Mathematics Times – July 2019

(^113)

I x dx 1  0 (3)

and  

1

2 0 2

1

I dx

a x



  (4)

Both the integrals are easy to evaluate. Indeed,

4 1

3

1

0

3 3

[ ]

4 4

x

x

I x





  (5)

and

1

2

0

[^1 ]

x

x

I

a x





 



(^)  
7.Sol:
8.Sol:
1 1 1
a a a a 1 1
  
  (6)
The data now reduces to a quadratic in a, viz.
a a  1 72i.e. a a^2  72 0 , whose
roots are 8 and -9.
7.Sol: Let
2 1 3
1 0 2
3 2 1
Q
 
 
 
 
then the given
equation is written interms of Q, we get
6
1
( K KT)
k
x P QP




now

6

1

T ( k KT T)

k

X PQP





We have (A B A B )T T T

6

1

T ( K KT T)

k

X P QP X



  

Let

1

1

1

R

 

 

 

  

, then

6

1

K KT [ KT ]

k

XR P QP R P R R



  

6 6

k 1 K K 1 K

P QR P QR

 

  

  

But

6

1

2 2 2

2 2 2

2 2 2

k PK

 

 

 

 

 and

6

3

6

QR

 

 

 

  

, then

2 2 2 6 30

2 2 2 3 30 30

2 2 2 6 30

XR R

     

      

     

         

^  30

now

6

1

Trace Trace K KT

k

X P QP



  

 

 



6

1

Trace(K KT) 6(Trace ) 18

k

P QP Q



  

i.e.,

1 1

1 30 1

1 1

X

   

   

   

     

^

1

( 30 ) 1 30 0

1

X I O X I

 

     

 

  

 (^) X I 30 is non-invertible
8.Sol: (b,c,d) 2
sin
( ) ; 0
x
f x x
x

 
2
4
cos( ) 2 sin
'( )
x x x x
f x
x
   

(^44)
cos tan
cos ( 2tan ) 2 2
x x x
x x x x x
x x
  
  
  
  
 
1
ln 2 ,2
2
 n n 
 
the derivative goes from
positive to negative. Therefore, it’s a point of
maxima.
Similarly, in
3
ln 2 1,2
2
 n n  
 
there is a
minima.
So, the first minima 1 1
3
:1
2
y y  and the first
maxima 1 1
5
: 2
2
x x  and tan 1 1
2
y
y

 
and tan( ) 1 1
2
x
x

 