Teachers' Copy
Test Name: RELATION AND
Board: CBSE Standard: 12th Subject: Mathematics
FUNCTION
Time: 2 Minutes Worksheet
Section B
3 Marks
Attempt any 5 questions out of 10.
1: Check whether the relation R in the set Z of integers defined as R = {(a, b) : a + b is divisible by 2} is reflexive, symmetric or transitive. Write the equivalence
class containing 0 i.e. [0].
(i) Reflexive:
Since, a + a = 2a which is even
∴ (a, a) ∈ R ∀ a ∈ Z
Hence R is reflexive.
(ii) Symmetric:
If (a, b) ∈ R, then a + b = 2λ
⇒ b + a = 2μ
⇒ (b, a) ∈ R. Hence R is symmetric.
(iii) Transitive:
If (a, b) ∈ R and (b, c) ∈ R
then a + b = 2λ ...(i)
and b + c = 2μ ...(ii)
Adding (i) and (ii) we get
a + 2b + c = 2(λ + μ)
⇒ a + c = 2(λ + μ – b)
⇒ a + c = 2k
where λ + μ – b = k
⇒ (a, c) ∈ R
Hence R is transitive
[0] = {...–4, –2, 0, 2, 4...}
2: Prove that the relation R on Z, defined by R = {(x, y) : (x – y) is divisible by 5} is an equivalence relation.
For reflexive
x – x = 0, for every x ∈ Z is divisible by 5 ⇒ (x, x) ∈ R
For symmetric
(x, y) ∈ R ⇒ x – y is divisible by 5 ⇒ y – x is divisible by 5
⇒ (y, x) ∈ R ⇒ R is symmetric
For transitive
Let (x, y) ∈ R and (y, z) ∈ R
(x, y) ∈ R ⇒ x – y = 5λ ...(i)
(y, z) ∈ R ⇒ y – z = 5μ ...(ii)
adding (i) and (ii), x – z = 5 (λ + μ) = 5k
⇒ (x, z) ∈ R ⇒ R is transitive
Hence R is an equivalence relation.
3: Show that the relation R on R defined as R = {(a, b) : a ≤ b}, is reflexive, and transitive but not symmetric.
Clearly a ≤ a ∀ a ∈ R ⇒ (a, a) ∈ R ⇒ R is reflexive.
For transitive:
Let (a, b) ∈ R and (b, c) ∈ R, a, b, c ∈ R
⇒ a ≤ b and b ≤ c ⇒ a ≤ c ⇒ (a, c) ∈ R
⇒ R is transitive.
For non-symmetric:
Let a = 1, b = 2, As 1 ≤ 2 ⇒ (1, 2) ∈ R
but 2 ≰ 1 ⇒ (2, 1) ∉ R
⇒ R is non-symmetric.
4: 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.
Reflexive:
R is reflexive, as 1 + a.a = 1 + a2 > 0 ⇒ (a, a) ∈ R
∀a∈R
Symmetric:
If (a, b) ∈ R
then, 1 + ab > 0
⇒ 1 + ba > 0
⇒ (b, a) ∈ R
Hence, R is symmetric.
Transitive:
Let a = – 8, b = –1, c = 1
Since, 1 + ab = 1 + (– 8)(– 1) = 9 > 0
⇒ (a, b) ∈ R
also, 1 + bc = 1 + (−1) ( ) = > 0
1 1
2 2
⇒ (b, c) ∈ R
But, 1 + ac = 1 + (−8) (
1
) = −3 < 0
2
Hence, R is not transitive.
5: Show that the relation R in the set N × N defined by (a, b) R (c, d) if a2 + d2 = b2 + c2 ∀ a, b, c, d ∈ N, is an equivalence relation.
Let (a, b) ∈ N × N
then,
a2 + b2 = a2 + b2
∴ (a, b) R (a, b)
Hence R is reflexive.
Let (a, b), (c, d) ∈ N × N be such that
(a, b) R (c, d)
2 2
⇒ a + d = b + c
2 2
2 2
⇒ c + b = d + a
2 2
⇒ (c, d) R (a, b)
Hence, R is symmetric.
Let (a, b), (c, d), (e, f) ∈ N × N be such that
(a, b) R (c, d), (c, d) R (e, f).
2 2 2 2
⇒ a + d = b + c ...(i)
and c2 + f2 = d2 + e2 ...(ii)
Adding eqn. (i) and (ii),
2 2 2 2 2
⇒ a + d + c + f = b + c + d + e
2 2 2
2 2 2 2
⇒ a + f = b + e
⇒ (a, b) R (e, f)
Hence, R is transitive
Since, R is reflexive, symmetric and transitive.
Therefore, R is an equivalence relation.
Section C
5 Marks
Attempt any 5 questions out of 10.
1: Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible
outcomes of the throw every time belongs to set {1, 2, 3, 4, 5, 6}. Let A be the set of players while B be the set of all possible outcomes.
A = {S, D}, B = {1, 2, 3, 4, 5, 6}
Raji wants to know the number of functions from A to B. How many number of functions are possible?
If a set P has m elements and set Q has n elements, then the number of functions possible from P to Q is nm.
So, number of functions from A to B = 62
2: Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible
outcomes of the throw every time belongs to set {1, 2, 3, 4, 5, 6}. Let A be the set of players while B be the set of all possible outcomes.
A = {S, D}, B = {1, 2, 3, 4, 5, 6}
Raji wants to know the number of relations possible from A to B. How many numbers of relations are possible?
As the total number of Relations that can be defined from a set P to Q is the number of possible subsets of P×Q.
If n(P)= m and n(Q) = n then n(P × Q)=mn and the number of subsets of P × Q = 2mn.
So, number of relations possible from A to B = 22×6 = 212
If n(A) = p and n(B) = q then n(A × B) = pq and the number of subsets of A × B = 2pq.
3: Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible
outcomes of the throw every time belongs to set {1, 2, 3, 4, 5, 6}. Let A be the set of players while B be the set of all possible outcomes.
A = {S, D}, B = {1, 2, 3, 4, 5, 6}
Let R be a relation on B defined by R = {(1, 2), (2, 2), (1, 3), (3, 4), (3, 1), (4, 3), (5, 5)}. Is R an equivalence relation?
OR
Show that a relation, R : B → B be defined by R = {(x, y) : y is divisible by x} is a reflexive and transitive but not symmetric.
R = {(1, 2), (2, 2), (1, 3), (3, 4), (3, 1), (4, 3), (5, 5)}
R is not reflexive. (3, 3) R
R is not symmetric.
Because, for (1, 2) R there
(2, 1) R.
R is not transitive.
Because for all element of B there does not exist, (a, b) (b, c) R and (a, c) R.
OR
R is reflexive, since every element of B i.e.,
B = {1, 2, 3, 4, 5, 6} is divisible by itself.
i.e., (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) R
further, (1, 2) R
but (2, 1) ∉ R
Moreover,
(1, 2), (2, 4) R
⇒
(1, 4) R
⇒
R is transitive.
Therefore, R is reflexive and transitive but not symmetric.
4: An organization conducted bike race under two different categories-Boys and Girls. There were 28 participants in all. Among all of them, finally
three from category 1 and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his
college project.
let B = {b1 , b2 , b3 } and G = {g1 , g2 }, where B represents the set of Boys selected and G be the set of Girls selected for the final race.
Based on the above information, answer the following questions:
How many relations are possible from B to G ?
Number of relations = 2mn = 22×3 = 26 = 64
5: An organization conducted bike race under two different categories-Boys and Girls. There were 28 participants in all. Among all of them, finally
three from category 1 and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his
college project.
let B = {b1 , b2 , b3 } and G = {g1 , g2 }, where B represents the set of Boys selected and G be the set of Girls selected for the final race.
Based on the above information, answer the following questions:
Among all the possible relations from B to G, how many functions can be formed from B to G ?
Number of functions from B to G = 23 = 8
6: An organization conducted bike race under two different categories-Boys and Girls. There were 28 participants in all. Among all of them, finally
three from category 1 and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his
college project.
let B = {b1 , b2 , b3 } and G = {g1 , g2 }, where B represents the set of Boys selected and G be the set of Girls selected for the final race.
Based on the above information, answer the following questions:
Let R : B → B be defined by R = {(x, y) : x and y are students of the same sex}. Check if R is an equivalence relation.
OR
A function f : B → G be defined by f = {(b1 , g1 ), (b2 , g2 ), (b3 , g1 )}.
Check if f is bijective. Justify your answer.
R = {(x, y) : x and y are students of same sex.}
Since x and x are of the same sex
So (x, x) ∈ R for all x
∴ R is reflexive
If x and y are of the same sex then y and x are also of the same sex
∴ R is symmetric
If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R
Then x and z will be of the same sex
∴ R is transitive
Sine R is reflexive, symmetric and transitive
∴ R is an equivalence relation.
OR
Given
R = {(b1 , g1 ), (b2 , g2 ), (b3 , g1 )}
Since b1 and b3 have the same image g1
∴ R is not injective
Since all elements of G has a pre-image
∴ R is surjective
7: Show that the function f : ℝ → {x ∈ ℝ : –1 < x < 1) defined by
x
f (x) =
1 + |x|
, x ∈ ℝ is one-one and onto function.
We have,
\[f\left( x \right)=\left\{ \begin{align}
& \frac{x}{1+x},\,\,\text{if}\,x\ge 0 \\
& \frac{x}{1-x},\,\,\text{if }x<0 \\
\end{align} \right.\]
Now, we consider the following cases.
Case 1: When x ≥ 0 ,
we have
x
f(x) =
1 + x
Injectivity: let x, y ∈ ℝ+ ∪ {0} such that f (x) = f (y), then
⇒
x y
=
1 + x 1 + y
x + xy = y + xy
x=y
So, f is injective function.
Surjectivity : when x ≥ 0 , we have
f(x) =
x
≥ 0
1 + x
and f(x) =
x
f (x) = 1 − < 1
1 + x
as ≥ 0
Let y ∈ [0, 1), thus for each y ∈ [0, 1) there exists
y
x = ≥ 0
1 + y
such that
y
y 1−y
f ( ) = = y
y
1 + y 1 +
1−y
So, f is onto function on [0,∞) to [0, 1).
Case 2: When x < 0 ,
we have
x
f (x) =
1 − x
Injectivity: Let x, y ∈ ℝ– i.e., x, y < 0 , such that f (x) = f (y), then
x y
⇒ =
1 − x 1 − y
x – xy = y – xy
x=y
So, f is injective function.
Surjectivity: x < 0 ,
we have
x
f (x) = < 0
1 − x
also,
x 1
f (x) = = −1 + > −1
1 − x 1 − x
–1 < f (x) < 0.
Let y ∈ (1, 0) be an arbitrary real number and there exists
y
x = < 0
1 + y
Such that, f(x) =
y
y 1 +y
f ( ) = = y.
y
1 + y 1 −
1+ y
So, for y ∈ (–1, 0), there exists
y
x = < 0
1 + y
such that f(x) = y .
Hence, f is onto function on (–∞, 0) to (–1, 0).
Case 3: Injectivity
Let x > 0 & y < 0 such that
f(x) = f(y)
⇒
x x
=
1 + x 1 − y
⇒ x – xy = y + xy
⇒ x – y = 2xy, here LHS > 0 but RHS < 0 , which is inadmissible.
Hence, f (x) ≠ f (y) when x ≠ y.
Hence f is one-one and onto function.
8: A function f : R – {–1, 1} → R is defined by:
x
f (x) =
2
x − 1
(i) Check if f is one-one.
(ii) Check if f is onto.
Show your work.
(i) Assumes f(x) = f(y) and evaluates the same as:
x y
=
2 2
x − 1 y − 1
⇒ x(y2 – 1) = y(x2 – 1)
⇒ xy2 – x – yx2 + y = 0
⇒ (y – x)(xy + 1) = 0
Uses the above step to conclude that
= y or xy = –1.
Takes a pair of numbers x and y such that xy = –1 to show that f is not one-one.
For example, for
1 2
x = and y = − 2, f (x) = −
2 3
and
2
f (x) = −
3
(ii) Equates f(x) to y and solves the same to express x in terms of y as:
x
= y
2
x − 1
⇒ x = yx2 – y
⇒ yx2 – x – y = 0
⇒
2
1 ± √1 − 4y
x =
2y
Since 1 + 4y2 > 0, real root exists and also they are not ± 1
⇒
2
1 ± √1 − 4y
x =
2y
∈R – {-1, + 1}
Writes that for any y ∈ R (codomain), there exists x ∈ R – {–1, 1} (domain) such that f(x) = y. Hence concludes that f is onto.
9: Consider f : R+ → [– 5, ∞) given by f(x) = 9x2 + 6x – 5 where R+ is the set of all non-negative real numbers. Prove that f is one-one and onto
function.
One-one: Let x1, x2 ∈ R+ such that f(x1) = f(x2)
9x12 + 6x1 – 5 = 9x22 + 6x2 – 5
⇒ 9(x12 – x22) + 6(x1 – x2) = 0
⇒ (x1 – x2) {9(x1 + x2) + 6} = 0
⇒ x1 – x2 = 0
or 9x1 + 9x2 + 6 = 0
which is not possible
⇒ x1 = x2
∴ f is one-one.
Onto: Let y = 9x2 + 6x – 5
⇒ 9x2 + 6x – (5 + y) = 0
⇒
−6 ± √36 + 4 (9) (5 + y)
x =
2 (9)
−6 ± √36 √1 + 5 + y
=
18
6 (−1 ± √y + 6)
−1 ± √y + 6
x = =
6 (3) 3
Now, x ∈ R+ ⇒ x ≥ 0 and so
−1 − √y + 6
x =
3
is rejected
−1 + √y + 6
x =
3
Now x ≥ 0
−1 + √y + 6
⇒ ≥ 0
3
⇒ √y + 6 ≥ 1
≥1⇒y+6≥1
⇒y≥–5
∴ Rf = {y : y ∈ [– 5, ∞)} = codomain of f.
∴ f is onto.
Hence, f is one one and onto function.
10: A function f : [– 4, 4] → [0, 4] is given by f(x) =
2
√16 − x
. Show that f is an onto function but not one-one function. Further, find all possible values of 'a' for which f(a) =
√7
