Detailed Solutions MATHEMATICS - I D-01/DC-01 December 2004
Q1(a)
Ans. C
Q1(b)
Given
= 
=
=
Ans.
B
Q1(c)
=
=
Ans. A
Q1(d)
Since
are
roots of 
Now
Ans.
B
Q1(e)
Ans. C
Q1(f)
Ans.
A
Q1(h) The equation of line perpendicular to
is 
This passes through (-3,2)
=
=
=
From (i), required equation is
= 
Ans.
C
Q2(a) Let
, For 
which
is divisible by 7 . Let 
,where k is a positive integer
i.e.
is
divisible by 7. We have to show that this relation is true for 
Here
is
divisible by 7 and 7 itself divisible by 7. Thus P(k+1) is divisible by 7.
Hence result is true for 
,But it is true for 
. Thus it is true
for every positive integer
Q2(b) Let 
and 
Now
Q3(a) 
=
Q3(b) 
Cubic both sides
̃
or
̃
Q4(a)
When DABC is an acute angled
triangle.Draw perpendicular CD from C on AB
In
D CAD, we have 
In DCBD, we have
In DCBD,
Q.4 (b) Let 
L.H.S.
a Sin(B-C)+ b Sin (C-A)+c Sin(A-B)
= K Sin A Sin(B-C)+K Sin B Sin(C-A)+K Sin C Sin(A-B)
= K[Sin (B+C) Sin (B-C)+Sin(C+A) Sin(C-A)+Sin(A+B) Sin(A-B)]
= K(Sin2 B-Sin2C+ Sin2C -Sin2A+Sin2A-Sin2B]
= K(0)
= 0 = RHS
Q.5.a. Let A= (1,1) ,B= (3,5) ,C= (a,b)
The given points are collinear if x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0
Q.5.b. A line
through point (4,5) is 
This makes angle 450
with the line
whose slope is
-2. Therefore.
tan
450 =
or 
The
required lines are 
Q.6(a) Given circle is x2+ y2-4x-6y-9=0. Its center is (-f,-g) =(2,3)
The equation of circle whose center is (2,3) and radius r is (x-2)2+(y-3)2= r2
It passes through (-4,5) ̃(-4-2)2+(5-3)2 =r2 ̃ 36+4 = r2 ̃ r2 = 40
Required Circle is (x-2)2+(y-3)2=40
Q6(b) y2-8y-x+19=0 ̃ (y-4)2 =(x-3) ..............(1)
Let
Y = y-4, X= x-3 (1) becomes Y2=X, which is a parabola.
Here 4a=1 ̃ 
Vertex: Vertex = (X=0, Y=0) ̃ (x-3=0, y-4=0) ̃ (x=3, y=4) So, Vertex = (3,4)
Focus: (X=a,
Y=0) ̃ 
̃
Directrix:
Equation of directorix is X= -a ̃
x-3= -
̃ x = 
Q.7(a) 
= 
= 3.1 = 3
Q.7(b) Let a be any real number, then there exists an integer k such that k-1£ a £ k,
Case1: a ¹k-1
(LHL
at x=a) = 
(RHL
at x=a) = 
and
f(a)=k-1. Thus
so,
f(x) is continuous at x=a.
Case2: a=k-1
Now
while
Thus f(x) is not continuous at point a=k-1. Thus f(x) continuous at all points x¹ an
integer while it is discontinuous at integer points.
Q.8(a)
Let
be
the semi-vertical angle of a cone of given slant height 
. Then, CO=Cos, OA= sin. Let V be the
volume of the cone.
Then
For
maximum or minimum V, 
Thus
V is maximum when tan 
Q.8(b) The equation of given curve is y=x2-9..............(1)
This cuts the x-axis at the point where y=0 ̃ x2-9=0 ̃ x=3
Point of contact = (3,0) Differentiating (1) w.r. to x, we get
Equation of tangent at (3,0) is y-0 =6 (x-3) ̃ y -6x+18=0
Evaluation
of normal at (3,0) is y-0= 
Q.9(a) Let In = 
=
=
=
=
=
Q.9(b) Let I=
………………….. (1)
=
=
……(2)
Now
2.I
=
=
= 
̃ I = 
Q10(a) A rough sketch of the parobola is shown in Fig.
Let
S(a,o) be the focus and 
, be the latus rectum of the
parabola y2=4ax. The required area is 
Since the curve is symmetric about
x-axis. So, required area = 2 area (
)
Here, we slice the area
into vertical strips. For the
approximating rectangle shown in fig. we have length =y, width = Dx
Area
= y Dx = 
Since the approximating rectangle can move between x=0 and x=a
\Required area = 2 Area =2
= 
Q10 (b)Volume
of solid 
Q11(a) 
This is homogeneous equation
Let y=vx
= v+x
̃\v+x
̃ v+x
̃
x
=
= 
Separating the variables
Intigrating both sides
. Let
1-2v-v2 = t on l.h.s. ̃
(-2-2v)dv=dt
̃ (1+v)dv= 
\
̃
̃ 
= (1-2v-v2)-½
= 
̃ 
Q11 (b) 
̃
I.F.
= 
Solution is
