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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