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GOVERNMENT OF KARNATAKA

KARNATAKA STATE PRE-UNIVERSITY EDUCATION EXAMINATION BOARD

II YEAR PUC EXAMINATION JUNE/JULY 2017

SCHEME OF VALUATION

Subject Code: 35 (NS) Subject: Mathematics (NS)

Instructions:

a) Answer by alternate method should be valued and suitably awarded.

b) All answers (including extra, struck off and repeated) should be valued. Answers with maximum marks awarded must be considered.

c) If the question numbers for the answers are wrong or question numbers are not written, write the correct question numbers with respect to the answers by circling it and value them.

Qn No PART A Marks

1 e 4 1

2 0x 1

3 11

2

2 1

A

1

4 2 2 x 1

5

22 cos dy

x xdx

1

6 1. sin3

3 G I x C

1

7

The unit vector in the direction of a is ˆ| |

aa

a

1( 2 )

6 i j k

1

8 The direction cosines of y-axis are 0,1,0 OR cos ,cos0,cos

2 2

OR 0, 1,0. OR cos ,cos ,cos2 2

.

1

9 Any point in the feasible region that gives the optimal value (maximum or

minimum) of the objective function is called an optimal solution linear

programing problem.

1

10 P(A B) P(B| A) P(A) 0.4 0.8 0.32 . 1

PART B 11 Let z C be an arbitrary element. As g : B C is onto, there exists y B such

that g y z .

OR For y B there exists x A such that f x y , since f : A B is onto.

1

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OR Writing: f and g are onto f A B and g B C

Therefore for all z C there exists x A such that g f x g f x g y z .

Hence g f : A C is onto.

OR Showing g f A C and conclusion.

1

12 Taking 1tan x OR tanx 1

Getting : 2

1

2

1cos

1

x

x

12 tan x . 1

13 Getting :

3 3 2       5 5 5

sin sin sin

OR 1sin sin x x, x ,2 2

1

Getting: 1 13 2 2sin   sin   .

5 5 5sin sin

1

14

Writing: Area of the triangle is

1 0 11

6 0 12

4 3 1

1

Getting: 15

2

1

15 Getting: log log(sin )y x x OR v vd d

u u vlogudx dx

1

Getting: cos

sin log(sin )sin

xdy xx x x

dx x OR

coslog(sin )

sin

dy xy x x

dx x

1

16 Getting: 2 3 cos

dy dyy

dx dx

1

Getting: 2

(3 cos )

dy

dx y

OR 2

cos 3

dy

dx y

1

17

Getting: 25

4

dy xm

dx y

1

Getting the points (0,5) and (0, 5) . 1

18

Getting: 2

tan 1.

tan cos

xG I dx

x x OR 2tan

. sectan

x

G I x dxx

OR

21. sec

tan G I x dx

x

OR Put tan x t and 1

. G I dtt

OR d tan x

G.Itan x

1

Getting:

. 2 tan G I x C 1

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19 Getting:

2 3

1 2.

( 1) ( 1)

xG I e dxx x

1

Getting: 2

1.

( 1)

xG I e Cx

1

20 Writing: Order is 4

1

Writing: Degree is not defined. 1

21 Getting: 2 2| | | | 8x a 1

Getting: | | 3x . 1

22

Getting : ˆ3 1 5 44

1 1 1

i j

i j k

a b k

1

Getting the area of a parallelogram is | ⃗ ⃗⃗| √ 1

23 Getting: 1 3 2 6b i j k and

1 2 2 b i j k

OR If is the angle between given lines then 1 2

1 2

cos| || |

b b

b b

OR 19

cos21

1

Getting: 1 19cos

21

.

1

24 P at least one of A and B P B(A ) P A P B )P A B(

OR P A B 1 P A B

1

Getting :

LHS 1 P A P B 1

PART C

25 Writing: R 1,2 , 2,3 , 3,4 , 4,5 , 5,6

1

Showing R is not reflexive by giving any counter example: as 1,1   R .

OR a a 1 a,a R

1

Showing R is not symmetric by giving any counter example as 1,2 R but

2,1 R .

OR If a,b R , i.e., b a 1 then a b 1 , i.e., b,a R

1

26

Writing: 1 2 3tan

1 2 3 4

x x

x x

1

Getting: 26 5 1 0 x x 1

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Getting: 1

6x

1

27 Writing: A IA OR

1 2 1 0

2 1 0 1A

1

Applying 2 2 12R R R and getting 1 2 1 0

0 5 2 1A

1

Getting: –1

1 2

5 5

2 1

5 5

A

.

1

28

Getting: 21 1sin sec

2 2tan

2

dx ta t

tdt

OR 1

sin

2sin cos2 2

dxa t

t tdt

1

Getting: cosdy

a tdt

1

Getting:2

costan

cos

sin

dy a tt

dx ta

t

1

29 Writing: The function 2( )f x x is continuous in [2,4]and differentiable in

(2,4)

1

Getting: ( ) 2f x x OR (2) 4 (4) 16f and f OR ( ) ( ) 16 4

64 2

f b f a

b a

1

Writing the conclusion: c 3 (2,4) . Therefore Mean Value Theorem verified. 1

30 Writing: 2 2(+ 15 – )S x x 1

Getting: 4 – 30dS

xdx

1

Writing: Required numbers are 15 15

,2 2

1

31 Getting:

1 2

1 2

I dx

x x

1

Getting: . log | 1| 2log| 2 | G I x x C 1+1

32 Put 1cos x OR

. cos G I d

1

Getting: G.I sin cos C

1

Getting: 2 1. 1 cos G I x x x C

1

33 Figure: 1

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OR

3

22 2

30

2 2

Area cos cos cosx dx x dx x dx

OR 2

0

Area 4 cos x dx

Getting: Area 3

222 3

022

sin [sin ] sinx x x

OR

/2

0Area 4 sin x

1

Getting: Area 4 square unit. 1

34 Getting:

2

2

dy x

dx y

1

Getting: 2 2y dy xdx OR 2 2 C y dy xdx OR

32

3

yx C

1

Required equation of curve is 3

2 53

yx .

1

35 Figure :

OR Writing: AP m

PB n

1

Writing: nAP mPB OR n OP OA m OB OP

1

Getting:

mb nar

m n

1

36 Writing: (1, 2, 4)AB OB OA x

OR (1,0, 3)AC OC OA

OR (3, 3, 2)AD OD OA

1

Writing: 0AB AC AD OR

1 2 4

1 0 3 0

3 3 2

x

1

Getting: 5.x 1

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37 Writing: 5 2 4 a i j k OR 2 3 N i j k OR ( ) 0 r a N

1

Getting: Vector equation [ (5 2 4 )] (2 3 ) 0 r i j k i j k 1

Getting: Cartesian equation 2 3 20 x y z 1

38 Writing:

1Probability that six occurP(A) s

6

OR 5

Probability that six does not occur( )6

P A s

1

Writing: Let E: ‘man reports that six’ 3

P(E | A)4

OR 1

P(E | A )4

OR

P A P E | AP A | E

P A P E | A P A P E | A

1

Getting Required probability: 3

P(A | E)8

. 1

PART D

39 Defining g : ,

  3  ( )   

4

yg y ,

OR Defining g : ,      3

  ( )   4

xg x ,

OR Writing 4 3 y x      3

     4

yx .

1

Showing gof x x . 1

Getting f g y y ,

OR f og x x .

1

Stating g f I and f g I and conclusion 1

Writing: 1 y   3

4f

 y

 

OR

1 x   3

4f

 x

 

OR 1f g

1

ALTERNATE METHOD

Let 1 2x , x be such that 1 2f x f x 1

Getting : 1 2x x . Therefore f is one-one. 1

Let y then 4x f xy 3 y     3

 x   4

OR Showing: Every element of is the image of the element of .

1

Proving Onto: y ,y    3

4

y     3such that f y

4

.

OR Writing: f ( ) or Range is the codomain, hence f is onto.

1

Writing: 1 x   3

4f

 x

  OR

1 y   3

4f

 y

 

1

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40

Finding:

9

12

30

AC

1

Finding:

1

8

2

BC

1

Finding:

0 7 8

5 0 10

8 6 0

A B

OR

10

20

28

AC BC

1

Finding:

10

20

28

A B C

1

Conclusion: A B C AC BC 1

41

Let

1 1 1

0 1 3

1 2 1

A

,

x

X y

z

and

6

11

0

B

OR

Writing: 1 1X A B adjA B

A

OR 1

X adjA BA

1

Getting: 9A 1

Getting:

7 3 2

( ) 3 0 3

1 3 1

adj A

Note: If any 4 cofactors are correct award 1 mark

2

Getting: 1 ;  2        3x y and z . 1

42 Getting: 1

3sin(log ) 4cos(log )y

x x

x x

1

Writing: 1y 3sin(log ) 4cos(log ) x x x 1

Getting: 2 1

3cos(log ) 4sin(log )y 1

x xx y

x x

1

Getting: 2

2 1y [3cos(log ) 4sin(log )] x xy x x 1

Getting: 2

2 1y 0. x xy y

1

43 Writing: 312 / s

dvcm

dt OR

1

6h r

1

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Writing: 21

3V r h OR

312V h 1

Getting: 236dv dh

hdt dt

1

Getting:

2

12 36 4dh

dt

1

Getting:

1/

48

dhcm s

dt

1

44

Getting: I =∫√ 2 2

2 2

2

2

xa x x xdx

a x

1

Getting: I 2 2 2 2 2

2 2

1

x a x a x dx a dx

a x

1

Getting: 2 2 2 2 22 log I x a x a x a x 1

Getting: 2 2 2

2 2log2 2

x a x aI x a x C

1

Getting: 2

21 1log 1

2 2

x xI x x C

1

45 Figure:

1

Area

2

2

0

4 (2 )A x x dx

OR Area 1

4 (area of the circle with radius 2 ) – (area of the triangle in the 1

st

quadrant).

1

Area

22

2 1

0

44 sin 2

2 2 2 2

x x xx x

[For integrating each 24 x and (2 ) x carries 1 mark]

OR 21 1Area 2 2 2

4 2

1+1

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Getting: Area 2 square units. 1

46

Getting:

22

dx x y

dy y y OR

12

dxx y

dy y

1

Writing: 1

2P and Q yy

1

Getting:

1

log. .1

dy

yye eI Fy

1

Writing general solution ( . .) ( . .)x I F Q I F dy C

OR 1 1

2x y d y Cy y

1

Getting: 2 . x

y Cy

1

47 Drawing any appropriate figure and explanation

1

Writing: AP r a and AB b a are collinear vectors. OR AP AB 1

Getting vector form ( ) r a b a 1

Writing: 1 1 1  ˆOA a x i y j z k , 2 2 2  ˆ OB b x i y j z k

and ˆOP r xi y j zk .

1

Getting: Cartesian form 1 1 1

2 1 2 1 2 1

x x y y z z

x x y y z z

.

1

48 Writing:

110,

2 n p and

1

2q

1

Writing:

10– 1010 101 1

(2 2 2

)1

x x

x xP X x C C

1

Getting:

1010

6 10

1 10! 1 1056

2 4!(10 4)!( )

5122

P X C

1

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Getting: ( 6) ( 6) ( 7) ( 8) ( 9) ( 10) P X P X P X P X P X P X 1

Getting: 193

( 6)512

P x 1

PART E 49 a Graph : Representation of both lines carries 1 mark and shading feasible region

carries 1 mark

2

Writing corner points O(0, 0), A (0, 4), E(2, 3) and D(4, 0) 1

Getting:

Sl.

No.

Corner

points

Corresponding

value of 3 9 z x y

1. O (0, 0) 0

2. A (0, 4) 16

3. E (2, 3) 6

4. D (4, 0) 12

1+1

Conclusion: minimum value of z is 12 at the point (4, 0).

Indicating Minimum value in the above table.

1

49 b Applying 1 1 2R R R , and 2 2 3R R R , we get

2 2

2 2

2

0

0

1

a b a b

LHS b c b c

c c

1+1

Taking    a b and b c as common from 1 2 R and R respectively,

2

0 1

0 1

1

a b

LHS a b b c b c

c c

1

Expanding along 1C we get

   0 0 1 LHS a b b c b c a b

1

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     a b b c c a RHS

50 a

Writing: 0

– 0

a a

a a

f x dx f x dx f x dx

1

Putting x t in the first integral on the RHS, then dx dt

Also, x a t a and 0 0x t

1

Getting: 0 0

a a a

a

f x dx f x dx f x dx

1

If  f x is even, then ( ) ( )f x f x

– 0 0 0

2

a a a a

a

f x dx f x dx f x dx f x dx

1

If  f x is odd, then ( ) ( )f x f x

– 0 0

0

a a a

a

f x dx f x dx f x dx .

1

Writing: 3 cosx x x is an odd function and

2

–2

3(   ) 0 x xcosx dx

1

50 b Getting: (0) 0f

1

Getting: 1RHL 1

Showing: (0)RHL f 1

Conclusion: ( )f x is not continuous at 0,x R .

Thus, for no value of , f is continuous at 0x

1