**SECTION A**

**Q 1. Read the following passage and answer the questions that follow it**:

So great is our passion for doing things for ourselves, that we are becoming increasingly less dependent on specialized labor. NO one can plead ignorance of a subject any longer, for three are countless do-it yourself publications. Armed with the right tools and materials, newly-weds gaily embark on the task of decorating their own fireplaces, laying-out their own gardens, building garages and making furniture. Some really keen enthusiasts go so far as to make their own record players and a radio transmitters. Shops cater for the do-it yourself craze not only by running special advisory services for novices, but by offering consumers bits and pieces which they can assemble at home. Such things provide an excellent outlet for pent up creative energy, but unfortunately not all of us are born handymen.

Wives tend to believe that their husbands are infinitely resourceful and versatile. Even husbands who can hardly drive a nail in straight are supposed to be born electricians, carpenters, plumbers and mechanics. When lights fuse, furniture gets rickety, pipes get clogged, or vacuum cleaners fail to operate, wives automatically assume that their husbands will somehow put things right. The worst thing about the do-it yourself game is that sometimes husbands live under the delusion that they can do anything even when they have been repeatedly proved wrong. It is a question of pride s much as anything else.

**Q 1.Read the following passage carefully and answer the questions that follow:**

People travelling long distances frequently have to decide whether they would prefer to go by land, sea or air. Hardly anyone can positively enjoy sitting in a train for more than a few hours. Train compartments soon get cramped and stuffy. It is almost impossible to take your mind off the journey. Reading is only a partial solution, for the monotonous rhythm of the wheels clicking on the rails soon lulls you to sleep. During the day, sleep comes in snatches. At night when you really wish to go to sleep, you rarely manage to do so. If you are lucky enough to get a couchette, you spend half the night staring at the small blue light in the ceiling, or fumbling to find your passport when you cross a frontier. Inevitably you arrive at your destination almost exhausted.

Long car journeys are even less pleasant, for it is quite impossible even to read. On motor-ways you can at least travel fairly safely at high speeds, but more often than not, the greater part of the journey is spent on narrow, bumpy roads which are crowded with traffic . By comparison, trips by sea offer a great variety of civilized comforts. You can stretch your legs on the spacious decks, play games, swim, meet interesting people and enjoy good food-always assuming, of course, that the sea is calm. If it is not, and you are likely to get sea-sick, no form of transport could be worse. Even if you travel in ideal weather, sea-journeys take a long time. Relatively few people are prepared to sacrifice up to a third of their holidays for the pleasure of travelling on a ship.

Aeroplanes have the reputation of being dangerous and even hardened travellers are intimidated by them. They also have the grave disadvantage of being the most expensive form of transport. But nothing can match them for speed and comfort. Travelling at a height of 30,000 feet, far above the clouds, and at over 500 miles an hour is an exhilarating experience. You do not have to devise ways of taking your mind off the journey, for an aeroplane gets you to your destination rapidly. For a few hours, you settle back in a deep armchair to enjoy the flight. The real escapist can watch a free film show and sip a hot or cold drink on some services. But even when such refreshments are not available, there is plenty to keep you occupied. An aeroplane offers you an unusual breath taking view of the world. You soar effortlessly over high mountains and deep valleys. You really see the shape of the land. If the landscape is hidden from view, you can enjoy the extraordinary sight of unbroken clouds, plains that stretch out for miles before you, while the sun shines brilliantly in a clear sky. The journey is so smooth that there is nothing to prevent you from reading or sleeping. However you decide to spend your time, one thing is certain: you will arrive at your destination fresh and uncrumpled. You will not have to spend the next few days recovering from a long and arduous journey.

**Q1) Using elementary row transformations, find the inverse of the matrix**

A = 4 5

3 4

Ans1) Here

A =

4 5

3 4

and I =

1 0

0 1

Now A = IA

=>

4 5

3 4=1 0

0 1

A=> Applying R1 (1/4)R1

1 5/4

3 4=1/4 0

0 1

A=> Applying R2 R2 – 3R1

1 5/4

0 1/4

=

1/4 0

-3/4 1

A=> Applying R2 4R2

1 5/4

0 1=1/4 0

-3 4

A=> Applying R1 R1 – (5/4)R2

1 0

0 1

=

4 -5

-3 4

AA-1 = 4 -5

-3 4

Q2) For the matrices A and B, verify that (A, B)’ = A’B’ where

A = 1 2

3 4

B = 4 5

1 2

Ans2) Here

A = 1 2

3 4

B = 4 5

1 2= 6 9

16 23= 6 16

9 23

R.H.S. = B’A’ =4 1 1 3 = 6 16 = L.H.S.

5 2 2 4 9 23

**Q2) Find the work done by the force = + 2 + , acting on a particle, if the particle is displaced from the point with position vector 2 + + to the point with position vector 3 + 2 + 4.**

Ans2) Here = + 2 +

call A(2 + + ) and B(3 + 2 + 4)

… = – = (3 + 2 + 4) – (2 + + )

= + + 3

… work done = .

= ( + 2 + ) . ( + + 3)

= 1 . I + 2 . I + I . 3 = 6

**Q4) For the matrices A and B, verify that (AB’ ) = A’B’ where**

A = 1 3

2 4

B = 1 4

2 5

Ans4) Here

A =

1 3

2 4

B =

1 4

2 5

L.H.S. =

7 19

10 28= 7 10

18 28

R.H.S. = B’A’ =

1 2

4 5

1 2

3 4

=

7 10

18 28= L.H.S

**Q7) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 3 or by 4?**

Ans7) Let P(3), P(4), P(3 and 4), P(3 or 4) respectively denote the Probs of the sum of the no.s on the two faces be divisible by 3, 4, 3 and 4, 3 or 4 then

P(3) = 12/16 = 1/3

… the favourable Pairs are

(1, 2); (1, 5); (2, 1); (2, 4); (3, 3); (3, 6); (4, 2); (4, 5); (5, 1); (5, 4); (6, 3); (6, 6)

P(4) = 9/36 = 1/4

… the favourable Pairs are

(1, 3); (2, 2); (2, 6); (3, 1); (3, 5); (4, 4); (5, 3); (6, 2); (6, 6)

P(3 and 4) = 1/36

(only one pair, (6, 6) is common)

P(3 or 4) = P(3) + P(4) – P(AB)

= 1/3 + 1/4 – 1/36 = (12 + 9 – 1)/36 = 5/9

**Q1) Find from first principle the derivative of (x2 + 1)/x w.r.t. x**

Ans1) Let y = (x2 + 1)/x = x + 1/x – (i)

Let x be a small incremental in the value of x and y be the corresponding incremental in the value of y, then

y + y = x + x + 1/(x + x) – (ii)

Now (ii) – (i)

=> y = x + 1/(x + x) – 1/x

y = x – x/(x(x + x)) => y/x = 1 – 1/(x(x + x))

… dy/dx =

y/x

=

(1 – 1/(x(x + x))

= 1 – 1/x2

= (x2 – 1)/x2

**Q2) Evaluate :**

1 – cos 5x

1 – cos 6x

Ans2)

1 – cos 5x =

1 – cos 6x 2sin2 (5x/2)

2sin2 (3x)

(5/2)2[sin (5x/2)/(5x/2)]2

32 [sin 3x/3x]2 = 25/36

**Q3) Using differentials, find the approximate value of 29**

Ans3) To find 29

Let x = 27 and x = 2 then x + x = 27 + 2 = 29

Taking y = x1/3, we get dy/dx = 1/3 x-2/3

Also y + y = (x + x)1/3 = 291/3

where y = x1/3 = 271/3 = 3

… 3 + y = 291/3

Again y = dydx x = 1/3 x-2/3 . 2 = 2/3(27)-2/3 = 2/27

291/3 = 3 + y = 3 + (2/27) = 83/27

** Q1) In a group there are 2 men and 3 women 3 person are selected at random from this group. Find the probability that 1 man and 2 women or 2 men and 1 woman are selected**.

Ans1) Given Men (2) and women (3)

To be selected : 3 Person

Total no. of cases = 5C3

Now P(1 M and 2 W or 2 M and 1 W) = P(1 M and 2 W) + P(2 M and 1 W)

= (2C1 . 3C2)/5C3 + (2C2 . 3C1)/5C3 = (2 . 3)/10 + (1 . 3)/10 = 9/10

**Q6) Evaluate :**

xtan x

1 – cos 2x

Ans6)

xtan x =

1 – cos 2x xsin x

cos x . 2sin2 x

x =

sin x 1 = 1 .1/2 = 1/2

2cos x

**Q9) Evaluate**

/2 sin x dx

0 1 + cos2 x

Ans9) Let I =

/2 sin x dx

0 1 + cos2 x

Put cos x = t => sin x dx = -dt

Also x = 0 => t = 1 and x = /2 => t = 0

… I = 0 -dt = – [tan-1 t]10 = -[0 - /4] = /4

1 1 + t2

**15) Evaluate**

/3 sec x tan x dx

0 1 + sec2 x

Ans15)

/3 sec x tan x dx

0 1 + sec2 x

Put sec x = t

sec x tan xdx = dt

Now when x = /3 => t = 2

x = 0 => t = 1

2 dt = tan-1 (t)

1 1 + t2 2

1

tan-1 (2) – tan-1 (1)

= tan-1 (2) – /4

**Q24) Prove that**

b+c c+a a+b

q+r r+p p+q

y+z z+x x+y = 2 a b c

p q r

x y z

**Q2) Express the matrix**

A = 3 1

-4 -1

as the sum of a symmetric and a skew symmetric matrix.

Ans2) We know for a square matrix A. 1/2(A + A’ ) is sym and 1/2(A – A’ ) is skew-symmetric. Also

A = 1/2 (A + A’ ) + 1/2 (A – A’ )

Now

A =

3 1

-4 -1

=>A’ =

3 -4

1 -1

… 1/2 (A + A’ ) = 1/2

6 -3

-3 -2

1/2(A – A’ ) = 1/2

0 5

-5 0

Thus A =

0 -3/2

-3/2 -1

+

0 5/2

-5/2 0

where first one is symmetric and the second skew symmetric

**Q3) Evaluate**

lim

x->/2 (2x – )/cos x

Ans3)

lim

x->/2 (2x – )/cos x putting /2 – x = , as x -> /2 , -> 0

= lim

->0 (2(/2 – ) – )/cos (/2 – )

= lim

->0

-2/sin = -2 (as lim /Sin = 1)

->0

(i) All Questions are compulsory

(ii) Question number 1 to 15 are of 2 marks each

(iii) Question number 16 to 25 are of 4 marks each

(iv) Question number 26 to 30 are of 6 marks each

**Q1) A coin is tossed 12 times. What is the Probability of getting exactly 8 tails? (Marks 2)**

Ans1) When a coin is tossed, we have S = {H, T}

p = P (getting a tail) = 1/2

… q = 1 – p = 1/2

Here n = 12

Now the Probability Distribution

B(n, p) = B(12, 1/2)

i.e. (q + p)12 where q = p = 1/2

… P (X = = 12C8 . q12 – 8 . p8

= 12C4 . q4 p8 = 12×11x10×9 x (1/2)4 + 8

4

= 498 = 495

212 4096

**Q2) Three coins are tossed simultaneously. List the sample space for the event. (Marks 2)**

Ans2) Sample space S, of tossing 3 coins simultaneously is given by:-

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

**Q3) Two cards are drawn without replacement from a well shuffled pack of 52 cards. What is the probability that one is red queen and the other is a king of black colour? (Marks 2)**

Ans3) We know that there are 2 red queens and two kings of black colour in a pack of 52 cards.

P (drawing a red queen) = 2/52 = 1/26

Since the card is not replaced, after the firs card, we have 51 cards left.

… P (drawing a black king) = 2/51

But these two draws are interchangeable.

… The reqd. probability = 2 x 1/26 x 2/51 = 2/(13 x 51) = 2/663

**Q4) Find a unit vector perpendicular to both = 3 + – 2 and = 2 + 3 – (Marks 2)**

Ans4) Here, = 3 + – 2 and = 2 + 3 -

=> x =

3 1 -2

2 3 -1

= 5 – + 7

… = Unit vector | to and

= ( x )/[ x ] = (5 – + 7)/(52 + (-1)2 + 72)

= (5 – + 7)/(75) = 1/(53) (5 – + 7)