NCERT Exemplar - Relations and Functions (Part - 1)

# NCERT Exemplar - Relations and Functions (Part - 1) | Mathematics (Maths) Class 12 - JEE PDF Download

Q.1. Let A = {a, b, c} and the relation R be defined on A as follows:
R = {(a, a), (b, c), (a, b)}.

Then, write minimum number of ordered pairs to be added in R to make R reflexive and transitive.
Ans.
Here, R = {(a, a), (b, c), (a, b)}
for reflexivity; (b, b), (c, c) and for transitivity; (a, c)
Hence, the required ordered pairs are (b, b), (c, c) and (a, c)

Q.2. Let D be the domain of the real valued function f defined by

Then, write D.
Ans.

For real value of f(x), 25 – x2 ≥ 0
⇒ – x2 ≥ – 25 ⇒ x2 ≤ 25 ⇒ - 5 ≤ x ≤ 5
Hence, D ∈ - 5 ≤ x ≤ 5 or [- 5, 5]

Q.3. Let f, g : R → R be defined by f (x) = 2x + 1 and g (x) = x2 – 2, ∀ x ∈ R, respectively. Then, find gof.
Ans.
Here, f(x) = 2x + 1 and g(x) = x2 – 2
∴ g of = g[f(x)]
= [2x + 1]2 – 2 = 4x2 + 4x + 1 – 2 = 4x2 + 4x – 1
Hence, g of = 4x2 + 4x – 1

Q.4. Let f: R → R be the function defined by f (x) = 2x – 3 ∀ x ∈ R. write f –1.
Ans.
Here, f(x) = 2x – 3
Let f(x) = y = 2x – 3
⇒ y + 3 = 2x

Q.5. If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f -1.
Ans.
Let y = f(x) ∴ x = f – 1(y)
∴ If f = {(a, b), (b, d), (c, a), (d, c)}
then f – 1 = {(b, a), (d, b), (a, c), (c, d)}

Q.6. If f: R → R is defined by f (x) = x2 – 3x + 2, write f (f (x)).
Ans.
Here, f(x) = x2 – 3x + 2
∴ f [f(x)] = [f(x)]2 – 3f(x) + 2
= (x2 – 3x + 2)2 – 3(x2 – 3x + 2) + 2
= x4 + 9x2 + 4 – 6x3 + 4x2 – 12x – 3x2 + 9x – 6 + 2
= x4 – 6x3 + 10x2 – 3x
Hence, f [f(x)] = x4 – 6x3 + 10x2 – 3x

Q.7. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by
g (x) = αx + β, then what value should be assigned to α and β.
Ans.
Yes, g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function.
Here, g(x) = αx + β
For (1, 1), g(1) = α.1 + β
1 = α + β ... (1)
For (2, 3), g(2) = α.2 + β
3 = 2α + β ... (2)
Solving eqs. (1) and (2) we get, α = 2, β = – 1

Q.8. Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) {(x, y): x is a person, y is the mother of x}.
(ii){(a, b): a is a person, b is an ancestor of a}.
Ans.
(i) It represents a function. The image of distinct elements of x under f are not distinct. So, it is not injective but it is surjective.
(ii) It does not represent a function as every domain under mapping does not have a unique image.

Q.9. If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write fog.
Ans.
fog = f [g(x)]
= f [g(2)] = f(3) = 5
= f [g(5)] = f(1) = 2
= f [g(1)] = f(3) = 5
Hence, fog = {(2, 5), (5, 2), (1, 5)}

Q.10. Let C be the set of complex numbers. Prove that the mapping f: C → R given by f (z) = |z|, ∀ z ∈ C, is neither one-one nor onto.
Ans.
Here, f(z) = |z| ∀ z ∈ C
f(1) = |1| = 1
f(- 1) = |- 1| = 1
f(1) = f(- 1)
But 1 ≠ - 1
Therefore, it is not one-one.
Now, let f(z) = y = |z|. Here, there is no pre-image of negative numbers. Hence, it is not onto.

Q.11. Let the function f: R → R be defined by f (x) = cos x, ∀ x ∈ R. Show that f is neither one-one nor onto.
Ans.
Here, f(x) = cos x ∀ x ∈ R

Therefore, the given function is not one-one. Also, it is not onto function as no pre-image of any real number belongs to the range of cos x i.e., [–1, 1].

Q.12. Let X = {1, 2, 3}and Y = {4, 5}. Find whether the following subsets of X × Y are functions from X to Y or not.
(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}
(ii) g = {(1, 4), (2, 4), (3, 4)}
(iii) h = {(1, 4), (2, 5), (3, 5)}
(iv) k = {(1, 4), (2, 5)}.
Ans.
Here, given that X = {1, 2, 3}, Y = {4, 5}
∴ X × Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}
f is not a function because there is no unique image of each element of domain under f.
(ii) g = {(1, 4), (2, 4), (3, 4)}
Yes, g is a function because each element of its domain has a unique image.
(iii) h = {(1, 4), (2, 5), (3, 5)}
Yes, it is a function because each element of its domain has a unique image.
(iv) k = {(1, 4), (2, 5)}
Clearly k is also a function.

Q.13. If functions f: A → B and g: B → A satisfy g of = IA, then show that f is one-one and g is onto.
Ans.
Let x1, x2 ∈ gof
gof {f(x1)} = gof {f(x2)}
⇒ g(x1) = g(x2) [∵ gof = IA]
∴ x= x2
Hence, f is one-one. But g is not onto as there is no pre-image of A in B under g.

Q.14. Let f: R → R be the function defined by

Then, find the range of f.
Ans.

Range of cos x is [– 1, 1]

⇒ 2y – y cos x = 1 ⇒ y cos x = 2y – 1

Now – 1 ≤ cos x ≤ 1

Q.15. Let n be a fixed positive integer. Define a relation R in Z as follows: ∀ a, b ∈ Z, a R b if and only if a – b is divisible by n. Show that R is an equivalence relation.
Ans.
Here, ∀ a, b ∈ Z and a R b if and only if a - b is divisible by n.
The given relation is an equivalence relation if it is reflexive, symmetric and transitive.
(i) Reflexive:
a R a ⇒ (a - a) = 0 divisible by n
So, R is reflexive.
(ii) Symmetric:
a R b = b R a ∀ a, b ∈ Z
a - b is divisible by n (Given)
⇒ - (b - a) is divisible by n
⇒ b – a is divisible by n
⇒ b R a
Hence, R is symmetric.
(iii) Transitive:
a R b and b R c ⇔ a R c ∀ a, b, c ∈ Z
a - b is divisible by n
b - c is also divisible by n
⇒ (a - b) + (b - c) is divisible by n
⇒ (a - c) is divisible by n
Hence, R is transitive.
So, R is an equivalence relation.

Q.16. If A = {1, 2, 3, 4}, define relations on A which have properties of being:
(a) Reflexive, transitive but not symmetric
(b) Symmetric but neither reflexive nor transitive
(c) Reflexive, symmetric and transitive.
Ans.
Given that A = {1, 2, 3, 4}
∴ ARA = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), (2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)}
(a) Let R1 = {(1, 1), (2, 2), (1, 2), (2, 3), (1, 3)}
So, R1 is reflexive and transitive but not symmetric.
(b) Let R2 = {(2, 3), (3, 2)}
So, R2 is only symmetric.
(c) Let R3 = {(1, 1), (1, 2), (2, 1), (2, 4), (1, 4)}
So, R3 is reflexive, symmetric and transitive.

Q.17. Let R be relation defined on the set of natural number N as follows: R = {(x, y) : x ∈N, y ∈ N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.
Ans.
Given that x ∈ N, y ∈ N and 2x + y = 41
∴ Domain of R = {1, 2, 3, 4, 5, ..., 20}
and Range = {39, 37, 35, 33, 31, ..., 1}
Here, (3, 3) ∈ R
as 2 × 3 + 3 ≠ 41
So, R is not reflexive.
R is not symmetric as (2, 37) ∈ R but (37, 2) ∉R
R is not transitive as (11, 19) ∈ R and (19, 3) ∈ R but (11, 3) ∉ R.
Hence, R is neither reflexive, nor symmetric and nor transitive.

Q.18. Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
(a) An injective mapping from A to B
(b) A mapping from A to B which is not injective
(c) A mapping from B to A.
Ans.
Here, A = {2, 3, 4} and B = {2, 5, 6, 7}
(a) Let f: A → B be the mapping from A to B
f = {(x, y): y = x + 3}
∴ f = {(2, 5), (3, 6), (4, 7)} which is an injective mapping.
(b) Let g: A → B be the mapping from A → B such that
g = {(2, 5), (3, 5), (4, 2)} which is not an injective mapping.
(c) Let h: B → A be the mapping from B to A
h = {(y, x): x = y - 2}
h = {(5, 3), (6, 4), (7, 3)} which is the mapping from B to A.

Q.19. Give an example of a map
(i) Which is one-one but not onto
(ii) Which is not one-one but onto
(iii) Which is neither one-one nor onto.
Ans.
(i) Let f: N → N given by f(x) = x2
Let x1, x2 ∈ N then f(x1) = x12 and f(x2) = x22
Now, f (x1) = f(x2) ⇒ x12 =x22 ⇒ x12 -x22= 0 ⇒ (x1 + x2) (x1 - x2) = 0
Since x1, x2 ∈ N, so x1 + x2 = 0 is not possible.
∴ x1 – x2 = 0 ⇒ x1 = x2
∴ f(x1) = f(x2) ⇒ x1 = x2
So, f (x) is one to one function.
Now, Let f(x) = 5 ∈ N
then x2 = 5 ⇒ x = ± √5 ∉ N
So, f is not onto.

Hence, f (x) = x2 is one-one but not onto.

Since f (1) = f(2) but 1 ≠ 2,
So, f is not one-one.
Now, let y ∈ N be any element.
Then f (n) = y

⇒ n = 2y – 1 if y is even
n = 2y if y is odd or even

∴ Every y ∈ N has pre-image

∴ f is onto.
Hence, f  is not one-one but onto.
(iii) Let f: R → R be defined as f(x) = x2
Let x1 = 2 and x2 = - 2
f(x1) = x12 = (2)2= 4
f(x2) = x22 = (- 2)2 = 4
f(2) = f(- 2) but 2 ≠ - 2
So, it is not one-one function.
Let f (x) = – 2 ⇒ x2 = – 2 ∴ x = ± √-2 ∉ R
Which is not possible, so f is not onto.
Hence, f is neither one-one nor onto.

Q.20. Let A = R – {3}, B = R – {1}. Let f: A → B be defined by f (x) = (x-2)/(x-3) ∀ x ∈ A . Then show that f is bijective.
Ans.
Here, A ∈ R - {3}, B = R - {1}

Let x1, x2 ∈ f(x)
∴ f(x1) = f(x2)

So, it is injective function.

⇒ xy – 3y = x – 2
⇒ xy – x = 3y – 2
⇒ x(y – 1) = 3y – 2

⇒ f(x) = y ∈ B.
So, f (x) is surjective function.
Hence, f (x) is a bijective function.

Q.21. Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:
(i) f(x) = 2/x
(ii) g(x) = |x|
(iii) h(x) = x |x|
(iv) k(x) = x2

Ans.
(i) Given that – 1 ≤ x ≤ 1
Let x1, x2 ∈ f(x)

So, f(x) is one-one function.
Let f(x) = y = x/2 ⇒ x = 2y
For y = 1, x = 2 ∉ [-1, 1]
So, f (x) is not onto. Hence, f(x) is not bijective function.
(ii) Here, g(x) = |x|
g(x1) = g(x2) ⇒ |x1| = |x2| ⇒ x1 = +x2
So, g (x) is not one-one function.
Let g(x) =y= |x| ⇒ x = + y ∉ A ∀ y ∈ A
So, g (x) is not onto function.
Hence, g (x) is not bijective function.
(iii) Here, h(x) = x|x|
h(x1) = h f(x2)
⇒ x1|x1| = x2|x2| ⇒ x1 = x2
So, h (x) is one-one function.
Now, let h(x) = y = x|x| = x2 or - x2

∴ h(x) is not onto function.
Hence, h (x) is not bijective function.
(iv) Here, k(x) = x2 k(x1) = k(x2)
⇒ x12 = x22 ⇒ x1 = ± x2
So, k (x) is not one-one function.
Now, let k(x) = y = x2 ⇒ x = ± √y
If y = – 1 ⇒ x = ± √- 1 ∉ A ∀ y ∈ A
∴ k(x) is not onto function.
Hence, k (x) is not a bijective function.

Q.22. Each of the following defines a relation on N:
(i) x is greater than y, x, y ∈ N
(ii) x + y = 10, x, y ∈ N
(iii) x y is square of an integer x, y ∈ N
(iv) x + 4y = 10 x, y ∈ N.

Determine which of the above relations are reflexive, symmetric and transitive.
Ans.
(i) x is greater than y, x, y ∈ N
For reflexivity x > x ∀ x ∈ N which is not true
So, it is not reflexive relation.

So, it is not symmetric relation.
For transitivity, x R y, y R z ⇒ x R z ∀x, y, z ∈ N ⇒ x > y, y > z ⇒ x > z
So, it is transitive relation.
(ii) Here, R = {(x, y) : x + y = 10 ∀ x, y ∈ N}
R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}
For reflexive: 5 + 5 = 10, 5 R 5 ⇒ (x, x) ∈ R
So, R is reflexive.
For symmetric: (1, 9) ∈ R and (9, 1) ∈ R
So, R is symmetric.
For transitive: (3, 7) ∈ R, (7, 3) ∈ R but (3, 3) ∉ R
So, R is not transitive.
(iii) Here, R = {(x, y) : xy is a square of an integer, x, y ∈ N}
For reflexive: x R x = x . x = x2 is an integer
[∴ Square of an integer is also an integer]
So, R is reflexive.
For symmetric: x R y = y R x ∀ x, y ∈ N
∴ xy = yx (integer)
So, it is symmetric.
For transitive: x R y and y R z ⇒ x R z
Let xy = k2 and yz = m2

which is again a square of an integer.
So, R is transitive.
(iv) Here, R = {(x, y) : x + 4y = 10, x, y ∈ N}
R = {(2, 2), (6, 1)}
For reflexivity: (2, 2) ∈ R
So, R is reflexive.
For symmetric: (x, y) ∈ R but (y, x) ∉ R (6, 1) ∈ R but (1, 6) ∉ R
So, R is not symmetric.
For transitive: (x, y) ∈ R but (y, z) ∉ R and (x, z) ∈ R
So, R is not transitive.

Q.23. Let A = {1, 2, 3, ... 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].
Ans.
Here, A = {1, 2, 3, ..., 9}
and R → A × A defined by (a, b) R (c, d) ⇒ a + d = b + c
∀ (a, b), (c, d) ∈ A × A
For reflexive: (a, b) R (a, b) = a + b = b + a ∀ a, b ∈ A which is true. So, R is reflexive.
For symmetric: (a, b) R (c, d) = (c, d) R (a, b)
L.H.S. a + d = b + c
R.H.S. c + b = d + a
L.H.S. = R.H.S. So, R is symmetric.
For transitive: (a, b) R (c, d) and (c, d) R (e, f) ⇔ (a, b) R (e, f)
⇒ a + d = b + c and c + f = d + e
a + d = b + c and d + e = c + f
(a + d) - (d + e) = (b + c) - (c + f)
a - e = b - f
a + f = b + e
(a, b) R (e, f)
So, R is transitive.
Hence, R is an equivalence relation.
Equivalent class of {(2, 5)} is {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

Q.24. Using the definition, prove that the function f : A → B is invertible if and only if f is both one-one and onto.
Ans.
A function f : X → Y is said to be invertible if there exists a
function g : Y → X such that gof = IX and fog = IY and then the inverse of f is denoted by
f -1.
A function f : X → Y is said to be invertible if f is a bijective function.

Q.25. Functions f, g : R → R are defined, respectively, by f (x) = x2 + 3x + 1, g (x) = 2x – 3, find
(i) fog
(ii) gof
(iii) fof
(iv) gog
Ans.
(i) fog ⇒ f[g(x)] = [g(x)]2 + 3[g(x)] + 1
= (2x – 3)2 + 3(2x – 3) + 1 = 4x2 + 9 – 12x + 6x – 9 + 1 = 4x2 – 6x + 1
(ii) gof ⇒ g[f(x)] = 2[x2 + 3x + 1] – 3 = 2x2 + 6x + 2 – 3 = 2x2 + 6x – 1
(iii) fof ⇒ f[f(x)] = [f(x)]2 + 3[f(x)] + 1
= (x2 + 3x + 1)2 + 3(x2 + 3x + 1) + 1
= x4 + 9x2 + 1 + 6x3 + 6x + 2x2 + 3x2 + 9x + 3 + 1
= x4 + 6x3 + 14x2 + 15x + 5
(iv) gog ⇒ g[g(x)] = 2[g(x)] – 3 = 2(2x – 3) – 3 = 4x – 6 – 3 = 4x – 9

Q.26. Let * be the binary operation defined on Q. Find which of the following binary operations are commutative
(i) a * b = a – b ∀ a, b ∈ Q
(ii) a * b = a2 + b2 ∀ a, b ∈ Q
(iii) a * b = a + ab ∀ a, b ∈ Q
(iv) a * b = (a – b)2 ∀ a, b ∈ Q
Ans.
(i) a * b = a – b ∈ Q ∀ a, b ∈ Q.
So, *  is binary operation.
a * b = a – b and b * a = b – a ∀ a, b ∈ Q
a - b ≠ b - a
So, *  is not commutative.
(ii) a * b = a2 + b2 ∈ Q,
so * is a binary operation. a * b = b * a
⇒ a2 + b2 = b2 + a2 ∀ a, b ∈ Q Which is true.
So, * is commutative.
(iii) a * b = a + ab ∈ Q, so * is a binary operation.
a * b = a + ab and b * a = b + ba
a + ab ≠ b + ba ⇒ a * b ≠ b * a ∀ a, b ∈ Q.
So, *  is not commutative.
(iv) a * b = (a – b)2 ∈ Q, so * is binary operation.
a * b = (a - b)2  and b * a = (b - a)2 a * b = b * a
⇒ (a - b)2 = (b - a)2 ∀ a, b ∈ Q.
So, *  is commutative.

Q.27. Let * be binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R.  Then the operation * is

(i) Commutative but not associative
(ii) Associative but not commutative
(iii) Neither commutative nor associative
(iv) Both commutative and associative
Ans.
(i): Given that
a * b = 1 + ab ∀ a, b ∈ R
and b * a = 1 + ba ∀ a, b ∈ R
a * b = b * a  = 1 + ab
So, * is commutative.
Now a * (b * c) = (a * b) * c ∀ a, b, c ∈ R
L.H.S. a * (b * c) = a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc
R.H.S. (a * b) * c = (1 + ab) * c = 1 + (1 + ab) . c = 1 + c + abc
L.H.S. ≠ R.H.S.
So, *  is not associative.
Hence, * is commutative but not associative.

The document NCERT Exemplar - Relations and Functions (Part - 1) | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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## Mathematics (Maths) Class 12

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## FAQs on NCERT Exemplar - Relations and Functions (Part - 1) - Mathematics (Maths) Class 12 - JEE

 1. What are relations and functions in mathematics?
Ans. Relations and functions are fundamental concepts in mathematics. A relation is a set of ordered pairs, where each ordered pair consists of an input and an output. A function, on the other hand, is a special type of relation where each input has exactly one corresponding output. In other words, a function assigns a unique value to every input.
 2. How do you determine if a relation is a function?
Ans. To determine if a relation is a function, we need to check if each input has only one corresponding output. One way to do this is by using the vertical line test. If any vertical line intersects the graph of the relation at more than one point, then the relation is not a function. Another way is to examine the set of ordered pairs and check if there are any repeated inputs. If there are repeated inputs with different outputs, then the relation is not a function.
 3. What is the domain and range of a function?
Ans. The domain of a function is the set of all possible input values or the x-values. It represents the values for which the function is defined. The range, on the other hand, is the set of all possible output values or the y-values. It represents the values that the function can take. In other words, the domain is the set of inputs that the function can accept, and the range is the set of outputs that the function can produce.
 4. How do you find the inverse of a function?
Ans. To find the inverse of a function, we need to interchange the input and output variables. Let's say we have a function f(x). The inverse of f(x) is denoted as f^(-1)(x). To find it, we first replace f(x) with y. Then we interchange x and y in the equation and solve for y. The resulting equation gives us the inverse function. It is important to note that not all functions have inverses, and for a function to have an inverse, it must pass the horizontal line test.
 5. What are the different types of functions?
Ans. There are several types of functions in mathematics. Some commonly studied types include: - Linear functions: These functions have a constant rate of change and can be represented by a straight line. - Quadratic functions: These functions have a squared term and can be represented by a parabola. - Exponential functions: These functions have a constant base raised to a variable exponent. - Logarithmic functions: These functions are the inverse of exponential functions and involve the logarithm of a variable. - Trigonometric functions: These functions involve ratios of the sides of a right triangle and are commonly used in trigonometry and geometry.

## Mathematics (Maths) Class 12

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