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KENDRIYA VIDYALAYA SANGATHAN, CHENNAI REGION
CLASS XII-COMMON PRE-BOARD EXAMINATION-2013-2014
Answer key (Mathematics)
Section A
1. � = 13 1 mark 2. � + � = 6 1 mark 3. �� �� = 3 1 mark
4. − �� 1 mark
5. 0 1 mark 6. 0 1 mark 7. 5 1 mark 8. � + � + � = 13 1 mark 9. 110 1 mark 10. 66 1 mark
Section B 11. Proving Reflexive 1 mark
Proving Symmetric 1 mark
Proving Transitive 1 ½ mark
Conclusion ½ mark 12. tan�� �13� + tan�� �15� + tan�� �17� + tan�� �18� ⇒ tan�� !"#!$��!"%!$+tan��
!&#!'��!&%!' 1 mark
⇒ tan�� ()+ tan�� ��� 1 mark
⇒ tan�� *&# "!!��*&% "!! 1 mark
⇒ tan�� 1
⇒ �( 1 mark
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OR
tan�� +,��,�-. + tan�� +,#�,#-. = +/(. ⇒ tan�� +,��,�-. + tan�� +,#�,#-. = tan�� 1
⇒ tan�� +,��,�-. = tan�� 1 − tan�� +,#�,#-. 1 mark
By applying formula on the R.H.S.
⇒ tan�� +,��,�-. = tan�� + �-,#�. 1 mark
Applying tan both sides and solving
� = ± �√- 2 mark
13. ����) = 2� + �3-4
⇒ log2����)3 = log2� + �3-4 ⇒ 1389�� + 7 log � = 20 log2� + �3 1 mark
Differentiating with respect to x
⇒ ��:�),,2,#:3 = +��:�),:2,#:3. ;:;, 2 mark
⇒ ;:;, = :
, 1 mark
OR
Let � 2 = <9=2> ⇒ cos�� �- = 2> ½ mark
Let y = tan�� √�#,B�√��,B√�#,B#√��,B
⇒ y = tan�� √�#CDE-F�√��CDE-F√�#CDE-F#√��CDE-F
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⇒ y = tan��2��GHIF�#GHIF3 ⇒ � = 2�( − >3 1 ½ mark
⇒ � = 2�( − �- cos�� �-3
Let � = cos�� �- 1 mark
⇒ � = +�( − �- �. ⇒ ;:;J = − �- 1 mark
14. For calculating LHL = 8 1 ½ mark
For calculating RHL = 8 1 ½ mark
For calculating K = 8 1 mark
15. ∫ LM2NOI(,�(3��PQN(, � = ∫ LM2-NOI-,PQN-,�(3- CDEB -, � 1 ½ mark
= ∫ �,[STU2� − 2 sec- 2�] � 1 ½ mark
= �,STU2� + < 1 mark
16. �√�- + 1 = lnX√�- + 1 − �Y Differentiating with respect to x
⇒ �+ �-√,B#�. 2� + √�- + 1 ;:;, = �
√,B#��, + �-√,B#� 2� − 1. 1 mark
⇒ ,:√,B#�+ √�- + 1 ;:;, = ,�√,B#�
X√,B#�YX√,B#��,Y
⇒ ,:√,B#�+ √�- + 1 ;:;, =− �
√,B#� 1 mark
⇒ √�- + 1 ;:;, =− 2�#,:3√,B#� 1 mark
⇒ 2�- + 13 ;:;, + �� + 1 = 0 1 mark
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17. ∫ |�=[U\�|�"B��
= ∫ |�=[U\�|���� + ∫ |�=[U\�|�"B� 1 mark
= ∫ �=[U\����� − ∫ �=[U\��"B� 1 mark
On integrating both integrals on right-hand side, we get
= ]− ,PQN�,� + NOI�,�B ^��
� − ]− ,PQN�,� + NOI�,�B ^�
"B 1 mark
= �� + �
�B 1 mark
18. ∫ + �_E`a" ,NOI2,#b3. �
= ∫ + �_E`a" ,2NOI,PQNb#PQN,NOIb3. �
= ∫ + �_E`a* ,2PQNb#PQG,NOIb3. �
= ∫ + PQNLPB,_2PQNb#PQG,NOIb3. � 2 marks
On substitution of <9=c + <9S�=[Uc = S = ∫ +− �
NOIb√G. S = − -√G
NOIb + d 1 mark
On substitution of t S = <9=c + <9S�=[Uc
= −+ -NOIb.eE`a2,#b3NOI, + d 1 mark
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19. Tf = g + i + jk,mnf = 2g + 4i − 5jk,<f = og + 2i + 3jk pHnf%Xqnf#PfYrqnf#Pfr p = √2 ------ (i) 1 mark
mnf + <f = 22 + o3g + 6i − 2jk rmnf + <fr = √o- + 4o + 44 ------ (ii) 1 mark
TfΧXmnf + <fY = t g i jk1 1 12 + o 6 −2t = −8g + 24 + o3i + 24 − o3jk------ (iii) 1 mark
By equation (i), (ii) & (iii)
u�vw#2(#x3y#2(�x3zk√xB#(x#(( u = √2 On solving we will get
o = 1 1 mark
OR
Tf + mnf + <f = 0 ⇒ XTf + mnfY- = 2−<f3- 1 mark
XTf + mnfY. XTf + mnfY = <f. <f ½ mark
|Tf|- + rmnfr- + 2Tf. mnf = |<f|- ½ mark
By Substitution of values
9 + 25 + 2Tf. mnf = 49 ⇒Tf. mnf = �{
- 1 mark
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⇒ |Tf|rmnfr<9=> = 152 By Substitution of values
<9=> = �- ⇒ > = 60° 1 mark
20. ,#�) = :#�
�} = J#�� TU ,��� = :�{
�- = J�)�
T1nnnnf = −g − i − jk T2nnnnf = 3g + 5i + 7jk m1nnnnf = 7g − 6i + jk m2nnnnf = g − 2i + jk T2nnnnf − T1nnnnf = 4g + 6i + 8jk 1 mark
m1nnnnfΧm2nnnnf = −4g − 6i − 8jk rm1nnnnfΧm2nnnnfr = √116 1 mark
Shortest distance = pXH-nnnnnf�H�nnnnnfY.Xq�nnnnnf%q-nnnnnfYrq�nnnnnf%q-nnnnnfr p 1 mark
= u− ��}√��}u
= √116 1 mark
OR
Equation of plane passing through 22,1, −13is
T2� − 23 + m2� − 13 + <2� + 13 = 0 ----- (i) ½ mark
(i) Passes through 2−1,3,43
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⇒ −3T + 2m + 5< = 0----- (ii) 1 mark
(i) is perpendicular to � − 2� + 4� = 10 ⇒ T − 2m + 4< = 0----- (iii) 1 mark
On solving (ii) and (iii)
T = 18, m = 17, < = 4 1 mark
From (i) equation of plane
18� + 17� + 4� − 49 = 0 ½ mark
21. Let X is the random variable denoting the number of selected scouts, X
takes values 0, 1, 2. ½ mark
~2� = 03 = B��B $��B = �v-({ ½ mark
~2� = 13 = X B��!% "��!Y $��B = �-4-({ ½ mark
~2� = 23 = "��B $��B = v)-({ ½ mark
Now mean = ∑2~O�O3 = -�(-({ 1 mark
Relevant Value 1 mark
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x
Y
2
22. t � � + � � + 2�� + 2� � � + �� + � � + 2� � t Applying �1 ⇒ �1 + �2 + �3 = t32� + �3 32� + �3 32� + �3� + 2� � � + �� + � � + 2� � t 1 mark
= 32� + �3 t 1 1 1� + 2� � � + �� + � � + 2� � t 1 mark
Applying d1 ⇒ d1 − d3, d2 ⇒ d2 − d3 = 32� + �3 t0 0 1� −� � + �� 2� � t 1 mark
Expanding along R1 we get
= 9�-2� + �3 1 mark
23. Diagram ½ mark
Volume of the tank = 8�� = 2��----- (i) ½ mark
Let C be the cost of making the tank
d = 70�� + 45Χ222� + 2�3 d = 70�� + 1802� + �3 1 mark
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From equation (i)
d = 70�. 4� + 1802� + 4�3 d = 280 + 180 +� + (
,. ----- (ii) 1 mark
;�;, = 180 +1 − (
,B. 1 mark
For maxima and minima, ;�;, = 0
⇒ 180 +1 − (,B. = 0
⇒ � = ± 2 as x≠-2, x=2
Now ;B��,B = 180Χ v,"
�-d��-�,�- = 180 > 0 ⇒ d[=�[U[���TS� = 2
By equation (ii)
d,�- = �=. 1000
OR
1 mark
Area of ∆��d = �- Χ��Χ�d = �
- Χ2m=[U>2T − T<9=>3 = Tm=[U>21 − <9=>3-----(i) 1 mark
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;�;F = Tm2sin- > + cos > − cos- >3 1 mark
For maxima and minima ;�;F = 0
Tm2sin- > + cos > − cos- >3 = 0 <9=> − <9=2> = 0 <9=> = <9=2> 2> = 2U\ ± > > = U\ ± F-----(ii) As > ∈ 20, \3 by equation (ii)
> = \ − >2
> = -�� 1 mark
-�>- = Tm22=[U2> − =[U>3 ];B�;FB^F�B�" < 0 ⇒ �is maximum. 1 mark
By equation (i)
���� = �√�( Tmsq.unit 1 mark
24. Let x, y, z be the amount of prize to be awarded in the field of
agriculture, education & social science respectively. The given situation
Can be written in the matrix form as:
�� = �
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Where � = �10 5 1515 10 51 1 1 �, � � ����� , � � �
70000550006000 � �� � � ⇒ � � ���� 1 mark
|�| � 75 1 mark
T�� � � 5 10 �125�10 �5 1755 �5 25 �
��� � H;��|�| � �
){ �5 10 �125�10 �5 1755 �5 25 � 2 marks
����� � 175 �
5 10 �125�10 �5 1755 �5 25 � �70000550006000 �
����� � �200010003000�
⇒ � � 2000, � � 1000, � � 3000 1 mark
Value based relevant answer 1 mark
25.
1 mark
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Point of intersection of given curves is � = �- 1 mark
Required Area = 2 �∫ √4�� + ∫ e��(,B( �"B!B!B4 � 1 ½ mark
= 2.2 ]�"B^4!B + 2 �,-e��(,B( + �
v sin�� +-,� .�!B"B 1 ½ mark
= ��v + √-
} − �( sin�� +��. 1 mark
OR
� = ∫ 2|� − 1| + |� − 2| + |� − 3|3�(4 ---- (i)
� = ∫ 2|� − 1|3�(4 + ∫ 2|� − 2|3�(4 + ∫ 2|� − 3|3�(4 1 mark
Let �� = ∫ 2|� − 1|3� = ∫ 21 − �3��4 + ∫ 2� − 13�(�(4
= − �- [21 − �3-]4� + �- [2� − 13-]�(
= 5 1 ½ mark
�- = 2|� − 2|3� = 22 − �3�-4 + 2� − 23�(
-(4
= − �- [22 − �3-]4- + �- [2� − 23-]-(
= 4 1 ½ mark
�� = ∫ 2|� − 3|3� = ∫ 23 − �3��4 + ∫ 2� − 33�(�(4
= − �- [23 − �3-]4� + �- [2� − 33-]�(
= 5 1 ½ mark
By equation (i)
� = �� + �- + �� = 5 + 4 + 5 = 14 ½ mark
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26. Let x and y be the number of pieces of type A and B manufactured per
week respectively. If z is the profit then, � = 80� � 120�
Maximize z subject to the constraint
9� � 12� ¡ 180 ⇒ 3� � 4� ¡ 60----(i)
� � 3� ¡ 30----(ii)
� ¢ 0, � ¢ 0----(iii) 2 Mark
2 mark
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Corner Points � = 80� + 120�
0(0,0)
A(20,0)
B(12,6)
C(0,10)
0
1600
1680 ← maximum
1200
1 mark
Relevant answer 1 mark
27. Let the event b defined as ¤� =The examinee guesses the answer ¤- =The examinee copies the answer ¤� =The examinee knows the answer � =The examinee answers correctly ½ marks
~2¤�3 = �} , ~2¤-3 = �
� , ~2¤�3 = 1 − +�}+ ��. = ���v 1 mark
~ + �¥!. = �( (Out of 4 choices 1 is correct) ½ marks
~ + �¥B. = �v ½ marks
~ + �¥". = 1 (If the answer is known it is always correct) ½ marks
~ +¥"� . =Required
~ +¥"� . = ¦2¥"3.¦+ §".¦2¥!3.¦+ §!.#¦2¥B3.¦+ §B.#¦2¥"3.¦+ §". 1 mark
On substitution
~ +¥"� . = ���( 1 mark
Yes the probability of copying is less than other probability. 1 mark
28. The given planes are 2� + � − 6� − 3 = 0---- (i) 5� − 3� + 4� + 9 = 0---- (ii) Equation of the plane passing through the intersection of (i) and (ii) 2� + � − 6� − 3 + o25� − 3� + 4� + 93 = 0 ---- (iii) 2 mark
Given that plane (iii) is parallel to ,��- = :��
( = J�{( ⇒ 22 + 5o3. 2 + 21 − 3o3. 4 + 24o − 63. 4 = 0 1 mark
On solving o = v) 1 ½ mark
On substitution of o in (iii) equation of plane 54� − 17� − 10� + 51 = 0 1 ½ mark
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29. �-� + �2� + �3� = 0
⇒ ;:;, = − ,:#:B,B ----- (i) ½ mark
Let � = ©� ⇒ ;:;, = © + � ;ª;, ½ mark
By equation (i)
© + � ;ª;, = − ª,B#ªB,B,B ⇒ ;ªªB#-ª = − ;,, 1 mark
∫ + ;ªªB#-ª. = ∫ +− ;,, . ⇒ �- ∫ +�ª − �
ª#-. © = ∫ +− ;,, . ⇒ log|©| − log|© + 2| = −2 log|�| + d 1 mark
⇒ log uª,Bª#-u = 89�j
ª,Bª#- = j 1 mark
Putting © = :, we get �-� = j2� + 2�3 1 mark
For Particular solution j = ��
Therefore Particular Solution is 3�-� = � + 2� 1 mark
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