Archive

Archive for the ‘Maths’ Category

Maths-1996 -Set I

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

Maths-1996 -Set II

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

Maths-1997 -Set I

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

Maths-1997 -Set II

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

Maths-1997 -Set III

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

Maths-1998-Set II

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

Maths-1999-Set I

(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 = 8) = 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)

Maths-1999-Set II

Q1) If y = cot x, show that d2y/dx2 + 2y dy/dx = 0 (Marks 2)
Ans1) Here y = cot x
differentiating both sides w.r.t. x:-
dy/dx = -cosec2x
differentiating again w.r.t. x:-
d2y/dx2 = -d/dx (cosec2x)
d2y/dx2 = -2cosecx(-cosec x . cot x ) = +2cosec2x . cot x
or d2y/dx2 = -2 dy/dx . y
or d2y/dx2 + 2y dy/dx = 0.

Q3) If
A = 2 2
-3 1
4 0 B = 6 2
1 3
0 4

find the matrix C such that A + B + C is a zero matrix. (Marks 2)
Ans3) Here
A = 2 2
-3 1
4 0 B = 6 2
1 3
0 4

we want to find C such that A + B + C = 0
< => C = 0 – A – B = -A – B
= – 2 2
-3 1
4 0 – 6 2
1 3
0 4
= -2 -2
3 -1
-4 0 + -6 -2
-1 -3
0 -4
= -8 -4
2 -4
-4 -4

Q5) Construct a 3 x 2 matrix whose elements in the ith row and the jth coloumn are given by:-
aij = (3i + j)/2 (Marks 2)

Ans5) We want a 3 x 2 matrix
=> 1 i 3 and 1 j 2
where aij = (3i + j)/2
… a11 = (3 + 1)/2 = 2; a12 = (3 + 2)/2 = 5/2;
a21 = (6 + 1)/2 = 7/2; a22 = (6 + 2)/2 = 4;
a31 = (9 + 1)/2 = 5; a32 = (9 + 2)/2 = 11/2
… The reqd matrix is :-

a11 a12
a21 a22
a31 a32
=

2 5/2
7/2 4
5 11/2

Maths-2000-Set I

Q1) Evaluate
{(1 – Cos x)/x2}

Ans1)
{(1 – Cos x)/x2} = (2Sin2x/2)/x2
= 2/4 (Sin(x/2)/(x/2))2 =2/4 . 12 = 1/2 (as Sinx/x=1)

Q2) A particle moves along a straight line so that S = t . Show that the acceleration is negative and is proportional to the cube of velocity.
Ans2) Here the equation of motion is
S = t
differentiating both sides w.r.t. ‘t’
v = dS/dt = 1/2t = 1/2S ( S = t )
or v = 1/2S
differentiating again w.r.t. ‘t’
accn = dv/dt = -1/2s2 ds/dt
= -2/4s2 ds/dt
= -(2/(1/v2)). v = -2v3
accn is negative and is proportional to cube of velocity.

Q3) Evaluate
Cos4xdx

Ans3) let
I = Cos4xdx = (Cos2x)2dx
= ((1 + Cos 2x)/2)2dx = 1/4 (1 + Cos22x + 2Cos 2x)dx
=1/4 [1 + (1 + Cos 4x)/2 + 2Cos 2x]dx
= 1/4[x + x/2 + (Sin 4x)/8 + (2Sin 2x)/2] + C
= 3x/8 + (Sin 4x)/32 + (1/4)Sin 2x + C

Q4) Evaluate
(1 + tan x)/(x + log sec x) dx

Ans4) Let
I = (1 + tan x)/(x + log sec x) dx
= log | x + log sec x | + C
[ as f'(x)/f(x) dx = log | f(x) | + C ]

Pages: 1 2 Next