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
Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
Exercise 1.3
1. Let
: {1, 3, 4}
{1, 2, 5} and
: {1, 2, 5}
{1, 3} be given by
= {(1, 2), (3, 5), (4, 1)} and
= {(1, 3), (2, 3), (5, 1)}. Write down
Ans. = {(1, 2), (3, 5), (4, 1)} and
= {(1, 3), (2, 3), (5, 1)}
Now, and
and
Hence, {(1, 3), (3, 1), (4, 3)}
2. Let
and
be functions from R to R. Show that:
Ans. (a) To prove:
L. H. S. = =
=
= R. H. S.
(b) To prove:
L. H. S. = =
=
= R. H. S.
3. Find
and
, if:
(i) and
(ii) and
Ans. To find: and
(i) and
and
=
=
(ii) and
and =
4. If
show that
for all
What is the inverse of 
Ans. Given:
L.H.S. = =
=
=
= = R.H.S.
Now,
Hence inverse of
5. State with reason whether following functions have inverse:
(i) : {1, 2, 3, 4}
{10} with
= {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) : {5, 6, 7, 8}
{1, 2, 3, 4} with
= {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) : {2, 3, 4, 5}
{7, 9, 11, 13} with
= {(2, 7), (3, 9), (4, 11), (5, 13)}
Ans. (i) = {(1, 10), (2, 10), (3, 10), (4, 10)}
It is many-one function, therefore has no inverse.
(ii) = {(5, 4), (6, 3), (7, 4), (8, 2)}
It is many-one function, therefore has no inverse.
(iii) = {(2, 7), (3, 9), (4, 11), (5, 13)}
is one-one onto function, therefore,
has an inverse.
6. Show that
R given by
is one-one. Find the inverse of the function
Range
Ans. Part I: R given by
Let , then
and
When then
is one-one.
Part II: Let Range of
for some
in
As
is onto.
Therefore,
7. Consider
: R
R given by
Show that
is invertible. Find the inverse of
Ans. Consider : R
R given by
Let R, then
and
Now, for , then
is one-one.
Let Range of
is onto.
Therefore, is invertible and hence,
.
8. Consider
given by
Show that
is invertible with the inverse
of
given by
where
is the set of all non-negative real numbers.
Ans. Consider and
Let R
, then
and
is one-one.
Now
as
is onto.
Therefore, is invertible and
.
9. Consider
given by
Show that
is invertible with
Ans. Consider and
Let R
, then
and
Now, then
is one-one.
Now, again
=
=
=
=
=
is onto.
Therefore, is invertible and
.
10. Let
be an invertible function. Show that
has unique inverse.
(Hint: Suppose and
are two inverses of
Then for all
Use one-one ness of
).
Ans. Given: be an invertible function.
Thus is 1 – 1 and onto and therefore
exists.
Let and
be two inverses of
Then for all
Y,
The inverse is unique and hence
has a unique inverse.
11. Consider
: {1, 2, 3}
given by
and
Find
and show that
Ans. , then it is clear that
is 1 – 1 and onto and therefore
exists.
Also, and
Hence,
12. Let
be an invertible function. Show that the inverse of
is
, i.e., 
Ans. Let be an invertible function.
Then is one-one and onto
X where
is also one-one and onto such that
and
Now, and
13. If
: R
R given by
then
is:
(A)
(B)
(C)
(D)
Ans. : R
R and
=
= =
=
Therefore, option (C) is correct.
14. Let
: R –
R be a function defined as
The inverse of
is the map
: Range of
given by:
(A)
(B)
(C)
(D)
Ans. Given: : R –
R and
Now, Range of
Let
Therefore, option (B) is correct.