# 2015 Summer-17301-Maths III-Model Answer Sheet

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### Model answer sheet Summer 17301 summer 2015 text only

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Summer 2015 Examination
Subject & Code: Applied Maths (17301) Model Answer Page No: 1/26
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
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 2/26
Que.
No.
Sub.
Marks
1)
a)
Ans.
b)
Ans.
c)
Ans.
Attempt any TEN of the following:
At which point on the curve 2
y x x = − 3 the slope of the tangent
is -5.
( ) ( )
( )
2
2
3
3 2
But tangent is parallel to x-axis.
3 2 5
4
3 4 4 4
the point is 4, 4 .
y x x
dy
x
dx
x
x
y
= −
∴ = −
∴ − = −
∴ =
∴ = − = −
∴ −
-------------------------------------------------------------------------------------
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
∴ = −
∴ = −
∴ = −
=
∴ − = =
∴ =
= = − <
∴ =
-------------------------------------------------------------------------------------
Evaluate: 3
sin cos x xdx ∫
3
3
4
4
Put sin
sin cos
cos
4
sin
4
x t
x xdx
xdx dt
t dt
t
c
x
c
=
∴ =
=
= +
= +

½
½
½
½
½
½
½
½
½
½
½
½
2
2
2
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 3/26
Que.
No.
Sub.
Marks
1)
d)
Ans.
e)
Ans.
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 ( )( )
1
3 2
dx
x x + + ∫
( )( )
( )( )
( )
( )( )
( ) ( )
1
3 2
1
3 2 3 2
1
1 *
1 1 1
3 2 3 2
1 1
3 2
log 3 log 2
I dx
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.
OR
( )( )
2
2
2 2
1
3 2
1
5 6
1
25 25 5 6
4 4
1
5 1
2 2
I dx
x x
dx
x x
dx
x x
dx
x
=
+ +
=
+ +
=
+ + − +
=
        + −    

½
½
½+½
½
½
½+½
½
½
2
2
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Subject & Code: Applied Maths (17301) Page No: 4/26
Que.
No.
Sub.
Marks
1)
f)
Ans.
g)
Ans.
h)
Ans.
5 1
1 2 2 log
1 5 1
2
2 2 2
2
log
3
x
c
x
x
c
x
 
+ −  
= +       + +      
  +
= +     +
-------------------------------------------------------------------------------------
Evaluate
log 2 2
0
e x
e dx ∫
log 2 2
log 2 2
0
0
2log 2 0
2
2 2
4 1
2 2
3
1.5
2
e
x
x e
e dx
e e
or
 
=    
= −
= −
=

-------------------------------------------------------------------------------------
Find the area between the line y x = 2 and x =1 and x = 3 .
3 3
1 1
3
2
3
2
1
1
2
2
2
2
3 1
8
y dx x dx
x
or x
⋅ = ⋅
 
= ⋅        
= −
=
∫ ∫
-------------------------------------------------------------------------------------
Find the order and degree of the following equation:
2
2
1 0 d y dy
dx dx
+ + =
Order = 2
2
2
2
For degree,
1
Degree = 2
d y dy
dx dx
    = +
 

½
½
½
½
1
½
½
½
½
1
1
2
2
2
2
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 5/26
Que.
No.
Sub.
Marks
1)
i)
Ans.
j)
Ans.
k)
Ans.
Form a differential equation if 2
y ax = 4 .
2
2
4
2 4
2
2 2
2 0 0.
2
y ax
dy
y a
dx
dy
y y x
dx
dy dy
y x or x y
dx dx
dy dy y
or x y or
dx dx x
=
∴ =
∴ = ⋅
∴ = =
− = − =
-------------------------------------------------------------------------------------
From a pack of 52 cards one card is drawn at random. Find the
probability of getting a king.
( )
( )
( ) ( )
( )
52
4
4
52
1
0.077
13
n n S
m n A
n A
p p A
n S
or
= =
= =
∴ = =
=
=
-------------------------------------------------------------------------------------
An unbiased coin is tossed 5 times. Find the probability of
3 2
5
3
1 1 0.5 1 0.5
2 2
5
1 1
2 2
5
0.3125
16
n r n r
r
p q p
Here n
p C p q
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.
1
½
½
½
½
½
½
½
1
½
2
2
2
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 6/26
Que.
No.
Sub.
Marks
1)
2)
l)
Ans.
a)
Ans.
b)
Ans.
A die is thrown, find the probability of getting an odd number.
( )
( )
( )
( )
6
3
3
6
1
0.5
2
n n S
m n A
n A
p
n S
or
∴ = =
∴ = =
∴ =
=
=
-------------------------------------------------------------------------------------
Attempt any four of the following:
Find the equation of tangent and normal to the curve
y = x(2 – x) at (2, 0).
( )
( )
( )
( )
2
2 2
the slope of tangent at 2, 0 is
2 4 2
the equation of tangent is
0 2 2
2 4 2 4
1 1 the slope of normal
2
the equation of normal is
1
0 2
2
2 2 2 2 0 2 2
y x x
dy
x
dx
m
y x
y x or x y
m
y x
y x or x y or x y
= −
∴ = −

= − = −

− = − −
∴ = − + + =
∴ = − =

− = −
∴ = − − − = − + + = 0
-------------------------------------------------------------------------------------
Find radius of curvature of the curve 3 3 x a y a = = cos , sin θ θ at
4
π
θ =
3
2
cos
3 cos sin
x a
dx
a
d
θ
θ θ
θ
=
∴ = −
½
½
½
½
1
½
½
½
½
½
½
½
2
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 7/26
Que.
No.
Sub.
Marks
2)
c)
Ans.
3
2
2
2
2
2
2
2
2
2
2
2
2
sin
3 sin cos
/ 3 sin cos
tan
/ 3 cos sin
&
1
sec
3 cos sin
,
4
tan 1
4
1
sec
4
3 cos sin
4 4
4 2
3
1
y a
dy
a
d
dy dy d a
dx dx d a
d y d dy d
dx d dx dx
a
at
dy
dx
d y and
dx a
a
dy
dx
θ
θ θ
θ
θ θ θ
θ
θ θ θ
θ
θ
θ
θ θ
π
θ
π
π
π π
ρ
=
∴ =
∴ = = = −

 
= ×    
= − ×

∴ =
= − = −
= ×
=
      +         ∴ =
( )2
2
2
3
2
3
2
1 1
4 2
3
3
2
d y
dx a
a
  + −  
=
=
-------------------------------------------------------------------------------------
Find the maximum and minimum value of 15 ³ ² 18
2
y x x x = − + .
2
2
2
2
15 ³ ² 18
2
3 15 18
6 15
, 0
3 15 18 0
2, 3
y x x x
dy
x x
dx
d y
x
dx
dy For stationary values
dx
x x
x
= − +
∴ = − +
∴ = −
=
∴ − + =
∴ =
½
½
½
½
½
½
½
½
½
½+½
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 8/26
Que.
No.
Sub.
Marks
2)
d)
Ans.
e)
Ans.
( )
( ) ( ) ( )
( )
( ) ( ) ( )
2
2
2
2
2, 6 2 15 3 0
2, max
15 2 ³ 2 ² 18 2 14
2
3, 6 3 15 3 0
3, min
15 3 ³ 3 ² 18 3 13.5
2
d y At x
dx
At x y has imum value and it is
y
d y At x
dx
At x y has imum value and it is
y
= = − = − <
∴ =
= − + =
= = − = >
∴ =
= − + =
-------------------------------------------------------------------------------------
Evaluate ( 1)
cos ²( )
x
x
e x dx
xe
+

( )
( )
( )
2
2
( 1) 1
cos ²( )
1
cos
sec
tan
tan
x
x
x x
x
x
x
Put xe t
e x dx xe e dx dt
xe
e x dx dt
dt
t
tdt
t c
xe c
=
+
∴ + ⋅ =
∴ + =
=
=
= +
= +

-------------------------------------------------------------------------------------
Evaluate
2
2
sec
3tan 2 tan 5
x
dx
x x − −

2
2 2
2
2 2
2
2 2
sec tan
3tan 2 tan 5 sec
1
3t 2 t 5
2 5 3t 2 t 5 3 t t
3 3
2 1 1 5 3 t t
3 9 9 3
1 4 3 t
3 3
x Put x t
dx
x x xdx dt
dt
=
− − ∴ =
=
− −
  ∴ − − = − −    
 
= − + − −    
     
= − −              

½
½
½
½
1
1
½
1
½
½
1
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 9/26
Que.
No.
Sub.
Marks
2)
f)
Ans.
2 2
2 2
1
1 4 3 t
3 3
1 1
3 1 4
t
3 3
1 4
t
1 1 3 3 log
3 4 1 4
2 t
3 3 3
5
t
1 3
log
8 t 1
1 3t 5 log
8 3t 3
1 3tan 5 log
8 3tan 3
I dt
dt
c
c
c
x
c
x
∴ =             − −      
=
        − −    
 
− −  
= ⋅ ⋅ +       − +      
 

 
= ⋅ +  
+    
  −
= ⋅ +     +
  −
= ⋅ +     +

-------------------------------------------------------------------------------------
Evaluate
1
sin
1 ²
x xdx
x

( ) ( )
( ) ( )
( )
1
1
1 1
sin
sin 1
1 ² 1 ²
sin
sin
sin sin
cos cos 1
cos cos
cos sin
sin cos sin
Put x t
x xdx dx dt
x x
Also x t
t t dt
d
t tdt tdt t dt
dt
t t t dt
t t t dt
t t t c
x x x c

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

∫ ∫ ∫

-------------------------------------------------------------------------------------
½
1
½
½
1
½
½
1
½
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 10/26
Que.
No.
Sub.
Marks
3)
a)
Ans.
b)
Ans.
Attempt any four of the following:
Evaluate
/ 2
0 9 4 ²
dx
x
π

The given problem cannot be solved within the given limits
because for the integrating the given function within the
prescribed limits the function must be well defined on the given
interval. For example at
2
x
π
= ,
1
9 4 ² − x
is a non-real number
and hence the function is not defined on the interval 0, 4
  π
  .
-------------------------------------------------------------------------------------
Evaluate
/3
/ 6
sin
sin cos
x
dx
x x
π
π
+

[ ]
/3
/ 6
/3
/ 6
/3
/ 6
/3
/ 6
/3
/ 6
Re / 2
sin sin cos
sin cos
& cos sin
cos
cos sin
sin cos 2
sin cos
2 1
2
3 6
12
place 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
π
π
π
π π
π
→ −
= ∴ →
+

∴ =
+
+
∴ =
+
∴ = ⋅
∴ =
= −
∴ =

OR
( )
( ) ( )
/3
/ 6
/3
/ 6
/3
/ 6
sin
sin cos
sin 2
sin cos 2 2
cos
cos sin
x
I dx
x x
x
dx
x x
x
I dx
x x
π
π
π
π
π
π
π
π π
=
+

=
− + −
∴ =
+

4
1
1
½
½
½
½
½
½
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 11/26
Que.
No.
Sub.
Marks
3)
c)
Ans.
[ ]
/3
/ 6
/3
/ 6
/3
/ 6
sin cos 2
sin cos
2 1
2
3 6
12
x x I dx
x x
I dx
I x
I
π
π
π
π
π
π
π π
π
+
∴ =
+
∴ = ⋅
∴ =
= −
∴ =

-------------------------------------------------------------------------------------
Find the area bounded by two curves 2 2 y x x y = = and
( )
( )
2 2
2
2
2 1
1
2
0
1
3
3/ 2
0
3
3/ 2
Given and
0, 1

2

3 3
2 1 1 0
3 3
1
0.333
3
b
a
y x x y
x x
x x
A y y dx
x x dx
x
x
or
= =
∴ =
∴ = =
= −
= −    
 
= −    
 
= ⋅ − −    
=

OR
( ) 2 1
1
2
0
1
3
3/ 2
0
3
3/ 2
0, 1

2

3 3
1 2 1 0
3 3
1
0.333
3
1
0.333
3
b
a
x x
A y y dx
x x dx
x
x
or
the area or
∴ = =
= −
= −    
 
= −    
 
= − ⋅ −    
= − −
∴ =

1
½
½
½
½
½ + ½
1
1
½
½
½ + ½
1
1
½
½
4
4
4
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Subject & Code: Applied Maths (17301) Page No: 12/26
Que.
No.
Sub.
Marks
3)
d)
Ans.
e)
Ans.
Solve ( )
2 3 3 xy dy x y dx − + = 0 given y = 0 when x = 1.
( )
( )
2 3 3
3 3
2
3 3 3
2 2 2
3
2
2
2
3
3
3
3
3
0
Put
1
1
1
1
log
3
log
3
1 & 0, 0
log
3
xy dy x y dx
dy x y
dx xy
dy dv
y vx v x
dx dx
dv v x vx
v x
dx xv x v
dv v
x v
dx v
dv
x
dx v
v dv dx
x
v
x c
y
x c
x
At x y c
y
x
x
− + =
+
∴ =
= ∴ = +
+ +
∴ + = =
+
∴ = −
∴ =
∴ =
∴ = +
∴ = +
= = =
∴ =
∫ ∫
-------------------------------------------------------------------------------------
Solve the differential equation ( )2 dy 2
x y a
dx
+ =
2 2
2
2
2 2 2
2 2
2
2 2
1 1
1
1
1
Put x y v
dy dv dy dv
or
dx dx dx dx
dv
v a
dx
dv a
dx v
dv a a v
dx v v
v
dv dx
a v
+ =
∴ + = = −
  ∴ − =    
∴ − =
+
∴ = + =
 
∴ =     +
1
½
½
½
½
½
½
½
½
½
½
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 13/26
Que.
No.
Sub.
Marks
3)
f)
Ans.
2
2 2
2
2 2
2 1
1 1
1
1
tan
tan tan
v
dv dx
a v
a
dv dx
a v
v
v a x c
a a
x y x y
x y a x c or y a c
a a

− −
 
∴ =     +
 
∴ − =     +
  ∴ − ⋅ = +    
    + +
∴ + − = + − =        
∫ ∫
∫ ∫
-------------------------------------------------------------------------------------
Solve dy 2
x y x
dx
− =
2
1
log
1
1
and
1
1 1
1.
pdx
dx
x
x
dy
x y x
dx
dy
y x
dx x
P Q x
x
IF e
e
e
x
y IF Q IF dx c
y x dx c
x x
y
dx c
x
y
x c
x

− =
∴ − ⋅ =
∴ = − =
∴ = ∫

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

-------------------------------------------------------------------------------------
½
1
½
1
1
1
1
4
4
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Subject & Code: Applied Maths (17301) Page No: 14/26
Que.
No.
Sub.
Marks
4)
a)
Ans.
Attempt any Four of the following:
Evaluate
5 3
3 3
1
9
9 3
x
dx
x x

− + + ∫
[ ]
5 3
3 3
1
5 3
3 3
1
5 3 3
3 3
1
5
1
5
1
Re 6
9
9 3
9 3 & 3 9
3
3 9
9 3 2
9 3
2 1
2
2 5 1
2
place 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
I
→ −

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

OR
( )
( ) ( )
[ ]
5 3
3 3
1
5 3
3 3 1
5 3
3 3
1
5 3 3
3 3
1
5
1
5
1
9
9 3
9 6
9 6 6 3
3
3 9
9 3 2
9 3
2 1
2
2 5 1
2
x
I dx
x x
x
I 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: 15/26
Que.
No.
Sub.
Marks
4)
b)
Ans.
c)
Ans.
Evaluate
4cos ² 9sin ²
dx
x x +

2
2
2
1
1
4cos ² 9sin ²
/ cos ²
4cos ² 9sin ²
cos ²
sec tan
4 9 tan ² sec
4 9 t ²
4
9 t ²
9
1
9 2
t ²
3
1 1
tan
9 2 / 3 2 / 3
1 3tan
tan
6 2
dx
x x
dx x
x x
x
xdx Put x t
x xdx dt
dt
dt
dt
t
c
x
c

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

-------------------------------------------------------------------------------------
Using integration find the area of the ellipse
2 2
2 2 1
x y
a b
+ =
( )
2 2
2 2
2
2 2
2
2
2 2 2
2
2 2
2 2
0
2 2
0
1
1
0 0 . ., ,
4
4
a
a
x y
a b
x
y b
a
b
y a x
a
b
y a x
a
Now y gives a x i e x a a
A ydx
b
a x dx
a
+ =
 
∴ = −    
∴ = −
∴ = −
= − = = −
∴ =
= −

½
½
1
1
1
1
1
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 16/26
Que.
No.
Sub.
Marks
4)
d)
Ans.
( )
2
2 2 1
0
2
1
2
4 sin
2 2
4
0 sin 1 0
2
4
2 2
a
b x a x
a x
a a
b a
a
b a
a
ab
π
π

   
= ⋅ − +        
 
= + −    
 
= ⋅    
=
-------------------------------------------------------------------------------------
Solve ( ) ( )
2 2 2 2 2 6 3 2 0 x xy y dx x xy y dy + − + − + =
( )
2 2
2 2
tan
2 2 2
3 2 3
2
3
3 2 2
2 6
6 2
3 2
6 2
.
2 6
2. 6
3 2 3
2
3
3 3
y cons t terms free from x
M x xy y
M
x y
y
N x xy y
N
x y
x
the equation is exact
Mdx Ndy c
x xy y dx y dy c
x x y
y y x c
y
or x x y y x c
= + −

∴ = −

= − +

∴ = −

+ =
+ − + =
∴ + ⋅ − + =
+ − + =
∫ ∫
∫ ∫
OR
( ) ( )
2 2 2 2
2 2
2 2
2 2 2 2
2 2 2 2
2 6 3 2 0
2 6
3 2
2 6
3 2
x xy y dx x xy y dy
dy x xy y
dx x xy y
dy dv Put y vx v x
dx dx
dv x vx v x
v x
dx x vx v x
+ − + − + =
+ −
∴ = −
− +
= ∴ = +
+ −
∴ + = −
− +
1
½
½
1
½
½
1
1
1
½
4
4
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Subject & Code: Applied Maths (17301) Page No: 17/26
Que.
No.
Sub.
Marks
4)
e)
Ans.
( ) ( )
2
2
2
2
2 2
2
2 3
2
2
2 3
2
2 3
2
2 3
2 6
3 2
2 6
3 2
2 6 3 2
3 2
2 9 3
3 2
3 2
2 9 3
3 2
2 9 3
1 9 6 3
3 2 9 3
dv v v
v x
dx v v
dv v v
x v
dx v v
dv v v v v v
x
dx v v
dv v v v
x
dx v v
v v dx dv
v v v x
v v dx dv
v v v x
v v
v v v
+ −
∴ + = −
− +
+ −
∴ = − −
− +
− + − − − +
∴ =
− +
− − + −
∴ =
− +
  − +
∴ =     − − + −
  − +
∴ =     − − + −
  − + −
∴− 
− − + −
∫ ∫
( ) 2 3
2 3
1
log 2 9 3 log
3
1
log 2 9 3 log
3
dx dv
x
v v v x c
y y y
x c
x x x
= 

∴− − − + − = +
      ∴− − − ⋅ + − = +            
∫ ∫
-------------------------------------------------------------------------------------
Solve ( )
2 2 1 0 + − = x dy x ydx
( )
( )
2 2
2
2
2
2
2
1
1
1 0
1
0
1
1
1
1 1 1
1
log tan
log tan
x dy x ydx
x
dy dx
y x
x
dy dx c
y x
dy dx c
y x
y x x c
or y x x c

+ − =
∴ − =
+
∴ − =
+
  ∴ − − =     +
∴ − − =
− + =
∫ ∫
∫ ∫
-------------------------------------------------------------------------------------
½
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: 18/26
Que.
No.
Sub.
Marks
4)
5)
f)
Ans.
a)
Ans.
Show that y x = sin log ( ) is solution of differential equation
2
2
2
0
d y dy
x x y
dx dx
+ + =
( )
( )
( )
( )
2
2
2
2
2
2
2
2
sin log
1
cos log
cos log
1
sin log
0
y x
dy
x
dx x
dy
x x
dx
d y dy
x x
dx dx x
d y dy
x x y
dx dx
d y dy
x x y
dx dx
=
∴ = ⋅
∴ =
∴ + = − ⋅
∴ + = −
∴ + + =
-------------------------------------------------------------------------------------
Attempt any Four of the following:
The probability that A can shoot at a target is 5
7
and B can
shoot at the same target is 3
5
. (A and B shoot independently.)
Find the probability that
i) The target is not shot at all.
ii) The target is shot by at least one of them.
( ) ( )
( ) ( )
( )
( ) ( )
5 5 2 ' 1
7 7 7
3 3 2 ' 1
5 5 5
) target is not shot ( '& ')
' '
2 2
7 5
4
0.114
35
P A P A
P B P B
i P P A B
P A P B
or
= ∴ = − =
= ∴ = − =
=
= ⋅
= ⋅
=
1
1
1
1
½
½
1
½
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 19/26
Que.
No.
Sub.
Marks
5)
b)
Ans.
) at least1shoot 1 (target is not shot) ( )
4
1 1 0.114
35
31 0.886
35
ii P p
or
or
= −
= − −
=
-------------------------------------------------------------------------------------
Note for Numerical Problems: For practical purpose, generally
the values of fractional numbers are truncated up to 3 decimal
points by the method of rounded-off. Thus the solution is taken up
to 3 decimal points only. If answer is truncated more than 3
decimal points, the final answer may vary for last decimal points.
Thus 31/35 is actually 0.885714285 but can be taken as 0.886. Due
to the use of advance calculators, such as modern scientific nonprogrammable
calculators, the step 31/35 may not written by the
students and then directly the answer 0.012 is written. In this case,
no marks to be deducted.
-------------------------------------------------------------------------------------
If 30% of the bulbs produced are defective, find the probability
that out of 4 bulbs selected,
a) one is defective,
b) at the most two are defective.
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
[ ]
1 3 4
1
0 4 1 3 2 2 4 4 4
0 1 2
3 1 4 0 4 4
3 4
30 0.3
100
0.7
4
)
0.3 0.7
0.412
) 0 1 2
0.3 0.7 0.3 0.7 0.3 0.7
0.24 0.412 0.265
0.916
1 3 4
1 0.3 0.7 0.3 0.7
1 0.076 0.008
0.916
n r n r
r
p
q
Here n
i p C p q
C
ii p p p p
C C C
OR
p p p
C C

= =
∴ =
=
=
=
=
= + +
= + +
= + +
=
= − +    
= − +    
= − +
=
1
½
1
1
1
1
OR
1
1
4
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 20/26
Que.
No.
Sub.
Marks
5)
c)
Ans.
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.
-------------------------------------------------------------------------------------
In a certain examination 500 students appeared, mean score is
68 and S. D. is 8. Assuming data is normally distributed, find
the number of students scoring,
a) less than 50
b) more than 60.
(Given that area between Z = 0 to Z = 2.25 is 0.4878 and area between
Z = 0 & Z = 1 is 0.3413.)
( ) ( )
( )
( )
( ) ( )
( ) ( )
68 8 500
50 68 ) 2.25
8
50 2.25
2.25
0.5 0 2.25
0.5 0.4878
0.0122
.
500 0.0122
6.1 . ., 6
60 68 ) 1
8
60 1
1 0 0
0.3413 0.5
0.8413
.
Given x N
x x i z
p x p z
p z
p z
no of students N p
i e
x x ii z
p x p z
p z p z
no of
σ
σ
σ
= = =
− −
= = = −
∴ ≤ = ≤ −
= ≤
= − ≤ ≤
= −
=
∴ = ⋅
= ×
=
− −
= = = −
∴ ≤ = − ≤
= − ≤ ≤ + ≤
= +
=

500 0.8413
420.65 . ., 421
students N p
i e
= ⋅
= ×
=
-------------------------------------------------------------------------------------
½
½
½
½
½
½
½
½ 4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 21/26
Que.
No.
Sub.
Marks
5)
d)
Ans.
e)
Ans.
Evaluate
0
5 4cos
dx
x
π
+

2
2 2
Put tan
2
2 1 and cos
1 1
x
t
dt t dx x
t t
=
− ∴ = =
+ +

x 0 π
t 0 ∞
2 2
0 0
2
2
0
2 2
0
1
0
1 1
1 2
5 4cos 1 1
5 4
1
1
2
9
1
2
3
1
2 tan
3 3
2
tan tan 0
3

dx dt
x t t
t
dt
t
dt
t
t
π ∞

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

2

3 2

3
π
π
 
=    
=
-------------------------------------------------------------------------------------
Evaluate 2
3 4
x
dx
x x + − ∫
( )( )
( ) ( )
2

3 4
1 4
1/ 5 4 / 5 (Please refer next note)
1 4
1 4 log 1 log 4
5 5
x
dx
x x
x
dx
x x
dx
x x
x x c
+ −
=
− +
 
= +     − +
= − + + +

½
½
½
½
1
½
½
2
1+1
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 22/26
Que.
No.
Sub.
Marks
5)
6)
f)
Ans.
a)
Ans.
Note: To find the partial fractions, traditional partial fraction method
is generally used. But apart from this direct method of partial
fraction is also allowed here.
-------------------------------------------------------------------------------------
Solve log log dy
x x y z x
dx
+ =
(Considering z as constant.)
( )
( )
1
log log log
2
2
log log
1
log
1
and
log
log
log log
1
Put log
log t
log
2
log
log
2
dx pdx x x x
dy
x x y z x
dx
dy z
y
dx x x x
z
P Q
x x x
IF e e e x
y IF Q IF dx c
z
y x x dx c
x
x t dx dt
x
y x z dt c
t
y x z c
x
y x z c
+ =
∴ + =
∴ = =

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

-------------------------------------------------------------------------------------
Attempt any Four of the following:
If ( ) ( ) 1 2 , '
2 3
P A P B = = and ( ) 2
3
P A B ∪ = , find P A B ( ' ' ∩ ) and
( ) P A
B
.
( )
( )
1
2
2 1 1
3 3
P A
P B
=
= − =
1
1
½
½
½
½
½
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 23/26
Que.
No.
Sub.
Marks
6)
b)
Ans.
c)
( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( )
( )
) ' ' '
1
2
1
3
1
0.333
3
)
1 1 2
2 3 3
1
0.167
6
1
6
1
3
1
0.5
2
i P A B P A B
P A B
or
ii P A B P A P B P A B
or
P A B
P A
B P B
or
∩ = ∪
= − ∪
= −
=
∩ = + − ∪
= + −
=

∴ =
=
=
-------------------------------------------------------------------------------------
If the probability that an electric motor is defective is 0.01, what
is the probability that the sample of 300 electric motors will
contain exactly 5 defective motors? ( 3
e 0.0498 −
= )
( )
( )
3 5
0.01 300
0.01 300 3
!
3
5
5!
0.101
m r
p n
m np
e m
p r
r
e
p

= =
∴ = = × =

=
⋅ ∴ =
=
-------------------------------------------------------------------------------------
Fit a Poisson distribution for the following observation:
x 20 30 40 50 60 70
f 8 12 30 10 6 4
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: 24/26
Que.
No.
Sub.
Marks
6)
Ans.
d)
Ans.
x f xy
20 8 160
30 12 360
40 30 1200
50 10 500
60 6 360
70 4 280
70 2860
( ) 40.857
2860 40.857
70
!
40.857
!
m r
r
mean m
e m
p
r
e
r

∴ = =
∴ =
=
-------------------------------------------------------------------------------------
A metal wire 36 m 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
2
2
2 2 36 18
18
18 18
18 2
2
, 0
18 2 0
9
9, 2 0
9, max
18 9
x y or x y
y x
But area A xy x x x x
dA
x
dx
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
∴ + = + =
∴ = −
= = − = −
∴ = −
∴ = −
=
∴ − =
∴ =
= = − <
∴ =
= − =
1
2
1
½
1
½
½
½
½
½
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject & Code: Applied Maths (17301) Page No: 25/26
Que.
No.
Sub.
Marks
6)
e)
Ans.
Find the equation of tangent to the curve 1 1
x y , 1
t t
= = − , when
t=2.
( )
( )
1 1 , 1
1
1
2, 0.5 0.5
1
,
0.5 1 0.5
0.5 0.5
1 0
x y
t t
y x
dy
dx
at t x and y
and slope m
the equationis
y b m x a
y x
y x
x y
= = −
∴ = −
∴ = −
∴ = = =
= −

− = −
∴ − = − −
∴ − = − +
∴ + − =
OR
( )
( )
2 2
2
2
1 1 , 1
1 1
1
1
1
2, 0.5 0.5
1
,
0.5 1 0.5
0.5 0.5
1 0
x y
t t
dx dy and
dt t dt t
dy
dy dt t
dx dx
dt t
dy
dx
at t x and y
and slope m
the equationis
y b m x a
y x
y x
x y
= = −
∴ = − =
∴ = =

∴ = −
∴ = = =
= −

− = −
∴ − = − −
∴ − = − +
∴ + − =
-------------------------------------------------------------------------------------
1
½ + ½
1
1
½
½
½ + ½
1
1
4
4
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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_____________________________________________________________________________________________________
Subject & Code: Applied Maths (17301) Page No: 26/26
Que.
No.
Sub.
Marks
6)
f)
Ans.
Find the area between the parabola 2
y x = −4 and the x-axis.
( )
( ) ( )
2
2
2
2
2
2
3
2
3
3
3 3
4 . ., 0
4 0
2, 2
4
4
3
2 2
2 2
3 3
32 10.667
3
b
a
y x and x axis i e y
x
x
A ydx
x dx
x
x
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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½ + ½
½
1
½
1 4