Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 1 Relations and Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Maths Chapter 1 Important Extra Questions Relations and Functions

### Relations and Functions Important Extra Questions Very Short Answer Type

Question 1.

If R = {(x, y) : x + 2y = 8} is a relation in N, write the range of R.

Solution:

Range of R = {1, 2, 3}.

[∵ When x = 2, then y = 3, when x = 4, then y = 2, when x = 6, then y = 1 ]

Question 2.

Show that a one-one function :

f{1, 2, 3} → {1, 2, 3} must be onto. (N.C.E.R.T.)

Solution:

Since ‘f’ is one-one,

∴ under ‘f’, all the three elements of {1, 2, 3} should correspond to three different elements of the co-domain {1, 2, 3}.

Hence, ‘f’ is onto.

Question 3.

What is the range of the function f(x) = \(\frac{|x-1|}{x-1}\) ? (C.B.S.E. 2010)

Solution:

When x > 1,

than f(x) = \(\frac{x-1}{x-1}\) = 1.

When x< 1,

than f(x) = \(\frac{-(x-1)}{x-1}\) = -1

Hence, R_{f} = {-1, 1}.

Question 4.

Show that the function f : N → N given by f(x) = 2x is one-one but not onto. (N.C.E.R.T.)

Solution:

Let x_{1}, x_{2} ∈ N.

Now, f(x_{1}) = f(x_{2})

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2}

⇒ f is one-one.

Now, f is not onto.

∵ For 1 ∈ N, there does not exist any x ∈ N such that f(x) = 2x = 1.

Hence, f is ono-one but not onto.

Question 5.

If f : R → R is defined by f(x) = 3x + 2 find f(f(x)). C.B.S.E. 2011 (F))

Solution:

f(f(x)) = 3 f(x) + 2

= 3(3x + 2) + 2 = 9x + 8.

Question 6.

If f(x) = \(\frac{x}{x-1}\) , x ≠1 then find fof. (N.C.E.R.T)

Solution:

Question 7.

If f: R → R is defined by f(x) = (3 -x^{3})^{1/3}, find fof(x)

Solution:

fof(x) = f(f(x)) = (3-(f(x))^{3})^{1/3}

= (3 – ((3 – x^{3})^{1/3})^{3})^{1/3}

= (3 – (3 – x^{3} ))^{1/3}= (x^{3})^{1/3} = x.

Question 8.

Are f and q both necessarily onto, if gof is onto? (N.C.E.R.T.)

Solution:

Consider f: {1, 2, 3, 4} → {1, 2, 3, 4}

and g : {1, 2, 3,4} → {1,2.3} defined by:

f(1) = 1, f(2) = 2, f(3) = f(4) = 3

g (1) = 1, g (2) = 2, g (3) = g (4) = 3.

∴ gof = g (f(x)) {1, 2,3}, which is onto

But f is not onto.

[∵ 4 is not the image of any element]

### Relations and Functions Important Extra Questions Short Answer Type

Question 1.

Let A be the set of all students of a Boys’ school. Show that the relation R in A given by:

R = {(a, b): a is sister of b} is an empty relation and the relation R’ given by :

R’ = {(a, b) : the difference between heights of a and b is less than 3 metres} is an universal relation. (N.C.E.R.T.)

Solution:

(i) Here R = {(a, b): a is sister of b}.

Since the school is a Boys’ school,

∴ no student of the school can be the sister of any student of the school.

Thus R = Φ Hence, R is an empty relation.

(ii) Here R’ = {(a,b): the difference between heights of a and b is less than 3 metres}.

Since the difference between heights of any two students of the school is to be less than 3 metres,

∴ R’ = A x A. Hence, R’ is a universal relation.

Question 2.

Let f : X → Y be a function. Define a relation R in X given by :

R = {(a,b):f(a) = f(b)}.

Examine, if R is an equivalence relation. (N.C.E.R.T.)

Solution:

For each a ∈ X, (a, a) ∈ R.

Thus R is reflexive. [∵ f (a) = f(a)]

Now (a, b) ∈ R

⇒ f(a) = f(b)

⇒ f(b) = f (a)

⇒ (b, a) ∈ R.

Thus R is symmetric.

And (a, b) ∈ R

and (b, c) ∈ R

⇒ f(a) = f(b)

and f(b) = f(c)

⇒ f(a)= f(c)

⇒ (a, c) ∈ R.

Thus R is transitive.

Hence, R is an equivalence relation.

Question 3.

Let R be the relation in the set Z of integers given by:

R = {(a, b): 2 divides a – b}.

Show that the relation R is transitive. Write the equivalence class [0]. (C.B.S.E. Sample Paper 2019-20)

Solution:

Let 2 divide (a – b) and 2 divide (b – c), where a,b,c ∈ Z

⇒ 2 divides [(a – b) + (b – c)]

⇒ 2 divides (a – c).

Hence, R is transitive.

And [0] = {0, ± 2, ± 4, ± 6,…].

Question 4.

Show that the function :

f : N → N

given by f(1) = f(2) = 1 and f(x) = x -1, for every x > 2 is onto but not one-one. (N.C.E.R.T.)

Solution:

Since f(1) = f(2) = 1,

∴ f(1) = f(2), where 1 ≠ 2.

∴ ‘f’ is not one-one.

Let y ∈ N, y ≠ 1,

we can choose x as y + 1 such that f(x) = x – 1

= y + 1 – 1 = y.

Also 1 ∈ N, f(1) = 1.

Thus ‘f ’ is onto.

Hence, ‘f ’ is onto but not one-one.

Question 5.

Find gof and fog, if:

f : R → R and g : R → R are given by f (x) = cos x and g (x) = 3x^{2}. Show that gof ≠ fog. (N. C.E.R. T.)

Solution:

We have :

f(x) = cos x and g(x) = 3x^{2}.

∴ gof (x) = g (f(x)) = g (cos x)

= 3 (cos x)^{2} = 3 cos^{2} x

and fog (x) = f(g (x)) = f(3x^{2}) = cos 3x^{2}.

Hence, gof ≠ fog.

Question 6.

If f(x) = \(\frac{4 x+3}{6 x-4}\) , x ≠ \(\frac{2}{3}\) find fof(x)

Solution:

We have: \(\frac{4 x+3}{6 x-4}\) …(1)

∴ fof(x) — f (f (x))

Question 7.

Let A = N x N be the set of ail ordered pairs of natural numbers and R be the relation on the set A defined by (a, b) R (c, d) iff ad = bc. Show that R is an equivalence relation.

Solution:

Given: (a, b) R (c, d) if and only if ad = bc.

(I) (a, b) R (a, b) iff ab – ba, which is true.

[∵ ab = ba ∀ a, b ∈ N]

Thus, R is reflexive.

(II) (a, b) R (c,d) ⇒ ad = bc

(c, d) R (a, b) ⇒ cb = da.

But cb = be and da = ad in N.

∴ (a, b) R (c, d) ⇒ (c, d) R (a, b).

Thus, R is symmetric.

(III) (a,b) R (c, d)

⇒ ad = bc …(1)

(c, d) R (e,f)

⇒ cf = de … (2)

Multiplying (1) and (2), (ad). (cf) – (be), (de)

⇒ af = be

⇒ (a,b) = R(e,f).

Thus, R is transitive.

Thus, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

Question 8.

Let f: R → R be the Signum function defined as :

and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0,1] ?

Solution:

For x ∈ (0,1].

And (gof) (x) = g(f(x)) = g(1)

[∵ f(x) = 1 ∀ x > 0]

= [1] = 1

⇒ (gof) (x) = 1 ∀ x ∈ (0, 1] …(2)

From (1) and (2), (fog) and (gof) do not coincide in (0, 1].

### Relations and Functions Important Extra Questions Long Answer Type 1

Question 1.

Show that the relation R on R defined as R = {(a, b):a ≤ b}, is reflexive and transitive but not symmetric.

Solution:

We have : R = {(a, b)} = a ≤ b}.

Since, a ≤ a ∀ a ∈ R,

∴ (a, a) ∈ R,

Thus, R reflexive.

Now, (a, b) ∈ R and (b, c) ∈ R

⇒ a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a, c) ∈ R.

Thus, R is transitive.

But R is not symmetric

[∵ (3, 5) ∈ R but (5, 3) ∉ R as 3 ≤ 5 but 5 > 3]

Question 2.

Prove that function f : N → N, defined by f(x) = x^{2} + x + 1 is one-one but not onto. Find inverse of f: N → S, where S is range of f.

Solution:

Let x_{1}, x_{2} ∈ N.

Now, f(x_{1}) = f(x_{2})

Thus, f is one-one.

Let y ∈ N, then for any x,

f(x) = y if y = x^{2} + x + 1

Now, for y = \(\frac{3}{4}\) , x = \(-\frac{1}{2}\) ∉ N.

Thus, f is not onto.

⇒ f(x) is not invertible.

Since, x > 0, therefore, \(\frac{\sqrt{4 y-3}-1}{2}\) > 0

⇒ \(\sqrt{4 y-3}\) > 1

⇒ 4y – 3 > 1

⇒ 4y > 4

⇒ y > 1.

Redefining, f : (0, ∞) → (1, ∞) makes

f(x) = x^{2} + x + 1 on onto function.

Thus, f (x) is bijection, hence f is invertible and f^{-1} : (1, ∞) → (1,0)

f^{-1}(y) = \(\frac{\sqrt{4 y-3}-1}{2}\)

Question 3.

Let A = (x ∈Z : 0 ≤ x ≤ 12}.

Show that R = {(a, b) : a, b ∈ A; |a – b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2]. (C.B.S.E 2018)

Solution:

We have:

R = {(a, b): a, b ∈ A; |a – b| is divisible by 4}.

(1) Reflexive: For any a ∈ A,

∴ (a, b) ∈ R.

|a – a| = 0, which is divisible by 4.

Thus, R is reflexive.

Symmetric:

Let (a, b) ∈ R

⇒ |a – b| is divisible by 4

⇒ |b – a| is divisible by 4

Thus, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a – b| is divisible by 4 and |b – c| is divisible by 4

⇒ |a – b| = 4λ

⇒ a – b = ±4λ ………….(1)

and |b – c| = 4µ

⇒ b – c = ± 4µ ………….(2)

Adding (1) and (2),

(a-b) + (b-c) = ±4(λ + µ)

⇒ a – c = ± 4 (λ + µ)

⇒ (a, c) ∈ R.

Thus, R is transitive.

Now, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

(ii) Let ‘x’ be an element of A such that (x, 1) ∈ R

⇒ |x – 1| is divisible by 4

⇒ x – 1 = 0,4, 8, 12,…

⇒ x = 1, 5, 9, 13, …

Hence, the set of all elements of A which are related to 1 is {1, 5, 9}.

(iii) Let (x, 2) ∈ R.

Thus |x – 2| = 4k, where k ≤ 3.

∴ x = 2, 6, 10.

Hence, equivalence class [2] = {2, 6, 10}.

Question 4.

Prove that the function f: [0, ∞) → R given by f(x) = 9x^{2} + 6x – 5 is not invertible. Modify the co-domain of the function f to make it invertible, and hence find f^{-1}. (C.B.S.E. Sample Paper 2018-19)

Solution:

Let y∈ R.

For any x, f(x) = y if y = 9x^{2} + 6x – 5

⇒ y = (9x^{2} + 6x + 1) – 6

= (3x + 1)^{2} – 6

For y = – 6 ∈ R, x = \(-\frac{1}{3}\) ∉ [0, ∞).

Thus, f(x) is not onto.

Hence, f(x) is not invertible.

We redefine,

f: [0, ∞) → [-5, ∞),

which makes f(x) = 9x^{2} + 6x – 5 an onto function.

Now, x_{1}, x_{2} ∈ [0, ∞) such that f(x_{1}) = f(x_{2})

⇒ (3x_{1} + 1)^{2} = (3x_{2} + 1)^{2}

⇒[(3x_{1} + 1)+ (3x_{2} + 1 )][(3x_{1} + 1)- (3x_{2} + 1)]

⇒ [3(x_{1} + x_{2}) + 2][3(x_{1} – x_{2})] = 0

⇒ x_{1} = x_{2}

[∵ 3(x_{1} + x_{2}) + 2 > 0]

Thus, f(x) is one-one.

∴ f(x) is bijective, hence f is invertible

and f^{-1}: [-5, ∞) → [0, ∞)

f^{-1} (y) = \(\frac{\sqrt{y+6}-1}{3}\)

Question 5.

Check whether the relation R in the set R of real numbers, defined by :

R = {(a, b): 1 + ab > 0}, is reflexive, symmetric or transitive. (C.B.S.E. Sample Paper 2018-19)

Solution:

R = {(a, b): 1 + ab> 0}.

Reflexive:

Now, 1 + a.a = 1 + a^{2} > 0

⇒ (a, a) ∈ R ∀ a ∈ R.

Thus, R is reflexive.

Symmetric:

Let (a, b) ∈ R.

Then 1 + ab > 0

⇒ 1 + ba > 0

⇒ (b, a) ∈ R.

Thus, R is symmetric.

Transitive:

Take a = -8,b = -1, c = \(\frac { 1 }{ 2 }\)

Now, 1 + ab = 1 + (-8) (-1)

= 9 > 0

⇒ (a, b) ∈ R

and, 1 + bc = 1 + (-1)(\(\frac { 1 }{ 2 }\) )

= 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) > 0

⇒ (b, c) ∈ R.

But 1 + ac – 1 + (-8) (\(\frac { 1 }{ 2 }\) )

= 1 – 4 = -3 < 0

⇒ (a, c) ∉ R.

Thus, R is not transitive.

Hence, R is reflexive, symmetric but not transitive.

Question 6.

Let T be the set of all triangles in a plane with R, a relation in T given by :

R = {(T_{1}, T_{2}): T_{1} is congruent to T_{2}}.

Show that R is an equivalence relation. (N.C.E.R.T.)

Solution:

We have:

R = {(T_{1}, T_{2}): T_{1} is congruent to T_{2}}.

Now (T_{1}, T_{2}) ∈ R.

[ ∵ Every triangle is congruent to itself]

Thus R is reflexive.

(T_{1}, T_{2}) ∈ R

⇒ T_{1} is congruent to T_{2}

⇒ T_{2} is congruent to T_{1}

(T_{2}, T_{1}) ∈ R.

Thus R is symmetric.

(T_{1}, T_{2}) ∈ R and (T_{2}, T_{3}) ∈ R

⇒ T_{1} is congruent to T_{2} and T_{2} is congruent to T_{3}

⇒ T_{1} is congruent to T_{3}

(T_{1}, T_{3}) ∈ R.

Thus R is transitive.

Hence, R is an equivalence relation.

Question 7.

If R_{1}, and R_{2} are equivalence relations in a set A, show that R_{1} ∩ R_{2} is also an equivalence relation. (N.C.E.R.T.)

Solution:

Since R_{1} and R_{2} are equivalence relations, [Given]

∴ (a, a) ∈ R_{1}

and (a, a) ∈ R_{2} ∀ a ∈ A

⇒ (a, a) ∈ R_{1} ∩ R_{2} ∀ ∈ G

Thus R_{1} ∩ R_{2} is reflexive.

Now (a, b) ∈ R_{1} ∩ R_{2}

⇒ (a, b) ∈ R_{1}

and (a, b) ∈ R_{2}

⇒ (b, a) ∈ R_{1}

and (b, a) ∈ R_{2}

⇒ (b, a) ∈ R_{1} ∩ R_{2}

Thus R_{1} n R2 is symmetric.

And (a, b) ∈ R_{1} ∩ R_{2}

and (b, c) ∈ R_{1} ∩ R_{2}

⇒ (a, c) ∈ R_{1}

and (a, c) ∈ R_{2}

⇒ (a, c) ∈ R_{1} ∩ R_{2}

Thus R_{1} ∩ R_{2} is transitive.

Hence, R_{1} ∩ R_{2} is an equivalence relation.

Question 8.

Show that f: R-{2} → R -{1} defined by:

f(x) = \(\frac{x}{x-2}\) is one-one.

Also, if g : R -{1} → R -{2} as g{x) = \(\frac{2 x}{x-1}\) find gof.

Solution:

We have: f(x) = \(\frac{x}{x-2}\) and g(x) = \(\frac{2 x}{x-1}\)

Question 9.

Let f: N → N be a function defined as :

f(x) – 9x^{2} + 6x – 5.

Show that f: N → S, where S is the range of f, is invertible. Find the inverse of f and hence find f^{-1} (43) and f^{-1} (163). (C.B.S.E. 2016)

Solution:

We have : f : N → S,

fix) = 9x^{2} + 6x – 5.

Now f(x_{1}) = f(x_{2})

⇒ 9x_{1}^{2} + 6x_{1} – 5 = 9 x_{2}^{2} + 6x_{2} – 5

⇒ 9(x_{1}^{2} – x_{2}^{2}) + 6 (x_{1} – x_{2}) = 0

⇒ (x_{1} – x_{2}) [9x_{1} + 9x_{2}+ 6] = 0

⇒ x_{1} = x_{2} [x_{1}, x_{2} ∈ N ]

⇒ f is one-one.

Let y ∈ S be an arbitrary number.

Now y = f(x)

⇒ y = 9x^{2}+ 6x – 5

⇒ y = (3x + 1)^{2} – 6

⇒ \(\sqrt{y+6}\) = 3x + 1

⇒ x = \(\frac{\sqrt{y+6}-1}{3}\) ∈ N

⇒ x = f^{-1} (y).

Since f is one-one and onto

⇒ f is invertible.

Question 10.

Let f: A → B be a function defined as f(x) = \(\frac{2 x+3}{x-3}\) where A = R – {3} and B = R – {2}.

Is the function ‘f ’one-one and onto ?

Is ‘f’ invertible? If yes, then find its inverse.

Solution:

Let x_{1}, x_{2} ∈ A = R – {3}.

Now, f(x_{1}) = f(x_{2})

⇒ \(\frac{2 x_{1}+3}{x_{1}-3}=\frac{2 x_{2}+3}{x_{2}-3}\)

⇒ (2x_{1} + 3) (x_{2}– 3) = (2x_{2} + 3) (x_{1} – 3)

⇒ 2x_{1}x_{2} – 6x_{1} + 3x_{2 }– 9 = 2x_{1} x_{2} – 6x_{2} + 3x_{1} – 9

⇒ – 6x_{1} + 3x_{2} = – 6x_{2} + 3x_{1}

⇒ 9x_{1} = 9x_{2}

⇒ x_{1} = x_{2}

Thus, ‘f’ is one-one.

Let y ∈ R- {2}.

Let y = f(x_{0}).

Then \(\frac{2 x_{0}+3}{x_{0}-3}\) = y

⇒ 2x_{0} + 3 = x_{0}y – 3y

⇒ x_{0}(y – 2) = 3(y + 1)

⇒ x_{0} = \(\frac{3(y+1)}{y-2}\)

Now , y ∈ R – {2 } ⇒ \(\frac{2 x_{0}+3}{x_{0}-3}\) ∈ R – {2}

Thus, ‘f’ is onto.

Hence, ‘f’ is one-one onto and consequently ‘f’ is invertible.

Also y = \(\frac{2 x+3}{x-3}\)

xy – 3y = 2x + 3

x(y- 2) = 3(y +1)

x = \(\frac{3(y+1)}{y-2}\)

f^{-1} (y) = \(\frac{3(y+1)}{y-2}\)

Hence, f^{-1} (x) = \(\frac{3(x+1)}{x-2}\) for all x ∈ R – {2}.

Question 11.

Let A = R – {2} and B = R – {1}. If f: A → B is a function defined by :

f(x) = \(\frac{x-1}{x-2}\) Show that f is one one and onto.

Hence, find f^{-1}.

Solution:

(i) One-one : Let x_{1} x_{2} ∈ R – {2} such that

f(x_{1}) = f(x_{2})

⇒ \(\frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}\)

⇒ x_{1}x_{2} – 2x_{1} – x_{2} + 2 = x_{1}x_{2} – 2x_{2} – x_{1} + 2

⇒ x_{1} = x_{2}

Thus,f one-one.

Onto : Let f(x) = y.

Thus \(\frac{x-1}{x-2}\) = y ⇒ x = \(\frac{2 y-1}{y-1}\)

∴ Range of f = R – {1}

= Co-domain of B

Thus f is onto.

(ii) f^{-1}(y) = \(\frac{2 y-1}{y-1}\)

Hence f^{-1}(x) = \(\frac{2 x-1}{x-1}\)

Question 12.

Show that the relation S on the set:

A = {x ∈ Z : 0 ≤ x ≤ 12} given by S = {(a, b) :a,b e Z, |a – b| is divisible by 3} is an equivalence relation.

Solution:

Reflexive :

(a, a) ∈ S

⇒ |a – a| i.e., | 0 | is divisible by 3.

Thus, S is reflexive.

Symmetric:

(a, b) ∈ S ⇒ | a – b | is divisible by 3

⇒ |b – a| is divisible by 3

⇒ (b, a) ∈ S.

Thus, S is symmetric.

Transitive :

Let (a, b) ∈ S and (b, c) ∈ S.

Thus |a – b| = 3m and |b – c| = 3n

⇒ a – b = ± 3m and b – c = ± 3 n.

∴ (a – c) = 3(± m ± n)

⇒ a – c is divisible by 3

⇒ | a – c | is divisible by 3

⇒ (a, c) ∈ S.

Thus, S is transitive.

Hence, S is an equivalence relation.

### Relations and Functions Important Extra Questions Long Answer Type 2

Question 1.

Let A = (1,2,3,…, 9} and R be the relation in A x A defined by (a, b) R (c, d) if: a + d = b + c for (a, b), (c, d) in A x A.

Prove that R is an equivalence relation. Also obtain the equivalence class {(2,5)}. (C.B.S.E. 2014)

Solution:

(i) We have : (a, b) R (c, d)

⇒ a + d = b + c on the set A = {1, 2, 3,…, 9}.

(a) (a, b) R (a, b)

⇒ a + b = b + a, which is true.

[∵ a + b = b + a ∀ a, b ∈ A]

Thus R is reflexive.

(b) (a, b) R (c, d)

⇒ a + d = b + c

(c, d) R (a, b)

⇒ c + b = d + a.

But c + b = b + c

and d + a = a + d ∀ a, b, c, d ∈ A.

∴ (a, b) R(c, d) = (c, d) R (a, b).

Thus R is symmetric.

(c) (a,b)R(c,d)

⇒ a + d = b + c ∀ a, b, c, d ∈ A ……..(1)

(c,d)R(e,f)

⇒ c + f = d + e ∀ c, d, e,f ∀ A ……………. (2)

∴ Adding (1) and (2),

(a + d) + (c +f) = (b + c) + (d + e)

⇒ a + f = b + e

⇒ (a, b) R (e, f).

Thus R is transitive.

Hence, the relation R is an equivalence relation.

(ii) {(2, 5)} = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

[∵ 2 + 4 = 5 + 1; etc.]

Question 2.

Let N denote the set of all natural numbers and R be the relation on N x N defined by : (a, b) R(c,d) is ad(b + c) = bc(a + d).

Show that R is an equivalence relation. (C.B.S.E. 2015)

Solution:

We have : (a, b) R (c, d)

⇒ ad (b + c) = bc (a + d) on N.

(i) {a, b) R (a, b)

⇒ ab(b + a) = ba (a + b)

⇒ ab(a + b) = ab (a + ft),

which is true.

Thus R is reflexive.

(ii) (a, b) R (c, d)

⇒ ad(b + c) = be (a + d)

⇒ bc(a + d) = ad (b + c)

⇒ cb(d + a) = da (c + b)

[∵ bc = cb and a + d = d + a;

etc. ∀ a, b, c, d ∈ N]

⇒ (cb) R (a,b).

Thus R is symmetric

(iii) Let (a , b) R (c, d) and (c, d) R ( e, f)

∴ ad (b + c) = bc (a + d)

and cf(d + e) = de(c + f)

⇒ be(a+f) = af(b+e)

⇒ af(b+e) = be(a+f)

⇒ (a, b) R (e,f).

Thus R is transitive.

Hence, R is an equivalence relation.

Question 3.

Show that the relation in the set A = {x:x ∈ W, 0 ≤ x ≤ 12} given by R = {(a, b) : (a – b) is a multiple of 4} is an equivalence relation. Also find the set of all. elements related to 2.

Solution:

We have :

R = {(a, b): (a- b) is a multiple of 4},

where a, b ∈ A = {x: x ∈ W, 0 ≤ x ≤ 12}.

(a) For any a ∈ A, we have :

(a – a) = 0, which is a multiple of 4

⇒ (a, a) ∈ R ∀ a ∈ A.

Thus R is reflexive.

(b) Let a, b ∈ A.

Now (a, b) ∈ R

⇒ (a – b) is a multiple of 4

⇒ (b – a) is a multiple of 4

⇒ (b, a) ∈ R

Thus R is symmetric.

(c) Let a, b, c ∈ A.

Now (a, b) ∈ R

and (b, c) ∈ R

⇒ (a – b) is a multiple of 4 and (b – c) is a multiple of 4

⇒ (a – b) + (b – c) is a multiple of 4

⇒ (a – c) is a multiple of 4

⇒ (a, c) ∈ R.

Thus R is transitive.

Hence, R is an equivalence relation.

Question 4.

Let R be the relation defined in the set A = {1,2,3,4,5,6,7} by :

R = {(a, b) : both a and b are either odd or even}.

Show that R is an equivalence relation. Further, show that all the elements of the subset:

(1, 3, 5, 7} are related to each other and all the elements of the subset {2,4,6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2,4,6}. (N.C.E.R.T.)

Solution:

We have:

R = {{a, b): both a, b are either odd or even}.

(i) Let a ∈ A.

Both a and a are either odd or even.

∴ (a, a) ∈ R.

Thus R is reflexive.

(ii) Let (a, b) ∈ R

⇒ both a and b are either odd or even

⇒ both b and a are either odd or even

⇒ (b, a) ∈ R.

Thus R is symmetric.

(iii) Let {a, b) ∈ R and (b, c) ∈ R.

∴ Both a, b and both b, c are either odd or even

⇒ both a, c are either odd or even

⇒ (a, c) ∈ R.

Thus R is transitive.

Hence, R is an equivalence relation.

Further all elements of {1,3,5,7} are related to each other.

[∵ All elements of this subset are odd] Similarly all elements of {2,4,6} are related to each other.

[∵All elements of this subset are even] But no element of {1, 3, 5, 7} is related to any element of {2, 4, 6}.

[∵Elements of {1, 3, 5, 7] are odd while elements of [2,4,6] are even]

Question 5.

Show that f : N → N given by :

is both one-one and onto. (A.I.C.B.S.E. 2012)

Solution:

One-One.

Here we discuss the following possible cases :

(i) When x_{1} is odd and x_{2} is even.

Here f(x_{1}) = f(x_{2})

⇒ x_{1} + 1 = x_{2} – 1

⇒ x_{2} = x_{1} = 2, which is impossible.

(ii) When x_{1} is even and x_{2} is odd.

Here f(x_{1}) =f(x_{2})

⇒ x_{1} – 1 = x_{2} + 1

⇒ x_{1} – x_{1} = 2, which is impossible.

(iii) When x_{1} and x_{2} are both odd.

Here f(x_{1}) = f(x_{2})

⇒ x_{1} + 1 = x_{2} + 1

⇒ x_{1} = x_{2}

∴ ‘f’ is one-one.

(iv) When x_{1} and x_{2} are both even.

Here f(x_{1}) = f(x_{2})

⇒ x_{1} – 1 = x_{2} – 1

⇒ x_{1} = x_{2}

∴ ‘ f’ is one-one.

Onto. Let ‘x’ be an arbitrary natural number. When x is an odd natural number, then there exists an even natural number (x + 1) such that: f(x + 1) = (x + 1) – 1 = x.

When x is an even natural number, then there exists an odd natural number (x – 1) such that:

f(x – 1) = (x – 1) + 1 = x.

∴ Each x ∈ N has its pre-image in N.

Thus ‘f ’ is onto.

Hence, ‘f ’ is both one-one and onto.

Question 6.

Show that the function f : R → R defined by:

\(\frac{x}{x^{2}+1}\) ∀ x ∈ R is neither one-one nor onto. Also, if g : R → R is defined g(x) = 2x -1, find fog (x). (C.B.S.E. 2018)

Solution:

We have : f(x) = \(\frac{x}{x^{2}+1}\)

(i) One-one, f(x_{1}) = f(x_{2})

⇒ \(\frac{x_{1}}{x_{1}^{2}+1}=\frac{x_{2}}{x_{2}^{2}+1}\)

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

⇒ x_{1}x_{2} (x_{2} – x_{1}) = x_{2} – x_{1}

x_{1}x_{2} = 1

x_{1} = \(\frac{1}{x_{2}}\)

x_{1} ≠ x_{2}

Thus, f is not one-one.

Onto:

⇒ f(x) = y

⇒ \(\frac{x}{x^{2}+1}\) = y

⇒ x = yx^{2} + y

⇒ x^{2}y + y – x = 0

⇒ x can not be expressed in y.

Thus, f is not onto.

Hence, f is neither one-one nor onto.

(ii) Since g (x) = 2x – 1,

∴ fog (x) = f(g (x)) = f(2x – 1)

= \(\frac{2 x-1}{(2 x-1)^{2}+1}=\frac{2 x-1}{4 x^{2}-4 x+2}\)

Question 7.

Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation.

(C.B.S.E. 2018 C)

Solution:

(x – x) = 0 is divisible by 3 for all x ∈ Z.

So, (x, x) ∈ R

:. R is reflexive.

(x – y) is divisibile by 3 implies (y – x) is divisible by 3.

So (x, y) ∈ R implies (y – x) ∈ R, x. y ∈ Z

⇒ R is symmetric.

(x – y) is divisibile by 3 and (y – z) is divisible by 3.

So (x – z) = (x – y) + (y + z) is divisible by 3.

∴ (x, z) ∈ R ⇒ R is transitive.

Hence, R is an equivalence relation.

Question 8.

Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by:

f(x) = [latyex]=\left(\frac{x-2}{x-3}\right)[/latex]. Show that ‘f’ is one-one and onto and hence find f^{-1}. (C.B.S.E. 2012)

Solution:

Let x_{1}, x_{2} ∈ R-{3}.

Now f(x_{1}) = f(x_{2})

⇒ \(\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}\)

⇒ (x_{1} – 2) (x_{2} – 3) = (x_{1} – 3) (x_{2} – 2)

⇒ x_{1}x_{2} – 3x_{1} – 2x_{2} + 6 = ⇒ x_{1}x_{2} – 2x_{1} – 3x_{2} + 6

⇒ x_{1} = x_{2}

Thus ‘f’ is one-one.

Let y ∈ R – {1}.

Let y = f(x_{0}).

Then \(\frac{x_{0}-2}{x_{0}-3}\) = y

x_{0} – 2 = x_{0}y – 3y

x_{0}(y-1) = 3y – 2

x_{0} = \(\frac{3 y-2}{y-1}\)

Now y ∈ R – {1}.

Thus ‘f’ is onto.

Hence, ‘f’ is one-one and onto function

⇒ ‘f’ is invertible.

Also y = \(\frac{x-2}{x-3}\)

⇒ xy — 3y = x – 2

⇒ x (y — 1) = 3y – 2

x = \(\frac{3y-2}{y-1}\)

f^{-1} (y) = \(\frac{3 y-2}{y-1}\)

Hence, f^{-1} (x) = \(\frac{3 x-2}{x-1}\)– for all x ∈ R-{1}.

Question 9.

Let Y = {n^{2}: n ∈ N} ⊂ N. Consider f : N → Y as f(n) = n^{2}. Show that if is invertible. Find the inverse of ‘f. (N.C.E.R.T.)

Solution:

Let y ∈ Y, where y is arbitrary.

Here y is of the form n^{2}, for n∈ N

⇒ n = √y.

This motivates a function :

g : Y → N, defined by g (y) = √y .

Now gof(n ) = g (f(n)) = g (n^{2}) = √n^{2}= n

and fog (y) =f(g (y))

= f (√y) = (√y)^{2} = y

Thus gof = I_{N}

and fog = I_{Y}.

Hence, ‘ f ’ is invertible with f^{-1} = g.

Question 10.

Let f : W → W be defined by:

Show that ‘f ’ is invertible. Find the inverse of ‘f ’. (Here ‘ W’ is the set of whole numbers) (A.I.C.B.S.E. 2015)

Solution:

We have : f : W → W defined by :

f is one-one.

When n_{1} and n_{2} are both odd,

then f(n_{1}) = f(n_{2})

⇒ n_{1} – 1 = n_{2} – 1

⇒ n_{1} = n_{2}

When n_{1} and n_{2} are both even, then

f(n_{1}) = f(n_{2})

⇒ n_{1} + 1 = n_{2} + 1

⇒ n_{1}= n_{2}

∴ In both cases,

f(n_{1}) = f(n_{2})

⇒ n_{1 }= n_{2}

When n_{1}is odd and n_{2} is even,

then f(n_{1}) = n_{1} – 1, which is even

and f(n_{2}) = n_{2} + 1, which is odd.

∴ (n_{1}) ≠ (n_{2})

⇒ f(n_{1}) ≠ f(n_{2})

Similarly when n_{1} is even and n_{2} is odd,

∴ (n_{1}) ≠ (n_{2})

⇒ f(n_{1}) ≠ f(n_{2})

In each case, ‘f’ is one-one.

f is onto.

When n is odd whole number, then there exists an even whole number

n – 1 ∈ W such that

f(n- 1) = (n -1) + 1 = n.

When n is even whole number, then there exists an odd whole number n + 1 ∈ W such that

f(n +1) = (n + 1) -1 = n.

Also f(1) = 0 ∈ W.

∴ each number of W has its pre-image in W.

Thus f ’ is onto.

Hence, ‘f’ is one-one onto

⇒ ‘f’ is invertible.

To obtain f^{-1}.

Let n_{1}, n_{2} ∈ W

such that f(n_{1}) = f(n_{2})

n_{1} + 1 = n_{2}, if n_{1} is even

n_{1} – 1 = n_{2}, if n_{1} is odd.

Hence f = f^{-1}