Mathematics Times – July 2019

now,

2

4 2 2

2 4 2 ( 4)( 4)

2 2 4

x x x

x x x Bx x

x x x



   



1 1 ;^22

R R R R

x x

  and R^3 R 3

x



2

4

1 2 2

4 4 4

2 1 2 1

4

2 2 1

x

B

x x x

x



     

  

  



as

1 2 2

2 1 2 5

2 2 1

x    B B

The desired order pair is 4,5 .

2.Sol: Given

3 1

(^15) 1 28
2
2 1
k k
k
k



5 13 46 0k k^2    or 5 13 66 0k k^2   
Now for 5 13 46 0k k^2   
(^23) ; 2.
5
 k k
 k 2 { k is an integer}
and for 5 13 66 0k k^2    has no real solution
exist.
for orthocenter we know BH AC
2 8 1
5 4


      
      
  2 1 (1)
also CH AB
2 8 1
2 3


     
    
3 8 10   (2)
Solving (1) and (2), we get
2,^1
2
 
ortho centre is
2,^1
2
 
 
3. Sol: Given 2 1 ; z (^) z i 3
3 1
2
 i
  is a cube root of unity
Now
2 2
2 7
1 1 1
1 1 3
1
  k
 
  
R R R R 1    2 3 1
2.Sol:

3. Sol:

4.Sol:

2 2

2 7

3 0 0

1 1 3

1

  k

 

  

 3 7    ^21   3 k

    1   k

i.e., k  1 2

(^)  z
4.Sol: Given
2 3
4 1
A
  
 
 
2 16 9
12 13
A
 
  
 
3 2 48 27
36 39
A
  
  
 
Also
24 36
12
48 12
A
  
 
 
3 12^2 48 27 24 36 72 63
36 39 48 12 84 51
A A
        
       
     