Exercise 5.1 Part-1
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NCERT Solutions class 12 Continuity & Differentiability
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 
.
NCERT Solutions class 12 Maths Exercise 5.1 Part-1
2. Examine the continuity of the function 
at 
Ans. Given:
Continuity at 
, 
= 
And 
Since 
, therefore, is continuous at .
NCERT Solutions class 12 Maths Exercise 5.1 Part-1
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.
NCERT Solutions class 12 Maths Exercise 5.1 Part-1
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 .
NCERT Solutions class 12 Maths Exercise 5.1 Part-1
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.
NCERT Solutions class 12 Maths Exercise 5.1 Part-1
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