Mathematics Times – July 2019

Since x y 1  1

tan ( ) tan ( )x 1  y 1

tan ( ) tan (x 1  y 1 1) and 1

5

2 1

2

  y

Notice the function tant is increasing in

5

2,

2

 

 

 

^ x y 1   1 1

Generalising: x yn n 1 similarlyy xn 1  n 1.

Since the function is differentiable, therefore

the points of maxima and minima occur

alternately.

Therefore, x y xn n 1  n 1

Now x x x yn 1  n (n 1  n 1 ) ( y xn 1  n) using

the foregoing proofs every quantity inside these

brackets is greater than 1. Therefore,

x xn 1  n 2.

SECTION-II
9.Sol: Given that

2 1

1

( 1)

( 1)2 2

2 0

2 4

n n

n n

n n

n n n

n

 





  





( 1)^2 ( 1)^2

0

2 8 4

n n n n n

   

^ n 0 or 4( 1) ( 1) 2 0n n n  n

i.e., n 4

now

4 4 4 5

1

0 1 0 5

k k

k k

C C

k



 



  



2 1 31^5

6.20

5 5



 

10.Sol: call the expression in the parentheses as
E. As a first measure of simplification we
replace the secants by cosines which we are
more comfortable with. Thus

 

10

0

1 1

4 7 7 1

cos cos

12 12 12 2

k

E

  k  k 



    

     

   



(1)

we notice that the denominator of each term

is of the form cos cos where

2



  .

As a result, cos sin and the product

cos sin  is a amenable to a simplification.

So,

10

0

1 1

2 7 7

2cos sin

12 12 12 2

k

E

  k  k

 

     

   

   



10

0

1 1

2 7

sin

6

k  k

 

  

 

 



 

10

0

1 1

(^2) sin 1
6
k k  
 
   
 
 

 (2)

As kis an integer, we can apply the (half)

periodicity of the sine function, whereby

(^) sink    1 sink  (3)
Using this and the fact that
1
sin
6 2
 
 
 
, E
further simplifies to
 
10
0 1
1 1
(^2) 1 sin
6
k k
E
  
 
  
 
 



 

(^101)
0
1
k
k



  

 

10

0

1

k

k

 (4)

This is a sum of 11 terms alternately 1 and -1,

the first term being 1. So the sum is 1. Hence

E 1.

The (unique) angle

SECTION-II
9.Sol:

10.Sol:

11.Sol:

3

,

4 4

 

  

 

with

sec 1 is 0.

11.Sol: Given that C a b R  , 

  

i.e., a b c ,

 

are coplanar