Mathematics Times – July 2019

  

2

0 1 0

cos 2sin sin cos 0 1 1 0

1 sin cos

x x x x

x x

    

 cos 2sin 0x x or sin cosx x

i.e.,

tan^1

2

x  or tan 1x

 It has two solutions.
8.Sol: Given A A I^2   5 7 0

i.e., A A I^2    5 7

 A A A A A IA.. ^1 5. ^1  7 ^1

 A I A 5 7 ^1

i.e.,^1  

(^15)
7
A  I A
Statement-I is not correct
Now A A A I^2 2 3^2  
A A I A A I5 7 2 3 ^2  
5 7 2 3 1A A A A^2  ^2  
3 5 7 10 A I A I  
5 20A I
 (^5) A I (^4) 
9.Sol: Let
2
2
2
1 2
2 3 1 3 3 3 12
2 3 2 1 2 1
x x x x
x x x x ax
x x x x
  
    
   
Put x 1, we get
0 0 3
2 3 0 12
2 3 3
a

    
 
     3(6 6) a (^12)    36 12 a
 a 24

10. Sol: Given the determinant

1 1

1 1

1 1

x

y

z

is non

negative. i.e.,

1 1

1 1 0

1 1

x

y

z



xyz x y z    2 0

using A.M-G.M inequality, we get

 

xyz     2 x y z xyz 3 1/3

(^) xyz 2 3 xyz1/3 0
Put xyz t^3
(^) t t^3   3 2 0
(^) t t (^2)  1 0^2 
(^) t  (^2) t^3   8

11. Sol: We know k adj A k A n n^1
    Here n A3;  5

i.e., 5 adj A 53 A^2

(^)   5 adjA 53 A^2  5 given
1
5
 A 

12. Sol: Let  

1 cos 1

sin 1 cos

1 sin 1

f



  



  



 1 sin cos cos sin cos    

 1 sin^2 ^1 

(^)  1 sin cos sin cos cos sin 1   ^2 ^2 
(^)  2 2sin cos cos 2  
(^)  2 sin 2 cos 2  (1)
Now, maximum value of f is
(^) 2 1 1 2 2^2   ^2
and minimum value of f is
8.Sol:
9.Sol:

10. Sol:
    11. Sol:
    12. Sol:

13.Sol:

14.Sol:

2 1 1 2 2.^2   ^2

13.Sol: Given B^2   0 B^3 0,...Bn 0

Now  

det 1 B^50  50 B

 

50 50 1 50 2 50 50

 C C B C B 0  1  2  ... C B 50  50 B

50 50

 C C B B 0  1  (^50) 
50
 C 0  1
14.Sol: Given
      
2
2 1 3 2
/ 2 1
(^1111) 1 3 4
2 2
r
r r r
n n a
n n n n
 
  
   