SECTION-
5.Sol:
meets the parabola y x^2 at twopoints
B 1,1 and also at F 1,1.
The region, say R 1 bounded by the three curves
is a part of the given set. It is easier obtain its
area, say by horizontal than by vertical
slicing. On R 1 , y varies from 1 to 4. For each
y 0 1, 4 , the intersection of R 1 with the
liney y 0 is the segment from y 0 ,1to
0
8
, 0
y
. It is shown by DE in the figure
above. Its length is^0
0
8
y
y
. Hence
4
0
(^10)
8
y dy
y
3 4
2
1
2
(8 log )
3
y
e
y
y y
16 2
8log 4
e 3 3
14
16 log 2
e 3
So A is correct.
SECTION-
5.Sol: Given an ellipse
2 2
: 2 2 1
x y
E
a b
(1)
let us first identify the rectangle, say R,of
maximum area inscribed in E and sides parallel
to the axes of E. If a bcos , sin is its
vertex in the first quadrant, the other vertices
are obtained by taking reflections into the axes.
So the sides of this rectangle are 2 cosa
and 2 sinb . Hence its area is
4 absin cos 2 ab 2 which is maximum
when
4
.
So, the rectangle R of maximum area has its
vertices at ,
2 2
a b
.
Next, we have to find an ellipse E'with sides
parallel to those of R and having the largest
area. Clearly, the marjor and minor axes
ofE'must be the lengthand the width of R,
i.e. 2 a and 2 brespectively. So,
2 2
' : 2 2 1
2 2
x y
E
a b
(2)
This construction of interlacing ellipses and
rectangles then goes on. The figure above
shows the construction ofR 1 from E 1 and
E 2 fromR 1.
The crucial observation is that the lengths of
the semi-major and semi-minor axes each form
a geometric progression with common
ratio
1
2
. We have a b 1 3, 1 2. Hence for
the n th ellipse En,we have
1
3
2
an n (3)
and ^1
2
2
bn n (4)
Note that the ratio of the major to the minor
axis of each ellipse is the same, as follows
3
2
.
Since the eccentricity depends only on this
ratio, (A) is immediately ruled out. Further this