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MATHEMATICS 2016 (A)

( NEW SYLLABUS )

πÂä∆Â

SCHEME OF VALUATION Subject Code : 35 (N/S)

Qn.

No.

Marks

Instructions :

1) For any alternate method, it should be valued and suitably

awarded.

2) All answers ( including extra, struck off and repeated ) should

be valued. Answer with maximum marks awarded must be

considered.

3) If the student had written wrong question number, write the

correct question number and be valued.

Code No. 35 (N/S) 2

Qn.

No.

Marks

PART - A

I. 1. Getting Answer =

!

"cot x " cosec!x + c 1

2. x = 6 or x = – 6 1

3. Writing

!

dy

dx= "

sin x

2 cos x 1

4. Writing

!

cos"

2= 0

1

5. Getting

!

OA = OB " AB = i#

" 3 j#

+ 3k#

1

6. Getting distance =

!

14

7

1

7.

!

x = 2 and y = 3 1

8. Getting

!

P(A" B) = 0 # 32 1

9. Not a binary operation 1

10. Definition 1

PART - B

II.

11. Getting

!

GE = tan"1

3

4" tan x

1 +3

4tan x

#

$

% % %

&

'

( ( (

=

!

tan"1 3

4" tan

"1tan x( ) = tan

"1 3

4" x

1

1

12. Writing

!

" = a b + c 1

b c + a 1

c a + b 1

Getting

!

" = 0

1

1

13. Getting

!

fog "1

2

#

$ %

&

' ( = 1

and

!

gof "1

2

#

$ %

&

' ( = 0

1

1

14. Putting x = cos θ and θ = cos

!

"1x

Proving the answer

1

1

15. Putting

!

x = cos " and getting y = 2

!

"

Getting

!

dy

dx=

"2

1" x2

1

1

3 Code No. 35 (N/S)

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Qn.

No.

Marks

16. Writing

!

y log x = x log a

Getting

!

y

x+ log x.

dy

dx= log a

and

!

dy

dx=

x log a " y

x log x

1

1

17. Getting

!

I =1

tan x+ tan x

" # $

% & '

( sec2

x dx

And

!

I = log|tan x|+tan2

x

2+ c

1

1

18. Writing

!

x = 27, "x = 2, x + "x = 25

!

f (x) = x1/3, f (x + "x) = (x + "x)

1/3

Getting

!

f /(x) =

1

3x2/3 and Answer = 2·926

1

1

19. Writing

!

"cos x dx

0

#

$

Getting Answer = 0

1

1

20. Writing

!

|a + b|2 = |a " b|2

And

!

|a|2 +|b|2 + 2(a . b) = |a|2 +|b|2 "2(a . b)

Getting

!

a . b = 0 and writing

!

a " to

!

b

1

1

21. Writing order = 4

Degree not defined

1

1

22. Getting

!

a . b = 1, |a|= 3, |b| = 3

Getting

!

cos" =1

3 and writing

!

" = cos#1 1

3

1

1

23. Getting

!

k =1

6

Getting

!

P(X " 2) = 1

1

1

24. Writing direction ratios 0, 1, 0

Equation of the line

!

x "1

0=

y "1

1=

z "1

0

1

1

Code No. 35 (N/S) 4

Qn.

No.

Marks

PART- C

III.

25. Writing LHS =

!

tan"1

1

2+

2

11

1"2

22

#

$

% % %

&

'

( ( (

+ tan"1 4

3

=

!

tan"1 3

4+ tan

"1 4

3

=

!

tan"1

3

4+ cot

"13

4=#

2= RHS

1

1

1

26. Writing A = IA

Getting any one non-diagonal element as zero

Getting the inverse =

!

1

5

3 1

"2 1

#

$ %

&

' (

1

1

1

27. Reflexive :

!

"a # A, (a $ a) = 0 which is

multiple of 4 :

!

(a, a) " R

Symmetric : Let

!

(a, b) " R #|a $ b| is multiple of 4

!

"|b # a| is multiple of

!

4 " (b, a) # R

Transitive : Let

!

(a, b) and ( b, c )

!

" R

!

" |a # b| and |b # c| are multiple of 4

!

" |a # b + b # c| = |a # c| is multiple of 4

!

" (a .c) # R

1

1

1

28. Writing

!

f (x) is continuous in [ 1, 3 ] and differentiable in ( 1, 3 )

!

f /(x) = 3x2

"10x " 3 or

!

f (3) = "27, f (1) = "7

and getting c = 7/3

Writing

!

c =7

3" (1, 3)

Note : If

!

c =7

3" (1, 3) is not written deduct one mark.

1

1

1

29. Getting

!

dx

d"= #3acos

2"sin "

!

dy

d"= 3asin

2"cos "

getting

!

dy

dx= "

sin #

cos #=

y

x3

1

1

1

5 Code No. 35 (N/S)

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Qn.

No.

Marks

30. Writing

!

P(E1) = P(E2) =1

2& P(A/ E1) = 1, P(A/ E2) =

1

2

Writing

!

P(E1/ A) =P(E1)P(A/ E1)

P(E1)P(A/ E1) + P(E2)P(A/ E2)

=

!

2

3

1

1

1

31. Writing

!

x

(x "1) (x " 2)=

A

x "1+

B

x " 2 and

!

x = A(x " 2) + B(x "1)

Getting A = – 1 and B = 2

Getting the answer =

!

" log(x "1) + 2 log(x " 2) + c

1

1

1

32. Getting

!

I =1

(t +1) (t + 2) dt"

=

!

1

t +1"

1

t + 2

# $ %

& ' (

) dt

=

!

log|t +1|" log|t + 2| + c

=

!

log|x2

+1|" log|x2

+ 2| + c

1

1

1

33. Writing

!

xy = 100 and

!

S = x + y is minimum

getting

!

ds

dx= 1"

100

x2

and

!

ds

dx= 0 " x = ± 10

getting

!

d2s

dx2

=200

x3

> 0 at x = 10

!

" x = 10 and y = 10

1

1

1

34. Finding the values x = 0 and x = 1

Writing area =

!

4x

0

1

" dx # 2x dx

0

1

"

Getting the area

!

=4

3"1 =

1

3 sq. unit

1

1

1

Code No. 35 (N/S) 6

Qn.

No.

Marks

35. Writing

!

(a " b) . (b " c) # (c " a){ }

Getting =

!

(a " b) . (b # c) " (b # a) + (c # a){ }

= 0

1

1

1

36. Getting

!

a2 " a1 = 2 i#

+ j#

" k#

and

!

b = 2 i"

+ 3 j"

+ 6k"

Writing distance

!

d = b " (a2 # a1)

|b|

=

!

9 i"

+14 j"

+ 4k"

49

!

d =293

7

1

1

1

37. Getting

!

a " b =

i#

j#

k#

1 2 2

3 2 6

= 8 i#

$ 4k#

Getting

!

a " b = 4 5, |a|= 3, |b|= 7

Getting

!

sin " = |a # b|

|a| |b| =

4 5

21

1

1

1

38. Writing

!

dy

dx=

x

y or

!

ydy = xdx

Getting general solution

!

ydy = xdx""

=

!

y2

2=

x2

2+ c

Putting x = 1, y = 1

!

" c = 0 and

Writing equation

!

x2= y2

1

1

1

7 Code No. 35 (N/S)

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Qn.

No.

Marks

PART - D IV.

39. Getting

!

AB =

"2 "6 12

4 12 "24

5 15 "30

#

$

% % % %

&

'

( ( ( (

Writing

!

A/ = "2 4 5 [ ]

!

B/

=1

3

"6

#

$

% %

&

'

( (

Getting

!

B/A

/=

"2 4 5

"6 12 15

12 "24 "30

#

$

% %

&

'

( (

Proving

!

(AB)/

= B/A

/

1 1 1 1 1

40. Writing

!

A =

2 "3 5

3 2 "4

1 1 "2

#

$

% % %

&

'

( ( (

, X =

x

y

z

#

$

% % %

&

'

( ( (

& B =

11

"5

"3

#

$

% % %

&

'

( ( (

Or

!

| A| = "1

Getting adj

!

A =

0 2 1

"1 "9 "5

2 23 13

#

$

% % %

&

'

( ( (

T

=

0 "1 2

2 "9 23

1 "5 13

#

$

% % %

&

'

( ( (

Writing

!

X = A"1B =1

| A| (adj A) B

Getting x = 1, y = 2 and z = 3

1 2 1 1

41. Getting

!

y = 4x2+12x +15 " x =

y # 6 # 3

2

Stating

!

g(y) =y " 6 " 3

2, y # S

or

!

g(x) = x " 6 " 3

2, x # S

Proving

!

gof (x) = 9(4x2+12x +15) = x and

writing

!

gof = IN

Proving

!

fog(y) = fy " 6 " 3

2

#

$ % %

&

' ( ( = y, y ) S

Or

!

fog(x) = fx " 6 " 3

2

#

$ % %

&

' ( ( = x, x ) S and writing

!

fog = IS

Writing

!

f "1(x) = x " 6 " 3

2or f "1 =

x " 6 " 3

2

1 1 1 1 1

Code No. 35 (N/S) 8

Qn.

No.

Marks

42. Figure

Writing

!

dx

dt= "3 cm/min and

!

dy

dt= 2 cm/min

Writing

!

P = 2x + 2y and getting

!

dP

dt= "2 cm/min

Writing

!

A = xy and getting

!

dA

dt= x

dy

dt+ y

dx

dt

Getting

!

dA

dt= 2 cm

!

2/min

1

1

1

1

1

43. Getting

!

dy

dx=

2sin"1 x

1" x2

Writing

!

1" x2 dy

dx= 2sin

"1 x

Getting

!

1" x2 d2y

dx2"

x

1" x2

dy

dx=

2

1" x2

Getting

!

(1" x2)d

2y

dx2" x

dy

dx= 2

1

1

1 + 1

1

44. Putting

!

x = a tan" # dx = asec2" d"

Getting

!

I =1

a1d"#

=

!

1

atan

"1 x

a

#

$ %

&

' ( + c

Writing

!

I =1

(x +1)2

+ 2

dx"

=

!

1

2

tan"1 x +1

2

#

$ %

&

' ( + c

1

1

1

1

1

9 Code No. 35 (N/S)

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Qn.

No.

Marks

45. Figure :

Getting equations of the sides

!

AB : y = 2(x "1), BC : y = 4 " x and AC : y =1

2(x "1)

Writing area =

!

2(x "1)dx + (4 " x)dx "1

2(x "1)dx

1

3

#2

3

#1

2

#

=

!

2 x

2

2" x

#

$ % %

&

' ( ( 1

2

+ 4x "x

2

2

#

$ % %

&

' ( ( 2

3

"1

2

x2

2" x

#

$ % %

&

' ( ( 1

3

Getting area =

!

3

2sq units

1 1

1 1 1

46. Figure :

Getting

!

AP " N and

!

AP . N = 0

Getting

!

( r " a ) . N = 0

Writing

!

a = x1 i"

+ y1 j"

+ z1 k"

, r = x i"

+ y j"

+ zk"

!

N = A i"

+ B j"

+ Ck"

and writing

!

[ (x " x1) i#

+ (y " y1) j#

+ (z " z1)k#

] [ A i#

+ B j#

+ Ck#

] = 0 Getting

!

A(x " x1) + B(y " y1) + C(z " z1) = 0

1

1

1

1

1

Code No. 35 (N/S) 10

Qn.

No.

Marks

47.

Writing

!

q =1

5, p =

4

5, n = 5

!

P(x = r) = 5Cr

4

5

"

# $ $

%

& ' '

r1

5

"

# $ $

%

& ' '

5(r

writing

!

P(x " 4) = P(x = 4) + P(x = 5)

getting

!

P(x " 4) =9.4

4

55

getting

!

P(x " 3) = 1# P(x $ 4) = 1#9.4

4

55

1 1 1 1 1

48. Writing

!

dx

dy+

x

y= ey

Getting

!

IF = e

1

ydy"

= elog

ey= y

Writing general solution

!

xy = eyydy"

Getting

!

xy = yey " eydy + c1#

!

xy = yey" ey

+ c

1

1

1 1

1

PART-E V.

49. a) Drawing graph

Getting corner points

!

A(60, 0), B(120, 0), C(60, 30) and

!

D(40, 20) Getting corresponding value of z at each corner point At A, Z = 300 At B, Z = 600, At C, Z = 600 and At D, Z = 400 Writing maximum value of Z is 600 at ( 120, 0 ) and ( 60, 30 ) Minimum value of Z is 300 at ( 60, 0 )

2 1 1 1 1

11 Code No. 35 (N/S)

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Qn.

No.

Marks

b) Writing

!

x"0

lim f (x) = f (o) = k

getting

!

x"0

lim

2sin2

x

2sin2 x

2

#

$ %

& %

'

( %

) %

= k

=

!

x"0

lim

sin2

x

x2

sin2

x

2

x2/4

# 4

$

%

& &

'

& &

(

)

& &

*

& &

= k

getting k = 4

1 1 1 1

50. a) Writing

!

I = f (x)dx = f (x)dx + f (x)dx

a

2a

"0

a

"0

2a

"

!

I = I1 + I2

getting

!

I2 = f (x)dx = f (2a " x)dx

0

a

#a

2a

#

writing

!

I = f (x)dx + f (2a " x)dx

0

a

#0

a

#

=

!

2 f (x) dx, f (2a " x) = f (x)

0

a

#

= 0,

!

f (2a " x) = " f (x)

writing

!

I = 2 cos5 xdx, f (2" # x) = f (x)

0

"

$

= 0,

!

f (" # x) = # f (x)

1 1

1 1 1 1

b) Getting LHS =

!

0 0 1

a " b b " c c

a3" b

3b

3" c

3c3

Getting LHS =

!

(a " b) (b " c)

0 0 1

1 1 c

a2

+ ab + b2

b2

+ bc + c2

c3

Getting LHS =

!

(a " b) (b " c) b2 + bc + c

2" a

2" ab " b

2[ ] Getting LHS =

!

(a " b) (b " c) (c " a) (a + b + c)

1 1 1 1