Mathematics Times – July 2019

7.Sol: It is easy to check (A) by looking at the

sign of f x'  for x  , 0. Note that

f x x x x x'   54 20 3  30 2   20 3 forx 0.

So f'         (^1)  5 20 30 20 3 2 0.
Hence fcannot be increasing on the entire
interval , 0. So (A) is false.
For (B) and (D), we first need to check if
f' 1 exists. This follows because for
0  x f x x1, '   2 1 which tends to 1
as x 1 . By the Lagrange MVT, the left
handed dervative of fatx 1 exists and
equals 1. The same is true for the right handed
derivative at x 1.
Since f x x x'   (^228)   7 1 asx 1 .
Sof'exists and equals 1 atx 1. Note also
that f x''  2 0 for 0  x 1. Sof' is
increasing on (0,1) by LMVT.
Butf x x''    4 8 0 for 1 x< 2. Sof'
changes its behaviour from increasing to
decreasing at x 1. Thereforef' has a
local maximum atx 1. That is, (B) is true.
However, as for differentiablility of f' at
x 1 , f x''  ^22 as x 1  while
f x x''    4 8 4 as x 1 . Hence
f'' 1 does not exist. So (D) is also true.
Finally, for (C), since f 0  1 but
f x  (^)  as x , by the intermediate
value Property of continuous functions,
f, 0 includes ,1. On the other
hand, for x3, f x'  1 logx   2  1 0.
So f is strictly increasing on 3,and tends
to  as x. Also  
1
3
3
f . Hence by
the IVP again, all points in
1
,
3
 
 
 
are in
the range of f. Since  
7.Sol:
8.Sol:
1
,1 ,
3
      R
 
fis onto.
8.Sol:  , are roots of x x^2   1 , then
2 2
2
( r r ) ( r r)
a ar r
   
 
 

  
 

(   r^2 r) ( r^2 r)
 
   


1 1
1
r r
ar
 
 
 


 

^ a a ar 2  r 1 r
i.e.,
2 2
1 2 2 2
n
r a a a ar n n
 
    

   

 

    an 2 ( ) an 2 1

Now^11

1

10 10

10

n n

n n n

n n

a

 

 

 

  



    

   

    



 



10 10

1 1

10 10

 

 

 



 





10 10

(10 )(10 ) 89

 

 

1 1

1 1

10 10 12

10 10 1 1 89

10 10

n n n

n n n n

b a a

 

 

 

 

 



   

 

 

Further, b a an n 1 n 1

In other words, the number bnsatisfy a very

known Recurrence relation called the

Fibonacci relation.

i.e.,

( n^1 n^1 ) ( n^1 n^1 )

bn

   

 

    



