Class... · CBSE Examination Paper, Delhi-2014 Time allowed: 3 hours Maximum marks: 100 ¡ General...
Transcript of Class... · CBSE Examination Paper, Delhi-2014 Time allowed: 3 hours Maximum marks: 100 ¡ General...
CBSE ExaminationPaper, Delhi-2014
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 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 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. You may ask for logarithmic tables, if required.
SET–I
SECTION–A
Question numbers 1 to 10 carry 1 mark each.
1. Let * be a binary operation, on the set of all non-zero real numbers, given by a bab
* =5
for all a,
b RÎ - { }.0 Find the value of x, given that 2 5 10* ( * ) .x =
2. If sin sin cos- -+æèç ö
ø÷ =1 11
51x , then find the value of x.
3. If 23 4
5
1
0 1
7 0
10 5x
yé
ëê
ù
ûú +
é
ëê
ù
ûú =
é
ëê
ù
ûú, find (x – y).
4. Solve the following matrix equation for x x O: [ ] .11 0
2 0-
é
ëê
ù
ûú =
5. If 2 5
8
6 2
7 3
x
x=
- , write the value of x.
6. Write the antiderivative of 31
xx
+æèç
öø÷.
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7. Evaluate: dx
x9 20
3
+ò .
8. Find the projection of the vector $ $ $i j k+ +3 7 on the vector 2 3 6$ $ $i j k- + .
9. If a®
and b®
are two unit vectors such that a b® ®
+ is also a unit vector, then find the angle
between a®
and b®
.
10. Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the
plane r®
. ($ $ $)i j k+ + = 2.
SECTION–BQuestion numbers 11 to 22 carry 4 marks each.
11. Let A = {1, 2, 3, …, 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for(a, b), (c, d) in A × A. Prove that R is an equivalence relation. Also obtain the equivalence class[(2, 5)].
12. Prove that cotsin sin
sin sin; ,- + + -
+ - -
æ
èç
ö
ø÷ = Îæ
èç1 1 1
1 1 20
4
x x
x x
xx
p öø÷.
OR
Prove that 21
5
5 2
72
1
8 4
1 1 1tan sec tan- - -æèç ö
ø÷ +
æ
èç
ö
ø÷ + æ
èç ö
ø÷ =
p.
13. Using properties of determinants, prove that
2 2
2 2
2 2
3
y y z x y
z z z x y
x y z x x
x y z
- -
- -
- -
= + +( ) .
14. Differentiate tan - -æ
è
çç
ö
ø
÷÷
121 x
x with respect to cos ( ),- -1 22 1x x when x ¹ 0.
15. If y xx= , prove that d y
dx y
dy
dx
y
x
2
2
21
0-æèç
öø÷ - = .
16. Find the intervals in which the function f x x x x( ) = - - +3 4 12 54 3 2 is
(a) strictly increasing
(b) strictly decreasing
OR
Find the equations of the tangent and normal to the curve x a= sin 3 q and y a= cos 3 q at
qp
=4
.
17. Evaluate:sin cos
sin . cos
6 6
2 2
x x
x xdx
+ò
OR
Evaluate: ( )x x x dx- + -ò 3 3 182
502 Xam idea Mathematics–XII
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18. Find the particular solution of the differential equation e y dxy
xdyx 1 02- + = , given that
y = 1 when x = 0.
19. Solve the following differential equation:
( )xdy
dxxy
x
2
21 2
2
1- + =
-
20. Prove that, for any three vectors a b c® ® ®
, ,
[ , , ] [ , , ]a b b c c a a b c® ® ® ® ® ® ® ® ®
+ + + = 2
OR
Vectors a b® ®
, and c®
are such that a b c® ® ® ®
+ + = 0 and | | ,| |a b® ®
= =3 5 and | |c®
= 7. Find the angle
between a®
and b®
.
21. Show that the lines x y z+
=+
=+1
3
3
5
5
7 and
x y z-=
-=
-2
1
4
3
6
5 intersect. Also find their
point of intersection.
22. Assume that each born child is equally likely to be a boy or a girl. If a family has two
children, what is the conditional probability that both are girls given that
(i) the youngest is a girl?
(ii) atleast one is a girl?
SECTION–C
Question numbers 23 to 29 carry 6 marks each.
23. Two schools P and Q want to award their selected students on the values of Discipline,
Politeness and Punctuality. The school P wants to award x each, y each and z each for
the three respective values to its 3, 2 and 1 students with a total award money of 1,000.
School Q wants to spend 1,500 to award its 4, 1, and 3 students on the respective values (by
giving the same award money for the three values as before). If the total amount of awards
for one prize on each value is 600, using matrices find the award money for each value.
Apart from the above three values, suggest one more value for awards.
24. Show that the semi-vertical angle of the cone of the maximum volume and of given slant
height is cos .-1 1
3
25. Evaluate: dx
x16
3
+ò cot/
/
p
p
26. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the
circle x y2 2 32+ = .
27. Find the distance between the point (7, 2, 4) and the plane determine by the points A(2, 5, –3),
B(–2, –3, 5) and C ( , , ).5 3 3-
OR
Find the distance of the point ( , , )- - -1 5 10 from the point of intersection of the line
r®
= - + + + +2 2 3 4 2$ $ $ ( $ $ $)i j k i j kl and the plane r i j k®
- + =.($ $ $) .5
Examination Papers – 2014 503
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28. A dealer in rural area wishes to purchase a number of sewing machines. He has only 5,760
to invest and has space for at most 20 items for storage. An electronic sewing machine cost
him 360 and a manually operated sewing machine 240. He can sell an electronic sewing
machine at a profit of 22 and a manually operated sewing machine at a profit of 18.
Assuming that he can sell all the items that he can buy, how should he invest his money in
order to maximise his profit? Make it as a LPP and solve it graphically.
29. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack threecards are drawn at random (without replacement) and are found to be all spades. Find theprobability of the lost card being a spade.
OR
From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by onewith replacement. Find the probability distribution of number of defective bulbs. Hence findthe mean of the distribution.
SET–II
Only those questions, not included in Set I, are given.
9. Evaluate: cos (sin ) .-ò
1 x dx
10. If vectors a®
and b®
are such that, | | ,| |a b® ®
= =32
3 and a b
® ®´ is a unit vector, then write the
angle between a®
and b®
.
19. Prove the following using properties of determinants:
a b c a b
c b c a b
c a c a b
a b c
+ +
+ +
+ +
= + +
2
2
2
2 3( )
20. Differentiate tan -
-
1
21
x
x with respect to sin ( )- -1 22 1x x .
21. Solve the following differential equation:
cosec x ydy
dxx ylog + =2 2 0.
22. Show that the lines 5
4
7
4
3
5
-
-=
-=
+
-
x y z and
x y z-=
-=
-8
7
2 8
2
5
3 are coplanar.
28. Evaluate: x x
x xdx
tan
sec ..
cosec0
p
ò
29. Prove that the semi-vertical angle of the right circular cone of given volume and least curved
surface area is cot .-1 2
SET–III
Only those questions, not included in Set I and Set II, are given.
9. Evaluate: e x x dxx (sin cos ) ./
-ò0
p 2
504 Xam idea Mathematics–XII
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10. Write a unit vector in the direction of the sum of the vectors a®
= + -2 2 5$ $ $i j k and
b i j k®
= + -2 7$ $ $.
19. Using properties of determinants, prove the following:
x xy xz
xy y yz
xz yz z
x y z
2
2
2
2 2 2
1
1
1
1
+
+
+
= + + + .
20. Differentiate tan - + -æ
è
çç
ö
ø
÷÷
121 1x
x with respect to sin ,-
+
æ
èç
ö
ø÷
1
2
2
1
x
x when x ¹ 0.
21. Find the particular solution of the differential equation dy
dx
x x
y y y=
+
+
( log )
sin cos
2 1 given that
y =p
2 when x = 1.
22. Show that lines r i j k i j®
= + - + -($ $ $) ( $ $)l 3 and r i k i k®
= - + +( $ $) ( $ $)4 2 3m intersect. Also find
their point of intersection.
28. Evaluate: x x x
x xdx
sin cos
sin cos
/
4 40 +ò
p 2
.
29. Of all the closed right circular cylindrical cans of volume 128 p cm3, find the dimensions ofthe can which has minimum surface area.
Examination Papers – 2014 505
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SET–I
SECTION–A
1. Given 2 5 10* ( * )x =
Þ 25
5*
x ´ = 10 Þ 2 * x = 10
Þ2
510
´=
xÞ x =
´10 5
2Þ x = 25.
2. Given sin(sin cos )- -+ =1 11
51x
Þ sin cos sin- - -+ =1 1 11
5x 1 Þ sin cos- -+ =1 11
5 2x
p
Þ sin -1 1
5 = - -p
2
1cos x Þ sin sin- -=1 11
5x Þ x =
1
5.
3. Given 2 3
5
4 1
0 1x
yé
ëê
ù
ûú +
é
ëê
ù
ûú =
7
10
0
5
é
ëê
ù
ûú
Þ6
10
8
2
1
0 1
7
10
0
5x
yé
ëê
ù
ûú +
é
ëê
ù
ûú =
é
ëê
ù
ûú Þ
7
10
8
2 1
7
10
0
5
+
+
é
ëê
ù
ûú =
é
ëê
ù
ûú
y
x
Equating we get 8 0+ =y and 2 1 5x + =
Þ y = -8 and x = 2 Þ x y- = + =2 8 10
4. Given [ ]x 11
2
0
0-
é
ëê
ù
ûú = 0
Þ [ ] [ ]x - =2 0 0 0
Þ x - =2 0 Þ x = 2
5. Given 2 5
8
6 2
7 3
x
x=
-
Þ 2 40 18 142x - = - -( ) Þ 2 40 322x - =
Þ 2 722x = Þ x2 36= Þ x = ± 6
6. Antiderivative of 31
31
xx
xx
dx+æèç
öø÷ = +
æèç
öø÷ò
= + òò31
xdxx
dx = +ò ò-3 1 2 1 2x dx x dx/ /
=+
+- +
++ - +
31
21
1
21
1 2 1 1 2 1
./ /x x
c
= ´ + +32
323 2 1 2x x c/ /
= + +2 23 2x x c/
506 Xam idea Mathematics–XII
SolutionsSolutionsSolutions
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7. Let Idx
x=
+ò
9 20
3
=+
òdx
x32 20
3
= é
ëêù
ûú-1
3 3
1
0
3
tanx
= -- -1
31 01 1[tan ( ) tan ( )] = -é
ëêù
ûú1
3 40
p =
p
12
8. Let a i j k®
= + +$ $ $3 7
b i j k®
= - +2 3 6$ $ $
Now projection of a®
on b®
= a b
b
® ®
®
.
| |
=+ + - +
- +
($ $ $).( $ $ $)
| $ $ $|
i j k i j k
i j k
3 7 2 3 6
2 3 6
=- +
+ - +
2 9 42
2 3 62 2 2( ) = =
35
49
35
7 = 5.
9. | | ( ).( )a b a b a b® ® ® ® ® ®
+ = + +2
Þ 1 22 2 2= + +® ® ® ®
| | . | |a a b b [Q | |a b® ®
+ = 1]
Þ 1 1 2 1= + +® ®a b. [Q a
® and b
® are unit vector, hence | |a
® = 1 and | |b
® = 1]
Þ 1 2 2= +® ®a b. Þ a b
® ®= -.
1
2
Þ | |.| |cosa b® ®
= -q1
2, where q is angle between a
® and b
®
Þ 1.1 cos q = -1
2[Q | | | |a b
® ®= = 1]
Þ cosq = -1
2
Þ cos cosqp
= -3
Þ cos cosq pp
= -æèç ö
ø÷
3
Þ cos cosqp
=2
3Þ q
p=
2
3.
10. Since, the required plane is parallel to plane r i j k®
+ + =.($ $ $) 2
\ Normal of required plane is normal of given plane.
Þ Normal of required plane = $ $ $.i j k+ +
\ Required vector equation of plane
{ r ai bj ck i j k®
- + + + + =( $ $ $)}.($ $ $) 0
Examination Papers – 2014 507
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SECTION–B
11. Given, R is a relation in A × A defined by
( , ) ( , )a b R c d a d b cÛ + = +
(i) Reflexivity: " Îa b A,
Q a b b a+ = + Þ ( , ) ( , )a b R a b
So, R in reflexive.
(ii) Symmetry: Let (a, b) R (c, d)
Q ( , ) ( , )a b R c d Þ a d b c+ = +
Þ b c d a+ = + [Q a, b, c, d ÎN and N is commutative under addition]
Þ c b d a+ = +
Þ ( , ) ( , )c d R a b
So, R is symmetric.
(iii) Transitivity: Let ( , ) ( , )a b R c d and ( , ) ( , )c d R e f
Now, ( , ) ( , )a b R c d and ( , ) ( , )c d R e f Þ a d b c+ = + and c f d e+ = +
Þ a d c f b c d e+ + + = + + +
Þ a f b e+ = +
Þ ( , ) ( , ).a b R e f
R is transitive.
Hence, R is an equivalence relation.
2nd Part: Equivalence class: [( , )] {( , ) : ( , ) ( , )}2 5 2 5= Î ´a b A A a b R
= Î ´ + = +{( , ) : }a b A A a b5 2
= Î ´ - ={( , ) : }a b A A b a 3
= {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}1 4 2 5 3 6 4 7 5 8 6 9
12. LHS =+ + -
+ - -
æ
èçç
ö
ø÷÷
-cotsin sin
sin sin
1 1 1
1 1
x x
x x, x Îæ
èç ö
ø÷0
4,
p
=+ + -
+
-cot(cos / sin / ) (cos / sin / )
(cos / sin
12 22 2 2 2
2
x x x x
x x x x/ ) (cos / sin / )2 2 22 2- -
æ
è
çç
ö
ø
÷÷
=+ + -
+ -
-cotcos / sin / cos / sin /
cos / sin / cos
1 2 2 2 2
2 2
x x x x
x x x x/ sin /2 2+
æ
èç
ö
ø÷
=æ
èç
ö
ø÷
-cotcos /
sin /
1 2
2
x
x = =-cot (cot / )1 2
2x
x = R.H.S.
OR
L.H.S. = æèç ö
ø÷ +
æ
èç
ö
ø÷ + æ
èç ö
ø÷- - -2
1
5
5 2
72
1
8
1 1 1tan sec tan
= æèç ö
ø÷ + æ
èç ö
ø÷
ìíî
üýþ
+æ
èç
ö
ø÷
- - -21
5
1
8
5 2
7
1 1 1tan tan sec
508 Xam idea Mathematics–XII
Given
04
02 8
20
80
< <
Þ < <
Þ Îæèç ö
ø÷ Ì
é
ë
êêêêêê
ù
û
úúúúúú
x
x
x
p
p
pp, ( , )
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Downloaded from www.studiestoday.com
=+
-
ì
íï
îï
ü
ýï
þï
+æ
èç
ö
ø÷ -- -2
1
5
1
8
11
5
1
8
5 2
711 1
2
tan.
tan [ sec tan ]Q - -= -1 1 2 1x x
= + -- -2
13
4039
40
50
4911 1tan tan = ´ +- -2
13
40
40
39
1
49
1 1tan tan
= æèç ö
ø÷ + æ
èç ö
ø÷- -2
1
3
1
7
1 1tan tan =´
- æèç ö
ø÷
æ
è
çççç
ö
ø
÷÷÷÷
+ æèç ö
ø÷- -tan tan1
2
12
1
3
11
3
1
7 Q2
2
1
1 1
2tan tan- -=
-
é
ëê
ù
ûúx
x
x
=
æ
è
ççç
ö
ø
÷÷÷
+ æèç ö
ø÷- -tan tan1 1
2
38
9
1
7 = ´æ
èç ö
ø÷ + æ
èç ö
ø÷- -tan tan1 12
3
9
8
1
7 = æ
èç ö
ø÷ + æ
èç ö
ø÷- -tan tan1 13
4
1
7
=+
- ´
æ
è
ççç
ö
ø
÷÷÷
-tan 1
3
4
1
7
13
4
1
7
= ´æèç ö
ø÷-tan 1 25
28
28
25 = -tan ( )1 1 =
p
4 = R.H.S.
13. L.H.S. D =
- -
- -
- -
2 2
2 2
2 2
y y z x y
z z z x y
x y z x x
Applying R R2 3« then R R1 2« , we have
D =
- -
- -
- -
x y z x x
y y z x y
z z z x y
2 2
2 2
2 2
Applying R R R R1 1 2 3® + + , we have
D =
+ + + + + +
- -
- -
x y z y z x z x y
y y z x y
z z z x y
2 2
2 2
Taking out ( )x y z+ + from first row, we have
D = + + - -
- -
( )x y z y y z x y
z z z x y
1 1 1
2 2
2 2
Applying C C C1 1 2® - and C C C2 2 3® - , we have
D = + + + + - + +
+ + - -
( ) ( )x y z y z x y z x y
z x y z x y
0 0 1
2
0
Expanding along first row, we have
D = + + + +( ) ( )x y z x y z 2 = + +( )a b c 3 = R.H.S.
Examination Papers – 2014 509
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14. Let ux
x=
-æ
è
çç
ö
ø
÷÷
-tan 121
and v x x= --cos ( )1 22 1
We have to determine du
dv
Let x x= Þ = -sin sinq q 1
Now, u =-æ
è
çç
ö
ø
÷÷
-tansin
sin
121 q
q
Þ u = æèç ö
ø÷-tan
cos
sin
1 q
q Þ u = -tan (cot )1 q
Þ u = -æèç ö
ø÷
é
ëêù
ûú-tan tan1
2
pq Þ u =
pq
2–
Þ u x= -p
2
1– sin Þdu
dx x=
-0
1
1 2–
Þdu
dx x=
-–
1
1 2
Again, v x x= --cos ( )1 22 1
Q x = sin q
\ v = --cos ( sin sin )1 22 1q q
Þ v = -cos ( sin . cos )1 2 q q
Þ v = -cos (sin )1 2q
Þ v = -æèç ö
ø÷
æèç
öø÷
-cos cos1
22
pq
Þ v = -p
q2
2
Þ v x= - -p
22 1sin
Þdv
dx x= -
-0
2
1 2Þ
dv
dx x= -
-
2
1 2
\du
dv
du
dxdv
dx
x
x
= = -
-
–
–
1
12
1
2
2
=1
2.
[Note: Here the range of x is taken as - < <1
2
1
2x ]
15. Given y xx=
Taking logarithm of both sides, we get
log . logy x x=
510 Xam idea Mathematics–XII
Q - < <1
2
1
2x Þ sin sin sin-æ
èç ö
ø÷ < < æ
èç ö
ø÷
pq
p
4 4
Þ - < <p
qp
4 4
Þ - < <p
qp
22
2
Þ p
qp
22
2> >– –
Þ pp
q> æèç ö
ø÷ >
22 0–
Þ p
q p p2
2 0 0– ( , ) [ , ]æèç ö
ø÷ Î Ì
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Downloaded from www.studiestoday.com
Differentiating both sides, we get
Þ 1 1
y
dy
dxx
xx. . log= + Þ
dy
dxy x= +( log )1 …(i)
Again differentiating both sides, we get
d y
dxy
xx
dy
dx
2
2
11= + +. ( log ).
Þd y
dx
y
x y
dy
dx
dy
dx
2
2
1= + . . [From (i)]
Þd y
dx
y
x y
dy
dx
2
2
21
= +æèç
öø÷ Þ
d y
dx y
dy
dx
y
x
2
2
21
0-æèç
öø÷ - =
16. Given f x x x x( ) = - - +3 4 12 54 3 2
Þ ¢ = - -f x x x x( ) 12 12 243 2 Þ ¢ = - -f x x x x( ) ( )12 22
Þ ¢ = - + -f x x x x x( ) { }12 2 22 Þ ¢ = - + -f x x x x x( ) { ( ) ( )}12 2 1 2
Þ ¢ = - +f x x x x( ) ( )( )12 2 1 … (i)
For critical points
¢ =f x( ) 0 Þ 12 2 1 0x x x( )( )- + =
Þ x = -0 1 2, , (critical points)
These critical points divide the real number line into 4 disjoint intervals ( , ), ( , ), ( , )- ¥ - -1 1 0 0 2
and ( , )2 ¥ .
For ( , )- ¥ - 1
¢ =f x( ) +ve × –ve × –ve × –ve = –ve [From (i)]
Þ f(x) is decreasing in ( , )- ¥ - 1
For (– 1, 0)
¢ =f x( ) +ve × –ve × –ve × +ve = +ve
Þ f(x) is increasing in (–1, 0)
For (0, 2)
¢ =f x( ) +ve × +ve × –ve × +ve = –ve
Þ f(x) is decreasing in (0, 2).
For (2, ¥)
¢f x( ) = +ve × +ve × +ve × +ve = +ve
Þ f(x) is increasing in (2, ¥).
Hence, f(x) is strictly increasing in (–1, 0) U (2, ¥)
and f(x) is strictly decreasing in ( , ) ( , ).- ¥ - 1 0 2U
OR
Q x a= sin 3 q and y a= cos 3 q
Þdx
da
qq q= 3 2sin . cos and
dy
da
qq q= -3 2cos sin
Examination Papers – 2014 511
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Þdy
dx
dy
ddx
d
a
a= =
-q
q
q q
q q
3
3
2
2
cos . sin
sin . cos = - cotq
Þdy
dx= - cotq
Þ Slope of tangent to the given curve at qp
qp
= =é
ëêù
ûú =44
dy
dx = - cot
p
4 = -1.
Since for qp
=4
, x a= sin 3
4
p and y a= cos 3
4
p
Þ x a=æèç
öø÷
1
2
3
and y a=æèç
öø÷
1
2
3
Þ xa
=2 2
and ya
=2 2
i.e., co-ordinates of the point of contact = a a
2 2 2 2,
æèç
öø÷
\ Equation of tangent is
ya
xa
-æèç
öø÷ = - -
æèç
öø÷
2 21
2 2( ). Þ y
ax
a- = - +
2 2 2 2
Þ x ya
+ =2
Also slope of normal (at qp
=4
) = -1
slope of tangent = -
-=
1
11
\ Equation of normal is
ya
xa
-æèç
öø÷ = -
æèç
öø÷
2 21
2 2( ).
Þ ya
xa
- = -2 2 2 2
Þ y x- = 0
17. Let Ix x
x xdx=
+ò
sin cos
sin . cos
6 6
2 2
Þ Ix x
x xdx=
+ò
(sin ) (cos )
sin . cos
2 3 2 3
2 2
Þ Ix x x x x x
x=
+ - +(sin cos )(sin sin . cos cos )
sin . cos
2 2 4 2 2 4
2 2 xdxò
Þ Ix x x x
x xdx=
- +ò
sin sin . cos cos
sin . cos
4 2 2 4
2 2 = - +ò ò òtan cot2 2x dx dx xdx
Þ I x dx x x dx= - - + -ò ò(sec ) ( )2 21 1cosec
Þ I x dx x dx x x x c= + - - - +ò òsec2 2cosec = - - +tan cotx x x c3
512 Xam idea Mathematics–XII
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OR
Let I x x x dx= - + -ò ( )3 3 182 … (i)
Let x Ad
dxx x B- = + - +3 3 182( ) Þ x A x B- = + +3 2 3( ) … (ii)
Þ x Ax A B- = + +3 2 3( )
Equating the co-efficient, we get
2 1A = and 3 3A B+ = - Þ A =1
2 and 3
1
23´ + = -B
Þ A =1
2 and B = - - = -3
3
2
9
2
\ I x x x= + -æèç ö
ø÷ + -ò ò
1
22 3
9
23 182( ) dx [From (i) and (ii)]
I x x x dx x x dx= + + - - + -òò1
22 3 3 18
9
23 182 2( )
Þ I I I= -1
2
9
21 2 … (iii) where I x x x dx122 3 3 18= + + -ò ( )
and I x x dx22 3 18= + -ò
Now I x x x dx122 3 3 18= + + -ò ( )
Let x x z2 3 18+ - =
Þ ( )2 3x dz+ =
\ I z dzz
c z c1
1
21
1
3
211
21
2
3= =
++ = +ò
+
( )
Þ I x x c12
3
21
2
33 18= + - +( ) … (iv)
Again I x x dx22 3 18= + -ò
= + + æèç ö
ø÷ - -ò x x dx2
2
23
2
3
2
9
418. . = +æ
èç ö
ø÷ - æ
èç ö
ø÷ò x
3
2
9
2
2 2
I x x x x x x22 21
2
3
23 18
81
4 2
3
23 1= +æ
èç ö
ø÷ + - -
´+æ
èç ö
ø÷ + + -log 8
Þ I x x x x x x22 21
2
3
23 18
81
8
3
23 18= +æ
èç ö
ø÷ + - - +æ
èç ö
ø÷ + + - +log c2 … (v)
Putting the value of I 1 and I 2 in (iii), we get
I x x x x x x= + - - +æèç ö
ø÷ + - + +
1
33 18
9
4
3
23 18
729
16
3
2
23
2 2( ) log æèç ö
ø÷ + + -x x2 3 18 + c
where cc
c= -é
ëê
ù
ûú
122
9
2
Examination Papers – 2014 513
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18. We have,
e y dxy
xdyx 1 02- + =
Þ e y dxy
xdyx 1 2- = - Þ xe dxx = -
-
y
ydy
1 2
Þ x e dxy
ydy
I
x
IIò ò= -
-1 2
Þ xe e dxdt
t
x x- =ò ò1
2, where t y= -1 2 (Using I LATE on LHS)
Þ xe et
Cx x- =æ
èç
ö
ø÷ +
1
2 1 2
1 2/
/Þ xe e t Cx x- = +
Þ xe e y Cx x- = - +1 2 , where x RÎ is the required solution.
Putting y = 1 and x = 0
0e e C0 0 21 1- = - + Þ C = -1
Therefore required particular solution is xe e yx x- = - -1 12 .
19. The given differential equation is
( )xdy
dxxy
x
2
21 2
2
1- + =
-
Þdy
dx
x
xy
x+
-=
-
2
1
2
12 2 2( )… (i)
This is a linear differential equation of the form dy
dxPy Q+ = ,
where Px
x=
-
2
12 and Q
x=
-
2
12 2( )
\ I F e e dx e xPdx x x x. . ( )
/( ) log( )= ò = ò = = -- -2 2 1 2 1 2 1
y xx
x dx C( )( )
( )2
2
212
12 1- =
-´ - +ò [Using: y I F Q I F dx C( . .) .( . .)= +ò ]
\ y xx
dx C( )2
21
2
1- =
-+ò
Þ y x( )2 1 21
2- = ´ log
x
xC
-
++
1
1Þ y x
x
xC( ) log2 1
1
1- =
-
++
This is the required solution.
20. L.H.S. = + + +® ® ® ® ® ®
[ , , ]a b b c c a = + + ´ +® ® ® ® ® ®
( ).{( ) ( )}a b b c c a
= + ´ + ´ + ´ + ´® ® ® ® ® ® ® ® ® ®
( ).{ }a b b c b a c c c a
= + ´ + ´ + ´® ® ® ® ® ® ® ®
( ).{ }a b b c b a c a [Q c c® ® ®
´ = 0]
514 Xam idea Mathematics–XII
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= ´ + ´ + ´ + ´ +® ® ® ® ® ® ® ® ® ® ® ® ®a b c a b a a c a b b c b b.( ) .( ) .( ) .( ) .(
® ® ® ® ®´ + ´a b c a) .( )
= + + + +® ® ® ® ® ® ® ® ® ® ® ® ®
[ , , ] [ , , ] [ , , ] [ , , ] [ ,a b c a b a a c a b b c b b® ® ® ® ®
+, ] [ , , ]a b c a
= + + + + +® ® ® ® ® ®
[ , , ] [ , , ]a b c b c a0 0 0 0 [By property of scalar triple product]
= +® ® ® ® ® ®
[ , , ] [ , , ]a b c b c a
= +® ® ® ® ® ®
[ , , ] [ , , ]a b c a b c [By property of circularly rotation]
=® ® ®
2[ , , ]a b c
OR
a b c® ® ®
+ + = 0 Þ ( ) ( )a b c® ® ®
+ = -2 2
Þ ( ) . ( ) .a b a b c c® ® ® ® ® ®
+ + =
Þ | | | | . | |a b a b c® ® ® ® ®
+ + =2 2 22 Þ 9 25 2 49+ + =® ®a b.
Þ 2 49 25 9a b® ®
= - -.
Þ 2 15| || |cosa b® ®
=q Þ 30 15cos q =
Þ cos cosq = = °1
260 Þ q = 60°
21. Given lines arex y z+
=+
=+1
3
3
5
5
7…(i)
x y z-=
-=
-2
1
4
3
6
5…(ii)
Let two lines (i) and (ii) intersect at a point P( , , )a b g .
Þ ( , , )a b g satisfy line (i)
Þa b g
l+
=+
=+
=1
3
3
5
5
7(say)
Þ a l= -3 1, b l= -5 3, g l= -7 5 …(iii)
Again ( , , )a b g also lie on (ii)
\ a b g-
=-
=-2
1
4
3
6
5
Þ3 1 2
1
5 3 4
3
7 5 6
5
l l l- -=
- -=
- -
Þ3 3
1
5 7
3
7 11
5
l l l-=
-=
-
I II III
Examination Papers – 2014 515
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From I and II From II and III
3 3
1
5 7
3
l l-=
- 5 7
3
7 11
5
l l-=
-
Þ 9 9 5 7l l- = - Þ 25 35 21 33l l- = -
Þ 4 2l = Þ 4 2l =
Þ l =1
2Þ l =
1
2
Since, the value of l in both the cases is same
Þ Both lines intersect each other at a point.
\ Intersecting point = = - - -æèç ö
ø÷( , , ) , ,a b g
3
21
5
23
7
25 [From (iii)]
= --æ
èç ö
ø÷
1
2
1
2
3
2, ,
22. A family has 2 children,
then Sample space = S = { , , , },BB BG GB GG where B stands for Boy and G for Girl.
(i) Let A and B be two event such that
A = Both are girls = {GG}
B = the youngest is a girl = {BG, GG}
PA
B
P A B
P B
æèç ö
ø÷ =
( )
( )
I[ { }]Q IA B GG=
PA
B
æèç ö
ø÷ =
1
42
4
= 1
2
(ii) Let C be event such that
C = at least one is a girl = { , , }BG GB GG
Now P(A/C) = P A C
P C
( )
( )
I[Q A C GGI = { }]
=
1
43
4
=1
3.
SECTION–C
23. According to question
3 2 1000x y z+ + =
4 3 1500x y z+ + =
x y z+ + = 600
The given system of linear equations may be written in matrix form as AX B= where
A =
é
ë
êêê
ù
û
úúú
3 2 1
4 1 3
1 1 1
, X
x
y
z
=
é
ë
êêê
ù
û
úúú
and B =
é
ë
êêê
ù
û
úúú
1000
1500
600
Q AX B= Þ X A B= -1 … (i)
516 Xam idea Mathematics–XII
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Now for A-1
| | ( ) ( ) ( )A = = - - - + -
3 2 1
4 1 3
1 1 1
3 1 3 2 4 3 1 4 1
= - - + = - + = - ¹6 2 3 8 3 5 0
Hence, A-1 exists.
Also, A11
1 3
1 11 3 2= = - = - A12
4 3
1 14 3 1= - = - - = -( )
A13
4 1
1 14 1 3= = - = A21
2 1
1 12 1 1= - = - - = -( )
A22
3 1
1 13 1 2= = - = A23
3 2
1 1= - = - - = -( )3 2 1
A31
2 1
1 36 1 5= = - = A32
3 1
4 39 4 5= - = - - = -( )
A33
3 2
4 13 8 5= = - = -
\ Adj A =
- -
- -
- -
é
ë
êêê
ù
û
úúú
2 1 3
1 2 1
5 5 5
T
=
- -
- -
- -
é
ë
êêê
ù
û
úúú
2 1 5
1 2 5
3 1 5
\ Aadj A
A
- = =-
- -
- -
- -
é
ë
êêê
ù
û
úúú
1 1
5
2 1 5
1 2 5
3 1 5
( )
| |
Putting the value of X A B, ,-1 in (i), we get
x
y
z
é
ë
êêê
ù
û
úúú
= -
- -
- -
- -
é
ë
êêê
ù
û
úúú
1
5
2 1 5
1 2 5
3 1 5
1000
150. 0
600
é
ë
êêê
ù
û
úúú
= -
- - +
- + -
- -
é
ë
1
5
2000 1500 3000
1000 3000 3000
3000 1500 3000
êêê
ù
û
úúú
= -
-
-
-
é
ë
êêê
ù
û
úúú
1
5
500
1000
1500
Þ
x
y
z
é
ë
êêê
ù
û
úúú
=
é
ë
êêê
ù
û
úúú
100
200
300
Þ x y z= = =` ` `100 200 300, ,
i.e., 100 for discipline
200 for politeness and
300 for punctuality
One more value like sincerity or truthfulness can be awarded.
24. Let ABC be cone having slant height l and semi-vertical angle q.
If V be the volume of cone then.
V DC AD= ´ ´1
3
2.p = ´ ´p
q q3
2 2l lsin cos
Þ Vl
=p
q q3
2
3sin cos
Examination Papers – 2014 517
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ÞdV
d
l
q
p=
3
3 [ sin sin . cos ]- +3 22q q q
For maximum value of V.
dV
dq= 0
Þpl 3
3[ sin sin . cos ]- + =3 22 0q q q
Þ - + =sin sin . cos3 22 0q q q
Þ - - =sin (sin cos )q q q2 22 0
Þ sin q = 0 or 1 2 02 2- - =cos cosq q
Þ q = 0 or 1 3 02- =cos q
Þ q = 0 or cosq =1
3
Now d V
d
l2
2
3
3q
p= { sin . cos sin . cos cos }- - +3 4 22 2 3q q q q q
Þd V
d
l2
2
3
3q
p= { sin cos cos }- +7 22 3q q q
Þd V
d
2
20q q
ù
ûúú
=
=
+ve
andd V
d
2
2 1
3
q q
ù
ûúú
=
=cos
–ve [Putting cosq =1
3 and sin q = 1
1
3
2
3
2
-æèç
öø÷ = ]
Hence for cosq =1
3 or q =
æèç
öø÷
-cos ,1 1
3 V is maximum.
25. Let Idx
x=
+ò 1
6
3
cotp
p
=
+ò
dx
x
x1
6
3
cos
sinp
p
Ix
x xdx=
+òsin
sin cosp
p
6
3
… (i)
=
-æèç ö
ø÷
-æèç ö
ø÷ + -æ
èç ö
ø÷
òsin
sin cos
p
p
p
2
2 26
3 x
xx
xx
dx Q f x dx f a b x dx
a
b
a
b
( ) ( )ò ò= + -é
ë
êê
ù
û
úú
Ix
x xdx=
+òcos
sin cosp
p
6
3
… (ii)
518 Xam idea Mathematics–XII
BD
A
C
l
q
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Adding (i) and (ii), we get
2
6
3
Ix x
x xdx=
+
+òsin cos
sin cosp
p
= ò dxp
p
6
3
2
6
3I x= [ ]p
p
= -p p
3 6 =
-2
6
p p =
p
6
Þ I =p
12.
26. The given equations are
y x= ...(i)
and x y2 2 32+ = ...(ii)
Solving (i) and (ii), we find that the line and the circle meetat B ( , )4 4 in the first quadrant. Draw perpendicular BM tothe x-axis.
Therefore, the required area = area of the region OBMO +area of the region BMAB.
Now, the area of the region OBMO
= =ò ò0
4
0
4y dx x dx = =
1
282
0
4[ ]x ...(iii)
Again, the area of the region BMAB
= = -ò òy dx x dx4
4 2
4
4 2 232 = - + ´ ´é
ëêù
ûú-1
232
1
232
4 2
2 1
4
4 2
x xx
sin
= ´ + ´ ´æèç ö
ø÷ - - + ´ ´
æè
- -1
24 2 0
1
232 1
4
232 16
1
232
1
2
1 1sin sinçöø÷
= - + = -8 8 4 4 8p p p( ) . ...(iv)
Adding (iii) and (iv), we get the required area = 4p sq units.
27. The equation of plane determined by the points A B( , , ), ( , , )2 5 3 2 3 5- - - and C( , , )5 3 3- is
x y z–
– – –
– –
2 5 3
2 2 3 5 5 3
5 2 3 5 3 3
0
- +
- +
- +
= Þ
x y z– –
–
–
2 5 3
4 8 8
3 2 0
0
+
- =
Þ ( – ) { } – ( – ) { – } ( ) { }x y z2 0 16 5 0 24 3 8 24 0+ + + + + =
Þ 16 32 24 120 32 96 0x y z- + - + + =
Þ 16 24 32 56 0x y z+ + - =
Þ 2 3 4 7 0x y z+ + - = … (i)
Now the distance of point (7, 2, 4) to plane (i) is
2 7 3 2 4 4 7
2 3 42 2 2
´ + ´ + ´ -
+ + =
+ + -14 6 16 7
29 =
29
29 = 29 unit.
Examination Papers – 2014 519
B (4
, 4)
MOX¢ X
Y
Y¢
A (4 2
, 0)
y =
x
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OR
Given line and plane are
r i j k i j k®
= - + + + +( $ $ $) ( $ $ $)2 2 3 4 2l …(i)
r i j k®
- + =. ($ $ $) 5 …(ii)
For intersection point, we solve equations (i) and (ii) by putting the value of r®
from (i) in (ii).
[( $ $ $) ( $ $ $)] .($ $ $)2 2 3 4 2 5i j k i j k i j k- + + + + - + =l
Þ ( ) ( )2 1 2 3 4 2 5+ + + - + =l Þ 5 5+ =l Þ l = 0
Hence, position vector of intersecting point is 2 2$ $ $i j k- + .
i.e., coordinates of intersection of line and plane is (2, –1, 2).
Hence, Required distance = + + - + + +( ) ( ) ( )2 1 1 5 2 102 2 2 = 9 16 144+ + = 169 = 13 units
28. Suppose dealer purchase x electronic sewing machines and y manually operated sewingmachines. If Z denotes the total profit. Then according to question
(Objective function) Z = 22x + 18 y
Also, x y+ £ 20
360 240 5760x y+ £ Þ 9 6 144x y+ £
x y³ ³0 0, .
520 Xam idea Mathematics–XII
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We have to maximise Z subject to above constraint.
To solve graphically, at first we draw the graph of line corresponding to given inequationsand shade the feasible region OABC.
The corner points of the feasible region OABC are O(0, 0), A(16, 0), B(8, 12) and C(0, 20).
Now the value of objective function Z at corner points are obtained in table as
Corner points Z x y= +22 18
O(0, 0) Z = 0
A(16, 0) Z = ´ + ´ =22 16 18 0 352
B(8, 12) Z = ´ + ´ =22 8 18 12 392 Maximum
C(0, 20) Z = ´ + ´ =22 0 18 20 360
From table, it is obvious that Z is maximum when x = 8 and y = 12.
Hence, dealer should purchase 8 electronic sewing machines and 12 manually operatedsewing machines to obtain the maximum profit 392 under given condition.
29. Let E1, E2 , E E3 4, and A be event defined as
E1 = the lost card is a spade card.
E2 = the lost card is a heart card.
E3 = the lost card is a club card.
E4 = the lost card is diamond card.
and A = Drawing three spade cards from the remaining cards.
P E P E P E P E( ) ( ) ( ) ( )1 2 3 413
52
1
4= = = = =
PA
E
C
C1
123
513
220
20825
æ
èç
ö
ø÷ = = , P
A
E
C
C2
133
513
286
20825
æ
èç
ö
ø÷ = =
PA
E
C
C3
133
513
286
20825
æ
èç
ö
ø÷ = = , P
A
E
C
C4
133
513
286
20825
æ
èç
ö
ø÷ = =
Now, required probability = PE
A1æ
èç
ö
ø÷
PE
A1æ
èç
ö
ø÷ =
æ
èç
ö
ø÷
æ
èç
ö
ø÷ +
æ
èç
ö
ø÷
P E PA
E
P E PA
EP E P
A
E
( ).
( ). ( ).
11
11
22
+æ
èç
ö
ø÷ +
æ
èç
ö
ø÷P E P
A
EP E P
A
E( ). ( ).3
34
4
=´
´ + ´ + ´ +
1
4
220
208251
4
220
20825
1
4
286
20825
1
4
286
20825
1
4´
286
20825
=+ + +
220
220 286 286 286
=220
1078 =
10
49
Examination Papers – 2014 521
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OR
Let the number of defective bulbs be represented by a random variable X. X may have value0, 1, 2, 3, 4.
If p is the probability of getting defective bulb in a single draw then
p = =5
15
1
3
\ q = Probability of getting non defective bulb = 11
3
2
3- = .
Since each trial in this problem is Bernaulli trials, therefore we can apply binomialdistribution as
P X r C p qnr
r n r( ) . .= = - when n = 4
P X C( ) .= = æèç ö
ø÷ æ
èç ö
ø÷ =0
1
3
2
3
16
81
40
0 4
Now P X C( ) .= = æèç ö
ø÷ æ
èç ö
ø÷ = ´ ´ =1
1
3
2
34
1
3
8
27
32
81
41
1 3
P X C( )= = æèç ö
ø÷ æ
èç ö
ø÷ = ´ ´ =2
1
3
2
36
1
9
4
9
24
81
42
2 2
P X C( ) .= = æèç ö
ø÷ æ
èç ö
ø÷ = ´ ´ =3
1
3
2
34
1
27
2
3
8
81
43
3 1
P X C( )= = æèç ö
ø÷ æ
èç ö
ø÷ =4
1
3
2
3
1
81
44
4 0
Now probability distribution table is
X 0 1 2 3 4
P(X)16
81
32
81
24
81
8
81
1
81
Now mean E(X) = Sp xi i
= ´ + ´ + ´ + ´ + ´016
811
32
812
24
813
8
814
1
81
Mean = + + +32
81
48
81
24
81
4
81 = =
108
81
4
3.
SET–II
9. Let I x dx= -ò cos (sin )1
= -æèç ö
ø÷
æèç
öø÷
-ò cos cos1
2
px dx = -æ
èç ö
ø÷ò
p
2x dx
\ I dx xdx= -ò òp
2
= - +p
2 2
2
xx
c
522 Xam idea Mathematics–XII
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10. a b a b n® ® ® ®
´ =| || |sin $q
Þ | | || |.| |sin $|a b a b n® ® ® ®
´ = q
Þ 1 = 32
3´ sin $q n Þ 1 2= sin |$|q n
Þ 1 2= sin q [ |$| ]Q n = 1
Þ sin q =1
2Þ q = °30 .
19. L.H.S. D =
a b c a b
c b c a b
c a c a b
+ +
+ +
+ +
2
2
2
Applying R R R R R R1 1 2 2 2 3® - ® -; , we get
D =
+ + - + +
+ + - + +
+ +
a b c a b c
a b c a b c
c a c a b
( )
( )
0
0
2
Taking ( )a b c+ + common along R1 and R2 , we get
D = + +
-
-
+ +
( )a b c
c a c a b
2
1 1 0
0 1 1
2
= + + -
+ + +
( )a b c
c a c c a b
2
1 0 0
0 1 1
2
[Applying C C C2 2 1® + ]
Again applying C C C3 3 2® + , we get
D = + +
+ + +
( )
( )
a b c
c a c c a b
2
1 0 0
0 1 0
2
= + + × + +( ) ( )a b c a b c2 2 (Q determinant of triangular matrix is product of
its diagonal elements)
= + +2 3( )a b c
D = R.H.S.
20. Let ux
x=
-
æ
èçç
ö
ø÷÷
-tan 1
21 and v x x= --sin ( )1 22 1
We have to determine du
dv
Let x = sin q Þ q = -sin 1 x
Now, u =-
æ
èçç
ö
ø÷÷
-tansin
sin
1
21
q
q
Examination Papers – 2014 523
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Þ u = æèç ö
ø÷-tan
sin
cos
1 q
qÞ u = q
Þ u x= -sin 1
Þdu
dx x=
-
1
1 2
Again, v x x= --sin (1 22 1
Þ v = --sin ( sin sin1 22 1q q
Þ v = -sin ( sin cos )1 2 q q
Þ v = -sin (sin )1 2q Þ v = 2q
Þ v x= -2 1sin Þdv
dx x=
-
2
1 2
\du
dv
du
dxdv
dx
x
x
= = -
-
=
1
12
1
1
2
2
2
[Note: Here the range of x is taken as - < <1
2
1
2x ]
21. cosec x ydy
dxx ylog + =2 2 0
cosec x ydy
dxx y. log = - 2 2
Þlog .y dy
y
x dx
x2
2
= -cosec
Þ y y dy x x dx-ò ò= -2 2. log sin
Þ log . . [ ( cos ) ( cosyy
y
ydy x x x
- + - +
- +-
- += - - - -ò
2 1 2 12
2 1
1
2 12 x dx) ]ò
Þ - + = --ò ò
122 2
yy y dy x x x x dxlog cos cos
Þ = - +- +
= - -- +
ò1
2 12
2 12
yy
yx x x x x dxlog cos [ sin sin ]
Þ = - - = - + - +1 1
2 22
yy
yx x x x x clog cos sin ( cos )
Þ = - + = - - +1
1 2 22
yy x x x x x c(log ) cos sin cos .
524 Xam idea Mathematics–XII
Q - < <1
2
1
2x
Þ sin sin sin-æèç ö
ø÷ < < æ
èç ö
ø÷
pq
p
4 4
Þ - < <p
qp
4 4
Þ - < <p
qp
22
2
Þ 22 2
qp p
Îæèç ö
ø÷– ,
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22. Given lines are5
4
7
4
3
5
-
-=
-=
+
-
x y z Þ
x y z-=
-=
+
-
5
4
7
4
3
5… (i)
andx y z-
=-
--8
7
2 8
2
5
3 Þ
x y z-=
-=
-8
7
4
1
5
3… (ii)
We know that, x x
a
y y
b
z z
c
-=
-=
-1
1
1
1
1
1
and x x
a
y y
b
z z
c
-=
-=
-2
2
2
2
2
2
are coplanar iff
x x y y z z
a b c
a b c
2 1 2 1 2 1
1 1 1
2 2 2
- - -
= 0
Now
8 5 4 7 5 3
4 4 5
7 1 3
3 3 8
4 4 5
7 1 3
- - - -
- =
-
-
( )
= + + + + -3 12 5 3 12 35 8 4 28( ) ( ) ( )
= + -51 141 192 = -192 192 = 0
Hence lines (i) and (ii) are coplanar.
28. Let Ix x
x xdx= ò0
p tan
sec . cosec = ò0 1 1
px
x
x
x x
dx
.sin
cos
cos.
sin
I x x dx= ò02p
sin
= - -ò02p
p p( ) sin ( )x x dx [ ( ) ( ) ]Q0 0
a af x dx f a x dxò ò= -
I x dx x x dx= -ò ò0
2
0
2p pp sin sin Þ 2
22
0
2I x dx= òp p
sin
= -òp p
21 2
0( cos )x dx = -ò ò
p pp p
2 22
0 0dx x dxcos
[ ]= -é
ëêù
ûúp pp
p
2 2
2
200
xxsin
Þ 22
04
2 0I = - - -p
pp
p( ) (sin sin )
Þ 22
02
I = -p
Þ I =p 2
4.
29. Let r, h, q be radius, height and semi-vertical angle of cone having volume V.
If S be the surface area of cone then
S r h r= +p 2 2 Þ S r h r2 2 2 2 2= +p ( )
Þ S rV
rr2 2 2
2
2 4
29= +
æ
èç
ö
ø÷p
p
QV r h
hV
r
=
=
é
ë
êêêê
ù
û
úúúú
1
33
2
2
p
p
Examination Papers – 2014 525
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Þ SV
rr2
2
2
2 49= + p
Þd S
dr
V
rr
( )2 2
3
2 3184= - + p
For extremum value of S or S2 .
d S
dr
( )2
0=
Þ - + =18
4 02
3
2 3V
rrp Þ
184
2
3
2 3V
rr= p
Þ rV V6
2
2
2
2
18
4
9
2= =
p pÞ r
V3 3
2=
p
Againd S
dr
V
rr
2 2
2
2
4
2 25412
( )= + p
Þd S
drr
V
2 2
23 3
2
0( )é
ëêê
ù
ûúú
>
=p
i.e., For rV
S3 23
2=
p, or S is minimum.
Hence for minimum curve surface area
r r h3 23
2
1
3= æ
èç ö
ø÷
pp
Þ rr h3
2
2= Þ
r
h=
1
2
Þ tan q =1
2Þ cot q = 2
Þ q = -cot ( )1 2 .
SET–III
9. Let I e x x dxx= -ò (sin cos )
0
2
p
I e x x dxx= - -ò0
2
p
(cos sin ) = - + -ò e x x dxx (cos ( sin ))
0
2
p
= -[ cos ]e xx02
p
[ ]Q e f x f x dx e f x cx x( ( ) ( )) . ( )+ ¢ = +ò
526 Xam idea Mathematics–XII
h
r
A
q
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= - -[ . cos . cos ]e e
pp2 0
20
= - -[ ]0 1 = 1.
10. a b i j k i j k® ®
+ = + - + + -( $ $ $) ( $ $ $)2 2 5 2 7 = + -4 3 12$ $ $i j k
\ Required vector in the direction of 4 3 12$ $ $i j k+ -
=+ -
+ + -
4 3 12
4 3 122 2 2
$ $ $
( )
i j k =
+ -4 3 12
169
$ $ $i j k =
+ -4 3 12
13
$ $ $i j k
= + -4
13
3
13
12
13$ $ $i j k.
19. L.H.S. D =
+
+
+
x xy xz
xy y yz
zx zy z
2
2
2
1
1
1
Applying C C C C1 1 2 3® + + , we have
D =
+ + +
+ + + +
+ + + +
1
1 1
1 1
2
2
x x y z xy xz
y x y z y yz
z x y z zy z
( )
( )
( )
D = +
+
+ + + +
+
1
1 1
1 1
1
1
2
2
2
2
xy xz
y yz
zy z
x y z
x xy xz
y y yz
z zy z
( )
Changing row into column, we have
D = +
+
+ + + +
+
1 1 1
1
1
1
1
2
2
2
2
xy y zy
xz yz z
x y z
x y z
xy y zy
xz yz z
( )
For I determinant we apply, C C C1 1 2® - , C C C2 2 3® -
For II determinant we take out a from 1st column, we have
D = - - + -
- - - +
+ + +
0 0 1
1 1
1 1
12 2
2 2
xy y y zy zy
xz yz yz z z
x x y z
y z
y( ) y zy
z yz z
2
2
1
1
+
+
Expanding along first row, we have
D = - - - - - - + -1 1 1 12 2 2[( ) ( ) ( ) ( )]xy y yz z xz yz y zy
+ + + + + - - + - + -x x y z y z y z y yz y z y z y z( ) [{ ( ) ( ) } ( ) (1 1 12 2 2 2 2 2 2 y z z2 - )]
On solving, we have
D = + + +1 2 2 2x y z = R.H.S.
Examination Papers – 2014 527
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20. Let ux
x=
+ -æ
è
çç
ö
ø
÷÷
-tan 121 1
and vx
x=
+
æ
èç
ö
ø÷
-sin 1
2
2
1
Now ux
x=
+ -æ
è
çç
ö
ø
÷÷
-tan 121 1
Let x = tan q Þ q = -tan 1 x
\ x =+ -æ
è
çç
ö
ø
÷÷
-tantan
tan
121 1q
q
Þ u =-æ
èç ö
ø÷-tan
sec
tan
1 1q
q =
-æ
è
ççç
ö
ø
÷÷÷
-tan cossin
cos
1
11
q
=-æ
èç ö
ø÷-tan
cos
sin
1 1 q
q
Þ u =
æ
è
ççç
ö
ø
÷÷÷
-tansin
sin . cos
1
222
22 2
q
q q =
æ
è
ççç
ö
ø
÷÷÷
-tansin
cos
1 2
2
q
q = tan tan- æ
èç ö
ø÷1
2
qÞ u =
q
2
Þ u x= -1
2
1tan
\du
dx x=
+
1
2 1 2( )… (i)
Again vx
x=
+
æ
èç
ö
ø÷
-sin 1
2
2
1
Let x = tan q Þ q = -tan 1 x
\ v =+
æ
èç
ö
ø÷
-sintan
tan
1
2
2
1
q
q
Þ v = -sin (sin )1 2q Qsintan
tan2
2
1 2q
q
q=
+
é
ëê
ù
ûú
Þ v = 2q Þ v x= -2 1tan
\dv
dx x=
+
2
1 2… (ii)
Nowdu
dv
du
dxdv
dx
= [From (i) and (ii)]
=+
+
1
2 1
2
1
2
2
( )x
x
=+
´+1
2 1
1
22
2
( )x
x =
1
4.
528 Xam idea Mathematics–XII
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21. Given differential equation is
dy
dx
x x
y y y=
+
+
( log )
sin cos
2 1
Þ (sin cos ) ( log )y y y dy x x dx+ = +2 1
Þ sin cos logy dy y y dy x x dx x dxò ò ò ò+ = +2
Þ sin [ sin sin ] log .y dy y y y dy xx
x
xdxò ò ò+ - = -
é
ëêê
ù
ûúú
22
1
2
2 2
+ ò x dx
Þ sin sin sin logy dy y y y dy x x x dx x dx c+ - = - + +ò ò ò ò2
Þ y y x x csin log= +2 … (i)
It is general solution.
For particular solution we put y =p
2 when x = 1
(i) becomes p p
2 21 1sin . log= + c
p
2= c [ log ]Q 1 0=
Putting the value of c in (i), we get the required particular solution.
y y x xsin log= +2
2
p.
22. Given lines are
r i j k i j®
= + - + -($ $ $) ( $ $)l 3
r i k i k®
= - + +( $ $) ( $ $)4 2 3m
Given lines also may be written in cartesian form as
x y z-=
-
-=
+1
3
1
1
1
0…(i)
andx y z-
=-
=+4
2
0
0
1
3…(ii)
Let given lines (i) and (ii) intersect at point ( , , ).a b g
Þ Point ( , , )a b g satisfy equation (i)
Þa b g
l-
=-
-=
+=
1
3
1
1
1
0 (say)
Þ a l b l g= + = - + = -3 1 1 1, ,
Also, point ( , , )a b g satisfy equation (ii)
\a b g-
=-
=+4
2
0
0
1
3
Þ3 1 4
2
1
0
1 1
3
l l+ -=
- +=
- +
Examination Papers – 2014 529
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Þ3 3
2
1
00
l l-=
- +=
I II III
From I and III From II and III
3 1
20
( )l -= - + =l 1 0
3 3 0l - = l = 1
l = =3
31
The value of l in both cases are same. Hence both lines (i) and (ii) intersect at a point.
The co-ordinate of intersecting point is (4, 0, –1).
28. Let Ix x x
x xdx=
+ò
sin cos
sin cos
/
4 40
2p
Þ I
x x x
=
-æèç ö
ø÷ -æ
èç ö
ø÷ -æ
èç ö
ø÷
ò0
2
4
2 2 2p
p p p
p
/ . sin . cos
sin2 2
4-æèç ö
ø÷ + -æ
èç ö
ø÷x x
dx
cosp
By Property
0
a
ò ò= -
é
ë
êê f x dx f a x dx
a( ) ( )
0
Þ I
x x x
x xdx=
-æèç ö
ø÷
+ò0
2
4 4
2p
p/ cos . sin
cos sin
Þ Ix x
x xdx
x x x=
+-ò ò
pp p
20
2
4 40
2/ /cos . sin
sin cos
sin . cos
sin 4 4x xdx
+ cos
Þ Ix x dx
x xI=
+-ò
pp
20
2
4 4
/ sin . cos
sin cos
Þ 22
0
2
4 4I
x x dx
x x=
+ò
pp/ sin . cos
sin cos =
+ò
pp
2 10
2 4
4
/
sin . cos
cos
tan
x x
xdx
x
[Dividing numerator and denominator by cos4 x]
=´ +
òp
p
2 2
2
10
2 2
2 2
/ tan . sec
(tan )
x x dx
x
Let
If
tan ; tan . sec
, ; ,
2 22
0 02
x z x x dx dz
x z x z
= =
= = = = ¥p
\ 2I =+
¥
òp
4 102
dz
z = - ¥p
4
10[tan ]z = ¥ -- -p
401 1(tan tan )
530 Xam idea Mathematics–XII
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\ 2I = p p
4 20-æ
èç ö
ø÷ Þ I =
p 2
16
29. Let r h, be radius and height of closed right circular cylinder having volume 128p cm3.
If S be the surface area then
S rh r= +2 2 2p p
S rh r= +2 2p( )
S rr
r= +æ
èç
ö
ø÷2
1282
2p .
Q V r h
r h
hr
=
Þ =
\ =
é
ë
êêêêê
ù
û
úúúúú
p
p p
2
2
2
128
128
Sr
r= +æèç ö
ø÷2
128 2p
ÞdS
dr rr= - +
æ
èç
ö
ø÷2
1282
2p
For extreme value of S
dS
dr= 0
Þ 2128
2 02
p - +æ
èç
ö
ø÷ =
rr
Þ - + =128
2 02r
r
Þ 2128
2r
r= Þ r 3 128
2=
Þ r 3 64= Þ r = 4
Again d S
dr r
2
2 32
128 22=
´+
æ
èç
ö
ø÷p
Þd S
dr r
2
24
ù
ûúú
=
=
+ve
Hence, for r = 4 cm, S(surface area) is minimum.
Therefore, dimensions for minimum surface area of cylindrical can are
radius r = 4 cm and h = 128 128
168
2r= = cm.
z z z
Examination Papers – 2014 531
h
r
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