cbse maths xii marking scheme delhi
TRANSCRIPT
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Stri ctly Confidential __(For I nternal and Restr icted Use Only)
Senior School Certificate Examination
March 2014
Marking Scheme ---- Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General I nstructions :
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The
answers given in the Marking Scheme are suggested answers. The content is thus indicative.
If a student has given any other answer which is different from the one given in the Marking
Scheme, but conveys the meaning, such answers should be given full weightage.
2. Evaluation is to be done as per instructions provided in the marking scheme. It should not
be done according to one's own interpretation or any other consideration __Marking
Scheme should be strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question(s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
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QUESTION PAPER CODE 65/1/1
EXPECTED ANSWERS/VALUE POINTS
SECTION - A
1-10. 1. x = 25 2.5
1x = 3. 10 4. x = 2 5. x = + 6
6. 2x3/2+ 2 x + c 7.12
8. 5 9.
3
2
10. ( ) ( ) 0kjikcjbiar =++++
or
( ) cbakjir ++=++ 110 =10 m
SECTION - B
11. AAb)(a,
a + b = b + a (a, b) R (a, b) R is reflexive 1 m
For (a, b), (c, d) AA
If (a, b) R (c, d) i.e. a + d = b + c c + b = d + a
then (c, d) R (a, b) R is symmetric 1 m
For (a, b), (c, d), (e, f) AA
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e a + f = b + e
then (a, b) R (e, f) R is transitive 1 m
R is reflexive, symmetric and transitive
hence R is an equivalance relation m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} m
Q. No. Marks
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= 3211 RRRR
2x2xzyx
yxz2z2z
zyxzyxzyx
++++++++
1 m
= ( ) ;
zyxzyxzyx
zyx02z
00zyx
++++++
++
133
122
CCC
CCC
2 m
= (x + y + z) .{0 .(x + y + z) + (x + y + z)2
} = (x + y + z)3
1 m
14. let ( ) xcos!!cosx,x12xcosv,x
x1tanu 121
2
1 ===
=
( ) xcos!!tantan!cos
!cos1tanu 11
2
1 ===
= 1 m
( ) ( )
=== !2
2
coscos!2sincos!cos1!cos2cosvand 1121
= xcos22
!2
2
1= 1 m
22 x1
2
dx
dv,
x1
1
dx
du==
1 m
2
1
2
x1
x1
1
dv
du2
2== 1 m
( In case, If x = sin ! then answer is2
1)
15. y = xx
log y = x log x, Taking log of both sides m
dx
dy
y
1 = log x + 1, Diff. w r t x 1 m
,x
1
dx
dy
y
1
dx
yd
y
12
22
2
=
Diff. w r t x 1 m
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0x
y
dx
dy
y
1
dx
yd2
2
2
=
m
16. (x)f = 12 x3 12 x2 24 x = 12 x (x + 1) (x 2) 1+ m
(x)f > 0, ),2()0,1( Ux ++ 1 m
(x)f < 0, )2,0()1,( U x 1 m
f(x) is strictly increasing in ),2()0,1( U m
and strictly decreasing in )2,0()1,( U
OR
Point at
=22
a,
22
ais
4
! m
!cos!sin3ad!
dx!;sin!cos3a
d!
dy 22 == 1 m
slope of tangent at
4
!
2
2
4
!
!cos!sin3a
!sin!cos3a
dx
dyis
4
!
==
=
=
= 1
4
cot = 1 m
Equation of tangent at the point :
02
ayx
22
ax1
22
ay =+
= 1 m
Equation of normal at the point :
0yx
22
ax1
22
ay =
= m
17. ( ) ][
++
=+
dxxcosxsin
xcosx3sinx)cosxsin(xcosxsindx
xcosxsin
xcosxsin22
2222222
22
66
1 m
= dx3xcosxsin
122
1 0 2
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+
= dx3xcosxsin
xcosxsin22
22
m
( ) += dx3xcosecxsec 22 m
= tan x cot x 3x + c 1 m
(Accept 2 cot 2x 3x + c also)
OR
( ) + dx183xx3x 2
( ) +++= dx183xx29
dx183xx32x2
1 221 m
( ) ( )
++= dx
2
9
23x
2
9183xx
3
2
2
12
22
32
1 m
( )2
9183xx
3
12
32 +=
c183xx2
3xlog
8
81183xx
2
2
3x
22 +++++
+
1 m
or ( )8
9183xx
3
12
32 +=
{ c183xx2
3xlog
2
81183xx)32( 22 ++++++x
18.dy
y1
ydxxedy
x
ydxy1e
2
x2x ==1 m
Integrating both sides
= dyy1
2y
2
1dxxe
2
x
cy1exe 2xx += 1+1 m
For x = 0, y = 1, c = 1 1y11)(xe:issolution 2x = + m
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222
cba2ba
=++ 1 m
=++ b&abetweenanglebeing!49,!cosba2259 1 m
3
!
2
1
532
15!cos ==
= 1 m
21. let v5
6z
3
4y
1
2xu;
7
5z
5
3y
3
1x====
+=
+=
+
General points on the lines are
(3u 1, 5u 3, 7u 5) & (v + 2, 3v + 4, 5v + 6) 1 m
lines intersect if
3u 1 = v + 2, 5u 3 = 3v + 4, 7u 5 = 5v + 6 for some u & v 1 m
or 3u v = 3 ........... (1), 5u 3v = 7 .............. (2), 7u 5v = 11 ................... (3)
Solving equations (1) and (2), weget2
3v,
2
1u == m
Putting u & v in equation (3) , intersectlines112
35
2
17 =
m
Point of intersection of lines is :
2
3,
2
1,
2
11 m
22. let b2, g
2be younger boy and girl
and b1, g1be elder, then, sample space of two children is
S = {(b1, b
2), (g
1, g
2), (b
1, g
2), (g
1, b
2)} 1 m
A = Event that younger is a girl = {(g1, g
2), (b
1, g
2)}
B = Event that at least one is a girl = {(g1, g
2), (b
1, g
2), (g
1, b
2)}
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E = Event that both are girls = {(g1, g
2)}
(i) P(E/A) =2
1
(A)P
A)(EP=
I1 m
(ii) P(E/B) =3
1
(B)P
B)(EP=
I1 m
SECTION - C
23. Here
600zyx
15003zy4x
1000z2y3x
=++=++=++ 1
BXAor
600
1500
1000
z
y
x
111
314
123
=
=
| A | = 3 ( 2) 2 (1) + 1 (3) = 5 0 BAX 1= m
Co-factors are
5A,5A,5A
1A,2A,1A3A,1A,2A
333231
232221
131211
=========
1 m
=
600
1500
1000
513
521
512
5
1
z
y
x
x = 100, y = 200, z = 300 1 m
i.e. Rs. 100 for discipline, Rs 200 for politeness & Rs. 300 for punctuality
One more value like sincerity, truthfulness etc. 1 m
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24. For correct figure m
Let radius, height and slant height of cone be r, h & l
(constant),hr 222 ll=+ m
Volume of cone (V) = hr31 2 m
( ) ( )3222 hh3
hh
3
V ll == 1 m
( )22 h33
dh
dvl= 1 m
3h0
dh
dv l== m
03
23
2h2dh
vd2
2
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26. Correct Figure 1 m
The line and circle intersect
each other at x = + 4 1 m
Area of shaded region
( ) +=24
4
22
4
0
dxx24dx x 1 m
24
4
12
4
0
2
24
xsin16
2
x32x
2
x
++
= 1 m
= 8 + 484
=sq.units 1 m
27. Equation of plane through points A, B and C is
0
023
884
3z5y2x
=+
074z3y2xi.e.
05632z24y16x
=++=++ 3+1 m
Distance of plane from (7, 2, 4) =4169
7)4(4)2(3)7(2
++++ 1 m
= 29 1 m
OR
General point on the line is ( ) ( ) ( ) k2#2j4#1i3#2 +++++ 1 m
Putting in the equation of plane; we get
( ) ( ) ( ) 52#214#113#21 =++++ 1 m
0#= 1 m
Point of intersection is 2)1,(2,ork2ji2 + 1 m
Distance = ( ) ( ) ( ) 131691025112 222 ==+++++ 1 m
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Probability of defective bulb =3
1
15
5= m
Probability of a non defective bulb =3
2
3
11 = m
Probability distribution is :
81
4
81
24
81
48
81
320:P(x)x
81
1
81
8
81
24
81
32
81
16:P(x)
43210:x
m
m2
3
4
or81
108
P(x)xMean == 1 m
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QUESTION PAPER CODE 65/1/2
EXPECTED ANSWERS/VALUE POINTS
SECTION - A
1-10. 1. 5 2. ( ){ }( ) 0kjikcjbiar =++++ or
( ) cbakjir ++=++
3. 2x3/2+ 2 x + c 4. 10 5. x = 2
6. x = + 6 7.5
1x = 8. x = 25
9. c2
x
2
x2
+ 10.6
110 = 10 m
SECTION - B
11. (x)f = 12 x3 12 x2 24 x = 12 x (x + 1) (x 2) 1+ m
(x)f > 0, ),2()0,1( Ux ++ 1 m
(x)f < 0, )2,0()1,( U x 1 mf(x) is strictly increasing in ),2()0,1( U
m
and strictly decreasing in )2,0()1,( U
OR
Point at
=
22
a,
22
ais
4! m
!cos!sin3ad!
dx
!;sin!cos3ad!
dy 22
== 1 m
slope of tangent at
4
!
2
2
4
!
!cos!sin3a
!sin!cos3a
dx
dyis
4
!
==
=
=
= 14
cot = 1 m
Marks
1 0 2
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Equation of tangent at the point :
02
ayx
22
ax1
22
ay =+
= 1 m
Equation of normal at the point :
0yx22
ax1
22
ay =
= m
12. ( ) ][
++
=+
dxxcosxsin
xcosx3sinx)cosxsin(xcosxsindx
xcosxsin
xcosxsin22
2222222
22
66
1 m
= dx3xcosxsin1
22
+
= dx3xcosxsin
xcosxsin22
22
m
( ) += dx3xcosecxsec 22 m
= tan x cot x 3x + c 1 m
(Accept 2 cot 2x 3x + c also)
OR
( ) + dx183xx3x 2
( ) +++= dx183xx29
dx183xx32x2
1 221 m
( ) ( )
++= dx
2
9
23x
2
9183xx
3
2
2
12
22
32
1 m
( )2
9183xx
3
12
32 +=
c183xx2
3xlog
8
81183xx
2
2
3x
22 +++++
+
1 m
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or ( )8
9183xx
3
12
32 +=
{ c183xx2
3xlog
2
81183xx)32(
22 ++++++x
13. Given differential equation can be written as
( )222 1x2
y1x
2x
dx
dy=+ 1 m
Integrating factor = 1xee 21)(xlogdx
1
x222
==
x 1 m
cdx1)(x1)(x
2
1)(xyisSolution2
22
2
+= 1 m
cdx1x
121)(xy
2
2 +=
c1x
1xlog1)(xy 2 +
+= 1 m
14. y = xx
log y = x log x, Taking log of both sides m
dx
dy
y
1 = log x + 1, Diff. w r t x 1 m
,x
1
dx
dy
y
1
dx
yd
y
12
22
2
=
Diff. w r t x 1 m
0x
y
dx
dy
y
1
dx
yd2
2
2
=
m
15.
+
+
+=
+++
accbbaac,cb,ba m
=
+++
+
acccabcbba 1 m
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=
+
+
+
cbbacaabacba 1 m
+
+
acbabb
=
=
=
0abbcbbaca,aba
=
=
c,b,a2cba2 1 m
OR
=+=++ cba0cba m
222
ccba
=
=
+
m
222
cba2ba
=++ 1 m
=++ b&abetweenanglebeing!49,!cosba2259 1 m
3
!
2
1
532
15!cos ==
= 1 m
16. cot1
+++
xsin1xsin1
xsin1xsin1
= cot1
+
+
+
22
22
2
xsin
2
xcos
2
xsin
2
xcos
2
xsin
2
xcos
2
xsin
2
xcos
2 m
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2
x
2
xcotcot
2
xsin2
2
xcos2
cot 11 =
=
= 1 m
OR
+
+=
7
25sec
8
1tan
5
1tan2LHS 111
7
1tan
40
11
8
1
5
1
tan2 11 +
+= 1+ m
7
1tan
3
11
3
12
tan7
1tan
3
1tan2 1
2
111 +
=+= 1 m
RHS
4
)1(tan
25
25tan
7
1tan
4
3tan 1111 ====+= 1 m
17. AAb)(a,
a + b = b + a (a, b) R (a, b) R is reflexive 1 m
For (a, b), (c, d) AA
If (a, b) R (c, d) i.e. a + d = b + c c + b = d + a
then (c, d) R (a, b) R is symmetric 1 m
For (a, b), (c, d), (e, f) AA
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e a + f = b + e
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then (a, b) R (e, f) R is transitive 1 m
R is reflexive, symmetric and transitive
hence R is an equivalance relation m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} m
18. let b2, g
2be younger boy and girl
and b1, g
1be elder, then, sample space of two children is
S = {(b1, b
2), (g
1, g
2), (b
1, g
2), (g
1, b
2)} 1 m
A = Event that younger is a girl = {(g1, g
2), (b
1, g
2)}
B = Event that at least one is a girl = {(g1, g
2), (b
1, g
2), (g
1, b
2)}
E = Event that both are girls = {(g1, g
2)}
(i) P(E/A) =2
1
(A)P
A)(EP=
I1 m
(ii) P(E/B) =
3
1
(B)P
B)(EP=
I1 m
19. LHS =
2bacac)b(a2
b2acbc)b(a2
bac)b(a2
++++++++
++
3211 CCCC
Using,
++ 1 m
=
cba00
0cba0
bac)b(a2
++
++++
133122 RRR;RRR
Using,
2 m
= 2 (a + b + c) {(a + b + c)2
0} Expanding along C1
= 2 (a + b + c)3
= RHS 1 m
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20. let u = ( ) xsin!!sinx;x1x2sinv;x1
xtan 121
2
1 ===
( ) xsin!!tantan!sin1
!sintanu 11
2
1
===
= 1 m
( ) xsin2!2!2sinsinx12xsinv& 1121 ==== 1 m
22 x1
2
dx
dv,
x1
1
dx
du== 1 m
2
1
2
x1
x1
1
dv
du 2
2== 1 m
( In case, if x =2
1isanswerthen!cos )
21. dxxsinxdyy
ylogyx
dx
dyylogxcosec 2
2
22 == 1 m
Integrating both sides we get
[ ]+= dxxcosx2xcosxy
1
y
ylog 2 1+1 m
+= dxxsin1sin xx2xcosx 2 m
cxcos2xsinx2xcosxy
1
y
ylog 2 +++= m
22. Equations of lines are :
35z
14y
78x;
53z
47y
45x ==+== 1 m
Here,
3c1,b7,a;5c4,b4,a
5z4,y8,x;3z7,y5,x
222111
222111
============
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317
544
833
cba
cba
zzyyxx
222
111
121212
== 3(17) + 3 (47) + 8 ( 24) = 0 1+1 m
lines are co-planar m
SECTION - C
23. Let x and y be electronic and
manually operated sewing machines purchased respectively
L.P.P. is Maximize P = 22x + 18y m
subject to 360x + 240y < 5760
or 3x + 2y < 48
x + y < 20 2 m
x > 0, y > 0
For correct graph 2 m
vertices of feasible region are
A (0, 20), B(8, 12), C(16, 0) & O(0, 0)
P(A) = 360, P(B) = 392, P(C) = 352 m For Maximum P, Electronic machines = 8
1 m
Manual machines = 12
24. Let E1: Event that lost card is a spade
E2: Event that lost card is a non spade
m
A : Event that three spades are drawn without replacement from 51 cards
43
411)P(E,
41
5213)P(E 21 ==== 1 m
3
51
3
13
2
3
51
3
12
1C
C)P(A/E,
C
C)P(A/E == 1 m
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3
51
3
13
3
51
3
12
3
51
3
12
1
C
C
4
3
C
C
4
1
C
C
4
1
/A)P(E
+
=
1+1 m
49
10= 1 m
OR
X = No. of defective bulbs out of 4 drawn = 0, 1, 2, 3, 4 1 m
Probability of defective bulb =3
1
15
5= m
Probability of a non defective bulb =3
2
3
11 = m
Probability distribution is :
81
4
81
24
81
48
81
320:P(x)x
81
1
81
8
81
24
81
32
81
16:P(x)
43210:x
m
m2
3
4or
81
108P(x)xMean == 1 m
25. Here
600zyx
15003zy4x
1000z2y3x
=++=++=++ 1
BXAor
600
15001000
z
yx
111
314123 =
=
| A | = 3 ( 2) 2 (1) + 1 (3) = 5 0 BAX 1= m
Co-factors are
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5A,5A,5A
1A,2A,1A
3A,1A,2A
333231
232221
131211
=========
1 m
=
600
1500
1000
513
521
512
5
1
z
y
x
x = 100, y = 200, z = 300 1 m
i.e. Rs. 100 for discipline, Rs 200 for politeness & Rs. 300 for punctuality
One more value like sincerity, truthfulness etc. 1 m
26. Equation of plane through points A, B and C is
0
023
884
3z5y2x
=+
074z3y2xi.e.
05632z24y16x
=++=++ 3+1 m
Distance of plane from (7, 2, 4) =4169
7)4(4)2(3)7(2++
++ 1 m
= 29 1 m
OR
General point on the line is ( ) ( ) ( ) k2"2j4"1i3"2 +++++ 1 m
Putting in the equation of plane; we get
( ) ( ) ( ) 52"214"113"21 =++++ 1 m
0"= 1 m
Point of intersection is 2)1,(2,ork2ji2 + 1 m
Distance = ( ) ( ) ( ) 131691025112 222 ==+++++ 1 m
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V (volume) = h,r3
2[V is constant]
A = ( )22222222 hrrrAz,r +=== ll m
+=
42
2222
r
9vrr
+=
22
242
r
9vr 1 m
=
32
232
r
18v4r
dr
dz1 m
6
2
2
2
9vr0
dr
dz== m
0r
54v12r
dr
zd;
2
9vrAt
42
222
2
2
62
2
>
+== 1 m
corved surface area is minimum iff 262 9vr2 =
i.e. 24262 hrr2 =
OR
r2h = m
( )2cot#2r
h#cot 1=== m
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QUESTION PAPER CODE 65/1/3
EXPECTED ANSWERS/VALUE POINTS
SECTION - A
1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x3/2+ 2 x + c
5.5
1x = 6. x = 25 7. 5
8. ( ) ( ) 0kjikcjbiar =++++ 9. 1
or
( ) cbakjir ++=++
10. k13
12j
13
3i
13
4+ 110 = 10 m
SECTION - B
11.
+
+
+=
+++
accbbaac,cb,ba m
=
+++
+
acccabcbba 1 m
=
+
+
+
cbbacaabacba 1 m
+
+
acbabb
=
=
=
0abbcbbaca,aba
=
=
c,b,a2cba2 1 m
OR
Marks
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=+=++ cba0cba m
222
ccba
=
=
+
m
222
cba2ba
=++ 1 m
=++ b&abetweenanglebeing49,cosba2259 1 m
3
!
2
1
532
15cos ==
= 1 m
12. Given differential equation can be written as
( )222 1x2
y1x
2x
dx
dy=+ 1 m
Integrating factor = 1xee 21)(xlogdx
1
x222
==
x 1 m
cdx1)(x1)(x
2
1)(xyisSolution
2
22
2
+= 1 m
cdx1x
121)(xy
2
2 +=
c1x
1xlog1)(xy 2 +
+= 1 m
13. ( ) ][
++
=
+dx
xcosxsin
xcosx3sinx)cosxsin(xcosxsindx
xcosxsin
xcosxsin22
2222222
22
66
1 m
= dx3xcosxsin
122
+
= dx3xcosxsin
xcosxsin22
22
m
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( ) += dx3xcosecxsec 22 m
= tan x cot x 3x + c 1 m
(Accept 2 cot 2x 3x + c also)
OR
( ) + dx183xx3x 2
( ) +++= dx183xx29
dx183xx32x2
1 221 m
( ) ( )
++= dx
2
9
23x
2
9183xx
3
2
2
12
22
32
1 m
( )2
9183xx
3
12
32 +=
c183xx2
3xlog
8
81183xx
2
2
3x
22 +++++
+
1 m
or ( )8
9183xx
3
12
32 +=
{ c183xx2
3xlog
2
81183xx)32(
22 ++++++x
14. (x)f = 12 x3 12 x2 24 x = 12 x (x + 1) (x 2) 1+ m
(x)f > 0, ),2()0,1( Ux ++ 1 m
(x)f < 0, )2,0()1,( U x 1 m
f(x) is strictly increasing in ),2()0,1( U mand strictly decreasing in )2,0()1,( U
OR
Point at
=
22
a,
22
ais
4! m
1 0 2
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cossin3ad
dx;sincos3a
d
dy 22 == 1 m
slope of tangent at
4!
2
2
4! cossin3a
sincos3a
dx
dyis
4
!
==
=
=
= 14
!cot = 1 m
Equation of tangent at the point :
02
ayx
22
ax1
22
ay =+
= 1 m
Equation of normal at the point :
0yx22
ax1
22
ay =
= m
15. AAb)(a,
a + b = b + a (a, b) R (a, b) R is reflexive 1 m
For (a, b), (c, d) AA
If (a, b) R (c, d) i.e. a + d = b + c c + b = d + a
then (c, d) R (a, b) R is symmetric 1 m
For (a, b), (c, d), (e, f) AA
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e a + f = b + e
then (a, b) R (e, f) R is transitive 1 m
R is reflexive, symmetric and transitive
hence R is an equivalance relation m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} m
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,x
1
dx
dy
y
1
dx
yd
y
12
22
2
=
Diff. w r t x 1 m
0x
y
dx
dy
y
1
dx
yd2
2
2
=
m
18. let b2, g
2be younger boy and girl
and b1, g
1be elder, then, sample space of two children is
S = {(b1, b
2), (g
1, g
2), (b
1, g
2), (g
1, b
2)} 1 m
A = Event that younger is a girl = {(g1, g
2), (b
1, g
2)}
B = Event that at least one is a girl = {(g1, g
2), (b
1, g
2), (g
1, b
2)}
E = Event that both are girls = {(g1, g
2)}
(i) P(E/A) =2
1
(A)P
A)(EP=
I1 m
(ii) P(E/B) =3
1
(B)P
B)(EP=
I1 m
19. LHS =
zzzyzx
zyyyyx
zxyxxx
zyx
1
322
232
223
++
+
33
2211
RzR
RyR,RxR
1 m
=
1zzz
y1yy
xx1x
zyx
zyx
222
222
222
+
++
m
=
1zzz
y1yy
zyx1zyx1zyx1
222
222
222222222
++
+++++++++ 3211 RRRR ++ 1 m
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=
10z
01y
00zyx1
2
2
222 +++
;133
122
CCC
CCC
1 m
= 1 + x2+ y2 + z2= RHS (Expand along C1) m
20. let x = xtantan1=
xtan2
1
2
2
tantan
tan
1tan1tanu 11
2
1 ==
=
+= 1 m
( ) xtan222sinsintan1
tan2
sinv
11
2
1
===
+= 1 m
( ) 22 x12
dx
dv;
x12
1
dx
du
+=
+= 1 m
( ) 41
2
x1
x12
1
dv
du 2
2 =+
+
= 1 m
( In case, if x =4
1isanswerthencot )
21. Differential equation can be written as : (sin y + y .cos y) dy = x .(2 .log x + 1) dx 1 m
Integrating both sides we get
c2
x
4
xxlog
2
x2ycosysinyycos
222
++
=++ 1+1 m
cxlogxysiny
2
+=At x = 1 and
y = =2
!c,
2
! solution is : y sin y = x2log x +
2
!+ m
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22. General points on the lines are
( ) ( ) ( ) ( )k13"i2"4&kj#1i3#1 ++++ 1 m
lines intersect if
"somefor3).........(113"2);.........(0#1(1);........."243#1 ==+=+ 1 m
From (2) & (3) 0"1,# == m
substituting in equation (1)
Since, 1 + 3(1) = 4 + 2 (0) is true lines interset 1 m
Point of intersection is : 1)0,(4,ork
i
4 m
SECTION - C
23. Let x and y be electronic and
manually operated sewing machines purchased respectively
L.P.P. is Maximize P = 22x + 18y m
subject to 360x + 240y < 5760
or 3x + 2y < 48
x + y < 20 2 m
x > 0, y > 0
For correct graph 2 m
vertices of feasible region are
A (0, 20), B(8, 12), C(16, 0) & O(0, 0)
P(A) = 360, P(B) = 392, P(C) = 352 m
For Maximum P, Electronic machines = 81 m
Manual machines = 12
24. Let E1: Event that lost card is a spade
E2: Event that lost card is a non spade
m
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27. Here
600zyx
15003zy4x
1000z2y3x
=++=++=++ 1
BXAor
600
1500
1000
z
y
x
111
314
123
=
=
| A | = 3 ( 2) 2 (1) + 1 (3) = 5 0 BAX 1= m
Co-factors are
5A,5A,5A
1A,2A,1A
3A,1A,2A
333231
232221
131211
======
===1 m
=
600
1500
1000
513
521
512
5
1
z
y
x
x = 100, y = 200, z = 300 1 m
i.e. Rs. 100 for discipline, Rs 200 for politeness & Rs. 300 for punctuality
One more value like sincerity, truthfulness etc. 1 m
28. le I =( )
+=+2
!
0
44
2!
0
44dx
xsinxcos
xsinxcosx2
!
I;dxxcosxsin
xcosxsinx1 m
Adding we get, 2 I =( ) +=+
2!
0
22
22!
0
44dx
xtan1xsecxtan2
4!;dx
xcosxsinxcosxsin
2! 2 m
= ( )8
!tantan
4
! 22!
0
21 =
x 2 m
16
!I
2
= 1 m
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29. let r and h be the radius and height of the cylinder then,
Volume of cylinder (V) 128!hr! 2 = 22
r
128
r!
128!h == 1 m
Surface area of cylinder = )rh(r!2rh!2r2! 22 +=+ 1 m
=
+=
2
2
r
1282r2!
dr
ds
r
128r2!S 1 m
= 0dr
ds r3 = 64 or r = 4 m
012!
64
25622!
r
25622!
dr
sd;4rAt
32
2
>=
+=
+== 1 m
surface area is minimum at r = 4 cm ; h = 8 cm 1 m