CBSE MATHS XII MARKING SCHEME Delhi

8/12/2019 CBSE MATHS XII MARKING SCHEME Delhi

1/371

Stri ctly Confidential __(For I nternal and Restr icted Use Only)

Senior School Certificate Examination

March 2014

Marking Scheme ---- Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3

General I nstructions :

1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The

answers given in the Marking Scheme are suggested answers. The content is thus indicative.

If a student has given any other answer which is different from the one given in the Marking

Scheme, but conveys the meaning, such answers should be given full weightage.

2. Evaluation is to be done as per instructions provided in the marking scheme. It should not

be done according to one's own interpretation or any other consideration __Marking

Scheme should be strictly adhered to and religiously followed.

3. Alternative methods are accepted. Proportional marks are to be awarded.

4. In question(s) on differential equations, constant of integration has to be written.

5. If a candidate has attempted an extra question, marks obtained in the question attempted

first should be retained and the other answer should be scored out.

6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full

marks if the answer deserves it.

7. Separate Marking Scheme for all the three sets has been given.


2/372

QUESTION PAPER CODE 65/1/1

EXPECTED ANSWERS/VALUE POINTS

SECTION - A

1-10. 1. x = 25 2.5

1x = 3. 10 4. x = 2 5. x = + 6

6. 2x3/2+ 2 x + c 7.12

8. 5 9.

3

2

10. ( ) ( ) 0kjikcjbiar =++++

or

( ) cbakjir ++=++ 110 =10 m

SECTION - B

11. AAb)(a,

a + b = b + a (a, b) R (a, b) R is reflexive 1 m

For (a, b), (c, d) AA

If (a, b) R (c, d) i.e. a + d = b + c c + b = d + a

then (c, d) R (a, b) R is symmetric 1 m

For (a, b), (c, d), (e, f) AA

If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e

Adding, a + d + c + f = b + c + d + e a + f = b + e

then (a, b) R (e, f) R is transitive 1 m

R is reflexive, symmetric and transitive

hence R is an equivalance relation m

[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} m

Q. No. Marks


3/37


4/374

= 3211 RRRR

2x2xzyx

yxz2z2z

zyxzyxzyx

++++++++

1 m

= ( ) ;

zyxzyxzyx

zyx02z

00zyx

++++++

++

133

122

CCC

CCC

2 m

= (x + y + z) .{0 .(x + y + z) + (x + y + z)2

} = (x + y + z)3

1 m

14. let ( ) xcos!!cosx,x12xcosv,x

x1tanu 121

2

1 ===

=

( ) xcos!!tantan!cos

!cos1tanu 11

2

1 ===

= 1 m

( ) ( )

=== !2

2

coscos!2sincos!cos1!cos2cosvand 1121

= xcos22

!2

2

1= 1 m

22 x1

2

dx

dv,

x1

1

dx

du==

1 m

2

1

2

x1

x1

1

dv

du2

2== 1 m

( In case, If x = sin ! then answer is2

1)

15. y = xx

log y = x log x, Taking log of both sides m

dx

dy

y

1 = log x + 1, Diff. w r t x 1 m

,x

1

dx

dy

y

1

dx

yd

y

12

22

2

=

Diff. w r t x 1 m


5/375

0x

y

dx

dy

y

1

dx

yd2

2

2

=

m

16. (x)f = 12 x3 12 x2 24 x = 12 x (x + 1) (x 2) 1+ m

(x)f > 0, ),2()0,1( Ux ++ 1 m

(x)f < 0, )2,0()1,( U x 1 m

f(x) is strictly increasing in ),2()0,1( U m

and strictly decreasing in )2,0()1,( U

OR

Point at

=22

a,

22

ais

4

! m

!cos!sin3ad!

dx!;sin!cos3a

d!

dy 22 == 1 m

slope of tangent at

4

!

2

2

4

!

!cos!sin3a

!sin!cos3a

dx

dyis

4

!

==

=

=

= 1

4

cot = 1 m

Equation of tangent at the point :

02

ayx

22

ax1

22

ay =+

= 1 m

Equation of normal at the point :

0yx

22

ax1

22

ay =

= m

17. ( ) ][

++

=+

dxxcosxsin

xcosx3sinx)cosxsin(xcosxsindx

xcosxsin

xcosxsin22

2222222

22

66

1 m

= dx3xcosxsin

122

1 0 2


6/376

+

= dx3xcosxsin

xcosxsin22

22

m

( ) += dx3xcosecxsec 22 m

= tan x cot x 3x + c 1 m

(Accept 2 cot 2x 3x + c also)

OR

( ) + dx183xx3x 2

( ) +++= dx183xx29

dx183xx32x2

1 221 m

( ) ( )

++= dx

2

9

23x

2

9183xx

3

2

2

12

22

32

1 m

( )2

9183xx

3

12

32 +=

c183xx2

3xlog

8

81183xx

2

2

3x

22 +++++

+

1 m

or ( )8

9183xx

3

12

32 +=

{ c183xx2

3xlog

2

81183xx)32( 22 ++++++x

18.dy

y1

ydxxedy

x

ydxy1e

2

x2x ==1 m

Integrating both sides

= dyy1

2y

2

1dxxe

2

x

cy1exe 2xx += 1+1 m

For x = 0, y = 1, c = 1 1y11)(xe:issolution 2x = + m


7/37


8/378

222

cba2ba

=++ 1 m

=++ b&abetweenanglebeing!49,!cosba2259 1 m

3

!

2

1

532

15!cos ==

= 1 m

21. let v5

6z

3

4y

1

2xu;

7

5z

5

3y

3

1x====

+=

+=

+

General points on the lines are

(3u 1, 5u 3, 7u 5) & (v + 2, 3v + 4, 5v + 6) 1 m

lines intersect if

3u 1 = v + 2, 5u 3 = 3v + 4, 7u 5 = 5v + 6 for some u & v 1 m

or 3u v = 3 ........... (1), 5u 3v = 7 .............. (2), 7u 5v = 11 ................... (3)

Solving equations (1) and (2), weget2

3v,

2

1u == m

Putting u & v in equation (3) , intersectlines112

35

2

17 =

m

Point of intersection of lines is :

2

3,

2

1,

2

11 m

22. let b2, g

2be younger boy and girl

and b1, g1be elder, then, sample space of two children is

S = {(b1, b

2), (g

1, g

2), (b

1, g

2), (g

1, b

2)} 1 m

A = Event that younger is a girl = {(g1, g

2), (b

1, g

2)}

B = Event that at least one is a girl = {(g1, g

2), (b

1, g

2), (g

1, b

2)}


9/379

E = Event that both are girls = {(g1, g

2)}

(i) P(E/A) =2

1

(A)P

A)(EP=

I1 m

(ii) P(E/B) =3

1

(B)P

B)(EP=

I1 m

SECTION - C

23. Here

600zyx

15003zy4x

1000z2y3x

=++=++=++ 1

BXAor

600

1500

1000

z

y

x

111

314

123

=

=

| A | = 3 ( 2) 2 (1) + 1 (3) = 5 0 BAX 1= m

Co-factors are

5A,5A,5A

1A,2A,1A3A,1A,2A

333231

232221

131211

=========

1 m

=

600

1500

1000

513

521

512

5

1

z

y

x

x = 100, y = 200, z = 300 1 m

i.e. Rs. 100 for discipline, Rs 200 for politeness & Rs. 300 for punctuality

One more value like sincerity, truthfulness etc. 1 m


10/3710

24. For correct figure m

Let radius, height and slant height of cone be r, h & l

(constant),hr 222 ll=+ m

Volume of cone (V) = hr31 2 m

( ) ( )3222 hh3

hh

3

V ll == 1 m

( )22 h33

dh

dvl= 1 m

3h0

dh

dv l== m

03

23

2h2dh

vd2

2


11/3711

26. Correct Figure 1 m

The line and circle intersect

each other at x = + 4 1 m

Area of shaded region

( ) +=24

4

22

4

0

dxx24dx x 1 m

24

4

12

4

0

2

24

xsin16

2

x32x

2

x

++

= 1 m

= 8 + 484

=sq.units 1 m

27. Equation of plane through points A, B and C is

0

023

884

3z5y2x

=+

074z3y2xi.e.

05632z24y16x

=++=++ 3+1 m

Distance of plane from (7, 2, 4) =4169

7)4(4)2(3)7(2

++++ 1 m

= 29 1 m

OR

General point on the line is ( ) ( ) ( ) k2#2j4#1i3#2 +++++ 1 m

Putting in the equation of plane; we get

( ) ( ) ( ) 52#214#113#21 =++++ 1 m

0#= 1 m

Point of intersection is 2)1,(2,ork2ji2 + 1 m

Distance = ( ) ( ) ( ) 131691025112 222 ==+++++ 1 m


12/37


13/3713

Probability of defective bulb =3

1

15

5= m

Probability of a non defective bulb =3

2

3

11 = m

Probability distribution is :

81

4

81

24

81

48

81

320:P(x)x

81

1

81

8

81

24

81

32

81

16:P(x)

43210:x

m

m2

3

4

or81

108

P(x)xMean == 1 m


14/3714



SECTION - A

1-10. 1. 5 2. ( ){ }( ) 0kjikcjbiar =++++ or

( ) cbakjir ++=++

3. 2x3/2+ 2 x + c 4. 10 5. x = 2

6. x = + 6 7.5

1x = 8. x = 25

9. c2

x

2

x2

+ 10.6

110 = 10 m

SECTION - B

11. (x)f = 12 x3 12 x2 24 x = 12 x (x + 1) (x 2) 1+ m

(x)f > 0, ),2()0,1( Ux ++ 1 m

(x)f < 0, )2,0()1,( U x 1 mf(x) is strictly increasing in ),2()0,1( U

m

and strictly decreasing in )2,0()1,( U

OR

Point at

=

22

a,

22

ais

4! m

!cos!sin3ad!

dx

!;sin!cos3ad!

dy 22

== 1 m

slope of tangent at

4

!

2

2

4

!

!cos!sin3a

!sin!cos3a

dx

dyis

4

!

==

=

=

= 14

cot = 1 m

Marks

1 0 2


15/3715


02

ayx

22

ax1

22

ay =+

= 1 m


0yx22

ax1

22

ay =

= m

12. ( ) ][

++

=+

dxxcosxsin


xcosxsin

xcosxsin22

2222222

22

66

1 m

= dx3xcosxsin1

22

+

= dx3xcosxsin

xcosxsin22

22

m




OR

( ) + dx183xx3x 2

( ) +++= dx183xx29

dx183xx32x2

1 221 m

( ) ( )

++= dx

2

9

23x

2

9183xx

3

2

2

12

22

32

1 m

( )2

9183xx

3

12

32 +=

c183xx2

3xlog

8

81183xx

2

2

3x

22 +++++

+

1 m


16/3716

or ( )8

9183xx

3

12

32 +=

{ c183xx2

3xlog

2

81183xx)32(

22 ++++++x

13. Given differential equation can be written as

( )222 1x2

y1x

2x

dx

dy=+ 1 m

Integrating factor = 1xee 21)(xlogdx

1

x222

==

x 1 m

cdx1)(x1)(x

2

1)(xyisSolution2

22

2

+= 1 m

cdx1x

121)(xy

2

2 +=

c1x

1xlog1)(xy 2 +

+= 1 m

14. y = xx

log y = x log x, Taking log of both sides m

dx

dy

y

1 = log x + 1, Diff. w r t x 1 m

,x

1

dx

dy

y

1

dx

yd

y

12

22

2

=

Diff. w r t x 1 m

0x

y

dx

dy

y

1

dx

yd2

2

2

=

m

15.

+

+

+=

+++

accbbaac,cb,ba m

=

+++

+

acccabcbba 1 m


17/3717

=

+

+

+

cbbacaabacba 1 m

+

+

acbabb

=

=

=

0abbcbbaca,aba

=

=

c,b,a2cba2 1 m

OR

=+=++ cba0cba m

222

ccba

=

=

+

m

222

cba2ba

=++ 1 m

=++ b&abetweenanglebeing!49,!cosba2259 1 m

3

!

2

1

532

15!cos ==

= 1 m

16. cot1

+++

xsin1xsin1

xsin1xsin1

= cot1

+

+

+

22

22

2

xsin

2

xcos

2

xsin

2

xcos

2

xsin

2

xcos

2

xsin

2

xcos

2 m


18/3718

2

x

2

xcotcot

2

xsin2

2

xcos2

cot 11 =

=

= 1 m

OR

+

+=

7

25sec

8

1tan

5

1tan2LHS 111

7

1tan

40

11

8

1

5

1

tan2 11 +

+= 1+ m

7

1tan

3

11

3

12

tan7

1tan

3

1tan2 1

2

111 +

=+= 1 m

RHS

4

)1(tan

25

25tan

7

1tan

4

3tan 1111 ====+= 1 m

17. AAb)(a,





For (a, b), (c, d), (e, f) AA




19/3719




[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} m

18. let b2, g


and b1, g

1be elder, then, sample space of two children is

S = {(b1, b

2), (g

1, g

2), (b

1, g

2), (g

1, b

2)} 1 m


2), (b

1, g

2)}


2), (b

1, g

2), (g

1, b

2)}


2)}

(i) P(E/A) =2

1

(A)P

A)(EP=

I1 m

(ii) P(E/B) =

3

1

(B)P

B)(EP=

I1 m

19. LHS =

2bacac)b(a2

b2acbc)b(a2

bac)b(a2

++++++++

++

3211 CCCC

Using,

++ 1 m

=

cba00

0cba0

bac)b(a2

++

++++

133122 RRR;RRR

Using,

2 m

= 2 (a + b + c) {(a + b + c)2

0} Expanding along C1

= 2 (a + b + c)3

= RHS 1 m


20/3720

20. let u = ( ) xsin!!sinx;x1x2sinv;x1

xtan 121

2

1 ===

( ) xsin!!tantan!sin1

!sintanu 11

2

1

===

= 1 m

( ) xsin2!2!2sinsinx12xsinv& 1121 ==== 1 m

22 x1

2

dx

dv,

x1

1

dx

du== 1 m

2

1

2

x1

x1

1

dv

du 2

2== 1 m

( In case, if x =2

1isanswerthen!cos )

21. dxxsinxdyy

ylogyx

dx

dyylogxcosec 2

2

22 == 1 m

Integrating both sides we get

[ ]+= dxxcosx2xcosxy

1

y

ylog 2 1+1 m

+= dxxsin1sin xx2xcosx 2 m

cxcos2xsinx2xcosxy

1

y

ylog 2 +++= m

22. Equations of lines are :

35z

14y

78x;

53z

47y

45x ==+== 1 m

Here,

3c1,b7,a;5c4,b4,a

5z4,y8,x;3z7,y5,x

222111

222111

============


21/3721

317

544

833

cba

cba

zzyyxx

222

111

121212

== 3(17) + 3 (47) + 8 ( 24) = 0 1+1 m

lines are co-planar m

SECTION - C

23. Let x and y be electronic and

manually operated sewing machines purchased respectively

L.P.P. is Maximize P = 22x + 18y m

subject to 360x + 240y < 5760

or 3x + 2y < 48

x + y < 20 2 m

x > 0, y > 0

For correct graph 2 m

vertices of feasible region are

A (0, 20), B(8, 12), C(16, 0) & O(0, 0)

P(A) = 360, P(B) = 392, P(C) = 352 m For Maximum P, Electronic machines = 8

1 m

Manual machines = 12

24. Let E1: Event that lost card is a spade

E2: Event that lost card is a non spade

m

A : Event that three spades are drawn without replacement from 51 cards

43

411)P(E,

41

5213)P(E 21 ==== 1 m

3

51

3

13

2

3

51

3

12

1C

C)P(A/E,

C

C)P(A/E == 1 m


22/3722

3

51

3

13

3

51

3

12

3

51

3

12

1

C

C

4

3

C

C

4

1

C

C

4

1

/A)P(E

+

=

1+1 m

49

10= 1 m

OR

X = No. of defective bulbs out of 4 drawn = 0, 1, 2, 3, 4 1 m

Probability of defective bulb =3

1

15

5= m

Probability of a non defective bulb =3

2

3

11 = m

Probability distribution is :

81

4

81

24

81

48

81

320:P(x)x

81

1

81

8

81

24

81

32

81

16:P(x)

43210:x

m

m2

3

4or

81

108P(x)xMean == 1 m

25. Here

600zyx

15003zy4x

1000z2y3x

=++=++=++ 1

BXAor

600

15001000

z

yx

111

314123 =

=

| A | = 3 ( 2) 2 (1) + 1 (3) = 5 0 BAX 1= m

Co-factors are


23/3723

5A,5A,5A

1A,2A,1A

3A,1A,2A

333231

232221

131211

=========

1 m

=

600

1500

1000

513

521

512

5

1

z

y

x

x = 100, y = 200, z = 300 1 m



26. Equation of plane through points A, B and C is

0

023

884

3z5y2x

=+

074z3y2xi.e.

05632z24y16x

=++=++ 3+1 m

Distance of plane from (7, 2, 4) =4169

7)4(4)2(3)7(2++

++ 1 m

= 29 1 m

OR

General point on the line is ( ) ( ) ( ) k2"2j4"1i3"2 +++++ 1 m

Putting in the equation of plane; we get

( ) ( ) ( ) 52"214"113"21 =++++ 1 m

0"= 1 m

Point of intersection is 2)1,(2,ork2ji2 + 1 m

Distance = ( ) ( ) ( ) 131691025112 222 ==+++++ 1 m


24/37


25/3725

V (volume) = h,r3

2[V is constant]

A = ( )22222222 hrrrAz,r +=== ll m

+=

42

2222

r

9vrr

+=

22

242

r

9vr 1 m

=

32

232

r

18v4r

dr

dz1 m

6

2

2

2

9vr0

dr

dz== m

0r

54v12r

dr

zd;

2

9vrAt

42

222

2

2

62

2

>

+== 1 m

corved surface area is minimum iff 262 9vr2 =

i.e. 24262 hrr2 =

OR

r2h = m

( )2cot#2r

h#cot 1=== m


26/3726



SECTION - A

1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x3/2+ 2 x + c

5.5

1x = 6. x = 25 7. 5

8. ( ) ( ) 0kjikcjbiar =++++ 9. 1

or

( ) cbakjir ++=++

10. k13

12j

13

3i

13

4+ 110 = 10 m

SECTION - B

11.

+

+

+=

+++

accbbaac,cb,ba m

=

+++

+

acccabcbba 1 m

=

+

+

+

cbbacaabacba 1 m

+

+

acbabb

=

=

=

0abbcbbaca,aba

=

=

c,b,a2cba2 1 m

OR

Marks


27/3727

=+=++ cba0cba m

222

ccba

=

=

+

m

222

cba2ba

=++ 1 m

=++ b&abetweenanglebeing49,cosba2259 1 m

3

!

2

1

532

15cos ==

= 1 m

12. Given differential equation can be written as

( )222 1x2

y1x

2x

dx

dy=+ 1 m

Integrating factor = 1xee 21)(xlogdx

1

x222

==

x 1 m

cdx1)(x1)(x

2

1)(xyisSolution

2

22

2

+= 1 m

cdx1x

121)(xy

2

2 +=

c1x

1xlog1)(xy 2 +

+= 1 m

13. ( ) ][

++

=

+dx

xcosxsin


xcosxsin

xcosxsin22

2222222

22

66

1 m

= dx3xcosxsin

122

+

= dx3xcosxsin

xcosxsin22

22

m


28/3728




OR

( ) + dx183xx3x 2

( ) +++= dx183xx29

dx183xx32x2

1 221 m

( ) ( )

++= dx

2

9

23x

2

9183xx

3

2

2

12

22

32

1 m

( )2

9183xx

3

12

32 +=

c183xx2

3xlog

8

81183xx

2

2

3x

22 +++++

+

1 m

or ( )8

9183xx

3

12

32 +=

{ c183xx2

3xlog

2

81183xx)32(

22 ++++++x

14. (x)f = 12 x3 12 x2 24 x = 12 x (x + 1) (x 2) 1+ m

(x)f > 0, ),2()0,1( Ux ++ 1 m

(x)f < 0, )2,0()1,( U x 1 m

f(x) is strictly increasing in ),2()0,1( U mand strictly decreasing in )2,0()1,( U

OR

Point at

=

22

a,

22

ais

4! m

1 0 2


29/3729

cossin3ad

dx;sincos3a

d

dy 22 == 1 m

slope of tangent at

4!

2

2

4! cossin3a

sincos3a

dx

dyis

4

!

==

=

=

= 14

!cot = 1 m


02

ayx

22

ax1

22

ay =+

= 1 m


0yx22

ax1

22

ay =

= m

15. AAb)(a,





For (a, b), (c, d), (e, f) AA






[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} m


30/37


31/3731

,x

1

dx

dy

y

1

dx

yd

y

12

22

2

=

Diff. w r t x 1 m

0x

y

dx

dy

y

1

dx

yd2

2

2

=

m

18. let b2, g


and b1, g

1be elder, then, sample space of two children is

S = {(b1, b

2), (g

1, g

2), (b

1, g

2), (g

1, b

2)} 1 m


2), (b

1, g

2)}


2), (b

1, g

2), (g

1, b

2)}


2)}

(i) P(E/A) =2

1

(A)P

A)(EP=

I1 m

(ii) P(E/B) =3

1

(B)P

B)(EP=

I1 m

19. LHS =

zzzyzx

zyyyyx

zxyxxx

zyx

1

322

232

223

++

+

33

2211

RzR

RyR,RxR

1 m

=

1zzz

y1yy

xx1x

zyx

zyx

222

222

222

+

++

m

=

1zzz

y1yy

zyx1zyx1zyx1

222

222

222222222

++

+++++++++ 3211 RRRR ++ 1 m


32/3732

=

10z

01y

00zyx1

2

2

222 +++

;133

122

CCC

CCC

1 m

= 1 + x2+ y2 + z2= RHS (Expand along C1) m

20. let x = xtantan1=

xtan2

1

2

2

tantan

tan

1tan1tanu 11

2

1 ==

=

+= 1 m

( ) xtan222sinsintan1

tan2

sinv

11

2

1

===

+= 1 m

( ) 22 x12

dx

dv;

x12

1

dx

du

+=

+= 1 m

( ) 41

2

x1

x12

1

dv

du 2

2 =+

+

= 1 m

( In case, if x =4

1isanswerthencot )

21. Differential equation can be written as : (sin y + y .cos y) dy = x .(2 .log x + 1) dx 1 m

Integrating both sides we get

c2

x

4

xxlog

2

x2ycosysinyycos

222

++

=++ 1+1 m

cxlogxysiny

2

+=At x = 1 and

y = =2

!c,

2

! solution is : y sin y = x2log x +

2

!+ m


33/3733

22. General points on the lines are

( ) ( ) ( ) ( )k13"i2"4&kj#1i3#1 ++++ 1 m

lines intersect if

"somefor3).........(113"2);.........(0#1(1);........."243#1 ==+=+ 1 m

From (2) & (3) 0"1,# == m

substituting in equation (1)

Since, 1 + 3(1) = 4 + 2 (0) is true lines interset 1 m

Point of intersection is : 1)0,(4,ork

i

4 m

SECTION - C

23. Let x and y be electronic and

manually operated sewing machines purchased respectively

L.P.P. is Maximize P = 22x + 18y m

subject to 360x + 240y < 5760

or 3x + 2y < 48

x + y < 20 2 m

x > 0, y > 0

For correct graph 2 m

vertices of feasible region are

A (0, 20), B(8, 12), C(16, 0) & O(0, 0)

P(A) = 360, P(B) = 392, P(C) = 352 m

For Maximum P, Electronic machines = 81 m

Manual machines = 12

24. Let E1: Event that lost card is a spade

E2: Event that lost card is a non spade

m


34/37


35/37


36/3736

27. Here

600zyx

15003zy4x

1000z2y3x

=++=++=++ 1

BXAor

600

1500

1000

z

y

x

111

314

123

=

=

| A | = 3 ( 2) 2 (1) + 1 (3) = 5 0 BAX 1= m

Co-factors are

5A,5A,5A

1A,2A,1A

3A,1A,2A

333231

232221

131211

======

===1 m

=

600

1500

1000

513

521

512

5

1

z

y

x

x = 100, y = 200, z = 300 1 m



28. le I =( )

+=+2

!

0

44

2!

0

44dx

xsinxcos

xsinxcosx2

!

I;dxxcosxsin

xcosxsinx1 m

Adding we get, 2 I =( ) +=+

2!

0

22

22!

0

44dx

xtan1xsecxtan2

4!;dx

xcosxsinxcosxsin

2! 2 m

= ( )8

!tantan

4

! 22!

0

21 =

x 2 m

16

!I

2

= 1 m


37/37

29. let r and h be the radius and height of the cylinder then,

Volume of cylinder (V) 128!hr! 2 = 22

r

128

r!

128!h == 1 m

Surface area of cylinder = )rh(r!2rh!2r2! 22 +=+ 1 m

=

+=

2

2

r

1282r2!

dr

ds

r

128r2!S 1 m

= 0dr

ds r3 = 64 or r = 4 m

012!

64

25622!

r

25622!

dr

sd;4rAt

32

2

>=

+=

+== 1 m

surface area is minimum at r = 4 cm ; h = 8 cm 1 m

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CBSE MATHS XII MARKING SCHEME Delhi

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