Mathematics Times – July 2019

Now  ^3  ^2

3 9

'

2 2

f x x f x x   and

f x x''  9

i.e.,  ^2

9

' 2 2 18

2

f   and f'' 2 18 

 f' 2 f'' 2 f'' 2 f' 2 0 

6.Sol: Given
2 15 2 15
y x x   1   x x  1 

 

2 14

15 1 1 (^21)
dy x x x
dx x
 
     
  
 
2 14
15 1 1 (^21)
x
x x
x
 
     
  
2
(^15).
1
dy y
dx x
 

^21 ^15
x dy y
dx
  
after differentiating with respect to x, we get
2
2
2 1 2 15
1
x dy x d y dy
x dx dx dx
  

 
2 2 2
(^22)
15
1 15 1.
1
dy d y
x x x y
dx dx x
    

 225 y
7.Sol: Given that f x  is continuous in the interval
[0, )
 it is continuous at x1, 2
xlim 1 f x xlim 1 f (^1) 
i.e.,
 
2
0
2 1
limx
h
a
 a


 a^2  2
i.e.,a  2
nowxlim 2 f x xlim 2 f x f  ^2 
i.e.,  
2
0 3
2 4
lim
h 2
b b
a
 h



a 2 { Continuous at x 2 }
2 4^2
2 2
b b
a

 
i.e.,2 4 4b b^2  
   b b^2 2 2 0
b 1 3
b 1 3
Hence a b,  2,1 3 
8.Sol: Given f x  is differentiable at x 1
i.e.,
     
0 0
1 1 1 1
limh limh
f h f f h f
 h  h
   


(^1)   (^1)  
0
cos 1 cos 1
1 limh
h b b
h
 

   
  
i.e.,   2
1
1
1 1 b

 
 
 b  1
Now, limx 1 f x limx 1 f x 
xlim 1 1   h lim cos 1h^0 a ^1  h b
i.e.,  ^1 a cos 1^1  b
    1 a cos 0^1  
i.e.,^1
2
a

  
1
2
 a  
Now
6.Sol:
7.Sol:
8.Sol:
9.Sol:
2
2
a
b

9.Sol: Given that the function is continuous.
^ f x  is Continuous at x 0
 limx 0 f x  12