xii cbse 2016 maths solutions

8/18/2019 XII CBSE 2016 Maths Solutions

1/21

® CBSE XII EXAMINATION-2016

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CLASS-XII / (CBSE)

PAGE # 1

SOLUTION 2 15 16

CBSE 12 th Board (MATHEMATICS) SET-1

P.T.O.

mailto:[email protected]://www.pccpr.esonance.ac.in/


2/21


3/21





CLASS-XII / (CBSE)

PAGE # 2

Ans 1. A = !"#

$%&' ! !

! !

cossinsincos

0 < ( <2"

A + AT = 2 I2

!"#

$%&)!

"#

$%& '*!

"#

$%&' 10

012

cossinsincos

cossinsincos

! !

! !

! !

! !

!"#

$%&

!

!

cos200cos2

= !!"

#$$%

&20

02

2 cos ( = 2

cos ( =2

122 )

( = 4"

Ans 2. | 3 A | = k | A || 3 A | = 27 | A |

k = 27

Ans 3. for unique solution | A | + 0

023

112111

+'k

C2 , C 2 – C 1 ; C 3 , C 3

– C 1

0313

312001

+''

''k

expansion along R 1 – (k – 3) – 3 + 0 – k + 3 – 3 + 0k + 0

Ans 4. - . 05ˆˆˆ2 )''*/,

k j i r

in Cartesian form

2x + y – z – 5 = 0

2x + y – z = 5

1555

2 )'* z y x

P.T.O.

mailto:[email protected]://www.pccp.resonance.ac.in/


4/21





CLASS-XII / (CBSE)

PAGE # 3

P.T.O.

1552 / 5

)'

** z y x

Intercept cutt of on the axes 0 1 23

4 5 ' 5,5,

25

1)**c z

b y

a x

a =25

b = 5 c = – 5

a + b + c = 5/2

Ans 5. - . - .0ˆˆˆ3ˆ9ˆ3ˆ !)*'6** k j i k j i # $

03

931

ˆˆˆ

!)

' # $

k j i

- . - . - . 09ˆ27ˆ93ˆ !

)''*''* $ # $ # k j i 37 + 9 8 = 0 ––– (1) 27 – 7 = 0 ––– (2) – 8 – 9 = 0 ––– (3)by eq n (2) & (3) 7 = 27 8 = – 989 7 value satisfy the eq n (1)So 7 = 27, 8 = – 9

Ans 6. k j i a ˆˆˆ

4 *')!

, k j i b ˆˆ2

ˆ2 *')

!

- . - .k j i k j i b a ˆˆ2ˆ2ˆˆˆ4 *'**')* !!

= k j i ˆ

2ˆ

3ˆ

6 *'unit vector parallel to - .

|| b a b a

b a !!

!!!!

**)*

=4936

ˆ2

ˆ3

ˆ6

***' k j i

=49

ˆ2

ˆ3

ˆ6 k j i *'

= k j i ˆ

72ˆ

73ˆ

76 *'

Ans 7. tan – 1 (x – 1) + tan – 1x + tan – 1(x + 1) = tan – 13x

tan – 1(x – 1) + tan

– 1 (x+1) = tan – 13x – tan

– 1x

tan – 1

- .- .00 1 2

334 5

*''**'

11111

x x x x

= tan – 1 0

1 23

4 5

*'

231

3

x

x x

- . 22 312

112

x x

x x

*)

''x (1 + 3x 2 ) = x (2 – x2)

x (1 + 3x 2 – 2 + x 2) = 0



5/21





CLASS-XII / (CBSE)

PAGE # 4

P.T.O.

x (4x 2 – 1) = 0

x = 0 ; 4x 2 – 1 = 0

x2 =4

1

x = ±21

x = 0, ±21

OR

L.H.S.

0 1 23

4 5

''00 1

2334

5 '

' ''2

12

31

414

tan121

86tan

x x

x x x

00000

1

2

33333

4

5

'6

''*

''

''

'

22

3

22

3

1

414

12186

1

414

12186

tan

x x

x x x

x x

x x x

00 1

2334

5 '*''

'''''x x x x x

x x x x x 4)86()41()121(

)121(4)41()86(tan 322

2231

00 1

2334

5 '**''*'*'''

42422

35331

3224481241484328246

tanx x x x x x x x x x x

00 1 2334

5 ****'

181621632tan 24

351

x x x x x

;<;=>

;?;@A

****'

1816)1816(

2tan 2424

1

x x x x

x

tan – 1 2x

L.H.S. = R.H.S.

Ans 8. [10 3] !"#

$%&y x

= [145]

[3 10] !"#

$%&y x

= [180]

10x + 3y = 1453x + 10y = 180by solving the equations we getx = 10, y = 15but Typist charge 2 Rs. Per Page from a Poor student shyamamount taken by shyam = 2 × 5 = 10 Rs.but from another person, he take for



6/21





CLASS-XII / (CBSE)

PAGE # 5

P.T.O.

5 Pages = 15 × 5= 75 Rs.

amount differ by = 75 – 10= 65 Rs. Less.

sympathy are reflect this problem

Ans 9. at x = 0 function is continuous,)0()(lim)(lim

00f x f x f

x x ))

'* ,,

R.H.L.

= 11

1111lim)0(lim)(lim

000 ****6'*)*)

,,, * h b

h b h

h b h f x f

h x x

= 0lim,x - .11

11

**'*

bh h

bh

= 0lim,x - .11 ** bh h

bh

= 0lim,x 11 ** bh

b

=2b

f(0) = 2

L.H.L.

)(lim0 x f x ', = )0(lim0 h f h ',

= 0lim,h h

h h a '

'*'*0

)0sin()0()1sin(

= 0lim,h

- .h

h h a '

'**' )sin()1(sin

= 0lim,h h

h a )1(sin *+

h h sin2

= (a+1) + 2

a + 3 = 22b

= 2

a = – 1 b = 4

Ans 10. x cos (a+y) = cos y

x = )(coscos

y a y *

dy dx

=- .

)(cos)sin(cos)sin()cos(

2 y a y a y y y a

**'''6*



7/21





CLASS-XII / (CBSE)

PAGE # 6

P.T.O.

dy dx

= )(cos)sin(cos)cos(sin

2 y a y a y y a y

****'

" sin (A – B) = sin A cos B – cos A sin B

dy dx = )(cossin2 y a

a * So dx dy = a y a

sin)(cos

2

*

sina 22

dx y d

= 2cos(a + y) × ( – sin (a+y)) dx dy

sina 22

dx y d

+ 2cos(a + y) × (sin (a+y)) dx dy

= 0

sina 22

dx

y d + sin2 (a+y))

dx dy

= 0

OR

y = 00 1

2334

5 '''

54146sin

21 x x

y = 0 1 23

4 5 ''' 21 41

54

56

sin x x

y = - . 0 1 23

4 5 ''6' 21 21

54

53

2sin x x

sin – 1 p – sin

– 1 q = sin – 1 0 1

234 5 ''' 22 11 p q q p

p = 2x q = 54

y = sin – 1 2x – sin

– 154

dx dy

= 01.2)2(1

12

'6' x

dx dy

= 241

2

x '

Ans 11. y = x 3 + 2x – 4

1C dx

dy 0 1

234

5 = 3x 2 + 2 ..........(1)

eq n of tangent : y – y1 = m(x – x1) ......(2)

is B to x + 14y + 3 = 0

m =1C

dx dy 0

1 23

4 5

= 3x 2 + 2



8/21





CLASS-XII / (CBSE)

PAGE # 7

P.T.O.

14y = – x – 3

y =14

3'' x

m × 114

1 ')'

m = 14

3x2 + 12 = 14

3x2 = 12

x2 = 14 x = ± 2if x = 2 x = – 2

y = 2 3 + 2.2 – 4 y = – 8 – 4 – 4y = 8 + 4 – 4 y = – 16y = 8

P 1 (2, 8) P 2 ( – 2, – 16)

eq n of tangent at P 1 (2, 8)y – 8 = 14 (x – 2)y – 8 = 14x – 2814x – y = 20

eq n of tangent at P 2 ( – 2, – 16)

y + 16 = 14 (x + 2)14x – y = 16 – 2814x – y = – 12

Ans 12. I = dx x

e x x

C ''

3

2

)32(

)52(

= dx x x

e e x C !!"#

$$%

&'

''/'3

332

)32(2)32(

= dx x x e e x C !!"

#$$%

&'

''

/'32

332

)32(2

)32(1

Let f(x) = 2)32(1'x whose, 2x

– 3 = t

D 2dx = dt

I = 22

33 dt t

t e e t !"

#$%& '

C

= E F2

)()( 13 dt

t f t f e e t C *

= )(2

3

t f e e t // + C

= C x

e e x *

'/'

232

3

)32(1

2



9/21





CLASS-XII / (CBSE)

PAGE # 8

P.T.O.

= C x e x *

' 22

)32(2

OR

I = dx x x x x C **

**)2()1(

12

2

Let )2()1(1

2

2

****x x

x x = )1()2( 2 *

*** x

C Bx x

A

= )1()2()2)(()1(

2

2

******

x x x C Bx x A

= )1()2()2()2()(

2

2

*******

x x AC x C B x B A

G A + B = 1 ........(1)2B + C = 1 ........(2)2C + A = 1 ........(3)

D A = 1 – 2C ........(4)From (2)

B =2

1 C '........(5)

Put (4) and (5) in (1) we get.

1 – 2C +2

1 C ' = 1

D 2 – 4C + 1 – C = 2D 5C = 1

D C = 51

G A = 1 – 2C = 1 – 52

= 53

B =2

1 C ' =

251

1 ' =

254

= 52

G )1()2(

1

2

2

**

**

x x

x x

= )2(

5 / 3

*x + )1(51

52

2 *

*

x

x

G I = C ****

dx x x x x

)1()2(1

2

2

D I = dx x

x

x )1(51

52

)2(5 / 3

2 *

**

*C



10/21





CLASS-XII / (CBSE)

PAGE # 9

P.T.O.

D I = !"#

$%&

**

*** C C dx

x dx

x x

x 1

11

251

)2(log53

22

D I = - . C x x x ***** ' 12 tan1log5

1)2(log

5

3

Ans 13. I = C' *

2

2

2

51dx

x x ...........(1)

Using Property Cb

a dx x f )( = C '*

b

a dx x b a f )(

I =- .C

' '*'2

2

2

51dx

x x

D I = dx x

x

x

C' *

2

2

2

515

.......(2)

Add (1) and (2), we get

2I = C'

/**2

2

2x

x

dxx5151

= C'

2

2

2 dxx

=

2

2

3

3 '!!"

#$$%

&x

=3

1 [ 2 3 – ( – 2) 3 ]

= 31

[ 8 – ( – 8) ]

2I = 316

G I = 38

Ans 14. I = dx x x x 243)3( ''*CLet x+3 = 8 # *'' )43( 2x x

dx d

D x+3 = 8 ( – 2x – 4) + 7

D x+3 = – 28 x – 48 + 7

G – 28 = 1

D 8 = – 21

– 48 + 7 = 3

D – 4 321 )*0 1 23

4 5 ' #



11/21





CLASS-XII / (CBSE)

PAGE # 10

P.T.O.

D 2 + 7 = 3

D 7 = 1

G I = dx x x x x dx d 22 431)43(

21 ''!"

#$%& *'''C

= –21

C C ''*'''' dx x x dx x x x x dx d 222 4343)43(

= – 21

23

)43( 2 / 32x x ''+ dx x x C *''' 4443 2

= – 3

)43( 2 / 32x x ''+ dx x

C *' 2)2(7

= – 3

)43( 2 / 32x x ''+

22*x

2)2(7 *' x + 27

00 1 2

334 5 *'

7

2sin 1

x +C

Ans 15. dx dy

= – x x y x

sin1cos

**

Ddx dy

+ x x

y x

x sin1sin1

cos*

')* ...(1)

This is a linear differential equation with

x x

P sin1cos*) , Q = x

x sin1 *'

G I.F. = dx x x

e sin1cos*C

= e log(1+sinx)

= (1 + sinx)

Multiplying both sides of (i) by I.F. = 1+sinx , we get

(1 + sinx) dx dy

+ y cosx = – x

Integrating with respect to x, we get

y (1 + sinx) = C dx x *'C

D y =)sin1(2

2 2

x x C

*'

......(2)

Given that y = 1 when x = 0



12/21





CLASS-XII / (CBSE)

PAGE # 11

G 1 = )01(22

*C

D C = 1 ......... (3)

G Put (3) in (2) , we get

y = )sin1(22 2

x x

*'

Ans 16. 2ye x/y dx + (y – 2x e x/y) dy = 0

D dy dx

= y x y x

e y y e x

/

/

22 '

The given D.E. is a homogeneous differential equation.

G Put x = vy

dy dx

= v + y dy dv

v + y dy dv

= v v

e ve 2

12 '

D y dy dv

= v e

ve v

v

''2

12

D y dy dv

= – v e 21

D 2ye v dv = – dy

D 2e v dv = – dy y 1

D 2 C dv e v = – C dy y 1

D 2e v = – log | y | + log C

D 2e v = log y c

D 2e x/y = log y c

Given that at x = 0 , y = 1.

G1

log2 0 c

e )/

D C = e 2

G 2e x/y = log y e 2

P.T.O.



13/21





CLASS-XII / (CBSE)

PAGE # 12

D log y = – 2e x/y + 2

D y =y x e e

/ 22 '

Ans 17. A (4, 5, 1) B (0, – 1, – 1) C (3, 9, 4) D ( – 4, 4, 4)

G - .k j i AB ˆ2ˆ6ˆ4 '''),

- .k j i AC ˆ3ˆ4ˆ **'),

- .k j i AD ˆ3ˆˆ8 **'),

G !"#

$%& ,,,

ADACAB )318341264

'''

'''

= 1841

23831

63134

4 '''

'''

*''

= – 4 ( 12 + 3 ) + 6 ( – 3 + 24) – 2 (1+32)

= – 60 + 126 – 66= 0

G Four points A, B, C, D are coplannar.

Ans 18.

P

P'(a,b,c)

A (-1, 8, 4)

eq n of Line BC12

1

12

1

12

1

z z z z

y y y y

x x x x

'')

'')

''D

$ )'')

'*)'

43

21

20 z y x

general coordinates of P

P (2 8, – 28 – 1, – 48 ×3)D.R of AP (2 8 +1, – 28 – 9, – 48 – 1)AP B BC2 (28 +1) – 2 ( – 28 – 9) – 4 ( – 48 – 1) = 048 + 2 + 4 8 + 18 + 16 8 + 4 = 024 + 24 8 = 08 = – 1

P.T.O.



14/21





CLASS-XII / (CBSE)

PAGE # 13

P ( – 2, 1, 7)

Coordinates of foot of B ( – 2 , 1, 7)Coordinaties of image of A is P ’ (a, b, c) is

221

')'a

, a = –

3

12

8 )*b , b = – 6

72

4 )*c , c = 10

P ’ ( – 3, – 6, 10)

Ans 19. bag A = 4 white, 2 blackbag y = 3 white, 3 black

E1 = first bag selected

E2 = second bag selected

P(E 1) = 21

P(E 2) = 21

P(A / E 1) = )6*6 54

62

52

64

3016

P(A / E 2) = 53

63

53

63 6*6 = 30

18

P(E 2 / A) = ) / ()() / ()() / ()(

2211

22

E AP E P E AP E P E AP E P

*/

P(E 2 /A) =3018

21

3016

21

3018

21

6*6

6

= 181618* = 34

18=

179

OR

P( win ) = 121

363 )

P ( lose ) =1211

P ( Awins ) = *60 1 23

4 5 *66*

121

1211

121

1211

1211

121

4

..........

a =121

r =144121

by using formula of infinite G.P.

P.T.O.



15/21





CLASS-XII / (CBSE)

PAGE # 14

P (Awins) = 2312

144121

1

121

)'

Ans 20. X = larger of of three numbersX = 3, 4, 5, 6

P (x = 3) = 6 × 41

51

61 66 = 20

1

P (x = 4) = 18 × 41

51

61 66 = 20

3

P (x = 5) = 36 × 41

51

61 66 = 20

6

P (x = 6) = 60 × 41

51

61 66 = 20

10

Xi Pi PiXi PiXi 2

120

320

620

1020

320

1220

3020

6020

920

4820

15020

36020

3

4

5

6

Mean =2

2

20105 0

1 23

4 5 )H i i X P = 5.25

205672 )H i i X P

Var(X) = - .22 i i i i X P X P H'H

=20

567 –

2

20

105 0 1

234

5 = 0.787

Ans 21. (a, b) * (c, d) = ( a + c, b + d )(i) Commutative

(a, b) * (c, d) = (a+c, b+d)(c, d) * (a, b) = (c+a, d+b)for all, a, b, c, d I R* is commulative on A

P.T.O.



16/21





CLASS-XII / (CBSE)

PAGE # 15

(ii) Associative : ______ (a, b), (e, d), (e, f) I A{ (a, b) * (c, d) } * (e, f)= (a + c, b+d) * (e, f)

= - .f d b e c a **** )(,)(= - .)(),( f d b e c a ****= (a*b) * ( c+d, d+f )= (a * b) { (c, a) * (e, f) }is associative on A

Let (x, y) be the identity element in A.then,

(a, b) * (x, y) = (a, b) for all (a,b) I A(a + x, b+y) = (a, b) for all (a, b) I AAa + x = a, b + y = b for all (a, b) I AA

x = 0, y = 0(0, 0) I A(0, 0) is the identity element in A.

Let (a, b) be an investible element of A.(a, b) * (c, d) = (0, 0) = (c, d) * (a, b)(a+c, b+d) = (0, 0) = (c+a, d+b)a + c = 0 b + d = 0a = – c b = – dc = – a d = – b

(a, b) is an invertible element of A, in such a case the inverse of (a, b) is ( – a, – b)

Ans 22. y = % % %

'* cos2sin4

% d dy

= 1)cos2(

sin4)cos4()cos2(2

2

'*

**%

% % %

= 1)cos2(

sin4cos4cos82

22

'*

**%

% % %

= 1)cos2(4cos82 '

**%

%

= 2

2

)cos2(coscos4%

% %

*'

% d dy = 2)cos2( )cos4(cos %

% % ' '

for increasing% d

dy > 0

0 1 23

4 5 I

2,0 "

%

0 J cos J 1(2 + cos K)2 always greator than 0

P.T.O.



17/21





CLASS-XII / (CBSE)

PAGE # 16

So,% d

dy is increaring on !"

#$%&

2,0 "

OR

Volume of cone

= h r 2

31

"

= - . - .! ! " cossin31 2 ##

l

v

0BA

= ! ! " cossin31 23#

! d dv

=3

l 3" [ – sin 3 ( + 2sin ( cos ( × cos ( ]

=3

l ! " sin3( – sin 2 ( + 2 cos 2 ( )

for maximum or minimum

0)! d

dv

3sin3 ! " l

( – sin 2 ( + 2 cos 2 ( ) = 0

sin ( + 02 cos 2 ( = sin 2 (tan 2 ( = 2

tan ( = 2

32

1cos ( = 31

( = cos – 1 3

1

again diff. w.r.t. ( , we get

2

2

! d v d

= 31

L l3 cos 2 ( (2 – 7 tan 2 ( )

at cos ( =3

1

02

2

M! d v d

V is maximum when cos ( =3

1 or ( = cos – 1

3

1

P.T.O.



18/21





CLASS-XII / (CBSE)

PAGE # 17

Ans 23.

B (4, 3)(1, 2)C

A (2,-2)

x

y1

x1

Equation of line AB : -

y + 2 = )2(2

23 '* x

D 2y = 5x – 14

Equation of line BC : -

y – 3 = )4(31 'x

D 3y = x + 5Equation of line CA : -

(y – 2) = – 4 (x – 1)4x + y = 6

G ar (N ABC)

= C'*3

2 5142

dy y

– C '3

253 dy y – C'

'2

2 46

dy y

= 575

– 25

–

424

= 20130

2050120300 )''

=2

13 sq.units

Ans 24. - . 04ˆ3ˆ2ˆ )'*'/ k j i r !- . 05ˆˆˆ2 )***'/ k j i r !

- . - .O P 054ˆˆˆ

2ˆ

3ˆ

2ˆ

)*'**'/**'/ $ $ k j i r k j i r !!

D E F 054ˆ)3(ˆ)2(ˆ)21( )*'***'*'/ $ $ $ $ k j i r !

D (1 – 28)x + ( – 2 + 8)y + (3 + 8)z = – 58 + 4

D1

345

245

2145

)

**'*

*'*'*

'*'

$ $

$ $

$ $

z y x

G$

$ $

$ *'*')

'*'

245

2145

P.T.O.



19/21





CLASS-XII / (CBSE)

PAGE # 18

D 1 – 28 = – 2 + 8D – 38 = – 3D 8 ) Q

G Equation of the required plane – x – y + 4z = – 1x + y – 4z – 1 = 0

Vector eq n of the required Plane

D - . 01ˆ4ˆˆ )''*/ k j i r !

Ans 25. Z = 0.1x + 0.09 yx + y J 50000x R 20000y R 10000y J x

(0,50000)

( 2 0 0 0

0, 3 0 0

0 0 )

( 4 0 0 0 0, 1 0 0 0 0 )

(50000,0)(20000,0)

(20000,

10000)

(0,10000)

(20000,20000)

(25000,25000)

z = 0.1 x + 0.09y

P (20000, 10000)1

P (40000, 10000)2

P (2 000, 000)3 5 25

P (2 000, 000)4 0 20

2900

4900

4750

3800

When A invest 40000 & B invest 10000 P.T.O.



20/21





CLASS-XII / (CBSE)

PAGE # 19

Ans 26.3

2

2

2

)(2)(

)()(

z y x xyz x z xy zy

xy y z zx zy zx y x

**)*

**

L.H.S.

Multipiying R 1, R 2 and R 3 by z, x, y respectively

=222

222

222

)()(

)(1

x z y xy z y y x y z x z x y z x z y x z

xyz **

*

take common z, x, y from C 1, C 2, & C3

= 222222

222

)()(

)(

x z y y x y z x z z y x

xyz xyz

**

*

C1 , C1 – C 3 and C 2 , C2

– C 3

= 72

4 )*c

taking common x+y+z from C 1 & C2

= (x + y + z) 2

2

2

2

)(0

0

x z x z y x z y x x y z z z y x

*'''''*

'*

R3 , R3 – (R 1 + R 2)

= (x + y + z) 2 xz z x

x x y z z z y x

2220

02

2

'''*

'*

C1 , zC 1 , C 2 , xC3

= xz z y x 2

)( ** xz xz xz x x y z x

z z y x z

222)(0

0)(2

2

'' '*

'*

C1 , C1 + C 3 C2 , C 2 + C 3

=xz

z y x 2)( **

xz x y z x x z z y x z

200)(

)(22

22

**

P.T.O.



21/21





CLASS-XII / (CBSE)

PAGE # 20

taking z and x common from R 1 & R2

=xz x y z x z z y x

zx xz

z y x

200

)( 2 **

6**

expanasion along R 3= (x+y+z) 2 × 2xz - .xz y z y x '** )()(= (x+y+z) 2 × 2xz (xz + xy + yz + y 2 – xz)= (x+y+z) 2 × 2xz (xy + yz + y 2)= 2xyz (x + y + z) 3

OR

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G A3 – 6A 2 + 7A + K I3 = 0

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xii cbse 2016 maths solutions

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