Mathematics Times – July 2019

 ( ) sin^4 cos^4  (6)

and

 ( ) sin^4  cos^4  2 sin^2  cos^2 (7)

The problem is now reduced to minimising

these functions over 0, 2. As a first

measure of simplification we note that both of

them can be expressed in terms of sin^2 

cos^2 . This is obvious for (7). For (6),

rewrite sin^4 cos^4  1 2sin^2  cos^2.

As  varies over [0, 2 ], sin^2  cos^2 varies

from 0 to

1

4

. So, we get

(^) tsin^2  cos^2 (8)
then
1
min 1 2 : 0,
4
 t t
  
    
  
and
min 2 2 : 0,^1
4
 t t t
  
     
  
Clearly
1
2
. For , we maximise
2  t t^2 for 0,^1
4
t  
 

. As t and t^2 increasing

functions over this interval, the maximum

occurs at

1

4

t and equals

1 1 37

2

4 16 16

  .

Hence

37

16

 .

Hence

1 37 29

2 16 16

     .

3.Sol: Clearly S is the set of points outside the

circle, say C, of radius 5 centred at 2 - i

shown below.

For any complex number z z z 0 , 0  0 is real

while z z 0  0 is purely imaginary. Hence the

numerator of w 0 is real while its denominator

is purely imaginary. Therefore w 0 is purely

imaginary number of the form ki for some

k. Then depending upon the sign of k the

principle argument of w 0 will be either

2



or

2



. From the diagram it is clear without much

work that P x y( , 0 0 ) lies in the positive

quadrant inside the sqaure with vertices at 0

and 1. So both x y 0 , 0 lie in (0, 1). Hence

4     z z 0 0 4 2x 0 2 .Similarly,,

z z i y i 0     0 2 (2 0 2) and

2 y 0  2 2. So, w 0 is of the form

a

bi

where

a, b are positive. Hence 0

a

w

bi

  and

therefore its argument is

2



.

4.Sol:xy 8 represents a rectangular hyperbola

with one branch in the first and other in the

third quadrant. But as we have y 1 for all

points in the given set, we ignore the branch in

the third quadrant. y x^2 is a parabola with

vertex at O. It meets the hyperbola xy 8 only

at one point 2, 4. Call it A.

3.Sol:

4.Sol:

The line y 1 meets the hyperbola xy 8 on;

y at the point 1,8. Call it C. The line y 1