Skip to content
Related Articles

Related Articles

Improve Article

Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 1

  • Last Updated : 30 Apr, 2021

Question 1. Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = 1R.

Solution:

As, it is mentioned here

f : R → R be defined as f(x) = 10x + 7

To, prove the function one-one

Let’s take f(x) = f(y)



10x + 7 = 10y + 7

x = y

Hence f is one-one.

To, prove the function onto

y ∈ R, y = 10x+7

x = \frac{y-7}{10} ∈ R

So, it means for y ∈ R, there exists x = \frac{y-7}{10}

f(x) = f(\frac{y-7}{10}) = 10(\frac{y-7}{10})+7 = y - 7 + 7 = y



Hence f is onto.

As, f is one-one and onto. This f is invertible function.

Let’s say g : R → R be defined as g(y) = \frac{y-7}{10}

g o f = g(f(x)) = g(10x+7) = \frac{(10x+7)-7}{10} = \frac{10x}{10} = x

f o g = f(g(x)) = f(\frac{x-7}{10}) = 10(\frac{x-7}{10})+7 = x - 7 + 7 = x

Hence, g : R → R such that g o f = f o g = 1R.

g : R → R is defined as g(y) = \frac{y-7}{10}

Question 2. Let f : W → 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:

The function f is defined as

f(n)= \begin{cases} n-1, \hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} odd\\ n+1,\hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}



As, we know f is invertible, if and only if f is one-one and onto.

ONE-ONE

For the pair of number, we will deal with three cases:

Case 1: When both numbers p and q are odd numbers.

f(p) = p-1

f(q) = q-1

f(p) = f(q)

p-1 = q-1

p – q = 0

Case 2: When both numbers p and q are even numbers.

f(p) = p+1

f(q) = q+1

f(p) = f(q)

p+1 = q+1

p – q = 0

Case 3: When p is odd and q is even

f(p) = p-1

f(q) = q+1

f(p) = f(q)

p-1 = q+1

p – q = 2

Subtracting an odd number and even always gives a odd number, not even. Hence, the case 3 result is impossible.

So, the function f is one-one, for case 1 and case 2 only.

ONTO

Case 1: When p is odd number

f(p) = p-1

y = p-1

p = y+1

Hence, when p is odd y is even.

Case 2: When p is even number

f(p) = p+1

y = p+1

p = y-1

Hence, when p is even y is odd.

So, it means for y ∈ W, there exists p = y+1 and y-1 for odd and even value of p respectively.

Hence f is onto.

As, f is one-one and onto. This f is an invertible function.

Let’s say g : W → W be defined as g(y)= \begin{cases} y-1, \hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} odd\\ y+1,\hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}

f = g

Hence, The inverse of f is f itself



Question 3. If f : R → R is defined by f(x) = x2– 3x + 2, find f (f(x)).

Solution:

f(x) = x2– 3x + 2

f(f(x)) = f(x2– 3x + 2)

= (x2– 3x + 2)2 – 3(x2– 3x + 2) + 2

= x4 + 9x2 + 4 -6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2

f(f(x)) = x4 – 6x3 + 10x2 – 3x

Question 4. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = \frac{x}{1+|x|}   , x ∈ R is one one and onto function.

Solution:

As, it is mentioned here

f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = \frac{x}{1+|x|}  , x ∈ R

As, we know f is invertible, if and only if f is one-one and onto.



ONE-ONE

For the pair of number, we will deal with three cases:

Case 1: When both numbers p and p are positive numbers.

The function f is defined as

Case 1: When both numbers p and q are positive numbers.

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

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

f(p) = f(q)

\frac{p}{1+|p|} = \frac{q}{1+|q|}

\frac{p}{1+p} = \frac{q}{1+q}

p(1+q) = q(1+p)

p = q

Case 2: When number p and q are negative numbers.

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

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

f(p) = f(q)

\frac{p}{1+|p|} = \frac{q}{1+|q|}

\frac{p}{1-p} = \frac{q}{1-q}

p(1-q) = q(1-p)

p = q



Case 3: When p is positive and q is negative

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

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

f(p) = f(q)

\frac{p}{1+|p|} = \frac{q}{1+|q|}

\frac{p}{1+p} = \frac{q}{1-q}

p(1-q) = q(1+p)

p + q = 2pq

Here, RHS will be negative and LHS will be positive. Hence, the case 3 result is impossible.

So, the function f is one-one, for case 1 and case 2.

ONTO

Case 1: When p>0.

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

y = \frac{p}{1+p}

p = \frac{y}{1-y} (y≠1)

Case 2: When p <0

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

y = \frac{p}{1-p}

p = \frac{y}{1+y} (y≠-1)

Hence, p is defined for all the values of y, p∈ R



Hence f is onto.

As, f is one-one and onto. This f is an invertible function.

Question 5. Show that the function f : R → R given by f(x) = x3 is injective.

Solution:

As, it is mentioned here

f : R → R defined by f(x) = x3, x ∈ R

To prove f is injective (or one-one).

ONE-ONE

The function f is defined as

f(x) = x3

f(y) = y3

f(x) = f(y)

x3 = y3

x = y

The function f is one-one, so f is injective.

Question 6. Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.

(Hint : Consider f(x) = x and g (x) = | x |).

Solution:

Two functions, f : N → Z and g : Z → Z

Taking f(x) = x and g(x) = |x|

Let’s check, whether g is injective or not

g(5) = |5| = 5

g(-5) = |-5| = 5

As, we can see here that

Taking two integers, 5  and -5

g(5) = g(-5)

but, 5 ≠ -5

So, g is not an injective function.

Now, g o f: N → Z is defined as

g o f = g(f(x)) = g(x) = |x|

Now, as x,y∈ N

g(x) = |x|

g(y) = |y|

g(x) = g(y)

|x| = |y|

x = y (both x and y are positive)

Hence, g o f is an injective.

Question 7. Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto.

(Hint : Consider f(x) = x + 1 and g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}

Solution:

Two functions, f : N → N and g : N → N

Taking f(x) = x+1 and g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}

 As, f(x) = x+1

y = x+1

x = y-1



But, when y=1, x = 0. Which doesn’t satiny this relation f : N → N.

Hence. f is not an onto function.

Now, g o f: N → N is defined as

g o f = g(f(x)) = g(x+1)

When x+1=1, we have

g(x+1) = 1 (1∈ N)

And, when x+1>1, we have

g(x+1) = (x+1)-1 = x

y = x, which also satisfies x,y∈ N

Hence, g o f is onto.

Question 8. Given a non empty set X, consider P(X) which is the set of all subsets of X.

Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

Solution:

Given, A and B are the subsets of P(x), A⊂ B

To check the equivalence relation on P(X), we have to check

  • Reflexive

As, we know that every set is the subset of itself.

Hence, A⊂ A and B⊂ B

ARA and BRB is reflexive for all A,B∈ P(X)

  • Symmetric

As, it is given that A⊂ B. But it doesn’t make sure that B⊂ A.

To be symmetric it has to be A = B

ARB is not symmetric.

  • Transitive

When A⊂ B and B⊂ C

Then of course, A⊂ C

Hence, R is transitive.

So, as R is not symmetric. 

R is not an equivalence relation on P(X).

Question 9. Given a non-empty set X, consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Solution:

Given, P(X) × P(X) → P(X) is defined as A*B = A∩B ∀ A, B ∈ P(X)

This implies, A⊂  X and B ⊂  X

So, A∩X = A and B∩X = B ∀ A, B ∈ P(X)

⇒ A*X = A and B*X = B

Hence, X is the identity element for intersection of binary operator.

Question 10. Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.

Solution:

Onto function from the set {1,2,3,…..,n} to itself is just same as the permutations of n.

1×2×3×4×…….×n

Which is n!.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :