NCERT Solutions Class 12

NCERT Solutions Class 12

Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)

Exercise 5.1

1. Prove that the function  is continuous at  at  and at  

Ans. Given: 

Continuity at    = 5 (0) – 3 = 0 – 3 = – 3

And  5 (0) – 3 = 0 – 3 = – 3

Since , therefore,  is continuous at .

Continuity at    = 5 (– 3) – 3 = – 15 – 3 = – 18

And  5 (– 3) – 3 = – 15 – 3 = – 18

Since , therefore,  is continuous at 

Continuity at    = 5 (5) – 3 = 25 – 3 = 2

And  5 (5) – 3 = 25 – 3 = 22

Since , therefore,  is continuous at .


 

2. Examine the continuity of the function  at 

Ans. Given: 

Continuity at ,    = 

And  

Since , therefore,  is continuous at .


 

3. Examine the following functions for continuity:

(a) 

(b)  

(c) 

(d)  

Ans. (a) Given: 

It is evident that  is defined at every real number  and its value at  is 

It is also observed that 

Since , therefore,  is continuous at every real number and it is a continuous function.

(b) Given: 

For any real number , we get  

And 

Since , therefore,  is continuous at every point of domain of  and it is a continuous function.

(c) Given: 

For any real number , we get

And 

Since , therefore,  is continuous at every point of domain of  and it is a continuous function.

(d) Given: 

Domain of  is real and infinite for all real 

Here  is a modulus function.

Since, every modulus function is continuous, therefore,  is continuous in its domain R.


 

4.  Prove that the function  is continuous at  where  is a positive integer.

Ans. Given:  where  is a positive integer.

Continuity at ,   

And 

Since , therefore,  is continuous at .


5. Is the function  defined by  continuous at  at  at  

Ans. Given: 

At , It is evident that  is defined at 0 and its value at 0 is 0.

Then  and 

Therefore,  is continuous at .

At , Left Hand limit of 

Right Hand limit of 

Here 

Therefore,  is not continuous at .

At  is defined at 2 and its value at 2 is 5.

, therefore, 

Therefore,  is not continuous at .


 

Find all points of discontinuity of  where  is defined by: (Exercise 6 to 12)

6.  

Ans. Given: 

Here  is defined for  i.e., on  and also for  i.e., on 

 Domain of  is  = R

 For all  is a polynomial and hence continuous and for all is a continuous and hence it is also continuous on R – {2}.

Now Left Hand limit =  = 2 x 2 + 3 = 4 + 3 = 7

Right Hand limit =  = 2 x 2 – 3 = 4 – 3 = 1

Since   

Therefore,  does not exist and hence  is discontinuous at  (only)


7. 

Ans. Given: 

Here  is defined for  i.e., on  and for  and also for  i.e., on 

 Domain of  is  = R

 For all  is a polynomial and hence continuous and for all  is a continuous and hence it is also continuous and also for all . Therefore,  is continuous on R – 

It is observed that  and  are partitioning points of domain R.

Now Left Hand limit = 

Right Hand limit = 

And 

Therefore,  is continuous at .

Again  Left Hand limit = 

Right Hand limit = 

Since 

Therefore,  does not exist and hence  is discontinuous at  (only).


8. 

Ans. Given:  i.e.,  if  and  if 

   if  if  and  if 

It is clear that domain of  is R as  is defined for  and .

For all  is a constant function and hence continuous.

For all  is a constant function and hence continuous.

Therefore  is continuous on R – {0}.

Now Left Hand limit = 

Right Hand limit = 

Since 

Therefore,  does not exist and hence  is discontinuous at  (only).


9. 

Ans. Given: 

At  L.H.L. =  And 

R.H.L. = 

Since   L.H.L. = R.H.L. = 

Therefore,  is a continuous function.

Now, for  

 

Therefore,  is a continuous at 

Now, for  

Therefore,  is a continuous at 

Hence, the function is continuous at all points of its domain.


10. 

Ans. Given: 

It is observed that  being polynomial is continuous for  and  for all  R.

Continuity at  R.H.L. = 

L.H.L. = 

And 

Since   L.H.L. = R.H.L. = 

Therefore,  is a continuous at  for all  R.

Hence,  has no point of discontinuity.


11. 

Ans. Given: 

At  L.H.L. = 

R.H.L. = 

Since   L.H.L. = R.H.L. = 

Therefore,  is a continuous at 

Now, for   and 

 

Therefore,  is a continuous for all  R.

Hence the function has no point of discontinuity.


12. 

Ans. Given: 

At  L.H.L. = 

R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

Now,   for   and for  

Therefore,  is a continuous for all  R – {1}

Hence for all given function  is a point of discontinuity.


 

13. Is the function defined by  a continuous function?

Ans. Given: 

At  L.H.L. = 

R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

Now,   for   and for  

Therefore,  is a continuous for all  R – {1}

Hence  is not a continuous function.


Discuss the continuity of the function  where  is defined by:

14. 

Ans. Given: 

In the interval  

  is continuous in this interval.

At  L.H.L. =  and R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

At  L.H.L. =  and R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

Hence,  is discontinuous at  and 


15. 

Ans. Given: 

At  L.H.L. =  and R.H.L. = 

Since   L.H.L. = R.H.L.

Therefore,  is continuous at 

At  L.H.L. =   and R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

When  being a polynomial function is continuous for all 

When . It is being a polynomial function is continuous for all 

Hence  is a point of discontinuity.


16. 

Ans. Given: 

At  L.H.L. =  and R.H.L. = 

Since   L.H.L. = R.H.L.

Therefore,  is continuous at 

At  L.H.L. =  and R.H.L. = 

Since   L.H.L. = R.H.L.

Therefore,  is continuous at 

17. Find the relationship between  and  so that the function  defined by is continuous at  

Ans. Given: 

Continuity at 

Also

 

 

 

 


 

18. For what value of  is the function defined by continuous at  What about continuity at  

Ans. Since  is continuous at 

 

And 

Here, therefore should be L.H.L. = R.H.L.

 0 = 1, which is not possible.

Again Since  is continuous at 

 

And 

Here, therefore should be L.H.L. = R.H.L.

 

 


 

19. Show that the function defined by  is discontinuous at all integral points. Here  denotes the greatest integer less than or equal to  

Ans. For any real number  we use the symbol  to denote the fractional part or decimal part of  For example,

[3.45] = 0.45

[–7.25] = 0.25

[3] = 0

[–7] = 0

The function  : R  R defined by  is called the fractional part function. It is observed that the domain of the fractional part function is the set R of all real numbers and the range of the set [0, 1).

Hence given function is discontinuous function.


 

20. Is the function  continuous at  ?

Ans. Given: 

L.H.L. = 

R.H.L. = 

And 

Since   L.H.L. = R.H.L. = 

Therefore,  is continuous at 


 

21. Discuss the continuity of the following functions:

(a) 

(b)  

(c)  

Ans. (a) Let  be an arbitrary real number then 

 

 

Similarly, we have 

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(b) Let  be an arbitrary real number then 

 

 

Similarly, we have 

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(c) Let  be an arbitrary real number then 

 

 

Similarly, we have 

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.


 

22. Discuss the continuity of cosine, cosecant, secant and cotangent functions.

Ans. (a) Let  be an arbitrary real number then 

 

 = 

  for all  R

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(b)  and domain  I

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(c)  and domain  I

 

 = 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(d)  and domain  I

 

 = 

 = 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.


23. Find all points of discontinuity of  where .

Ans. Given: 

At  L.H.L. = 

R.H.L. = 

  is continuous at 

When  and  are continuous, then  is also continuous.

When  is a polynomial, then  is continuous.

Therefore,  is continuous at any point.


24. Determine if  defined by  is a continuous function.

Ans. Here,  = 0 x a finite quantity = 0

Also 

Since,   therefore, the function  is continuous at 


 

25. Examine the continuity of  where  is defined by .

Ans. At  L.H.L. = 

  = 

R.H.L. = 

  = 

And 

 

Therefore,  is discontinuous at 


 

Find the values of  so that the function  is continuous at the indicated point in Exercise 26 to 29.

26.  at  

Ans. Here,      

 

 

Putting  where 

 = 

 = 

 ……….(i)

And  ……….(ii)

  when  [Given]

Because  is continuous at 

 

 From eq. (i) and (ii),

 


27.  at  

Ans. Here, 

  and 

 

 

Now,   when , we have

Therefore,  is continuous at  when .


28.  at  

Ans. Here, 

And 

Also 

Since the given function is continuous at 

 

 

 

 


29.  at  

Ans. When  we have  which being a polynomial is continuous at each point 

And, when  we have  which being a polynomial is continuous at each point 

Now    

 ……….(i)

  …….(i)

Since function is continuous, therefore, eq. (i) = eq. (ii)

 

 

 


 

30. Find the values of  and  such that the function defined by  is a continuous function.

Ans.  For  function is  constant, therefore it is continuous.

For  function  polynomial, therefore, it is continuous.

For  function is  constant, therefore it is continuous.

For continuity at  

 

  ……….(i)

For continuity at  

 

 ……….(ii)

Solving eq. (i) and eq. (ii), we get

 and 


 

31. Show that the function defined by  is a continuous function.

Ans. Let  and , then

Now  and  being continuous it follows that their composite  is continuous.

Hence  is continuous function.


32. Show that the function defined by  is a continuous function.

Ans. Given:  ….(i)

 has a real and finite value for all  R.

 Domain of  is R.

Let  and 

Since  and  being cosine function and modulus function are continuous for all real 

Now,  being the composite function of two continuous functions is continuous, but not equal to 

Again,   [Using eq. (i)]

Therefore,  being the composite function of two continuous functions is continuous.


33. Examine that  is a continuous function.

Ans. Let  and , then

Now,  and  being continuous, it follows that their composite,  is continuous.

Therefore,  is continuous.


34. Find all points of discontinuity of  defined by  

Ans. Given: 

When    = 

When   

When   

 

At  L.H.L. = 

R.H.L. = 

And 

Therefore, at   is continuous.

At  L.H.L. =  

R.H.L. = 

And 

Therefore, at   is continuous.

Hence, there is no point of discontinuity.

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