Mathematics Times – July 2019

Now

 

2

0

1

lim 12

sin log 1

4

x

x

e

x x

k







    

     

2

0

1

lim.

sin

x

x

x

e k

k

x x

k



 

    

    

    

   

 

(^412)
cos 1^4
4
x
x

  
 
i.e.,   1.. 1 .4 12^2 k 
 k 3
10.Sol: We have    
1 1 '^10
2
f  f f  
 
i.e.,f ^1  f^1   c^2
Now
1 1
' 0
2 2
f     b
 
Now 2 b c    1 2 1
11.Sol: Given f x  is continuous at x
i.e.,limxf x f  
Now
 ^2  ^2  
lim 2 cos 1 lim 1 cos
x x 2 cos 1
x x
  x   x x
   
   
 
2
2
2cos 1
lim^2
4 2cos 1
2 2
x
x
  x x
 
   
 
 
1 1 1
2 1 1 4
  


12. Sol: Given

9 2

2 9

f  

 

Now

2

0 2 0 2

2sin^3

1 cos3 2

limx limx

x

x

f f

 x  x

 

    

   

   

 

2

0 2

3

sin

lim^92

(^23)
2
x
x
f
 x
  
  
  
 
   
    
 
9 2
2 9
f  
 
13.Sol: Given f x x ^2
Now  
     
0
0
' 0 lim
h 0
f h f f h
f
 h h

 

We know
f h 
h
h

h f h  h
h
   
 
limx 0 0
f h
 h
 
i.e.,f' 0 0 
 f x  is differentiable at x0, which implies
continuity
f x x x R ^2 ,  is continuous as well as
differentiable at x0.
14.Sol:
1
sin , 0
( )
0 , 0
x x
f x x
x
  
   
  
 

0
lim sin^1 0 (0)
x x x f
  
  
Hence f(x) is continuous function
g(0) 0 lim ( ) x 0 g x
L.H.D:
x 0
limg(0 h) g(0)
 h
 

2
x 0
1
h sin 0
lim h
 h
 
 
 

LHD = 0
R.H.D:
h 0
limg(o h) g(0)
 h
 
10.Sol:
11.Sol:

12. Sol:

13.Sol:

14.Sol:

L.H.D:

R.H.D:

15.Sol:

2

h 0

1

h sin 0

lim n

 h

 

 

 

RHD = 0

LHD = RHD

Hence g x  at x 0 is differentiable.

15.Sol: f x x x   ^25

Given g x  is inverse of f x 