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Course: Class 12 math (India) > Unit 1

Lesson 8: Solutions to select NCERT problems

Select problems from miscellaneous exercise

solutions NCERT selected problems

In this article we will look at solutions of a few selected problems from miscellaneous exercise on chapter 1 of NCERT.

Problem 1:

Let

f : W \to W

be defined as

f (n) = n - 1

, if

n

is odd and

f (n) = n + 1

, if

n

is even. Show that

f

is invertible. Find the inverse of

f

. Here,

W

is the set of all whole numbers.

Solution:

A function is invertible if it is both one-one and onto. First let's check the behaviour of

f

$n$ ‍	$0$ ‍	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍	$5$ ‍	$\dots$ ‍
$f (n)$ ‍	$1$ ‍	$0$ ‍	$3$ ‍	$2$ ‍	$5$ ‍	$4$ ‍	$\dots$ ‍

See the pattern in the second row. All whole numbers will appear, and each number will appear only once. Hence,

f

is both one-one and onto. So

f

is invertible.

To find the inverse, we need to form a function which takes us back from the second row to the first row. We can do that by subtracting

1

from odd numbers and adding

1

to even numbers in the second row. Therefore our inverse function will be

f^{- 1} (n) = {\begin{cases} n - 1 & if n is odd \\ n + 1 & if n is even \end{cases}

Incidentally,

f^{- 1}

is same as

f

Problem 2:

Show that the function

f : R \to {x \in R : - 1 < x < 1}

defined by

f (x) = \frac{x}{1 + | x |}

x \in R

is one-one and onto function.

Solution:

Let's check the behaviour of the function in detail.

As the modulus function is involved, we can break the function into two parts.

f (x) = {\begin{cases} \frac{x}{1 + x} & for x \geq 0 \\ \frac{x}{1 - x} & for x < 0 \end{cases}

First let us look at the case

x \geq 0

$x$ ‍	$0$ ‍	$0.5$ ‍	$1$ ‍	$5$ ‍	$20$ ‍	$1000$ ‍	$\dots$ ‍
$f (x)$ ‍	$0$ ‍	$0.33$ ‍	$0.5$ ‍	$0.83$ ‍	$0.95$ ‍	$0.999$ ‍	$\dots$ ‍

The function starts at

0

and keeps on increasing to reach values close to

1

x

increases. However, we never reach exactly

1

as the denominator always stays greater than the numerator. The graph will look like:

Now let us look at the case

x < 0

$x$ ‍	$0$ ‍	$- 0.5$ ‍	$- 1$ ‍	$- 5$ ‍	$- 20$ ‍	$- 1000$ ‍	$\dots$ ‍
$f (x)$ ‍	$0$ ‍	$- 0.33$ ‍	$- 0.5$ ‍	$- 0.83$ ‍	$- 0.95$ ‍	$- 0.999$ ‍	$\dots$ ‍

Again, we start from

0

. The function keeps on decreasing to reach values close to

- 1

. However, we never reach exactly

- 1

. The full graph will look like:

From the above see that no two

x

values lead to the same

y

value. Also,

- 1 < y < 1

. Thus,

f

is both one-one and onto.

Problem 3:

Show that the function

f : R \to R

given by

f (x) = x^{3}

is injective.

Solution:

Injective means that the function is one-one.

f

will be one-one if no two different inputs give the same output.

Let

x_{1}

and

x_{2}

be two different inputs to

f

having the same output.

\begin{aligned} f (x_{1}) & = f (x_{2}) \\ x_{1}^{3} & = x_{2}^{3} \end{aligned}

We cannot directly say

x_{1} = x_{2}

from the above. We need to factorize the expression.

\begin{aligned} x_{1}^{3} - x_{2}^{3} & = 0 \\ (x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2}) & = 0 \\ x_{1} - x_{2} = 0 or x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} & = 0 \end{aligned}

Now

x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} = 0

does not have a real solution.

We are left with

x_{1} = x_{2}

, meaning the two inputs are the same and our initial assumption is false. For the same output, we cannot have two different inputs.

f

is one-one.

Problem 4:

Find the number of all onto functions from the set

{1, 2, 3, …, n}

to itself.

Solution:

Let

X = {1, 2, 3, …, n}

. Let

f : X \to X

be an onto function.

Because

f

is onto, arrows must cover all the elements in the codomain.

n

arrows start from the left. If any two arrows end at the same element, some element in the codomain will be left out. Therefore all arrows must end at different elements, meaning

f

is one-one as well.

So, number of possible onto functions will be equal to the number of possible ways in which we can create a one-one and onto map between

X

and

X

Start creating the map from the first element, i.e.

1

. For this we have

n

options in codomain. Arrow starting from

1

may end at any of the

n

elements in codomain.

Next pick

2

. For this we will have

n - 1

options in the codomain. Similarly for the next element we will have

n - 2

options in the codomain and so on. For the last element, we will be left with only one option.

Total number of possible ways to create a one-one and onto map between

X

and

X

n \cdot (n - 1) \cdot (n - 2) \dots 2 \cdot 1 = n!

Problem 5:

Let

A = {1, 2, 3}

. Then number of relations containing

(1, 2)

and

(1, 3)

which are reflexive and symmetric but not transitive is

(A)

1

(B)

2

(C)

3

(D)

4

Solution:

A = {1, 2, 3}

. A relation on

A

is a subset of

A \times A

First, to make the relation reflexive we must have the pairs

(1, 1)

(2, 2)

and

(3, 3)

It is given that relation must contain

(1, 2)

and

(1, 3)

. To make the relation symmetric we must add the pairs

(2, 1)

and

(3, 1)

as well. So till now we have

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}

Here

R

is non-transitive.

(2, 1)

and

(1, 3)

are present in

R

but

(2, 3)

is not. So we have found one relation.

Can we make another relation satisfying the given criteria?

A \times A

has a total of

9

pairs. We have two pairs left:

(2, 3)

and

(3, 2)

which can be added to

R

If we add

(2, 3)

, we need to add

(3, 2)

to make the relation symmetric. But then

R

becomes equal to

A \times A

and hence an equivalence relation.

Thus, number of possible relations is

1

Problem 6:

Let

A = {1, 2, 3}

. Then number of equivalence relations containing

(1, 2)

(A)

1

(B)

2

(C)

3

(D)

4

Solution:

A = {1, 2, 3}

. A relation on

A

is a subset of

A \times A

. We need to find equivalence relations, i.e. relations which are reflexive, symmetric and transitive.

First, to make the relation reflexive we must have the pairs

(1, 1)

(2, 2)

and

(3, 3)

It is given that relation must contain

(1, 2)

. So we must add

(2, 1)

to make the relation symmetric. Till now we have

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

Here see that

R

is transitive. For example,

(1, 2)

(2, 1)

and

(1, 1)

are present in

R

. We cannot find any instance where transitivity fail. So we have found one equivalence relation.

Now

A \times A

has a total of

9

pairs and we have four pairs left:

\begin{aligned} (1, 3), (3, 1) \\ (2, 3), (3, 2) \end{aligned}

R

if we add

(1, 3)

, we need to add

(2, 3)

to make

R

transitive. And then we must add

(3, 1)

and

(3, 2)

to make

R

symmetric.

R

if we add

(2, 3)

, we need to add

(1, 3)

to make

R

transitive. And then we must add

(3, 2)

and

(3, 1)

to make

R

symmetric.

See that whatever we do, we will end up adding all four pairs to

R

to make an equivalence relation. So we have only one more possible equivalence relation.

To conclude, number of possible equivalence relations is

2

Problem 7:

Number of binary operations on the set

{a, b}

are

(A)

10

(B)

16

(C)

20

(D)

8

Recall the definition of a binary operation on set

X

. It is a function from

X \times X \to X

Here,

X = {a, b}

How many functions can we make here?

We have four elements in the domain. Each element can be mapped to two elements in the co-domain. So, we have two choices for each element in the domain.

Number of possible functions is

2 \times 2 \times 2 \times 2 = 2^{4} = 16

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