NCERT Solutions class 12 Maths Exercise 5.1 (Ex 5.1) Chapter 5 Continuity and Differentiability

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NCERT Solutions for Class 12 Maths Exercise 5.1 Chapter 5 Continuity and Differentiability – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 (Ex 5.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.1 (Ex 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 = 22
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 limx→kf(x)=limx→k(x−5)=k−5=f(k)limx→k⁡f(x)=limx→k⁡(x−5)=k−5=f(k)
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 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   i.e. on (−3,3)(−3,3) 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   limx→c(x2)=c2=f(c)limx→c⁡(x2)=c2=f(c)
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.
∴∴ for no value of λλ, f(x) is continuous at x = 0
Again Since   is continuous at 
=>limx→1−f(x)=f(−1)=>limx→1−⁡f(x)=f(−1)
=>limx→1−4x+1=>limx→1−⁡4x+1
=>limh→04(1−h)+1=>limh→0⁡4(1−h)+1
=>4×1+1=>4×1+1
=> 5
And 
Here, L.H.L. = R.H.L.
∴∴ for any value of λλ, f(x) is continuous at x = 1.

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

 

Ans. Let n ∈∈ II
Then,limx→n−[x]=n−1limx→n−⁡[x]=n−1
∴[x]=n−1 ∀ x∈[n−1,n]∴[x]=n−1 ∀ x∈[n−1,n]
and g(n) = n – n = 0  [∴[n]=n because n∈I][∴[n]=n because n∈I]
Now, limx→n−g(x)=limx→n−(x−[x])=limx→n−x−limx→n−[x]limx→n−⁡g(x)=limx→n−⁡(x−[x])=limx→n−⁡x−limx→n−⁡[x]
=n−(n−1)=1=n−(n−1)=1
Also, limx→n+g(x)=limx→n+(x−[x])limx→n+⁡g(x)=limx→n+⁡(x−[x])
=limx→n+x−limx→n+[x]=n−n=0=limx→n+⁡x−limx→n+⁡[x]=n−n=0
Thus, limx→n−g(x)≠limx→n+g(x)limx→n−⁡g(x)≠limx→n+⁡g(x)
Hence, g(x) is discontinuous at all integral points. 

20. Is the function   continuous at   ?

 

Ans. Given:   
L.H.L. = limx→π−(x2−sinx+5)limx→π−⁡(x2−sin⁡x+5) = limh→0[(π−h)2−sin(π−h)+5]=π2+5limh→0⁡[(π−h)2−sin⁡(π−h)+5]=π2+5
R.H.L. = limx→π+(x2−sinx+5)limx→π+⁡(x2−sin⁡x+5) = limh→0[(π+h)2−sin(π+h)+5]=π2+5limh→0⁡[(π+h)2−sin⁡(π+h)+5]=π2+5
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.  The given function f is f(x) 
={sinx−cosx,ifx≠0−1ifx=0{sin⁡x−cos⁡x,ifx≠0−1ifx=0
It is evident that f is defined at all point of the real line.
Let c be a real number.
Case I:
If c≠≠0, then f(c) = sin c – cos c
limx→cf(x)=limx→c(sinx−cosx)=sinc−cosc∴limx→cf(x)=f(c)limx→c⁡f(x)=limx→c⁡(sin⁡x−cos⁡x)=sin⁡c−cosc∴limx→c⁡f(x)=f(c)
Therefore, f is continuous at all point x, such that x≠≠ 0
Case II:
If c = 0, then f(0)=-1
limx→0−f(x)=limx→0(sinx−cosx)=sin0−cos0=0−1=−1limx→0+f(x)=limx→0an>(sinx−cosx)=sin0−cos0=0−1=−1∴limx→0−f(x)=limx→0+f(x)=f(0)limx→0−⁡f(x)=limx→0⁡(sin⁡x−cos⁡x)=sin⁡0−cos0=0−1=−1limx→0+⁡f(x)=limx→0⁡(sin⁡x−cos⁡x)=sin⁡0−cos0=0−1=−1∴limx→0−⁡f(x)=limx→0+⁡f(x)=f(0)
Therefore, f is continuous at x=0
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus, f is a continuous function.

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, limx→2−f(x)=limx→2−kx2limx→2−⁡f(x)=limx→2−⁡kx2=  k×22k×22 
 limx→2+f(x)=3limx→2+⁡f(x)=3 and 
Since f(x)f(x) is continuous at x=2x=2 [given]
Therefore, limx→2−f(x)=limx→2+f(x)=f(2)limx→2−⁡f(x)=limx→2+⁡f(x)=f(2)
 
 

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)
= 
limx→5−f(x)=limh→0f(5−h)=k(5−h)+1=5k−hk+1=5k+1limx→5−⁡f(x)=limh→0⁡f(5−h)=k(5−h)+1=5k−hk+1=5k+1  …….(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 f(x)=cos(x2)f(x)=cos⁡(x2) 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 f(−1)=−2×(−1)−1=1f(−1)=−2×(−1)−1=1
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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