NCERT Solutions Class 12
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NCERT Solutions-Mathematics
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 1)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 2)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 4)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 5)
- Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 1)
- Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 2)
- Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 3)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 1)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 3)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 4)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 1)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 2)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 3)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 4)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 5)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 6)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 7)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 2)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 3)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 4)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 6)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 7)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 8)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 9)
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NCERT Solutions-Chemistry
- Aldehydes, Ketones and Carboxylic Acids : NCERT Solutions – Class 12 Chemistry
- Alcohols, Phenols and Ethers : NCERT Solutions – Class 12 Chemistry
- Amines : NCERT Solutions – Class 12 Chemistry
- Biomolecules : NCERT Solutions – Class 12 Chemistry
- Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
- Chemistry in Everyday Life : NCERT Solutions – Class 12 Chemistry
- Coordination Compounds : NCERT Solutions – Class 12 Chemistry
- Electrochemistry : NCERT Solutions – Class 12 Chemistry
- General Principles and Processes of Isolation of Elements : NCERT Solutions – Class 12 Chemistry
- Haloalkanes and Haloarenes : NCERT Solutions – Class 12 Chemistry
- Polymers : NCERT Solutions – Class 12 Chemistry
- Surface Chemistry : NCERT Solutions – Class 12 Chemistry
- The d-and f-Block Elements : NCERT Solutions – Class 12 Chemistry
- The p-Block Elements : NCERT Solutions – Class 12 Chemistry
- The Solid State : NCERT Solutions – Class 12 Chemistry
- Solutions : NCERT Solutions – Class 12 Chemistry
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NCERT Solutions-Biology
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NCERT Solutions-Physics
- Electrostatic Potential And Capacitance : NCERT Solutions – Class 12 Physics
- Electric Charges And Fields : NCERT Solutions – Class 12 Physics
- Semiconductor Electronics: Materials, Devices And Simple Circuits : NCERT Solutions – Class 12 Physics
- Ray Optics And Optical Instruments : NCERT Solutions – Class 12 Physics
- Nuclei : NCERT Solutions – Class 12 Physics
- Moving Charges And Magnetism : NCERT Solutions – Class 12 Physics
- Magnetism And Matter : NCERT Solutions – Class 12 Physics
- Electromagnetic Induction : NCERT Solutions – Class 12 Physics
- Dual Nature Of Radiation And Matter : NCERT Solutions – Class 12 Physics
- Current Electricity : NCERT Solutions – Class 12 Physics
- Communication Systems : NCERT Solutions – Class 12 Physics
- Atoms : NCERT Solutions – Class 12 Physics
- Alternating Current : NCERT Solutions – Class 12 Physics
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.