Relations & Functions Miscellaneous Exercise NCERT Solutions Class 12 Math Chapter 1 free PDF Download

Class 12 NCERT Solutions- Mathematics Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1

Table of Contents

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=y−710∈Rx=10y−7​∈R

So, it means for y ∈ R, there exists x=y−710x=10y−7​

f(x)=f(y−710)=10(y−710)+7=y–7+7=yf(x)=f(10y−7​)=10(10y−7​)+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)=y−710g(y)=10y−7​

gof=g(f(x))=g(10x+7)=(10x+7)−710=10×10=xgof=g(f(x))=g(10x+7)=10(10x+7)−7​=1010x​=x

fog=f(g(x))=f(x−710)=10(x−710)+7=x–7+7=xfog=f(g(x))=f(10x−7​)=10(10x−7​)+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)=y−710g(y)=10y−7​

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)={n−1,nisoddn+1,nisevenf(n)={n−1,nisoddn+1,niseven

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)={y−1,yisoddy+1,yiseveng(y)={y−1,yisoddy+1,yiseven

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) = x1+∣x∣   1+∣xx​ , 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)=x1+∣x∣  f(x)=1+∣xx​ , 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)=p1+∣p∣f(p)=1+∣pp

f(q)=q1+∣q∣f(q)=1+∣qq

f(p) = f(q)

p1+∣p∣=q1+∣q∣1+∣pp​=1+∣qq

p1+p=q1+q1+pp​=1+qq

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

p = q

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

f(p)=p1+∣p∣f(p)=1+∣pp

f(q)=q1+∣q∣f(q)=1+∣qq

f(p) = f(q)

p1+∣p∣=q1+∣q∣1+∣pp​=1+∣qq

p1−p=q1−q1−pp​=1−qq

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

p = q

Case 3: When p is positive and q is negative

f(p)=p1+∣p∣f(p)=1+∣pp

f(q)=q1+∣q∣f(q)=1+∣qq

f(p) = f(q)

p1+∣p∣=q1+∣q∣1+∣pp​=1+∣qq

p1+p=q1−q1+pp​=1−qq

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)=p1+∣p∣f(p)=1+∣pp

y=p1+py=1+pp

p=y1−y(y≠1)p=1−yy​(y=1)

Case 2: When p <0

f(p)=p1+∣p∣f(p)=1+∣pp

y=p1−py=1−pp

p=y1+y(y≠−1)p=1+yy​(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)={x−1,x>11,x=1g(x)={x−1,x>11,x=1​

Solution:

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

Taking f(x) = x+1 and g(x)={x−1,x>11,x=1g(x)={x−1,x>11,x=1​

 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!.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.