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F-37 ver.1 dated 20-05-2013-Scheme of Evaluation (SoE) Page 1of 6

[WINTER, 2013] SCHEME OF EVALUATION

PROGRAM BCA REVISED 2012

SEMESTER ISUBJECT CODE &

NAME

BCA 1030- BASIC MATHEMATICS

CREDIT 4

BK ID B0947

MAX. MARKS 60

Q.No Question and Scheme of Evaluation Unit/Page No.

Marks TotaMark

1(i) LetA = {1, 2, 3, 4, 5, 6}and B = {2, 4, 6, 8}. Find A B and

B A.

(ii) In a group of 50people, 35speak Hindi, 25speak both English

and Hindi and all the people speak at least one of the two languages.

How many people speak only English and not Hindi ? How many

people speak English?

A(i) We have A B = {1, 3, 5}, as the only elements of A which do not

belong to B are 1, 3 and 5. Similarly, B A = {8}.

We note that A B B A

(ii) Let H denote the set of people speaking Hindi and E the set of people

speaking English. We are given that n(H E) = 50, n(H) = 35,

n(HE) = 25. Now

n(H E) = n(H) + n(E H).

So 50 = 35 + n(E H), i.e. , n(E H) = 15.

Thus, the number of people who speak only English but not Hindi is

15.

Also, n(H E) = n(H) + n(E) n(H E) implies50 = 35 + n(E) 25,

which gives n(E) = 40.

Hence, the number of people who speak English is 40.

2

8

10

2 (i) Express 792in radians and (7/12)c in degrees

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(ii) Prove that

cos

sin1

sin1

cos

1sectan

1sectan

A

(i)

radians180/xx

cradians 5/22180/)792(792

10512

1807)

12

7( c

(ii)

1sectan

1sectanLHS

22 sectan1

1sectan

tansecsectan 22

1sectan

tansectansecsectan

1sectan

tansec1sectan

cos

1sin

cos

1

cos

sinsectan

..(1)

Now

sin1cos

sin1sin1

cos

sin1

sin1cos

sin1 2

22

22

sin1cos

1cossin

sin1

cos

sin1cos

cos2

(2)

From (1)and (2)the result follows

2

8

10

3 (i) Define continuity of a point

(ii) Test the continuity of the function f where f is defined by

27

2|2|

2

xif

xifx

x

xf

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A (i)Let f be a function of a real variable defined in an open interval containing

a. Then f is continuous at a if afxfItax

(ii)Whenx < 2, |x

2|is negative. So|x

2| =

(x

2)

12x

2xxf

Whenx > 2, |x2|is positive. So |x2| = x 2, 12x

2xxf

f(2) = 7. Thus

21

27

21

xif

xif

xif

xf

As in the previous worked example,fis continuous for all a < 2and all a >

2.

11ItxfIt2x2x

11ItxfIt2x2x

As xfIt,xfItxfIt2x2x2x

does not exist. So f is continuous at all

points except 2.

2

8

10

4Solve

4xy

2xy

dx

dy

A

The given equation is4xy

2xy

dx

dy

[this is where

]b

b

a

a

--------- (1)

Puttingx = X + h, y = Y + k(h, k being constants)

So that dx = dX, dy = dY,(1) becomes

)4hk(XY

)2hk(XY

dX

dY

---------

(2)

Put k + h2 = 0and kh4 = 0,

So that, h =1, k = 3.

Therefore (2) becomes,XY

XY

dX

dY

..(3)

Which is homogeneous in X and Y.

10 10

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Put Y = vX,thendx

dvXv

dX

dY

Therefore (3) becomes,1v

1v

dX

dvXv

Or 1

21

1

1 2

v

vvv

v

v

dX

dvX

X

dXdv

vv

v

221

1

Integrating both the sides

c

X

dXdv

vv21

v22

2

1

2

Or

cXlogvv21log2

1 2

Or

c2XlogX

Y

X

Y21log 2

2

2

Or log (X2+ 2XY Y2) = 2c

Or X2+ 2XY y2 = e 2c= c

PuttingX = xh = x + 1,and Y = yk = y- 3,equation (4) becomes,

c3y3y1x21x 22

Which is the required solution.

5 A bag contains two red balls, three blue balls and five green balls.

Three balls are drawn at random. Find the probability that

a) the three balls are of different colours

b) two balls are of the same colour

c) all the three are of the same colour.

A Let E denote the given event..

a) We can choose one red ball in C (2, 1)ways, etc. So

3,10C1,5C1,3C1,2C

EP

4

13.2.1.

8.9.10

5.3.2

10 10

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b) Let E1, E2, E3denote the events that two balls among the three are red,

blue and green respectively.

15

13.2.1.

8.9.10

8.1

3,10C

1,8C2,2CEP 1

Similarly

407

3.2.1.8.9.10

7.3

3,10C

1,7C2,3CEP 2

125

3.2.1.8.9.10

5.

2.1

4.5

3,10C

1,5C2,5CEP 3

So P(E) = P(E1) + P(E2) + P(E3)12

5

40

7

15

1

120

50218

120

79

c) As there are only two red balls, the chosen three balls are of the same

colour only if they are all blue or green.

Let E1, E2 denote the events that the three balls are blue and green

respectively.

1201

3.2.1.8.9.10

1

3,10C

3,3CEP 1

121

3.2.1.8.9.10

3.2.1.

2.1

4.5

3,10C

3,5CEP 2

SoP(E) = P(E1) + P(E2)12

1

120

1

120

101

120

11

6 Solve: 2x + 3y + 4z = 20, x + y + 2z = 9, 3x + 2y + z = 10

AIn this example,

1023

911

2032

1103

291

4202

1210

219

4320

123

211

432

321

= 2(1 4) 3 (1 6) + 4(2 3) = 5

1= 20(1 4) 3(9 20) + 4 (18 10) = 5

2= 2(9 20) 20(1 6) + 4 (10 27) = 10

3= 2(10 18) 3 (10 27) + 20(2 3) = 15

Hence

10 10

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.3z,2y,1x 321

*A-Answer

NotePlease provide keywords, short answer, specific terms, and specific examples (wherever necessary)

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