95M-4
Sr. No. 4
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Marine Engineering at Operational Level
MATHEMATICS
CLASS IV
(Time allowed - 3 hours)
Morning Paper
INDIA (2001)
N.B. -
Total Marks 100
(1) All Questions are compulsory.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
x2
2
1. lf the normal at the end of latus rectum of the ellipse a
minor axis, then prove that e4 + e2 -1 = 0.

y2
b2
1
passes through an extremity of
2. Find the equation of the ellipse with its centre at origin axes along the co-ordinate axis and which
passes through the points (2,2) and (3,1).
3. Find the differential equation of the family of curves xy = Aex + Be-x + x2.
4. Eliminate the constant a,b from the relation asinx + bcosx = y to form a different equation.
5. Prove that the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 cuts the sphere x2 + y2 + z2 + 2u'x +
2v'y + 2w'z + d' = 0 in a great circle, if 2 ( uu' + v v' + ww' ) = 2r'2 + d + d' , where r' is the radius of
the second sphere.
6. A five digit number is formed by the digits 1,2,3,4,5 without repetition. Find the probability that the
number formed is divisible by 4.
7. (a) Prove that tan (/4 + /2) + tan (/4 - /2) = 2 sec 
(b) If A + B + C = 180O prove that –
sin2 A/2 + sin2 B/2 + sin2 C/2 = 1 – 2 sin B/2 sin C/2.
8. If  and  be the roots of ax2 + bx + c = 0, find the value of the following terms of the coefficients: (a)  - 
(b) 2 + 2
(c) 3 + 3
(d) 4 + 4.
9. Find the square root of x2 + y2 – 1
x - y – 9
2
2
y
x
j
y x
4
10. Integrate the following (a)
(x + 1/x) (x2 + 1/x2)dx
(b)
x2 cos 2 x dx
(c)
x3 tan4 x2 sec2 x4 dx
(d)
cos x – cos 2 x dx
1 – cos x
------------------------------X-------------------------------
95M-4
Sr. No. 4
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Marine Engineering at Operational Level
MATHEMATICS
CLASS IV
(Time allowed - 3 hours)
Morning Paper
INDIA (2001)
N.B. -
Total Marks 100
(1) All Questions are compulsory.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
Answers
Answer for Question No. 1
Ans : We know that the extemity of latus rectum in the lst quadrant is (ae, b2/a) normal at which is
x  ae
y  b2 / a

ae / a2
or ,
b2 / ab2
x  ae g

b
a
e
a
F
y
G
H
b2
a
I
J
K
lt passes through the extemicity of minor axis is the 2nd quadrant i.e., ( 0, - b)
0 , ae g

b
a
e

or, a2 e2 = ab
e4 
b2
a
2
F
b
G
H
a

b2
a
I
J
K
or, a2 - b2 = ab.
 e2 = b/a.
 1  e2 or e4  e2  1  0
Answer for Question No. 2
Ans : Let the equation of the ellipse be 4x2 + 13y2 = 1
This passes through the points ( 2, 2 ) and ( 3, 1)
Hence 4A + 4B = 1
and
9A+ B = 1
Solving these two equations, we get A = 3/ 32 , B = 5/32
Hence, equations of the ellipse is
3 2
5 2
x 
y 1
32
32
or, 3x2 + 5y2 = 32
Answer for Question No. 3
Ans : Differentiating w. r. t. , we get
x
dy
 y  Ae x  Be  x  2x
dx
Differentiating again, we get
x
d2 y
dx
2
2
dy
 4e x  Be  x  2  xy  x2  2
dx
Thus two variables A and B have been eliminated and we have got a differential equation of 2nd
x
order ,
d2 y
dx
2
2
dy
 xy  x2  2
dx
Answer for Question No. 4
d2 y
Ans : Differentiating twice we get
dx2
y
or,
d2 y
dx2
y0
Answer for Question No. 5
Ans : The equation of the plane of the circle through the given spheres is
2 ( u - v' ) x + 2 ( v -v' ) y+2 ( w - w') z + d - d' = 0
lf the circle is a great circle of the second sphere, then the centre ( -u' , - v', - w' ) lies on it.

- 2 ( u - u' ) u' - 2 ( v - v' ) v' - 2 ( w - w' ) w' + d - d' = 0,
or
2 ( uu' + v v' + ww' ) = 2 ( u'2 + v'2 + w'2 - d' ) + d + d'
or
2 ( uu' + v v ' + ww' ) = 2r'2 + d + d' )
Answer for Question No. 6
Ans : Total no. of numbers formed by the digits = 5! = 120
A number is divisible by 4 if the numbers placed in the last two digits is divisible by 4 . So that
last two digits can be 12, 24 , 32 or 52 . Corresponding to each of these 4 ways there are 3! = 6
ways .
 The favourable no. of ways
Answer for Question No. 8
(a)
(b)
(c)
(d)
b2 – 4ac / a
b2 – 2ac /a2
3abc – b3 / a3
b4 – 4ab2c + 2a2c2 / a4
Answer for Question No. 9
(a)   ( x/y – y/x +j/2)
Answer for Question No. 10
(a)
(b)
(c)
(d)
x4 /4 + x2/2 – log x – 1 / 2x2 + C
¼ (2x2 – 1) sin 2x + ½ x cos 2x + C
1/20 tan5 x4 + C
x + 2 sin x + C
= 4 x 6 = 24
Hence probability 
24
1

120
5