MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer
Subject Code:
17301
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.
1
Marking Scheme
Answer Attempt any TEN of the following:
a) Ans
20
2 Find the point on the curve y x 6 x 8 where the tangent is parallel to X-axis
02
y x2 6x 8
dy 2x 6 dx
½
tangent is parallel to X-axis
½
dy 0 dx 2x 6 0
½
x 3 y 32 6 3 8 1
½
Point is 3, 1 -------------------------------------------------------------------------------------------------------------b) Ans
Find the radius of curvature of the curve xy c at point c, c
02
xy c x
dy y0 dx Page No.01/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
1
b)
Subject Code:
17301 Marking Scheme
Answer
dy y dx x dy x dx y d2y 2 dx 2 x at c, c
½
dy c 1 dx c 2 c 1 c 2 d y 2 dx c2 c
½
3
dy 2 2 3 2 2 1 1 1 dx Radius of curvature 2 d2y c dx 2 c 2 2 2c 2
c) Ans
½
½
--------------------------------------------------------------------------------------------------------1 Evaluate dx sin 1 x 1 x 2 1 sin 1 x 1 x 2 dx put sin 1 x t
1
1 x2 1 dt t log t c
02
dx dt
½ ½ ½
log sin 1 x c
d) Ans
½
---------------------------------------------------------------------------------------------------1 Evaluate dx 3 4 2 3x
1 4
2 3x
3
02
dx
Page No.02/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
1
d)
Subject Code:
Marking Scheme
Answer
1 3
2 3x 4
2 3x
2 3x
17301
dx
½ 3 4
3 1 4
dx 1
1 2 3 x 4 1 . . c 1 3 3 4
1
3 1 4 1 4 2 3x 4 c 3
½
--------------------------------------------------------------------------------------------------e) Ans
Evaluate tan 1 xdx
tan
1
02
x.1dx
d tan 1 x dx tan x 1dx 1dx. dx x tan 1 x.x dx 1 x2 1 2x x tan 1 x dx 2 1 x2 1 x tan 1 x log 1 x 2 c 2
½
1
½ ½ ½
----------------------------------------------------------------------------------------------------------- 2
f)
02
Evaluate sin 3 xdx 0
2
Ans
sin
3
xdx
0
2
sin
2
x.sin xdx
0
2
1 cos x sin xdx
½
2
0
Put cos x t
when x 0 , t 1
sin xdx dt
when x
2
,t0
½
Page No.03/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
1
f)
Subject Code:
17301 Marking Scheme
Answer 0
1 t 2 dt 1 0
1 t 2 dt 1
½
0
t3 t 3 1
1 0 1 3 2 3
½
OR 2
sin
3
dx
½
0
2
0
3sin x sin 3 x dx 4
½
2
1 cos 3 x 3 cos x 4 3 0 1 1 0 03 4 3 1 8 . 4 3 2 3
½
½
-----------------------------------------------------------------------------------------------------------g)
Find the area of the region bounded by x2 16 y, y 1, y 4 and Y-axis in first 02
quadrant
Ans
b
A xdy a 4
4 y dy
½
1
4
3 y2 4 3 2 1
½
Page No.04/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
1
g)
Subject Code:
17301 Marking Scheme
Answer
8 3 3 4 2 12 3 8 7 3 56 3
½
½
---------------------------------------------------------------------------------------------------------------------
h) Ans
Determine the order and degree of x 2
02
d2y dy x my 2 dx dx
1
Order=2
1
Degree=1 -----------------------------------------------------------------------------------------------------------------------
i) Ans
Form the differential equation if y 4 x A .Where A is arbitrary constant 2
y 4 x A
dy 8 x A dx
1 dy y 4 8 dx
02
2
½
2
1 dy y4 64 dx
½ 2
½
2
dy 16 y dx j) Ans
½
----------------------------------------------------------------------------------------------------------------------A fair die is rolled. What is the probability that the number on the die is a prime number
02
S 1, 2,3, 4,5, 6 n s 6 A 2,3,5
½
n A 3 p A
n A 3 ns 6
½ [
1
1 or 0.5 2
-----------------------------------------------------------------------------------------------------------------------
Page No.05/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
1
k) Ans
Subject Code:
17301 Marking Scheme
Answer From 4 men and 2 women, 3 persons are chosen at random to form a committee. Find the probability that the committee consists of at least one person of either sex.
02
n S 6 c3 20
½
n A 4 c1 2 c2 4 c2 2 c1
1
16 16 4 p A or 0.8 20 5
½
---------------------------------------------------------------------------------------------------------------------
l)
A person fires 10 shorts at target. The probability that any shot will hit the target is
02
3 .Find the probability that the target is hit exactly 5 times. 5
Ans
3 3 2 Given : n 10, p , q 1 p 1 5 5 5 n r nr p r cr p q 5
3 2 p 5 c5 5 5 0.2007
5
1
10
1
-----------------------------------------------------------------------------------------------------------------------
m) Ans.
Evaluate
1 9 4x2
1 9 4x2 dx
OR
1 32 2 x
02
dx
2
dx
2x 1 sin 1 . c 3 2
1 dx 9 2 2 x 4 1 1 = dx 2 2 3 2 x 2 1 1 1 2 x x = sin 1 c sin c 2 2 3/ 2 3
1
1
-----------------------------------------------------------------------------------------------------------------------
Page No.06/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
1
n)
Subject Code:
17301 Marking Scheme
Answer e
Evaluate
02
1
x .log xdx 1
e
Ans
1
x .log xdx 1
Put log x t
when x 1 t log1 0
1 dx dt x
when x e
½
t log e 1
1
tdt
½
0 1
t2 2 0 1 2
½
½
----------------------------------------------------------------------------------------------------------Attempt any FOUR of the following:
2 a) Ans.
Evaluate tan x dx
16 04
Considering index as 3 tan 3 xdx tan 2 x tan xdx
½
sec 2 x 1 tan xdx
tan x sec 2 x tan x dx tan x sec 2 xdx tan xdx In first integral , put tan x t
1
sec 2 xdx dt
1
tdt log sec x c
1
t2 log sec x c 2 tan 2 x log sec x c 2
½
Note: If student attempted to solve the problem assuming any index value then consider it and reward appropriate marks to it. ---------------------------------------------------------------------------------------------Page No.07/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
2
b)
Subject Code:
Marking Scheme
Answer Evaluate
17301
04
log x dx x 2 log x 3 log x log x
Ans
x 2 log x 3 log x dx Put log x t
½
1 dx dt x t 2 t 3 t dt
½
t
consider
2 t 3 t t A 3 t B 2 t
½
A B 2t 3t
½
Put t 2 A 2 Put t 3
½
B3
2 t 3 t
c) Ans
t
t
2 3 2t 3t
dt
½
2 3 dt 2t 3t
2 t 3 t 2 log 2 t 3log 3 t c 2 log 2 log x 3log 3 log x c
½
-----------------------------------------------------------------------------------------------------------Find the equations of the tangent and normal to the ellipse 2 x 2 3 y 2 5 which is perpendicular to the line 3x 2 y 7 0
04
½
Slope of line 3x 2 y 7 0 is 3 m1 2 2 2x 3y2 5 dy 0 dx dy 4 x 2 x dx 6 y 3y 4 x 6 y
Page No.08/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
2
c)
Subject Code:
17301 Marking Scheme
Answer
2 x 3y Line and tangent are perpendicular
½
slope of tangent m2 m1.m2 1 3 2 x . 1 2 3y x y y x
½
2 x2 3 x 5 2
2 x 2 3x 2 5 x2 1 x 1 if x 1 y 1 point is 1, 1
½
point is 1,1
½
if x 1 y 1
Equation of tangent at (1, 1) is 2 x 1 3 2x 3y 5 0 y 1
½
Equation of tangent at ( 1,1) is 2 y 1 x 1 3 2x 3y 5 0
½
Equation of normal at (1, 1) 3 y 1 x 1 2 3x 2 y 1 0
½
Equation of normal at ( 1,1) is 3 x 1 2 3x 2 y 1 0 y 1
½
-----------------------------------------------------------------------------------------------------------------------
d) Ans
Find the radius of curvature for the curve x a cos3 , y a sin 3 at x a cos3
y a sin 3
dx 3a cos 2 sin d
dy 3a sin 2 cos d
4
04
½+½
Page No.09/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
2
d)
Subject Code:
17301 Marking Scheme
Answer
dy dy d 3a sin 2 cos dx dx 3a cos 2 sin d dy tan dx d2y d sec 2 2 dx dx 1 sec 2 dx d 1 sec 2 3a cos 2 sin
½
½
at
4
½
dy tan 1 dx 4 d2y 1 sec 2 dx 4 3a cos 2 sin 4 4 2 1 2 . 2 1 1 3a 2 2 2 1 3a 2 2 2
½
4 2 3a
dy 2 1 dx Radius of curvature d2y dx 2
1 1
2
3
3
2
2
½
4 2 3a ½
3a or 1.5 a 2
-----------------------------------------------------------------------------------------------------------------------
Page No.10/29
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
2
e)
Subject Code:
Marking Scheme
Answer A bullet is fired into a mud tank and penetrates 120t 3600t 2 meters in ‘t’ seconds after impact. Calculate maximum depth of penetration.
Ans
17301
04
y 120t 3600t 2 dy 120 7200t dt d2y 7200 0 Depth is maximum. dt 2 dy Let 0 dt 120 7200t 0 t
½ ½
½ 1
1 60
1 1 Maximum depth y 120 3600 60 60 1 meter
2
½ 1
----------------------------------------------------------------------------------------f) Ans
Evaluate
x
2
5x 4 dx 8 x 12
5x 4 dx 8 x 12 5x 4 5x 4 Consider 2 x 8 x 12 x 6 x 2
x
04
2
5x 4 A B x 6 x 2 x 6 x 2
½ ½
5x 4 A x 2 B x 6 Put x 6 26 13 4 2 Put x 2
A
6 3 B 4 2 13 3 5x 4 2 2 2 x 8 x 12 x 6 x 2 3 13 2 5x 4 2 dx 2 dx x 8 x 12 x6 x2
½ ½
½ ½
Page No.11/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
2
f)
Subject Code:
Marking Scheme
Answer I
17301
1
13 3 log x 6 log x 2 c 2 2
-----------------------------------------------------------------------------------------------------------------------
16
Attempt any FOUR of the following:
3
1
a)
Evaluate
2
3 1
Ans
2
3
04
1 dx 2 4 x 12 x 13
2
1 dx 4 x 12 x 13 2
2 1
1 2 1 dx 4 3 x 2 3 x 13 2 4
3x Third term 2 4. x 2
½
½
9 4
1
1 2 1 dx 4 3 x 2 3 x 9 9 13 2 4 4 4
½
1
1 2 1 dx 2 4 3 3 2 2 x 1 2
½
1
1
1 3 2 tan 1 x 2 3 4 2 1 1 3 1 3 3 tan 1 tan 1 4 2 2 4 2 2 1 tan 1 2 4
½
½
--------------------------------------------------------------------------------------------------------------------
b)
Evaluate
3
1 6
Ans
3
1 6
3
3
1 dx cot x
04
1 dx cot x
Page No.12/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
3
b)
Answer
I
3
Marking Scheme
3
1 dx cos x 6 1 3 sin x
17301
1 dx 3 cot x
1 6
Subject Code:
3
1 dx cos x 6 1 3 sin x
3
I
3
3
sin x dx sin x 3 cos x
3
6
I
3
3
sin x 3 6
3
3
I
sin x 2
3 sin x 3 cos x 2 2
6
3
dx
dx
3
cos x dx -------------------- 2 cos x 3 sin x
3
6
½
1
3 sin x 3 cos x 3 6 3 6
6
I
-------------------- 1
½
add (1) and (2)
II
2I
6
6
sin x 3 cos x dx 3 sin x 3 cos x
3 3
2I
6
3 3 sin x cos x + dx 3 sin x 3 cos x 3 cos x 3 sin x 3
3
3
1dx
½
½
6
2 I x 3
6
2I I
3
12
½
6 ½ Page No.13/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
3
c) [
Ans
Subject Code:
17301 Marking Scheme
Answer Find the area of the region enclosed between parabola y x 2 1 and the line y 2x 1
04
y x 2 1 and y 2 x 1 x2 1 2x 1 x2 2x 0 x x 2 0
1
x 0& x 2 b
Area y1 y2 dx
½
a
2
2 x 1 x 2 1 dx 0
2
2 x 1 x 2 1dx 0
2
2 x x 2 dx 0
2
2 x 2 x3 3 0 2
1 1
23 22 0 3 4 3
½
-----------------------------------------------------------------------------------------------------------------------
d) Ans
Solve x 2 .
dy x 2 xy y 2 dx
04
dy x 2 xy y 2 dx dy x 2 xy y 2 dx x2 Put y vx x2 .
dy dv v x dx dx
2 dv x x vx vx vx dx x2 dv x 2 vx 2 v 2 x 2 vx dx x2
½ 2
½
Page No.14/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
3
d)
Subject Code:
17301 Marking Scheme
Answer 2 2 dv x 1 v v vx dx x2 dv v x 1 v v2 dx dv x 1 v2 dx xdv 1 v 2 dx
1
1 1 dv dx 2 1 v x 1 1 1 v 2 dv x dx tan 1 v log x c
½
y tan 1 log x c x
½
1
-----------------------------------------------------------------------------------------------------------------------
e) Ans
Solve cos 2 x 2 y 1 2 cos 2 x 2 y 1 2
dy dx
04
dy dx
Put x 2 y v dy dv dx dx dv cos 2 v dx 1 dx dv cos 2 v solution is 1 2
dx sec
2
1
1
½
vdv
x tan v c
1
x tan x 2 y c
½
----------------------------------------------------------------------------------------------------------------------
f)
Solve 1 x 2
1 dy y e tan x dx
04
1
Ans
dy 1 e tan x y dx 1 x 2 1 x2 dy Comparing with Py Q dx Page No.15/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
3
f)
Subject Code:
Marking Scheme
Answer 1
½
1 e tan x P ,Q 1 x2 1 x2 Integrating factor e
17301
1
Pdx
e
1 x2 dx
e tan
-1
x
1
y.IF Q.IFdx 1
ye
tan 1 x
e tan x tan 1 x e dx 1 x2
e dx tan 1 x
ye tan
1
x
2
1 x2
Put tan 1 x t 1 dx dt 1 x2 ye tan ye tan
ye tan
1
1
1
x
x
x
Put e tan x t 1 1 e tan x . dx dt 1 x2 -1
OR
et dt e 2t dt 2
1
ye tan
1
ye tan
-1
2t
e c 2
e 2 tan 2
1
x
x
x
tdt
c
ye tan
x
½
2
t c 2
e tan -1 x
1
½
2
½
2
c
----------------------------------------------------------------------------------------------------------------------Attempt any FOUR of the following:
4 a)
Ans
10 x dx Evaluate 2 2 3 x 10 x 2 7 10 x dx --------------------(1) Let I 2 2 3 x 10 x 2 7 10 7 3 x I dx 2 2 7 3 x 10 7 3 x 3 2 7 x I dx ----------------------(2) 2 2 10 x x 3 2
7
16 04
1
Adding (1) and (2)
10 x 7 x2 dx 2 2 2 2 x 10 x 10 x x 3 3
7
2
II
1 Page No.16/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
a)
Subject Code:
Marking Scheme
Answer
10 x x 2 2I 2 dx 2 3 x 10 x
½
2
7
17301
7
2 I 1dx
½
3
2 I x 3
7
½
2I 7 3 4 2 I 2 I
½
----------------------------------------------------------------------------------------------------------------------
b)
1
[
04
Evaluate x.sin 1 xdx 0
1
Ans
x.sin
1
xdx
0
1 d sin 1 x dx sin x x.dx x.dx dx 0 0 0 1
1
1
½
1
x2 x2 1 sin x. . dx 2 0 2 1 x2
½
x 2 .sin 1 x 1 1 x 2 1 dx 2 2 0 1 x2
½
1 x 2 .sin 1 x 1 1 x 2 1 2 2 0 1 x 2 1 x2
dx
1 x 2 .sin 1 x 1 1 1 x2 2 2 0 1 x2
dx
1
1
1
x 2 .sin 1 x 1 x 12 1 x 2 sin 1 x sin 1 x 2 2 2 2 0
1
1
x 2 .sin 1 x x 1 1 x 2 sin 1 x 2 4 4 0 12.sin 1 1 1 0 sin 1 1 0 2 4
½
1 2 . 2 4 2 4 8 8
1 Page No.17/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
c)
Ans
Subject Code:
17301 Marking Scheme
Answer Find the area of the region in the first quadrant enclosed by the X-axis the line y x and the circle x 2 y 2 8
04
y x , x2 y 2 8 x2 x2 8 2x2 8 x2 4 x 2 In first quadrant x 0 to x 2
1
Due to symmetry about the line y=x the region in first quadrant of the circle is divided into two equal parts and hence can be integrated as follows: b
A y2 y1 a
A 0 2
x A 2 2 A 2
8
2
x 2 x dx
8
2
8
2
x2
2
2
½ 2
8
2
8
2
2
2
2 x 1 x sin 2 8 0
2 2 2 sin 2 0 8 1
1
½
A 2 4 2 4
1
OR y x , x2 y2 8 y2 4 y 2
In first quadrant
1
y 0 to y 2 b
A x2 x1 dy a
2 A 0
8
2
y 2 y dy
½ Page No.18/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
c)
Subject Code:
Marking Scheme
Answer
y A 2 2 A 2
8 8
2
2
y2
2
2
2
8
2
8 2
2
17301
2
y2 1 y sin 8 2 0
1
2 2 2 sin 2 0 8
½
1
A 2 4 2 4
1
OR
Find the area of the region in the first quadrant enclosed by the X-axis the line y x and the circle x 2 y 2 8
y x , x2 y 2 8 x2 x2 8 2 x2 8 x2 4 x 2 1
In first quadrant point of intersection is 2, 2 Let BM X - axis Required area A region OBMO A region BAMB 2
A region OBMO y dx 0 2
x dx 0 2
x2 2 2 0
1 Page No.19/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
c)
Subject Code:
Marking Scheme
Answer A region BAMB
04
2 2
17301
y dx
2 2 2
8 x 2 dx
2
x 2
8
2
x2
8
2
2 0 4sin 1 2 1 4 2 4sin 1 2 2 1
2
2 2
1 x sin 8 2
8
2
2
2
8 2
2
2 sin 8 1
2 2 4 4 2
1
Required area 2 2
d) Ans
1
----------------------------------------------------------------------------------------------------------------------Solve y 3 .sec 2 xdx 3 y 2 tan x sec 2 y dy 0
y 3 .sec 2 xdx 3 y 2 tan x sec 2 y dy 0 Comparing with Mdx Ndy c M y 3 .sec 2 x M 3 y 2 sec 2 x y M N y x
N=3 y 2 tan x sec 2 y N 3 y 2 sec 2 x x
1 1
D.E. is exact Solution is
y .sec 3
2
xdx sec 2 ydy c
1
y 3 tan x tan y c -----------------------------------------------------------------------------------------------------------------------
Page No.20/31
1
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
e)
Solve x y 1
Ans
Put x y 1 v dy dv 1 dx dx dy dv 1 dx dx dv v 2 1 1 dx dv 2 v2 v 1 dx dv v2 1 v2 dx v2 dv dx 2 1 v
Subject Code:
Marking Scheme
Answer 2
17301
04
dy 1 dx
1
1
solution is v2 1 v 2 dv dx
½
1 v2 1 1 v 2 dv dx 1 1 1 v 2 dv dx v tan 1 v x c
1 ½
x y 1 tan 1 x y 1 x c y 1 tan 1 x y 1 c ----------------------------------------------------------------------------------------------------------------------
f)
Verify that y em sin
1 x2
1
x
is a solution of differential equation
04
d2y dy x m2 y 0 2 dx dx Page No.21/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
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SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
Ans
Answer
Consider y em sin
1
Subject Code:
17301 Marking Scheme
x
1 dy 1 e m sin x .m dx 1 x2 dy my dx 1 x2
½
dy my dx Squaring, 1 x2
2
dy 1 x 2 m 2 y 2 dx
½ 2
dy d 2 y dy dy 1 x 2 2 x m 2 2 y 2 dx dx dx dx
1
2
2
2 dy dy dy 2 d y 1 x 2 x 2m 2 y 2 dx dx dx dx
1
d2y dy x m2 y 2 dx dx 2 d y dy 1 x 2 2 x m 2 y 0 dx dx 1 x 2
1
OR
1 x 2
d2y dy x m2 y 0 2 dx dx
Consider y e m sin
1
x
1 dy 1 e m sin x .m dx 1 x2 dy my dx 1 x2
dy my dx d 2 y dy 1 dy 1 x2 2 2 x m dx dx 2 1 x 2 dx 2 d2y dy dy 1 x2 x m 1 x2 2 dx dx dx 2 d y dy 1 x 2 2 x m my dx dx
½
1 x2
[[[[[
1
1 ½
Page No.22/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
4
Subject Code:
17301 Marking Scheme
Answer
d2y dy x m2 y 2 dx dx d2y dy 1 x 2 2 x m2 y 0 dx dx 1 x 2
1
----------------------------------------------------------------------------------------------------------------------
5
Attempt any FOUR of the following: a)
A Card is drawn at random from a well shuffled pack of 52 playing cards. A=event that the card drawn is not a spade. B=event that the card drawn is king. Verify that events A and B are independent.
Ans
16
04
n s 52 C1 52
½
n A C1 39 39
½
n B 4 C1 4 p A
n A 39 3 n s 52 4
p B
n B 4 1 n s 52 13
½
½
3 1 3 4 13 52 A B Event that the card drawn is a king of heart orof diamond or of club Consider p A p B =
n A B 3 C1 3 p A B
n A B 3 ns 52
p A B p A p B
1 ½ ½
A and B Independent events. 04 b)
1 Assuming that the probability of a fatal accident during the year is . 1200 Calculate the probability that in a factory employing 300 workers there will be at least two fatal accidents in a year. e 0.25 0.7788
Ans
Given n 300, p
1 1200
1
300 0.25 1200 p at least two fatal accidents 1 p 0 p 1
m np
1
Page No.23/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
5
b)
Subject Code:
17301 Marking Scheme
Answer
e0.25 .025 0 e0.25 .025 1 p at least two fatal accidents 1 0! 1! = 1 0.7788 0.7788 0.25
1
1
= 0.0265
-------------------------------------------------------------------------------------------------------------------
c)
In certain examination 500 students appeared.Mean score is 68 with S.D 8.Find the
04
number of students scoring i) less than 50 ii) more than 60 Ans
Given x 68 , 8 x 50
i) when z
xx
50 68 8
2.25
2
ii) when x 60 z
xx
60 68 8
2
1
NOTE : As the areas for the above problem are not given, the students cannot solve the problem completely. If students attempted to solve the problem and calculated upto value z. Full marks to be rewarded. ----------------------------------------------------------------------------------------------------------------------
d) Ans
Evaluate
1
4 5sin 2 x dx
04
1
4 5sin 2 x dx Put tan x t , dx
dt 2t , sin 2 x 2 1 t 1 t2
dt 1 t2 2t 4 5 2 1 t
1
Page No.24/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
5
d)
Answer
dt 4 4t 2 10t dt 2 4t 10t 4 dt 100 100 4t 2 10t 4 16 16 dt 2 2 5 3 2 t 2 2 5 3 2t 1 2 2 1 c log 3 5 3 2 2 2t 2 2 2 1 2t 1 log c 6 2t 4
Subject Code:
17301 Marking Scheme
1 2 tan x 1 log c 6 2 tan x 4 OR
½
½
1
½ ½
1
4 5sin 2 x dx Put tan x t , dx
dt 2t , sin 2 x 2 1 t 1 t2
dt 1 t2 2t 4 5 2 1 t dt 4 4t 2 10t dt 4t 2 10t 4 1 dt 4 t2 5 t 1 2 1 dt 4 t 2 5 t 25 1 25 2 16 16
1
½
Page No.25/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
5
d)
Subject Code:
17301 Marking Scheme
Answer
1 dt 4 5 2 3 2 t 4 4 5 3 t 1 1 4 4 log 3 5 3 4 2 t 4 4 4 1 2t 1 log 6 2t 4
½
1 2 tan x 1 log 6 2 tan x 4
½
1
½
---------------------------------------------------------------------------------------------------------------------
e)
2
Evaluate
0
2
Ans
sin 2 x dx 2 x
4 sin
04
sin 2 x dx 2 x
4 sin 0
2
0
2sin x cos x dx 4 sin 2 x
Put sin x t
when x 0
cos xdx dt
when x
1
0
2
t 0
1
t 1
2tdt 4 t2
consider, 2t
2t 2t 2 4t 2 t 2 t
A B 2t 2t
2 t 2 t 2t A 2 t B 2 t put t 2 A 1
½
put t 2 B 1
½ Page No.26/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
5
e)
Subject Code:
17301 Marking Scheme
Answer 2tdt 1 1 dt 2 4 t 2 t 2 t 0 0
1
1
1
log 2 t log 2 t 0 1
log 3 log1 log 2 log 2
1
log 3 log 4 OR 2
sin 2 x dx 2 x
4 sin 0
2
0
2sin x cos x dx 4 sin 2 x
Put sin x t
when x 0
cos xdx dt
when x
1
0
2
t0
1
t 1
2tdt 4 t2 1
0
2tdt 4 t2
log 4 t 2
1
1 0
log 4 1 log 4 02
1
2
1
log 3 log 4
-----------------------------------------------------------------------------------------------------------------------
f) Ans
ex Solve 2 x e x log y dx 1 dy 0 y comparing with Mdx Ndy 0 M 2 x e x log y
N=
ex 1 y
M e x N e x , y y x y M N y x given D.E.equation is exact
04
½ +½ 1
Page No.27/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
Subject Code:
17301 Marking Scheme
Answer
solution is
5
y cons tan t
Mdx
y cons tan t
2x e
terms free from x
x
Ndy c
log y dx 1dy c
1
2x2 e x log y y c 2 2 x e x log y y c
1 16
Attempt any FOUR of the following:
6 a)
A Husband and wife appeared in an interview for two vacancies in an office The Probability of husband's selection is
1 1 and that of wife's selection is 7 5
04
Find the probability that i) both of them are selected. ii) only one of them is selected. Ans
Given p H
1 7
p W
1 5
p H ' 1 p H
6 4 and p W ' 1 p W 7 5 1 1 1 i ) p H W p H p W 7 5 35 ' ' ii ) p H W p H W p H p W ' p H ' p W
1 1
1 4 6 1 = 7 5 7 5 2 7 b)
1 1
-----------------------------------------------------------------------------------------------------------A company manufactures electric motors. The probability that an electric motor is defective is 0.01. What is the probability that a sample of 300 electric motors will contain exactly 5 defective motors? e 0.0498 3
Ans
04
Given n 300, p 0.01 m np 300 0.01 3
1
3 5
e 3 5! 0.1008 P 5
2 1 Page No.28/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
6
c)
Subject Code:
17301 Marking Scheme
Answer In a test of 2000 electric bulbs it was found that the life of particular make was normally distributed with average life of 2040 hrs. and S.D.of 60 hrs. Estimate the number of bulbs likely to burn for
04
(i) between 1920hrs. and 2160 hrs. (ii) more than 2150 hrs. Ans
Given that A 2 0.4772 A 1.83 0.4664
x 2040, 60 i) x 1920, x 2160 1920 2040 2 60 x x 2160 2040 z 2 60 P(Between 1920 hrs. and 2160 hrs.) A 2 A 2 0.4772 0.4772 z
xx
½ 1
0.9544 Number of bulbs having life between 1920 hrs. and 2160 hrs. 0.9544 2000 1908.8 1909
½
ii) x 2150 x x 2150 2040 1.83 S .D 60 P More than 2150 hrs. 0.5 A 1.83 z
d) Ans
½
0.5 0.4664 0.0336 Number of bulbs having life more than 2150 hrs. 0.0336 2000 67.2 67
½
Find the maximum and minimum values of 2 x3 3x 2 36 x 10
04
y 2 x3 3x 2 36 x 10 dy 6 x 2 6 x 36 dx d2y 12 x 6 dx 2 dy 0 dx
1
½ ½
Page No.29/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
6
d)
Subject Code:
17301 Marking Scheme
Answer 6 x 2 6 x 36 0 or x 2 x 6 0 x 3 or x 2
1
for x 3 d2y 12 3 6 30 dx 2 y is minimum at x 3
½ ½
ymin 2 3 3 3 36 3 10 71 3
2
for x 2 d2y 12 2 6 30 dx 2 y is maximum at x 2
½
ymax 2 2 3 2 36 2 10 54
½
3
e)
2
-----------------------------------------------------------------------------------------------------------------------
Find the equation of the tangent and the normal to the curve 2 x xy 3 y 18 at 3,1 2
Ans
04
2
2 x 2 xy 3 y 2 18 at 3,1 dy dy 4x x y 6 y 0 dx dx dy dy 4x x y 6 y 0 dx dx dy y 4 x dx 6 y x
1
at 3,1 dy 1 4 3 11 dx 6 1 3 3 slope of tangent
11 3
½
Equation of tangent is 11 x 3 3 3 y 3 11x 33 y 1
1
11x 3 y 36 0 1 3 dy 11 dx Equation of normal is
slope of normal
½
Page No.30/31
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
SUMMER – 17 EXAMINATION Model Answer Q. No.
Sub Q. N.
6
e)
f) Ans
Subject Code:
17301
Answer
Marking Scheme
3 x 3 11 11y 11 3 x 9 3x 11y 2 0 -----------------------------------------------------------------------------------------------------------Find by integration the area of the ellipse 4 x 2 9 y 2 36
1
y 1
04
4 x 2 9 y 2 36 x2 y2 1 9 4 4 y2 9 x2 9 2 2 y 3 x2 3
½
b
Area 4 ydx a
1
3
2 2 4 3 x 2 dx 3 0 3
8 32 x 2 dx 30 8x 3 2
8 3 3 2
3 x 2
2
3
3 3 2
2
3
x sin 2 3 0
3
2
1
1
3 sin 1 0 2 3 2
½
8 9 0 sin 1 1 3 2 8 9 3 2 2 6
1
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. -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Page No.31/31
