Sample Question Paper (With value based questions)

Issued by CBSE for 2013 Examination

MathematicsClass–XII

Blue Print

S. No. Topics VSA SA LA Total

1. (a) Relations and Functions — 4(1) — —

(b) Inverse Trigonometric Functions 2(2) 4(1) — 10(4)

2. (a) Matrices 2(2) — 6(1) —

(b) Determinants 1(1) 4(1) — 13(5)

3. (a) Continuity and Differentiability 1(1) 12(3) — —

(b) Applications of Derivatives — — 6(1) 44(11)

(c) Integration — 12(3) — —

(d) Application of Integrals — — 6(1) —

(e) Differential Equations 1(1) — 6(1) —

4. (a) Vectors 2(2) 4(1) — —

(b) 3-dimensional Geometry 1(1) 4(1) 6(1) 17(6)

5. Linear Programming — — 6(1) 6(1)

6. Probability — 4(1) 6(1) 10(2)

Total 10(10) 48(12) 42(7) 100(29)

Note:

— Number of questions are given within brackets and marks outside the brackets.

— The Question Paper will include question(s) based on values to the extent of 5 marks.

Time allowed: 3 hours Maximum marks: 100

¡ General Instructions:

1. All questions are compulsory.

2. The question paper consists of 29 questions divided into three Sections A, B and C. Section Acomprises of 10 questions of one mark each; Section B comprises of 12 questions of four marks each;and Section C comprises of 7 questions of six marks each.

3. All questions in Section A are to be answered in one word, one sentence or as per the exactrequirement of the question.

4. There is no overall choice. However, internal choice has been provided in 4 questions of four markseach and 2 questions of six marks each. You have to attempt only one of the alternatives in all suchquestions.

5. Use of calculator is not permitted. You may ask for logarithmic tables, if required.

SECTION–A

Question numbers 1 to 10 carry 1 mark each.

1. Using principal values, write the value of 21

23

1

2

1 1cos sin- -+ .

2. Evaluate tan cos sin- -æèç ö

ø÷

é

ëêù

ûú1 12 2

1

2 .

3. Write the value of x y z+ + , if

1 0 0

0 1 0

0 0 1

1

1

0

é

ë

êêê

ù

û

úúú

é

ë

êêê

ù

û

úúú

= -

é

ë

êêê

ù

û

úúú

x

y

z

.

4. If A is a square matrix of order 3 such that |adj A| = 225, find |A’|

5. Write the inverse of the matrix cos sin

sin cos

q q

q q-

é

ëê

ù

ûú.

6. The contentment obtained after eating x-units of a new dish at a trial function is given by theFunction C(x) = x3 + 6x2 + 5x + 3. If the marginal contentment is defined as rate of change of(x) with respect to the number of units consumed at an instant, then find the marginalcontentment when three units of dish are consumed.

7. Write the degree of the differential equation d y

dx

d y

dx

dy

dx

2

2

2

22 1 0

æ

èçç

ö

ø÷÷ - - + = .

CBSE Sample Question Paper–2013(with Value Based Questions)

8. If a®

and b®

are two vectors of magnitude 3 and 2

3, respectively such that a b

® ®´ is a unit vector,

write the angle between a®

and b®

.

9. If a i j k®

= + -7 4$ $ $ and b i j k®

= + +2 6 3$ $ $, find the projection of a®

on b®

.

10. Write the distance between the parallel planes 2 3 4x y z- + = and 2 3 18x y z- + = .

SECTION–B

Question numbers from 11 to 22 carry 4 marks each.

11. Prove that the function f : N ® N, defined by f x x x( ) = + +2 1 is one – one but not onto.

12. Show that sin [cot {cos(tan )}]- - =+

+

1 12

2

1

2x

x

x

OR

Solve for x: 32

14

1

12

2

1

1

2

12

2

1sin cos tan- - -

+

æ

èç

ö

ø÷ -

-

+

æ

èç

ö

ø÷ +

-

x

x

x

x

x

x2 3

æ

èç

ö

ø÷ =

p .

13. Two schools A and B decided to award prizes to their students for three values: honesty(x), punctuality (y) and obedience (z). School A decided to award a total of `11,000 forthese three values to 5, 4 and 3 students, respectively, while school B decided to award`10,700 for these three values to 4, 3 and 5 students, respectively. If all the three prizestogether amount to `2,700 then

(i) Represent the above situation by a matrix equation and form Linear equations byusing matrix multiplication.

(ii) Is it possible to solve the system of equations so obtained using matrices?

(iii) Which value you prefer to be rewarded most and why?

14. If x a= -( sin )q q and y a= -( cos )1 q , find d y

dx

2

2.

15. If yx

x=

-

-sin 2

21, show that ( ) .1 3 02

2

2- - - =x

d y

dxx

dy

dxy

16. The function f x( ) is defined as f x

x ax b x

x x

ax b x

( )

,

,

,

=

+ + £ <

+ £ £

+ < £

ì

íï

îï

2 0 2

3 2 2 4

2 5 4 8

. If f x( ) is continuous on [0, 8],

find the values of a and b.

OR

Differentiate tan - + - -

+ + -

é

ë

êê

ù

û

úú

12 2

2 2

1 1

1 1

x x

x x with respect to cos-1 2x .

CBSE Sample Question Paper (iii)

17. Evaluate: x x

xdx

3

2

1

1

+ +

-ò .

OR

Evaluate: ex

xdxx ( – sin )

( cos ).

1

1 -ò

18. Evaluate: 2

1 22 2

x

x xdx

( )( ).

+ +ò

19. Evaluate: log( tan ) ,1

0

4

+ò x dx

p

using properties of definite integrals.

20. Let a i j k®

= + -4 5$ $ $, b i j k®

= - +$ $ $4 5 and c i j k®

= + -3$ $ $. Find a vector d®

which is perpendicular

to both a®

and b®

and satisfying d c® ®

=. 21.

21. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, –1, 2),B(5, 2, 4), and C(–1, –1, 6)

OR

Find the equation of the perpendicular drawn from the point P(2, 4, –1) to the line x y z+

=+

=-

-

5

1

3

4

6

9. Also, write down the coordinates of the foot of the perpendicular from

P to the line.

22. There is a group of 50 people who are patriotic out of which 20 believe in non violence.Two persons are selected at random out of them, write the probability distribution for theselected persons who are non violent. Also find the mean of the distribution. Explain theimportance of non violence in patriotism.

SECTION–C

Question numbers from 23 to 29 carry 6 marks each.

23. If A =

-

-

-

é

ë

êêê

ù

û

úúú

1

2

3

2

3

3

3

2

4

, find A-1. Hence solve the following system of equations:

x y z+ - = -2 3 4; 2 3 2 2x y z+ + = and 3 3 4 11x y z- - =

24. Find the equations of tangent and normal to the curve yx

x x=

-

- -

7

2 3( )( ) at the point where it

cuts the x-axis.

OR

Prove that the radius of the base of right circular cylinder of greatest curved surface areawhich can be inscribed in a given cone is half that of the cone.

25. Find the area of the region which is enclosed between the two circles x y2 2 1+ = and

( )x y- + =1 12 2 .

(iv) Xam idea Mathematics–XII

26. Find the particular solution of the differential equation:

( sin ) (tan )x y dy y dx- + = 0: given that y = 0 when x = 0.

27. Find the vector and Cartesian equations of the plane containing the two lines

r i j k i j k®

= + - + + +( $ $ $) ($ $ $)2 3 2 5l and r i j k i j k®

= + + + - +( $ $ $) ( $ $ $)3 3 2 3 2 5m .

28. A dealer in rural area wishes to purchase a number of sewing machines. He has only`5,760.00 to invest and has space for at most 20 items. An electronic sewing machine costshim `360.00 and a manually operated sewing machine `240.00. He can sell an ElectronicSewing Machine at a profit of `22.00 and a manually operated sewing machine at a profitof `18.00. Assuming that he can sell all the items that he can buy how should he invest hismoney in order to maximize his profit. Make it as a linear programming problem andsolve it graphically. Keeping the rural background in mind justify the ‘values’ to bepromoted for the selection of the manually operated machine.

29. In answering a question on a MCQ test with 4 choices per question, a student knows the

answer, guesses or copies the answer. Let 1

2 be the probability that he knows the answer,

1

4

be the probability that he guesses and 1

4 that he copies it. Assuming that a student, who

copies the answer, will be correct with the probability 3

4, what is the probability that the

student knows the answer, given that he answered it correctly?

Arjun does not know the answer to one of the questions in the test. The evaluationprocess has negative marking. Which value would Arjun violate if he resorts to unfairmeans? How would an act like the above hamper his character development in thecoming years?

OR

An insurance company insured 2000 cyclists, 4000 scooter drivers and 6000 motorbikedrivers. The probability of an accident involving a cyclist, scooter driver and a motorbikedriver are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with anaccident. What is the probability that he is a scooter driver? Which mode of transportwould you suggest to a student and why?

SECTION–A

1. 21

23

1

22

33

6

1 1cos sin- -+ = ´ + ´p p

=7

6

p1

2. tan cos sin tan cos- - -æèç ö

ø÷

é

ëêù

ûú= æ

èç ö

ø÷

1 1 12 21

22

3

p

= ´æèç ö

ø÷

-tan 1 21

2 = tan ( )- =1 1

4

p1

CBSE Sample Question Paper (v)

SolutionsSolutionsSolutions

3. Given

1 0 0

0 1 0

0 0 1

1

1

0

é

ë

êêê

ù

û

úúú

é

ë

êêê

ù

û

úúú

= -

é

ë

êêê

ù

û

úúú

x

y

z

Þ

x

y

z

é

ë

êêê

ù

û

úúú

= -

é

ë

êêê

ù

û

úúú

1

1

0

Þ x y z= = - =1 1 0, , Þ x y z+ + = - + =1 1 0 0 1

4. We know that |adj A| = | |A n- 1 and | | | |A A= ¢

\ | |¢ =-A 3 1 225 Þ | |¢ = =A 2 225 15 1

5. Let A = cos sin

sin cos

q q

q q-

é

ëê

ù

ûú

Then, |A| = cos sin2 2 1q q+ = and adj A =-é

ëê

ù

ûú

cos sin

sin cos

q q

q q

\ A- =-é

ëê

ù

ûú

1 cos sin

sin cos

q q

q q1

6. M X C X x x( ) ( )= ¢ = + +3 12 52

M X at X( )( )= = + +3 27 36 5 = 68 units 1

7. 2 1

8. a®

× b®

= | a®

| |b®

|sin q $n

Þ | a®

× b®

| = | a®

| |b®

||sin |q Þ 1 32

3= ´ sin q

Þ sin q =1

2or q

p=

61

9. Projection of a®

on b®

= b a

b

® ®

®=

+ -

+ +=

.

| |

14 6 12

2 6 3

8

72 2 21

10. Given parallel planes are

2 3 4x y z- + = ...(i) and 2 3 18x y z- + = ...(ii)

Let P(x y z1 1 1, , ) be a point on plane (i)

\ 2 3 41 1 1x y z- + =

Now the distance 'd' from P x y z( , , )1 1 1 to plane (ii) is given as

dx y z

=- + -

+ +

2 3 18

4 1 9

1 1 1

d =-

= =´

=4 18

14

14

14

14 14

1414

1

SECTION–B

11. f x x x( ) = + +2 1

Let x y N1 1, Î such that f x f y( ) ( )1 1= 1

(vi) Xam idea Mathematics–XII

\ x x y y12

1 12

11 1+ + = + + Þ - + - =x y x y12

12

1 1 0

Þ ( )( )x y x y1 1 1 1 1 0- + + =

Þ ( )x y1 1 0- = [As x y1 1 1 0+ + ¹ any x y1 1, Î N] 1

Þ x y1 1= Þ f is one-one function

Clearly f x x x( ) = + + ³2 1 3 for x NÎ 1/2

But f x( ) does not assume values 1 and 2 11

2

\ f N N: ® is not onto function

12. cos(tan ) cos cos- -=+

æ

èçç

ö

ø÷÷

=+

1 1

2 2

1

1

1

1x

x x1

cot sin( )

sin- - -

+

æ

èçç

ö

ø÷÷

=+

+ +

æ

è

çç

ö

ø

÷÷

=1

2

12

2

11

1

1

1 1x

x

x

x

x

2

2

1

2

+

+

æ

è

çç

ö

ø

÷÷

1

\ sin cot sin sin- -

+

æ

èçç

ö

ø÷÷

é

ë

êê

ù

û

úú

=+

+

æ

è

çç

1

2

12

2

1

1

1

2x

x

x

ö

ø

÷÷

é

ë

êê

ù

û

úú

=+

+

x

x

2

2

1

21+1

OR

Let x = tan q 1/2

LHS = 3 sin -1 2

14

1

12

2

12

2

tan

tancos

tan

tantan

q

q

q

q+

æ

èç

ö

ø÷ -

-

+

æ

èç

ö

ø÷ +- -

-

æ

èç

ö

ø÷

1

2

2

1

tan

tan

q

q

\ = 3 2 4 2 2 21 1 1sin (sin ) cos (cos ) tan (tan )- - -- +q q q 11

2

= 3 2 8 4 2 2 1´ - + = = -q q q q tan x 1

Þ 23 6

1 1tan tan- -= Þ =x xp p

Þ x =1

31

13. (i) The given situation can be represented as follows:

5 4 3

4 3 5

1 1 1

11000

10700

2700

é

ë

êêê

ù

û

úúú

é

ë

êêê

ù

û

úúú

=

é

ë

êx

y

zêê

ù

û

úúú

or 5 4 3 11000x y z+ + =

4 3 5 10700x y z+ + =

x y z+ + = 2700 11

2

(ii) Let A =

é

ë

êêê

ù

û

úúú

5 4 3

4 3 5

1 1 1

CBSE Sample Question Paper (vii)

Q | | ( ) ( ) ( )A = - - - +5 2 4 1 3 1 = - + + = - ¹10 4 3 3 0 1

Q A-1 exists, so equations have a unique solution. 1/2

(iii) Any answer of three values with proper reasoning which will be considered correct.1

[For example, I prefer the value "punctuality" for rewards, because a punctualstudents can study better.]

14. x a= -( sin )q q Þ dx

da a

qq

q= - =( cos ) sin1 2

2

2 1

y a= -( cos )1 q Þ dy

da a

qq

q q= =. sin sin cos2

2 21

\dy

dx

a

a

= =2

2 2

22

22

sin . cos

sin

cot

q q

qq

1/2

d y

dx

d

dx a a

2

2

2

2 2

1

2 2

1

22

1

22

1

4= - = - = -cosec co

q qq q

.sin

.sin

sec4

2

q1

1

2

15. yx

x=

-

-sin 1

21Þ 1 2 1- = -x y x. sin 1/2

Differentiating both sides w.r.t x, we get

12

2 1

1

1

2

2 2- +

-

-=

-x

dy

dx

y x

x x

( )1

\ ( )1 1 02- - - =xdy

dxxy 1/2

Differentiating again w.r.t x, we get

( ) .1 2 1 022

2- - - +

é

ëêù

ûú=x

d y

dxx

dy

dxy x

dy

dx1

Þ ( )1 3 022

2- - - =x

d y

dxx

dy

dxy 1

16. lim ( ) limx x

f x x ax b a b® - ® -

= + + = + +2 2

2 2 4

lim ( ) lim ( )x x

f x x® + ® +

= + =2 2

3 2 8 1

As f is continuous at x = 2 Þ 2 4 8a b+ + = Þ 2 4a b+ = ...(i) 1

Similarly as f is continuous at x = 4,

\ lim ( ) lim ( )x x

f x f x® - ® +

=4 4

Þ lim ( ) lim ( )x x

x ax b® - ® +

+ = +4 4

3 2 2 5 , 14 8 5= +a b ...(ii) 1

Solving (i) and (ii), we get a b= = -3 2, 1

(viii) Xam idea Mathematics–XII

OR

yx x

x x=

+ - -

+ + -

é

ë

êê

ù

û

úú

-tan 12 2

2 2

1 1

1 1 , Let x2 2= cos q 1

\ tancos cos

cos costan

cos- -+ - -

+ + -

é

ëê

ù

ûú =1 11 2 1 2

1 2 1 2

q q

q q

q q

q q

-

+é

ëêù

ûúsin

cos sin1

=-

+= -æ

èç ö

ø÷

é

ëêù

ûú= -- -tan

tan

tantan tan1 11

1

q

q

p

4q

p

4q 1

\ y x= - -p

4

1

2

1 2cos : Let z x= -cos 1 2

\ y z= -p

4

1

2Þ

dy

dz= -

1

2

1

2

1

2+

17. I = x x

xdx x

x

x xdx

3

2

1

1

2 1

1 1

+ +

-= +

+

- +

é

ëê

ù

ûúò ò ( )( )

1

Let 2 1

1 1 1 1

x

x x

A

x

B

x

+

- +=

-+

+( )( ) Þ A + B = 2, A – B = 1 Þ A B= =

3

2

1

2, 1

\ I = x dxdx

x

dx

x

xx x Cò ò ò+

-+

+= + - + + +

3

2 1

1

2 1 2

3

21

1

21

2

log( ) log( ) 1+1

OR

ex

xdx

ex x

xx

x

1

1

1 22 2

22

2

-

-æèç ö

ø÷ =

-é

ë

òsin

cos

( sin cos )

sin

êêê

ù

û

úúú

ò dx 1

= e

x x

xdx e

xx x1 2

2 2

22

1

2 22

2-æ

è

ççç

ö

ø

÷÷÷

= -òsin cos

sin

ccosec otx

dx2

æèç ö

ø÷ò 1/2

= - +ò òex

dx ex

dxx xcot2

1

2 2

2cosec 1/2

= - - æèç ö

ø÷ + òòcot . .

xe e

xdx e

xdxx x x

2 2

1

2

1

2 2

2 2cosec cosec 1

= - - + +òòex

ex

dx ex

dx Cx x xcot2

1

2 2

1

2 2

2 2cosec cosec 1

= - +ex

Cx cot2

18. I = 2

1 22 2

x

x xdx

( )( )+ +ò

Let x t2 = , 2x dx dt= Þ I = dt

t t( )( )+ +ò 1 21

CBSE Sample Question Paper (ix)

Let 1

1 2 1 2( )( )t t

A

t

B

t+ +=

++

+Þ A + B = 0

2 1A B+ = on solving, we get A = 1, B = -1 11

2

\ I = dt

t

dt

tt t C

+-

+= + - + +òò 1 2

1 2log| | log| | 1

= + - + +log| | log| |x x C2 21 2 1/2

19. Let I = log( tan )1

0

4

+ò x dx

p

...(i)

log tan

0

4

14

p

pò + -æ

èç ö

ø÷

é

ëêù

ûúx dx = log

tan

tan1

1

10

4

+-

+æèç ö

ø÷ò

x

xdx

p

1

or I = logtan

log log( tan )]2

12 1

0

4

0

4

+æèç ö

ø÷ = - +ò òx

dx x dx

p p

...(ii) 1

Adding (i) and (ii), we get

2I = log log24

2

0

4

dx

p

pò = 1

Þ I = p

82log 1

20. Let d xi yj zk®

= + +$ $ $

As d a® ®

^ and d b® ®

^ Þ d a® ®

=. 0 and d b® ®

=. 0 1

d a® ®

=. 0 Þ 4 5 0x y z+ - = and d b® ®

=. 0 Þ x y z- + =4 5 0 ...(i) 1

d c® ®

=. 21 Þ 3 21x y z+ - = ...(ii) 1/2

Solving (i) and (ii), we get x y z= = - = -7 7 7, , 1

\ d i j k®

= - -$ $ $7 7 1/2

21. The equation of plane passing through (3, –1, 2), (5, 2, 4) and (–1, –1, 6) is

x y z- + -

- + + - +

- - - + -

é

ë

êêê

ù

û

úúú

=

3 1 2

3 5 1 2 2 4

3 1 1 1 6 2

0 or

x y z- + -

-

é

ë

êêê

ù

û

úúú

=

3 1 2

2 3 2

4 0 4

0 2

Þ ( )( ) ( )( ) ( )( )x y z- - + + - =3 12 1 16 2 12 0

Þ 12 16 12 76x y z- + = or 3 4 3 19x y z- + = ...(i) 1/2

Length of ^ from (6, 5, 9) to (i) is

(x) Xam idea Mathematics–XII

18 20 27 19

3 4 3

6

342 2 2

- + -

+ += 1

1

2+

OR

Any point on the line x y z+

=+

=-

-=

5

1

3

4

6

9l is ( , ; )l l l- - - +5 4 3 9 6 1

1

2+

Let q a b g( , , ) is the foot of the perpendicular drawn from point P( , , )2 4 1- to the given linex y z+

=+

=-

-

5

1

3

4

6

9…(i)

Since, q a b, g( , ) be on line …(i)

\a b g

l+

=+

=-

-=

5

1

3

4

6

9( )say

\ a l b l g l= - = - = - +5 4 3 9 6, ,

Þ PQ i j k®

= - - + - - + - + +( )$ ( )$ ( ) $l l l5 2 4 3 4 9 6 1

= - + - + - +( )$ ( )$ ( ) $l l l7 4 7 9 7i j k

\ PQ^ line (i)

Þ ( ). ( ). ( )l l l- + - + - + - =7 1 4 7 4 9 7 9 0

Þ 98 98l = Þ l = 1 1

\ The pt. Q is (–4, 1, –3) 1/2

\ Equation of PQ is x y z-

=-

=+2

6

4

3

1

2 and foot of ̂ is ( , , )- -4 1 3 1

22. Let X denote the number of non-violent persons out of selected two. X can takevalues

0, 1, 2 non-violent 20: Violent patriotism: 30 1/2

P(X = 0) = 30 29

50 49

87

245

´

´= 1/2

P(X = 1) 30

50

20

49

20

50

30

49´ + ´ =

30 20 2

50 49

120

245

´ ´

´= 1/2

P(X = 2) = 20 19

50 49

38

245

´

´= 1/2

CBSE Sample Question Paper (xi)

P (2 , 4, 1)

x + 5

A Q B 1

y + 3

4=

z – 6

–9=

\ Probability distribution is

X 0 1 2

P(x) 87

245

120

245

38

245

Mean = 087

2451

120

2452

38

245

196

245´ + ´ + ´ = 1

Importance: In order to have a peaceful environment both the values are requiredpatriotism and non-violence because of patriotism with violence could be verydangerous. 1

SECTION–C

23. The given matrix is A =

-

- -

é

ë

êêê

ù

û

úúú

1 2 3

2 3 2

3 3 4

, | |A = – 6 + 28 + 45 = 67 ¹ 0 1

\ A-1 exists

Adj A =

-

-

- -

é

ë

êêê

ù

û

úúú

6 17 13

14 5 8

15 9 1

, Þ A- =

-

-

- -

é

ë

êêê

ù

û

úúú

1 1

67

6 17 13

14 5 8

15 9 1

21

2

The given system of equations can be written as AX = B

Where A =

1 2 3

2 3 2

3 3 4

-

- -

é

ë

êêê

ù

û

úúú

, X =

x

y

z

é

ë

êêê

ù

û

úúú

, B =

-é

ë

êêê

ù

û

úúú

4

2

11

1

X A B= =

-

-

- -

é

ë

êêê

ù

û

úúú

-é

ë

êêê

ù

û

ú-1 1

67

6 17 13

14 5 8

15 9 1

4

2

11úú

= -

é

ë

êêê

ù

û

úúú

= -

é

ë

êêê

ù

û

úúú

1

67

201

134

67

3

2

1

1

\ x y z= = - =3 2 1, , 1/2

24. The given curve cuts the x-axis at x = 7, and y = 0 1/2

yx

x x=

-

- +

7

5 62 Þ

dy

dx

x x x x

x x=

- + - - -

- +

( ) ( )( )

( )

2

2 2

5 6 7 2 5

5 61

1

2

dy

dxx( )

( ) ( )

( )at = =

- + -

- +=7

49 35 6 0

49 35 6

1

202

1

2

1

2+

\ Equation of tangent to the curve at (7, 0) is

y x- = -01

207( ) or x y- - =20 7 0 1

1

2

Equation of normal to the curve at (7, 0) is

y x x y- = - - = + - =0 20 7 20 140 0( ) 11

2

(xii) Xam idea Mathematics–XII

OR

Let x and r be radius of base of cylinder and cone respectively

Let OC = x VOB B DB, ~D D ¢ 1

\VO

B D

OB

DB¢= Þ ¢ =

-= ¢B D

h r x

rh

( )1

Let S be the curved surface area of cylinder.

\ S xh xh r x

r

h

rrx x= ¢ =

-é

ëê

ù

ûú = -2 2

2 2p pp( )

[ ] 1

\dS

dx

h

rr x

d s

dx

h

r= - =

-<

22

40

2

2

p p( ), 1

Þ S is maximum

dS

dx= 0 Þ r x= 2 1

\ S is is maximum when xr

=2

, i.e., when radius of base of cylinder is half the radius of base

of cone. 1

25. On solving the equations of the two circles, we get points of intersection as A1

2

3

2,

æ

èç

ö

ø÷ and

D1

2

3

2,

-æ

èç

ö

ø÷ 1

Area of shaded region = 2 (Area OABCO)

= - - + -

é

ë

êêêê

ù

û

úúúú

ò ò2 1 1 12

0

1

22

1

2

1

( )x dx x dx 1

=-

- - +-æ

èç ö

ø÷

é

ëêù

ûú+ --2

1

21 1

1

2

1

1 212 1

0

1

2 2( )( ) sin

xx

x xx +é

ëêù

ûú-1

2

11

2

1

sin x 2

CBSE Sample Question Paper (xiii)

Y

O B

Y’

Ö3 )( ,2

1

2A

X’ X

-Ö3 )( ,2

1

2D

2 2x + y = 1 2 2(x –1) + y = 1

a

O'

A B

B'A'

V

h'

r x

h

DCO

=-

+-æ

èç ö

ø÷ - -

ìíî

üýþ

+ - -- - -3

4

1

21 1

3

4

1 1 1sin sin ( ) sin ( ) sin - æèç ö

ø÷

ìíî

üýþ

1 1

21

= - - + + - -3

4 6 2 2

3

4 6

p p p p

= -æ

èç

ö

ø÷

2

3

3

2

p sq. units 1

26. The given differential equation can be written as

dx

dyy x y+ =(cot ) cos

I.F. = e e yy dy ycot log sin sinò = =

\ The solution is x sin y = sin cosy y dy C+ò 1

= +ò1

22sin y dy C 1

or x y y Csin cos=-

+1

42 1

It is given that y = 0, when x = 0 1

C - =1

40 Þ C =

1

4

\ x y y ysin ( cos ) sin= - =1

41 2

1

2

2 1

Þ 2x = sin y is the reqd. solution 1

27. Here a i j k1 2 3®

= + -$ $ $ and a i j k2 3 3 2®

= + +$ $ $ 1

b i j k1 2 5®

= + +$ $ $ and b i j k2 3 2 5®

= - +$ $ $

n b b

i j k® ® ®

= ´ =

-

é

ë

êêê

ù

û

úúú

1 2 1 2 5

3 2 5

$ $ $

= + -20 10 8$ $ $i j k 11

2

\ Vector equation of the required plane is ( ).r a n® ® ®

- =1 0 or r n a n® ® ® ®

=. .1

or r i j k i j k i j k®

+ - = + - + - =.( $ $ $) ( $ $ $).( $ $ $)20 10 8 2 3 20 10 8 40 + + =10 24 74 2

Þ r i j k®

+ - =.( $ $ $)10 5 4 37

The cartesian equation of plane is 10 5 4 37x y z+ - = 11

2

28. Suppose number of electronic operated machine = x and number of manually operatedsewing machines = y 1/2

(xiv) Xam idea Mathematics–XII

\ x y+ £ 20 ...(i)

and, 360 240 5760x y+ £ or 3 2 48x y+ £ ...(ii)

x y³ ³0 0,

To maximise Z x y= +22 18 2

Corners of feasible region are A P B( , ), ( , ), ( , )0 20 8 12 16 0

ZA = ´ =18 20 360 1/2

ZP = ´ + ´ =22 8 18 12 392

ZB = 352

\ Z is maximum at x = 8 and y = 12

\ The dealer should invest in 8 electric and 12 manually operated machines.

Keeping the ‘save environment’ factor in mind the manually operated machine should bepromoted so that energy could be saved. 1

29. Let A be the event that he knows the answer, B be the event that he guesses and C be theevent that he copies. 1/2

Then P A( ) =1

2, P B( ) =

1

4 and P C( ) =

1

41/2

Let X be the event that he has answered correctly.

Also, PX

AP

X

Bæèç ö

ø÷ = æ

èç ö

ø÷ =1

1

4, and P

X

Cæèç ö

ø÷ =

3

41

CBSE Sample Question Paper (xv)

2

Y

28

3x + 2y = 48

x + y = 20

X20 24 28

Y’

4 8 12 16

B(16, 0)

OX’

24 (0, 24)

8

16

12

A(0, 20)

20

4

P(8, 12)

Thus, required probability = PA

X

PX

AP A

PX

AP A P

X

BP

æèç ö

ø÷ =

æèç ö

ø÷ ´

æèç ö

ø÷ ´ + æ

èç ö

ø÷ ´

( )

( ) ( ) ( )B PX

CP C+ æ

èç ö

ø÷ ´

1

=´

´ + ´ + ´

1

21

1

21

1

4

1

4

1

4

3

4

=

+ +

=

1

21

2

1

16

3

16

2

31

If Arjun copies the answer, he will be violating the value of honesty in his character. Heshould not guess the answer as well as that may fetch him negative marking for a wrongguess. He should accept the question the way it is and leave it unanswered as cheatingmay get him marks in this exam but this habit may not let him develop an integrity ofcharacter in the long run. 2

OR

Let the events defined are

E1: Person chosen is a cyclist

E2: Person chosen is a scooter-driver

E3: Person chosen is a motorbike driver 1/2

A: Person meets with an accident 1/2

P(E1) = 1

6 , P(E2) =

1

3 , P(E3) =

1

21

PA

E1

æ

èç

ö

ø÷ = 0.01, P

A

E2

æ

èç

ö

ø÷ = 0.03, P

A

E3

æ

èç

ö

ø÷ = 0.15, P

E

A2æ

èç

ö

ø÷ = Required 1

PE

A

PA

EP E

PA

EP E P

A

E

2 22

11

2

æ

èç

ö

ø÷ =

æ

èç

ö

ø÷

æ

èç

ö

ø÷ +

æ

è

. ( )

. ( ) çö

ø÷ +

æ

èç

ö

ø÷. ( ) . ( )P E P

A

EP E2

33

=´

´ + ´ + ´

=´

= =

1

3

3

1001

6

1

100

1

3

3

100

1

2

15

100

600

100 52

6

52

3

261

Suggestion: Cycle should be promoted as it is good for 1/2

(i) Health 1/2

(ii) No pollution 1/2

(iii) Saves energy( no petrol) 1/2

(xvi) Xam idea Mathematics–XII

Value Based QuestionsMathematics–XII

1. Everyone wants to be a perfect ideal human being. Let us assume that dishonesty is one ofthe factors that affects our perfectness and perfectness has an inverse square relationshipwith dishonesty. For any value x of level of dishonesty we have a unique value y ofperfection.

(i) Write down the equation that relates y with x.

(ii) Is this relationship from x XÎ = ¥( , )0 to y Î ¥( , )0 , forms a function?

(iii) For what level of dishonesty one can achieve 1

4th level of perfection?

(iv) What will be the change in level of perfection when the level of dishonesty changes from 4 to 2?

Sol. (i) yx

x= ¹1

02

,

(ii) Yes

(iii) When y =1

4, we have

1

4

12

=x

Þ 4 2= x Þ x = ±2, but x can not be –ve

\ x = 2

(iv) When x = 4, y =1

16

When x = 2, y =1

4

\ Change in level of perfection = 1

4

1

16- =

3

16

2. A trust fund has ̀ 30,000 that is to be invested in two different types of bonds. The first bond pays 5% p.a. interest which will be given to orphanage and second bond pays 7% interestp.a. which will be given to financial benefits of the trust. Using matrix multiplication,determine how to divide `30,000 among two types of bonds if the trust fund obtains anannual total interest of `1800.

(i) What are the values reflected in the question?

(ii) Why is it required to help orphan children?

Sol. Let `x be invested in Ist bond, then `30,000 – x will be invested in IInd bond.

Total interest = `1800

Now, [ ] [ ]x x30000

5

1007

100

1800-

é

ë

êêê

ù

û

úúú

=

Þ5

10030000

7

1001800

xx+ - ´ =( )

Þ 5x + 210000 – 7x = 1800 × 100

Þ –2x + 210000 = 180000 Þ x = =30000

215000

\ For investment in IInd bond, amount = 30000 - 15000 = 15000.

So, equal amount is invested in both of the bonds.

(i) Values reflected are helping poor and needy children. Provided that the interestrate in financial benefits (IInd bond) is more than the Ist bond (money given toorphanage) trust decides to invest fund equally. It reflects that the motive of thetrust is not to only to earn the interest but also to help the needy orphan children.This charity should be a concern of every one.

(ii) The children living in orphanage are also talented and possess potential. If theyare given the proper brought up and opportunity, they can contribute to thedevelopment of the society and country and will become good citizens.

3. Of the students in a school; it is known that 30% has 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who has 100%attendance attain A grade and 10% irregular students attain A grade in their annualexamination. At the end of the year, one student is chosen at random from the schooland he has A grade. What is the probability that the student has 100% attendance?

(i) Write any two values reflected in this question.

(ii) Is regularity required only in school? Justify your answer

Sol. Let E1 : Student has 100% attendance

E2 : Student is irregular

A : Student attains A grade

A

E1

: Student attains A grade given that she has 100% attendance

A

E2

: Student attains A grade given that she is irregular

E

A1 : Student has 100% attendance given that she attains A grade

Using Bayes’ theorem

(xviii) Xam idea Mathematics–XII

V

A

L

U

E

B

A

S

E

D

Q

U

E

S

T

I

O

N

S

PE

A

P E PA

E

P E PA

EP E

11

1

11

2

æ

èç

ö

ø÷ =

æ

èç

ö

ø÷

æ

èç

ö

ø÷ +

( ).

( ). ( ).PA

E2

æ

èç

ö

ø÷

=´

´ + ´

30

100

70

10030

100

70

100

70

100

10

100

=´

´ + ´

30 70

30 70 70 10 =

+=

30

30 10

3

4

(i) Regularity and intelligence

(ii) Regularity is the value which is required at every stage of our life. In ourchildhood, during school education we can inculcate this value in our personality. Regularity increases our capabilities and makes us able to put the best of ourpotential. We are able to achieve certain targets due to regular efforts.

4. In a survey of 20 richest person of three residential society A, B, C it is found that insociety A, 5 believes in honesty, 10 in hard work, 5 in unfair means while in B, 5 inhonesty, 8 in hard in work, 7 in unfair means and in C, 6 in honesty, 8 in hard work, 6 inunfair means. If the per day income of 20 richest persons of society A, B, C are `32,500,30,500, 31,000 respectively, then find the per day income of each type of people bymatrix method.

(i) Which type of people has more per day income.

(ii) According to you, which type of person is better for country.

Sol. Let x, y, z be the per day income of person believing in honesty, hard work and unfairmeans, respectively. The given situation can be written in matrix form as

A X = B, Where

A X

x

y

z

B=

é

ë

êêê

ù

û

úúú

=

é

ë

êêê

ù

û

úúú

=

5 10 5

5 8 7

6 8 6

32500

3050. , 0

31000

é

ë

êêê

ù

û

úúú

Q AX B= Þ X A B= -1 ...(i)

Now for A-1

|A| =

5 10 5

5 8 7

6 8 6

= 5(48 – 56) – 10(30 – 42) + 5(40 – 48) = 40 ¹ 0

Also, C111 11

8 7

8 648 56 8= - = - = -+( ) ( )

Value Based Questions (xix)

V

A

L

U

E

B

A

S

E

D

Q

U

E

S

T

I

O

N

S

C121 21

5 7

6 630 42 12= - = - - =+( ) ( )

C131 31

5 8

6 840 48 8= - = - = -+( ) ( )

C212 11

10 5

8 660 40 20= - = - - = -+( ) ( )

C222 21

5 5

6 630 30 0= - = - = -+( ) ( )

C232 31

5 10

6 840 60 20= - = - - =+( ) ( )

C 313 11

10 5

8 770 40 30= - = - =+( ) ( )

C 323 21

5 5

5 735 25 10= - = - - = -+( ) ( )

C 333 31

5 10

5 840 50 10= - = - = -+( ) ( )

Adj (A) =

- -

-

- -

é

ë

êêê

ù

û

úúú

=

- -

-

- -

8 12 8

20 0 20

30 10 10

8 20 30

12 0 10

8 20

T

10

é

ë

êêê

ù

û

úúú

Aadj A

A

- =1 ( )

| |

=

- -

-

- -

é

ë

êêêêê

ù

û

úúúúú

8

40

20

40

30

4012

400

1

48

40

20

40

10

40

=

- -

-

- -

é

ë

êêêêê

ù

û

úúúúú

1

5

1

2

3

43

100

10

401

5

1

2

1

4

Putting the value of X, A-1, B in (i) we get

Þ

x

y

z

é

ë

êêê

ù

û

úúú

=

- -

-

- -

é

ë

êêêêê

ù

û

úúúú

1

5

1

2

3

43

100

1

41

5

1

2

1

4ú

é

ë

êêê

ù

û

úúú

.

32500

30500

31000

Þ

x

y

z

é

ë

êêê

ù

û

úúú

=

é

ë

êêê

ù

û

úúú

1500

2000

1000

Þ

x

y

z

é

ë

êêê

ù

û

úúú

=

é

ë

êêê

ù

û

úúú

2000

1000

3000

(xx) Xam idea Mathematics–XII

V

A

L

U

E

B

A

S

E

D

Q

U

E

S

T

I

O

N

S

Þ x y z= = =1500 2000 1000, ,

Hence, per day income of person who believe in honesty = `1,500

Per day income of person who believe in hard work = `2,000

Per day income of person who believe in unfair means = `1,000

(i) A person, who believe in hard work has more per day income.

(ii) A person, who believe in hard work and honesty, are better for country.

5. The male-female ratio of a village increases continuously at the rate proportional to theratio at any time. If the ratio of male-female of the villages was 1000 : 980 in 1999 and1000 : 950 in 2009, what will be the ratio in 2019?

(i) Why gender equality is value for society?

(ii) What should society do to reduce the male-female ratio to 1?

Sol. Let male-female ratio at any time be r.

Given dr

dtrµ Þ

dr

dtkr= where k is the constant of proportionality.

We have dr

rk dt=

Integrating both sides we get

log logr kt c= + where log c is the constant of integration.

Þ log logr c kt- = Þ logr

ckt=

\ r c e kt= ...(i)

Let us start reckoning time from the year 1999 for this problem.

So in 1999, t = 0 and r = =1000

980

50

49

Substituting in (i) we get

50

49

0= c e. Þ c =50

49

\ (i) becomes

r e kt=50

49...(ii)

Also in the year 2009, t = 10 and r = =1000

950

20

19

Substituting in (ii) we get

20

19

50

49

10= e k Þ e k10 98

95=

Substituting in (ii) we get

Value Based Questions (xxi)

V

A

L

U

E

B

A

S

E

D

Q

U

E

S

T

I

O

N

S

r e kt t

= = æèç

öø÷

50

49

50

49

98

95

10 10 10.( ) ...(iii)

In the year 2019, t = 20

\ r =50

49

98

95

50

49

98

95

98

95

20

10æèç

öø÷ = ´ ´

=´

´

100 98

95 951 085~ . ~ 1085 1000:

Thus in the year 2019, the male-female ratio will be 1085 : 1000

(i) Gender equity promotes economic growth, reduce fertility, child mortality andunder nutrition.

(ii) (a) Stop female-foeticide.

(b) Empower women to realise their rights.

(c) Provide special opportunities to women to come at par with men in all walks of life.

6. A window is in the form of rectangle surmounted by a semi-circular opening. Totalperimeter of the window is 10 m. What will be the dimensions of the whole opening toadmit maximum light and air?

(i) How having large windows help us in saving electricity and conservingenvironment?

(ii) Why optimum use of energy is required in the Indian context?

Sol. Let ABCED be required window having length 2x and width y. If A is the area of window. Then

A xy x= +21

2

2p

= - - +x x x x( )10 21

2

2p p

= - - +10 21

2

2 2 2x x x xp p

Given, Perimeter = 10 Þ 21

22 10x y y x+ + + =p

Þ 2 10 2y x x= - - p

= - -10 21

2

2 2x x xp = - +æèç

öø÷10 2

1

2

2x xp

Obviously, window will admit maximum light and air if its area A is maximum.

Now,dA

dxx= - +æ

èç

öø÷10 2 2

1

2p

(xxii) Xam idea Mathematics–XII

V

A

L

U

E

B

A

S

E

D

Q

U

E

S

T

I

O

N

S

y

E

D

A

C

B2x

For maxima or minima of A

dA

dx= 0

Þ 10 2 21

20- +æ

èç

öø÷ =x p Þ 10 4 0- + =x( )p

Þ x =+

10

4 pÞ

d A

dx

2

24= - +( )p < 0

Þ For maximum value of A, x =+

10

4 p and thus y =

+

10

4 p

Therefore, for maximum area i.e., for admitting maximum light and air,

Length of rectangular part of window = 220

4x =

+ p

Width = 10

4 + p

(i) Large windows allow more light during daytime and hence will reduce the use ofelectricity. Saving energy (electricity) helps in conserving environment aselectricity is produced by using natural resources which we should conserve forthe sake of future generation.

(ii) India is the 2nd most populated country in the world so have more consumers ofenergy but less sources of its production. Therefore, in Indian context energysaving is like energy production.

7. In a competition, a brave child tries to inflate a huge spherical balloon bearing slogansagainst child labour at the rate of 900 cubic centimeter of gas per second. Find the rate at which the radius of the balloon is increasing when its radius is 15 cm.

(i) Which values have been reflected in this question?

(ii) Why child labour is not good for society?

Sol. Let r be the radius and V be the volume of the balloon. Then

dV

dt= 900 cm3/sec

dr

dt= ?, when r = 15

V r=4

3

3p

Differentiating both sides w.r.t ‘t’

dV

dtr

dr

dt=

4

33 2p

900 4 15 2= p( )dr

dt Þ

dr

dt=

´ ´=

900

225 4

1

p p cm/sec

Value Based Questions (xxiii)

V

A

L

U

E

B

A

S

E

D

Q

U

E

S

T

I

O

N

S

(i) Three values reflected are bravery, sympathy for child labour and raising voiceagainst child labour.

(ii) We know that child labour is illegal and harmful to both society and country. Weshould spread awareness in society so that child labour should be abolished. Inthe childhood they should be sent to school for their education so that they cancontribute for the development of the society.

8. A manufacturing company makes two type of teaching aids A and B of mathematics ofclass XII. Each type of A requires 9 labour hours for fabricating and 1 labour hour forfinishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours forfinishing. For fabricating and finishing, the maximum labour hours available are 180and 30, respectively. The company makes a profit of `80 on each piece of type A and`120 on each piece of type B. How many pieces of type A and B should bemanufactured per week to get a maximum profit? What is the maximum profit perweek?

Is teaching aid necessary for teaching learning process? If yes, justify your answer.

Sol. Let x and y be the number of pieces of type A and B manufactured per weekrespectively. If Z be the profit then,

Objective function, Z = 80 120x y+

We have to maximize Z, subject to the constraints

9 12 180x y+ £ Þ 3 4 60x y+ £ ...(i)

x y+ £3 30 ...(ii)

x y³ ³0 0, ...(iii)

The graph of constraints are drawn and feasible region OABC is obtained, which isbounded having corner points O A B( , ), ( , ), ( , )0 0 20 0 12 6 and C ( , )0 10

(xxiv) Xam idea Mathematics–XII

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X40 50 60

Y’

15

x + 3y = 30

3x + 4y = 60

A(20, 0)

C(0, 10)

10 20 30OX’

10

5B(12, 6)

Now the value of objective function is obtained at corner point as

Corner point Z = 80 120x y+

O (0, 0) 0

A (20, 0) 1600

B (12, 6) 1680 Maximum

C (0, 10) 1200

Hence, the company will get the maximum profit of `1,680 by making 12 pieces of typeA and 6 pieces of type B of teaching aid.

Yes, Teaching aid is necessary for teaching learning process as

(i) it makes learning very easy.

(ii) it provides active learning.

(iii) students are able to grasp and understand concept more easily and in activemanner.

9. A village has 500 hectares of land to grow two types of plants, X and Y. Thecontribution of total amount of oxygen produced by plant X and plant Y are 60% and40% per hectare respectively. To control weeds, a liquid herbicide has to be used for X and Y at rates of 20 litres and 10 litres per hectare, respectively. Further no more than 8000 litres of herbicides should be used in order to protect aquatic animals in a pond whichcollects drainage from this land. How much land should be allocated to each crop so asto maximise the total production of oxygen?

(i) How do you think excess use of herbicides affect our environment?

(ii) What are the general implications of this question towards planting treesaround us?

Sol. Let plants X and Y be grown in x and y hectares.

So, x ³ 0 and y ³ 0

x y+ £ 500 ...(i)

Contribution of oxygen by the plants = 60% of x + 40% of y

zx y

x y= + = +6

10

4

100 6 0 4. .

Also, Amount of liquid herbicides required = ( )20 10x y+ litres

Given 20 10 8000x y+ £

Þ 2 800x y+ £ ...(ii)

The LPP for given problem is

Maximum, Z x y= +0 6 0 4. .

S.t. x y+ £ 500 ...(iii)

and 2 800x y+ £ ...(iv)

x y, ³ 0

Value Based Questions (xxv)

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Sketching a graph for the above LPP, we get the region shown in the figure

Solving x y+ = 500 and 2 800x y+ = , we get

B « ( , )300 200

Corner point Value of the optimizing function

(0, 0) Z = 0 + 0 = 0

(400, 0) Z = 0.6 × 400 + 0.4 × 0 = 240

(300, 200) Z = 0.6 × 300 + 0.4 × 200

= 180 + 80 = 260

(0, 500) Z = 0.6 × 0 + 0.4 × 500 = 200

Maximum Production of oxygen will be achieved when plant X is planted in 300hectare and plant Y is planted in 200 hectare.

(i) Excess herbicide will get absorbed in the soil and may contaminate the watersource also. Thus it can affect the whole ecosystem.

(ii) Care should be taken while planting trees that the variety of the plants is such thatthey provide more oxygen for our environment.

(xxvi) Xam idea Mathematics–XII

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100 200 X300O 400 500 600 700

(500, 0)

900

B

(0, 800)

x + y = 500

A(0, 500)

(400, 0) C(0, 0)

2x + y =

800

10. In shop A, 30 tin pure ghee and 40 tin adulterated ghee are kept for sale while in shop B, 50 tin pure ghee and 60 tin adulterated ghee are there. One tin of ghee is purchasedfrom one of the shop randomly and it is found to be adulterated. Find the probabilitythat it is purchased from shop B.

(i) How adulteration is dangerous for humanity?

(ii) What you can do against adulteration?

Sol. Let the event defined be as

E1 = Selection of shop A.

E2 = Selection of shop B.

A = Purchasing of a tin having adulterated ghee.

P E( )11

2= , P E( )2

1

2=

PA

E1

40

70

4

7

æ

èç

ö

ø÷ = = , P

A

E2

60

110

6

11

æ

èç

ö

ø÷ = =

PE

A2æ

èç

ö

ø÷ = required

PE

A

P E PA

E

P E PA

EP E

22

2

11

2

æ

èç

ö

ø÷ =

æ

èç

ö

ø÷

æ

èç

ö

ø÷ +

( ).

( ). ( ).PA

E2

æ

èç

ö

ø÷

=

+

=

+

=

1

2

6

111

2

4

7

1

2

6

11

3

112

7

3

11

21

43

.

. .

(i) Adulteration is dangerous as it is harmful for user’s health.

(ii) To prevent adulteration, we should spread awareness against it in society.

11. In a self-assessment survey 60% persons claimed that they never indulged incorruption, 40% persons claimed that they always speak truth and 20% say that theyneither indulged in corruption nor tell lies.

A person is selected at random out of this group.

(i) What is the probability that the person is either corrupt or tells lie?

(ii) If the person never indulged in corruption, find the probability that she/hetells, truth.

(iii) If the person always speaks truth find the probability that she/he claims to have never indulge in corruption.

(iv) What values have been discussed in this question?

Value Based Questions (xxvii)

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(v) Why is it must for all to practice values in our life?

Sol. Let A : Set of persons never indulged in corruption

B : Set of persons always speak truth

Then, P A( ) =60

100, P B( ) =

40

100

and P A B( )Ç =20

100

(i) P(Either A or B) = P A B P A P B P A B( ) ( ) ( ) ( )È = + - Ç

= + - = =60

100

40

100

20

100

80

100

4

5

(ii) PB

A

P A B

P A

æèç

öø÷ =

Ç= =

( )

( )

20

10060

100

1

3

(iii) PA

B

P A B

P B

æèç

öø÷ =

Ç= =

( )

( )

20

10040

100

1

2

(iv) The following values have been discussed

(a) We should never indulge in any type of corruption.

(b) We should never tell lies i.e., we should always speak truth.

(v) Values contribute to intellectual development, use of abilities, achieve creativity,personal development and development of society.

(xxviii) Xam idea Mathematics–XII

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