
m m
a b dx
dy b a b
dx a
b
y b
− −
− −
        + =    
    ∴ ⋅ + ⋅ ⋅ =        

        ⋅ + ⋅ ⋅ =    
∴ ⋅ + ⋅ ⋅ =
∴ = = −

− = − ( )
2
2
x a
a
ay ab bx ab
bx ay ab
x y
a b

∴ − = − +
∴ + =
∴ + =
½
½
½
½
1
½
½
1
½
½
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 27/28
Que.
No.
Sub.
Marks
6)
e)
Ans.
f)
Ans.
Divide 80 into two parts such that their product is maximum.
Let x, y be the numbers.
But x + y = 80 i. e., y = 80 - x
To maximize, p xy x x = = − (80 )
2
2
2
2
2
80
80 2
2
, 0
80 2 0 80 2
40
40, 2 0
40, max .
p x x
dp
x
dx
d p
dx
dp For stationary values
dx
x or x
x
d p At x
dx
At x p has imum value
∴ = −
∴ = −
∴ = −
=
∴ − = =
∴ =
= = − <
∴ =
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Two points A(1, 4) and B(9, 12) are on the parabola 2
y x =16 .
Show that the area enclosed between the chord AB and the
parabola is 16
3
.
1 4
9 1 12 4
1 4
3
x y
x y
y x
− −
=
− −
∴ − = −
∴ = +
Note: Students may use another form to find the equation of
this line, such as slope point form.
( )
( )
( ) ( )
2 1
9
1
9
2
3/ 2
1
2
3/ 2
4 3
2
4 3
3 2
8 9 8 1 9 3 9 3
3 2 3 2
16 5.333
3
b
a
A y y dx
x x dx
x
x x
or
∴ = −
= − −
 
= ⋅ − −    
   
= − − − − −        
=

1
½
½
½
½
½
½
1
1
1
½
½
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 27/28
Que.
No.
Sub.
Marks
6)
OR
( )
( )
( ) ( )
2 1
9
1
9
2
3/ 2
1
2
3/ 2
1 4
9 1 12 4
1 4
3
3 4
2
3 4
2 3
9 8 1 8 3 9 9 3
2 3 2 3
16 5.333
3
16 5.333
3
b
a
x y
x y
y x
A y y dx
x x dx
x
x x
or
Area A or
− −
=
− −
∴ − = −
∴ = +
∴ = −
= + −
 
= + − ⋅    
   
= + − − + −        
= − −
∴ =

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Important Note
In the solution of the question paper, wherever possible all
the possible alternative methods of solution are given for the
sake of convenience. Still student may follow a method other
than the given herein. In such case, FIRST SEE whether the
method falls within the scope of the curriculum, and THEN
ONLY give appropriate marks in accordance with the scheme
of marking.
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1
1
1
½
½ 4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 28/28
Area Under Standard Normal Curve
e. g., The value of P z (0 0.45 ≤ ≤ ) is 0.1736.