SECTION-
5.Sol:
meets the parabola y x^2 at twopointsB 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 eachy 0 1, 4 , the intersection of R 1 with theliney y 0 is the segment from y 0 ,1to08
, 0
y
. It is shown by DE in the figure
above. Its length is^0
08
y
y. Hence4
0(^10)
8
y dy
y
3 4
2
12
(8 log )
3y
e
yy 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 maximumwhen
4
.So, the rectangle R of maximum area has itsvertices 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 12 2x 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 commonratio1
2. We have a b 1 3, 1 2. Hence for
the n th ellipse En,we have 13
2an n (3)and ^12
2bn n (4)Note that the ratio of the major to the minoraxis of each ellipse is the same, as follows3
2.Since the eccentricity depends only on this
ratio, (A) is immediately ruled out. Further this