Mathematics Times – July 2019

x0,5 for which F x  0. Hence (D) is
correct.
2.Sol: If P, Q and R are collinear, then

1

1 1

PQ kPR

 

 

 

   

  

 

^

1 1

1

1



 

  



Therefore,  cannot take value of 1 and 0.

i.e., 0,1.

3.Sol: We have

2.Sol:

3.Sol:

4.Sol:

det ( ) det (^1 ) (det )(det )^1

det

R PQP P Q

P

    

 

 

det( )Q

i.e., det ( ) 48 4R   x^2

Consider option (a), we have x 1. That is,

det( ) 44 0R  

Therefore the equation

0

0

0

R







   

   

   

      

will have trival solution. i.e.,      0.

now consider option (b), in which we have

given, PQ QP

i.e., PQP QI Q^1  

^ R Q

 No value of x satisfies this condition

likewise consider option (c), we have given

2

det 0 4 0 8

5

x x

x x

 

 

 

 

(40 4 ) 8 x^2 

48 4 x^2

det ( ),R x R 

So option (c) is correct.

2 1 2/3

0 4 4/3

0 0 6

R

 

 

 

 

and finally consider option

(d), we have

1

(R I a6 ) 0

b

 

  

 

  

i.e.,

2

4 0

3

b

  a 



4

2 0

3

b

 a 

a 2 and b 3

^ a b  5

So option (d) is correct

finally consider option (d) which is.

4.Sol: rewrite the given function as

0

0

2 3

cos cos

2 2

( )

2 2

1 cos

2

n

k

n

k

k

n n

f n

k

n











      

     

       



   

    

    





0

0

2 3

( 1)cos

2 2

( )

( 1) cos 2 2

2

n

k

n

k

k

n

n n

f n

n k

n











      

     

       



    

    

   





sin ( 1)

( 1)cos^2 cos^3

(^2) sin 2
( )^2
sin ( 1)
( 1)^2 cos 2( 2)
sin 2( 2)
2
n
n n n
n n
f n n
n
n n n
n
n

 




    
       
     
       
    
  
    
      
    
     
    
 
i.e.,
( 1)cos cos
( )^22
( 1) 1
n
f n n n
n
     
    
      
 
( ) cos
2
f x
n
  
   
  
now consider option (a), we have
sin 7cos cos^1 sin 0
7

   
 