Mathematics Times – July 2019

i.e.,    ' 

1

g f x'

f x



1

g'(f (x))

f '(x)



1

g'(f (2))

f '(2)

 

1

g'(7)

3

 

16. Sol: Given  

2

2

sin , 0

sin , 0

x x

h x

x x

 



 

 

2

2

2 cos , 0

'

2 cos , 0

x x x

h x

x x x

 

 

 

We can see from the above function

h' 0 h' 0 h' 0 

Therefore h x'  is continuous at x 0

Now  

2 2 2

2 2 2

2 cos 2 sin , 0

''

2 cos 2 sin , 0

x x x x

h x

x x x x

    



    

Similarly, we can see that h'' 0 h'' 0 . 

Therefore h x''  is not continuous at x 0

So h x'  is not differentiable at x 0

17.Sol: Let P x a x b x c 1   12   1 1

P x a x b x c 2   2 2  2  2

P x a x b x c 3   3 2  3  3

When a a a b b b c c c1 2 3 1 2 3 1 2 3, , , , , , , , are real numbers.

 

2

1 1 1 1 1 1

2

2 2 2 2 2 2

2

3 3 3 3 3 3

2 2

2 2

2 2

a x b x c a x b a

A x a x b x c a x b a

a x b x c a x b a

    

    

 

    

Given      

T

B x A x A x  

 

12 1 1 22 2 2 32 3 3

1 1 2 2 3 2

1 2 3

2 2 2

2 2 2

a x b x c a x b x c a x b x c

B x a x b a x b a x b

a a a

       

    

 

 

2

1 1 1 1 1 1

2

2 2 2 2 2 2

2

3 3 3 3 3 3

2 2

2 2

2 2

a x b x c a x b a

a x b x c a x b a

a x b x c a x b a

    

    

 

    

From the above multiplication, we can see that the

degree of determinant of B x  cannot be less than

4.

18.Sol: Given f x  satisfies all condition of Rolle’ss

theorem in the interval 1,1

i.e.,f^1  f^1 f c' ^0

 2      a b 2 a b

  b 2

now

'^1 0 6^11 2 2 0

2 4 2

f           a  

     

1

2

 a

 2 a b   1

19.Sol: Given f x x      1 ; 1x x 3

Clearly f x  is not continuous at integral values

of x. i.e., x0,1,2,3.

But in statement -2 f x  is continuous at x 3

Statement 2 is false

20.Sol: Given that

  2

1

2

f u

u u



  and

 

1

1

u x

x





Now, it is clear from the given function u x  is

discontinuous at x 1 and

f u  is discontinuous for u 2,1

i.e.,

1 1

2

1 2

x

x

   

 and

1

1 2

1

x

x

  



 The given composite function is discontinuous

for

16. Sol:

17.Sol:

18.Sol:

19.Sol:

20.Sol:

21.Sol:

(^1) ,1,
2
x and 2.
21.Sol: fog f g x f x     1  
fog f g x f x     1  
    1 1 x 2
   ^1 x^1
Let fog x 
    y 1 x 1
, 0
, 0
x x
y
x x
 
 
 