EXAMINATION PAPERS – 2010 MATHEMATICS CBSE (Delhi) CLASS – XII Time allowed: 3 hours Maximum marks: 100 General Instructions: 1. All questions are compulsory. 2. The question paper consists of 29 questions divided into three Sections A, B and C. Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks each and Section C comprises of 7 questions of six marks each. 3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculator is not permitted. Set–I SECTION–A Question numbers 1 to 10 carry 1 mark each. 1. What is the range of the function fx x x () | | ( ) ? = - - 1 1 2. What is the principal value of sin - - æ L ç ö l ÷ 1 3 2 ? 3. If A = - æ L ç ö l ÷ cos sin sin cos a a a a , then for what value of a is A an identity matrix? 4. What is the value of the determinant 0 2 0 2 3 4 4 5 6 ? 5. Evaluate : log . x x dx ò 6. What is the degree of the following differential equation? 5 6 2 2 2 x dy dx dy dx y x æ L ç ö l ÷ - - = log Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com
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EXAMINATION PAPERS – 2010
MATHEMATICS CBSE (Delhi)CLASS – XII
Time allowed: 3 hours Maximum marks: 100
General Instructions:
1. All questions are compulsory.
2. The question paper consists of 29 questions divided into three Sections A, B and C. Section Acomprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks eachand Section C comprises of 7 questions of six marks each.
3. All questions in Section A are to be answered in one word, one sentence or as per the exactrequirement of the question.
4. There is no overall choice. However, internal choice has been provided in 4 questions of four markseach and 2 questions of six marks each. You have to attempt only one of the alternatives in all suchquestions.
5. Use of calculator is not permitted.
Set–I
SECTION–A
Question numbers 1 to 10 carry 1 mark each.
1. What is the range of the function f xx
x( )
| |
( )?=
-
-
1
1
2. What is the principal value of sin - -æ
èç
ö
ø÷
1 3
2 ?
3. If A =-æ
èç
ö
ø÷
cos
sin
sin
cos
a
a
a
a , then for what value of a is A an identity matrix?
4. What is the value of the determinant
0 2 0
2 3 4
4 5 6
?
5. Evaluate : log
.x
xdxò
6. What is the degree of the following differential equation?
5 62 2
2x
dy
dx
d y
dxy x
æ
èç
ö
ø÷ - - = log
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7. Write a vector of magnitude 15 units in the direction of vector $ $ $.i j k- +2 2
8. Write the vector equation of the following line:x y z-
=+
=-5
3
4
7
6
2
9. If 1 2
3 4
3 1
2 5
7 11
23
æ
èç
ö
ø÷
æ
èç
ö
ø÷ =
æ
èç
ö
ø÷
k, then write the value of k.
10. What is the cosine of the angle which the vector 2$ $ $i j k+ + makes with y-axis?
SECTION–B
11. On a multiple choice examination with three possible answers (out of which only one iscorrect) for each of the five questions, what is the probability that a candidate would get fouror more correct answers just by guessing?
12. Find the position vector of a point R which divides the line joining two points P and Q whose
position vectors are ( )2 a b® ®
+ and ( )a b® ®
- 3 respectively, externally in the ratio 1 : 2. Also,
show that P is the mid-point of the line segment RQ.
13. Find the Cartesian equation of the plane passing through the points A ( , , )0 0 0 and
B( , , )3 1 2- and parallel to the line x y z-
=+
-=
+4
1
3
4
1
7.
14. Using elementary row operations, find the inverse of the following matrix :
2 5
1 3
æ
èç
ö
ø÷
15. Let Z be the set of all integers and R be the relation on Z defined as R a b a b Z= Î{( , ) ; , , and ( )a b- is divisible by 5.} Prove that R is an equivalence relation.
16. Prove the following:
tan cos , ( , )- -=-
+
æ
èç
ö
ø÷ Î1 11
2
1
10 1x
x
xx
OR
Prove the following :
cos sin sin- - -æèç
öø÷ + æ
èç
öø÷ = æ
èç
öø÷
1 1 112
13
3
5
56
65
17. Show that the function f defined as follows, is continuous at x = 2, but not differentiable:
f x
x
x x
x
x
x
x
( )
,
,
,
=
-
-
-
< £
< £
>
ì
íï
îï
3 2
2
5 4
0 1
1 2
2
2
OR
Find dy
dx, if y x x x x= - - --sin [ ].1 21 1
148 Xam idea Mathematics – XII
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18. Evaluate : ex
xdxx
ò-
-
æ
èç
ö
ø÷
sin
cos.
4 4
1 4
OR
Evaluate : 1
1 2
2-
-òx
x xdx
( ).
19. Evaluate : p
p
/
/ sin cos
sin.
6
3
2ò+x x
xdx
20. Find the points on the curve y x= 3 at which the slope of the tangent is equal to the
y-coordinate of the point.
21. Find the general solution of the differential equation
x xdy
dxy
xxlog . log+ = ×
2
OR
Find the particular solution of the differential equation satisfying the given conditions:
dy
dxy x= tan , given that y = 1 when x = 0.
22. Find the particular solution of the differential equation satisfying the given conditions:
x dy xy y dx2 2 0+ + =( ) ; y = 1 when x = 1.
SECTION–C
Question numbers 23 to 29 carry 6 marks each.
23. A small firm manufactures gold rings and chains. The total number of rings and chainsmanufactured per day is atmost 24. It takes 1 hour to make a ring and 30 minutes to make achain. The maximum number of hours available per day is 16. If the profit on a ring is Rs. 300 and that on a chain is Rs 190, find the number of rings and chains that should bemanufactured per day, so as to earn the maximum profit. Make it as an L.P.P. and solve itgraphically.
24. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards aredrawn at random and are found to be both clubs. Find the probability of the lost card beingof clubs.
OR
From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random.Find the probability distribution of the number of defective bulbs.
25. The points A B( , , ), ( , , )4 5 10 2 3 4 and C ( , , )1 2 1- are three vertices of a parallelogram ABCD.Find the vector equations of the sides AB and BC and also find the coordinates of point D.
26. Using integration, find the area of the region bounded by the curve x y2 4= and the line
x y= -4 2.
Examination Papers – 2010 149
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OR
Evaluate: 0
p
ò +
x x
x xdx
tan
sec tan.
27. Show that the right circular cylinder, open at the top, and of given surface area andmaximum volume is such that its height is equal to the radius of the base.
28. Find the values of x for which f x x x( ) [ ( )]= - 2 2 is an increasing function. Also, find the
points on the curve, where the tangent is parallel to x-axis.
29. Using properties of determinants, show the following:
( )
( )
( )
( )
b c
ab
ac
ab
a c
bc
ca
bc
a b
abc a b c
+
+
+
= + +
2
2
2
32
Set-II
Only those questions, not included in Set I, are given.
3. What is the principal value of cos ?- -æ
èç
ö
ø÷
1 3
2
7. Find the minor of the element of second row and third column ( )a23 in the followingdeterminant:
2
6
1
3
0
5
5
4
7
-
-
11. Find all points of discontinuity of f, where f is defined as follows :
f x
x
x
x
x
x
x
( )
| | ,
,
,
=
+
-
+
£ -
- < <
³
ì
íï
îï
3
2
6 2
3
3 3
3
OR
Find dy
dx, if y x xx x= +(cos ) (sin ) ./1
12. Prove the following:
tan cos ,- -=-
+
æ
èç
ö
ø÷1 11
2
1
1x
x
x x Î( , )0 1
OR
Prove the following:
cos sin sin- - -æèç
öø÷ +
3æèç
öø÷ = æ
èç
öø÷
1 1 112
13 5
56
65
150 Xam idea Mathematics – XII
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14. Let * be a binary operation on Q defined by
a bab
* =3
5
Show that * is commutative as well as associative. Also find its identity element, if it exists.
18. Evaluate: 0 1
p
ò +
x
xdx
sin.
20. Find the equations of the normals to the curve y x x= + +3 2 6 which are parallel to the line
x y+ + =14 4 0.
23. Evaluate 1
3 23 2ò +( )x x dx as limit of sums.
OR
Using integration, find the area of the following region:
( , ) ;x yx y x y2 2
9 41
3 2+ £ £ +
ìíï
îï
üýï
þï
29. Write the vector equations of the following lines and hence determine the distance betweenthem:
x y z-=
-=
+1
2
2
3
4
6;
x y z-=
-=
+3
4
3
6
5
12
Set-III
Only those questions, not included in Set I and Set II, are given.
1. Find the principal value of sin cos .- --æèç
öø÷ + -æ
èç
öø÷
1 11
2
1
2
9. If A is a square matrix of order 3 and | | | |,3A K A= then write the value of K.
11. There are two Bags, Bag I and Bag II. Bag I contains 4 white and 3 red balls while another Bag II contains 3 white and 7 red balls. One ball is drawn at random from one of the bags and it isfound to be white. Find the probability that it was drawn from Bag I.
14. Prove that : tan ( ) tan ( ) tan ( ) .- - -+ + =1 1 11 2 3 p
OR
If tan tan ,- --
-
æ
èç
ö
ø÷ +
+
+
æ
èç
ö
ø÷ =1 11
2
1
2 4
x
x
x
x
p find the value of x.
17. Show that the relation S in the set R of real numbers, defined as S a b a b R= Î{( , ) : , and a b£ 3}
is neither reflexive, nor symmetric nor transitive.
19. Find the equation of tangent to the curve yx
x x=
-
- -
7
2 3( ) ( ) , at the point, where it cuts the
x-axis.
23. Find the intervals in which the function f given by
f x x x x( ) sin cos ,= - £ £0 2p
is strictly increasing or strictly decreasing.
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24. Evaluate 1
4 2ò -( )x x dx as limit of sums.
OR
Using integration find the area of the following region :
{( , ) :| | }x y x y x- £ £ -1 5 2
SOLUTIONS
Set–I
SECTION–A
1. We have given
f xx
x( )
| |
( )=
-
-
1
1
| |( ),
( ),x
x if x or x
x if x or x- =
- - > >
- - - < <
ìíî
11 1 0 1
1 1 0 1
(i) For x > 1, f xx
x( )
( )
( )=
-
-=
1
11
(ii) For x < 1, f xx
x( )
( )
( )=
- -
-= -
1
11
\ Range of f xx
x( )
| |
( )=
-
-
1
1 is { , }.-1 1
2. Let x = -æ
èç
ö
ø÷
-sin 1 3
2
Þ sin x = -3
2 Þ sin sinx = -æ
èç
öø÷
p
3Q
3
2 3=
é
ëê
ù
ûúsin
p
Þ x = -p
3
The principal value of sin - -æ
èç
ö
ø÷
1 3
2 is - ×
p
3
3. We have given
A =-é
ëê
ù
ûú
cos
sin
sin
cos
a
a
a
a
For the identity matrix, the value of A11 and A12 should be 1 and value of A12 and A21should be 0.
i.e., cos a = 1 and sin a = 0
As we know cos 0 1° = and sin 0 0° =
Þ a = °0
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4.
0 2 0
2 3 4
4 5 6
= 03
5
4
62
2
4
4
60
2
4
3
5- + (expanding the given determinant by R1)
= - 22
4
4
6
= - - =2 12 16 8( )
The value of determinant is 8.
5. We have givenlog x
xdxò
Let log x t= Þ 1
xdx dt=
Given integral = t dtò
= +t
c2
2=
(log )xc
2
2+
6. 5 62 2
2x
dy
dx
d y
dxy x
æ
èç
ö
ø÷ - - = log
Degree of differential equation is the highest power of the highest derivative. In above d y
dx
2
2 is
the highest order of derivative.
\ Its degree = 1.
7. Let A i j k®
= - +$ $ $2 2
Unit vector in the direction of A®
is $$ $ $
( ) ( ) ( )A
i j k=
- +
+ - +
2 2
1 2 22 2 2 = - +
1
32 2($ $ $)i j k
\ Vector of magnitude 15 units in the direction of A®
= 15 152 2
3$ ($ $ $)
Ai j k
=- +
= - +15
3
30
3
30
3$ $ $i j k
= - +5 10 10$ $ $i j k
8. We have given line asx y z-
=+
=-
-
5
3
4
7
6
2
By comparing with equationx x
a
y y
b
z z
c
-=
-=
-1 1 1 ,
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We get given line passes through the point ( , , )x x x1 2 3 i.e., ( , , )5 4 6- and direction ratios are ( , , )a b c i.e., (3, 7, –2).
Now, we can write vector equation of line as
A i j k i j k®
= - + + + -( $ $ $) ( $ $ $)5 4 6 3 7 2l
9.1 2
3 4
3 1
2 5
7 11
23
é
ëê
ù
ûú
é
ëê
ù
ûú =
é
ëê
ù
ûúk
LHS = 1 2
3 4
3 1
2 5
é
ëê
ù
ûú
é
ëê
ù
ûú
= ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) (
1 3 2 2
3 3 4 2
1 1 2 5
3 1 4
+
+
+
+ 5)
é
ëê
ù
ûú =
7 11
17 23
é
ëê
ù
ûú
Now comparing LHS to RHS, we get
\ k = 17
10. We will consider
a i j k®
= + +2$ $ $
Unit vector in the direction of a®
is $$ $ $
( ) ( ) ( )a
i j k=
+ +
+ +
2
2 1 12 2 2
=+ +
=+ +2
4
2
2
$ $ $ $ $ $i j k i j k
= + +2
2
1
2
1
2$ $ $i j k = + +
1
2
1
2
1
2$ $ $i j k
The cosine of the angle which the vector 2$ $ $i j k+ + makes with y-axis is 1
2
æèç
öø÷ .
SECTION–B
11. No. of questions = =n 5
Option given in each question = 3
p = probability of answering correct by guessing =1
3
q = probability of answering wrong by guessing = -1 p = 11
3
2
3- =
This problem can be solved by binomial distribution.
P r Cnr
n r r
( ) = æèç
öø÷
1æèç
öø÷
-2
3 3
where r = four or more correct answers = 4 or 5
(i) P C( )42
3
1
3
54
4
= æèç
öø÷
æèç
öø÷ (ii) P C( )5
1
3
55
5
= æèç
öø÷
154 Xam idea Mathematics – XII
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\ P P P= +( ) ( )4 5
= æèç
öø÷
æèç
öø÷ + æ
èç
öø÷
54
45
5
52
3
1
3
1
3C C
= æèç
öø÷ +
1é
ëêù
ûú=
´ ´ ´
é
ëêù
ûú1
3
10
3 3
1
3 3 3 3
11
3
4
= = ×11
2430 045
12. The position vector of the point R dividing the join of P and Q externally in the ratio 1 : 2 is
ORa b a b®® ® ® ®
=- - +
-
1 3 2 2
1 2
( ) ( )
=- - -
-
® ® ® ®a b a b3 4 2
1 =
- -
-= +
® ®® ®3 5
13 5
a ba b
Mid-point of the line segment RQ is
( ) ( )3 5 3
22
a b a ba b
® ® ® ®® ®+ + -
= +
As it is same as position vector of point P, so P is the mid-point of the line segment RQ.
13. Equation of plane is given by
a x x b y y c z z( ) ( ) ( )- + - + - =1 1 1 0
Given plane passes through (0, 0, 0)
\ a x b y c z( ) ( ) ( )- + - + - =0 0 0 0 …(i)
Plane (i) passes through (3, –1, 2)
\ 3 2 0a b c- + = …(ii)
Also plane (i) is parallel to the linex y z-
=+
-=
+4
1
3
4
1
7
a b c- + =4 7 0 …(iii)
Eliminating a b c, , from equations (i), (ii) and (iii), we get
x y z
3
1
1
4
2
7
0-
-
=
Þ x y z-
-- +
-
-=
1
4
2
7
3
1
2
7
3
1
1
40
Þ x y z( ) ( ) ( )- + - - + - + =7 8 21 2 12 1 0
Þ x y z- - =19 11 0 , which is the required equation
14. Given, A =é
ëê
ù
ûú
2 5
1 3
We can write, A IA=
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i.e., 2
1
5
3
1
0
0
1
é
ëê
ù
ûú =
é
ëê
ù
ûú A
1
1
2
3
1
0
1
1
é
ëê
ù
ûú =
-é
ëê
ù
ûú A [ ]R R R1 1 2® -
1
0
2
1
1
1
1
2
é
ëê
ù
ûú =
-
-é
ëê
ù
ûú A [ ]R R R2 2 1® -
1
0
0
1
3
1
5
2
é
ëê
ù
ûú =
-
-é
ëê
ù
ûú A [ ]R R R1 1 22® -
A- =-
-é
ëê
ù
ûú
1 3
1
5
2
15. We have provided
R a b a b Z a b= Î -{( , ) : , , and ( ) is divisible by 5}
(i) As ( )a a- = 0 is divisible by 5.
\ ( , )a a R a RÎ " Î
Hence, R is reflexive.
(ii) Let ( , )a b RÎ
Þ ( )a b- is divisible by 5.
Þ - -( )b a is divisible by 5. Þ ( )b a- is divisible by 5.
\ ( , )b a RÎ
Hence, R is symmetric.
(iii) Let ( , )a b RÎ and ( , )b c ZÎ
Then, ( )a b- is divisible by 5 and ( )b c- is divisible by 5.
( ) ( )a b b c- + - is divisible by 5.
( )a c- is divisible by 5.
\ ( , )a c RÎ
Þ R is transitive.
Hence, R is an equivalence relation.
16. We have to prove
tan cos ,- -=-
+
æ
èç
ö
ø÷1 11
2
1
1x
x
x x Î( , )0 1
L.H.S. = tan [ tan ]- -=1 11
22x x
=-
+
é
ë
êê
ù
û
úú
-1
2
1
1
12 2
2 2cos
( ) ( )
( ) ( )
x
x
=-
+
æ
èç
ö
ø÷ =-1
2
1
1
1cosx
x R.H.S. Hence Proved.
156 Xam idea Mathematics – XII
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OR
cos sin sin- - -æèç
öø÷ + æ
èç
öø÷ = æ
èç
öø÷
1 1 112
13
3
5
56
65
LHS =1æ
èç
öø÷ + æ
èç
öø÷
- -cos sin1 12
13
3
5
= æèç
öø÷ + æ
èç
öø÷
- -sin sin1 15
13
3
5Q cos sin- -æ
èç
öø÷ = æ
èç
öø÷
é
ëêù
ûú1 112
13
5
13
= ´ - æèç
öø÷ + ´ - æ
èç
öø÷
é
ë
êê
ù
û
úú
-sin 12 25
131
3
5
3
51
5
13
= ´ + ´é
ëêù
ûú=- -sin sin1 15
13
4
5
3
5
12
13
56
65 = RHS
LHS = RHS Hence Proved
17. We have given, f x
x
x x
x
x
x
x
( )
,
,
,
=
-
-
-
< £
< £
>
ì
íï
îï
3 2
2
5 4
0 1
1 2
2
2
At x = 2,(i) RHL LHL
= lim ( )x
f x® +2
=® -lim ( )
xf x
2
= lim ( )h
f h®
+0
2 =® -lim ( )
xf x
2
= lim { ( ) }h
h®
+ -0
5 2 4 = - - -®
lim { ( ) ( )}h
h h0
22 2 2
= 10 – 4 = 6 = lim {( ) ( )}h
h h®
- - -0
2 4 2 1 = 2 × 3 = 6
Also, f ( )2 = 2 2 2 8 2 62( ) - = - =
Q LHL = RHL = f ( )2
\ f x( ) is continuous at x = 2
(ii) LHD RHD
= lim( ) ( )
h
f h f
h®
- -
-0
2 2= lim
( ) ( )
h
f h f
h®
+ -
0
2 2
= lim[ ( ) ( )] ( )
h
h h
h®
- - - - -
-0
22 2 2 8 2= lim
[ ( ) ] ( )
h
h
h®
+ - - -
0
5 2 4 8 2
= lim[ )
h
h h h
h®
+ - - + -
-0
28 2 8 2 6= lim
h
h
h® 0
5
= limh
h h
h®
-
-0
22 7=
®lim ( )
h 05
= lim ( )h
h®
- +0
2 7 = 7 = 5
Examination Papers – 2010 157
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Q LHD ¹ RHD
\ f x( ) is not differentiable at x = 2
OR
We have given
y x x x x= - - --sin [ ].1 21 1
= - - --sin [ ( ) ]1 2 21 1x x x x
Þ y x x= -- -sin sin1 1
[using sin sin sin [ ]- - -- = - - -1 1 1 2 21 1x y x y y x
Differentiating w.r.t. x, we get
dy
dx x x
d
dxx=
--
-
1
1
1
12 2( )( )
=-
--
=-
--
1
1
1
1
1
2
1
1
1
2 12 2x x x x x x.
( )
18. ex
xdxx
ò-
-
æ
èç
ö
ø÷
sin
cos
4 4
1 4
=-æ
è
çç
ö
ø
÷÷ò e
x x
xdxx 2 2 2 4
2 22
sin cos
sin [sin sin cos4 2 2 2x x x= and 1 4 2 22- =cos sin ]x x
= -ò e x x dxx (cot )2 2 2cosec2
= -ò òcot .2 2 2x e dx e x dxx x cosec2
= - -ò[cot . ( ) . ]2 2 2x e x e dxx xcosec2 - ò2 2e x dxx cosec2
= + - òòcot . . .2 2 2 2 22x e x e dx x e dxx x xcosec cosec2 = +e x cx cot 2
OR
We have given
1
1 2
2-
-òx
x xdx
( ) =
-
-ò
1
2
2
2
x
x xdx
=-
-ò
x
x xdx
2
2
1
2 =
1
2
2 2
2
2
2ò-
-
æ
è
çç
ö
ø
÷÷
x
x xdx
=- + -
-ò
1
2
2 2
2
2
2
( ) ( )x x x
x xdx
= +-
-
æ
è
çç
ö
ø
÷÷ò
1
21
2
2 2
x
x xdx …(i)
158 Xam idea Mathematics – XII
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By partial fractionx
x x
x
x x
A
x
B
x
-
-=
-
-= +
-
2
2
2
2 1 2 12 ( )
x A x Bx- = - +2 2 1( ) …(ii)
Equating co-efficient of x and constant term, we get
2 1A B+ = and - = -A 2
Þ A B= = -2 3,
\x
x x x x
-
-= +
-
2
2
2 3
1 22
From equation (i)
1
1 2
1
21
1
2
2 3
1 2
2-
-= + +
-
æ
èç
ö
ø÷ò òò
x
x xdx dx
x xdx
( )
= + - - +1
2
3
41 2x x x clog| | log| |
19. Given integral can be written as
I = p
p / 3
/
sin cos
( sin )6 1 1 2ò+
- -
x x
xdx =
+
- -òp
p / 3
/
sin cos
( sin cos )6 21
x x
x xdx
Put sin cosx x t- =
so that, (cos sin )x xdt
dx+ =
when x =p
6, t = - = -sin cos
p p
6 6
1
2
3
2
when x =p
3, t = - = -sin cos
p p
3 3
3
2
1
2
Þ I = [ ]1
2
3
2
3
2
1
22
1
1
2
3
2
3
2
1
2
1-
- -
-
-
ò-
=dt
ttsin
= sin sin- --é
ëê
ù
ûú - -
é
ëê
ù
ûú
1 13
2
1
2
1
2
3
2
= sin sin- --é
ëê
ù
ûú + -
é
ëê
ù
ûú
1 13
2
1
2
3
2
1
2 = --2
1
23 11sin ( )
20. Let P x y( , )1 1 be the required point. The given curve is
y x= 3 …(i)
dy
dxx= 3 2
dy
dxx
x y
æ
èç
ö
ø÷ =
1 1
123
,
Examination Papers – 2010 159
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Q the slope of the tangent at ( ),x y y1 1 1=
3 12
1x y= …(ii)
Also, ( , )x y1 1 lies on (i) so y x1 13= …(iii)
From (ii) and (iii), we have
3 12
13x x= Þ x x1
213 0( )- =
Þ x1 0= or x1 3=
When x y1 130 0 0= = =, ( )
When x y1 133 3 27= = =, ( )
\ the required points are (0, 0) and (3, 27).
21. x xdy
dxy
xxlog log+ =
2
Þdy
dx x xy
x+ =
1 22log
...(i)
This is a linear differential equation of the form
dy
dxPy Q+ =
where Px x
=1
log and Q
x=
22
\ I.F. = òePdx
=ò
e x xdx
1
log
[Let log x tx
dx dt= \ =1
]
=ò
= =e e ttdt
t1
log = log x
\ y xx
x dx Clog log= +ò22
[ \ solution is y Q dx C( . .) ( . .) ]I F I F= +ò
Þ y x x x dx clog log .= +ò-2 2
I II
Þ y x xx
x
xdxlog log=
-
é
ëêê
ù
ûúú
--
é
ëêê
ù
ûúú
é
ëêê
ù
ûú
- -
ò21
1
1
1 1
ú+ C
Þ y xx
xx dx Clog
log= - +
é
ëê
ù
ûú +-
ò2 2
Þ y xx
x xClog
log= - -
é
ëê
ù
ûú +2
1
Þ y xx
x Clog ( log )= - + +2
1 , which is the required solution
160 Xam idea Mathematics – XII
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OR
dy
dxy x= tan Þ
dy
yx dx= tan
By integrating both sides, we get
dy
yx dx= òò tan .
log log|sec |y x C= + ...(i)
By putting x = 0 and y = 1 (as given), we get
log log (sec )1 0= + C
C = 0
\ (i) Þ log log|sec |y x=
Þ y x= sec
22. x dy y x y dx2 0+ + =( )
x dy y x y dx2 = - +( )
dy
dxy
x y
x= -
+( )
2
dy
dx
xy y
x= -
+æ
è
çç
ö
ø
÷÷
2
2…(i)
Putting y vx= and dy
dxv x
dv
dx= + in equation (i)
v xdv
dx
vx v x
x+ = -
+æ
è
çç
ö
ø
÷÷
2 2 2
2Þ v x
dv
dxv v+ = - +( )2
Þ x dv
dxv v= - -2 2
Þdv
v v
dx
x2 2+= - (by separating variable)
Þ1
2
12v v
dvx
dx+
= -ò ò (Integrating both sides)
Þ1
2 1 1
12v v
dvx
dx+ + -
= - òò
Þ1
1 1
12 2( )v
dvx
dx+ -
= - òò
Þ1
2
1 1
1 1log log log
v
vx C
+ -
+ += - +
Þ1
2 2log log log
v
vx C
+= - +
Examination Papers – 2010 161
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Þ log log logv
vx C
++ =
22 2
Þ log log log ,v
vx k
++ =
2
2 where k C= 2
Þ log logvx
vk
2
2+= Þ
vx
vk
2
2+=
Þ
y
xx
y
x
k×
+
=
2
2
Qy
xv=é
ëêù
ûú
Þ x y k y x2 2= +( ) …(ii)
It is given that y = 1 and x = 1, putting in (ii), we get
1 3= k Þ k =1
3
Putting k =1
3 in (ii), we get
x y y x2 1
32= æ
èç
öø÷ +( )
Þ 3 22x y y x= +( )
SECTION–C
23. Total no. of rings & chain manufactured per day = 24.
Time taken in manufacturing ring = 1 hour
Time taken in manufacturing chain = 30 minutes
One time available per day = 16
Maximum profit on ring = Rs 300
Maximum profit on chain = Rs 190
Let gold rings manufactured per day = x
Chains manufactured per day = y
L.P.P. is
maximize Z x y= +300 190
Subject to x y³ ³0 0,
x y+ £ 24
x y+ £1
216
Possible points for maximum Z are
(16, 0), (8, 16) and (0, 24).
At (16, 0), Z = + =4800 0 4800
162 Xam idea Mathematics – XII
32
28
24
20
16
12
8
4
0 4 8 12 16 20 24
x+y=24
x + y=16
(8,16)
(0,24)
Y
X
(16,0)
12
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At (8, 16), Z = + =2400 3040 5440 ¬ Maximum
At (0, 24), Z = + =0 4560 4560
Z is maximum at (8, 16).
\ 8 gold rings & 16 chains must be manufactured per day.
24. Let A E1 1, and E2 be the events defined as follows:
A : cards drawn are both club
E1: lost card is club
E2 : lost card is not a club
Then, P E( )113
52
1
4= = , P E( )2
39
52
3
4= =
P A E( / )1 = Probability of drawing both club cards when lost card is club = 12
51
11
50´
P A E( / )2 = Probability of drawing both club cards when lost card is not club = 13
51
12
50´
To find : P E A( / )1
By Baye’s Theorem,
P E A( / )1 = P E P A E
P E P A E P E P A E
( ) ( / )
( ) ( / ) ( ) ( / )1 1
1 1 2 2+
=´ ´
´ ´ + ´ ´
1
4
12
51
11
501
4
12
51
11
50
3
4
13
51
12
50
= 12 11
12 11 3 13 12
´
´ + ´ ´ =
11
11 39
11
50+=
OR
There are 3 defective bulbs & 7 non-defective bulbs.
Let X denote the random variable of “the no. of defective bulb.”
Then X can take values 0, 1, 2 since bulbs are replaced
p P D= =( )3
10 and q P D= = - =( ) 1
3
10
7
10
We have
P XC C
C( )= =
´=
´
´=0
7 6
10 9
7
15
72
30
102
P XC C
C( )= =
´=
´ ´
´=1
7 3 2
10 9
7
15
71
31
102
P XC C
C( )= =
´=
´ ´
´=
12
1 3 2
10 9
1
15
70
32
02
\ Required probability distribution is
X 0 1 2
P x( ) 7/15 7/15 1/15
Examination Papers – 2010 163
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25. The points A B( , , ), ( , , )4 5 10 2 3 4 and C ( , , )1 2 1- are three vertices of parallelogram ABCD.