# 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Miscellaneous Exercise

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## Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Miscellaneous Exercise

Question 1.
Let f : R → R be defined as f (x) = 10x + 7. Find the function g : R → R such that
gof = fog = IR.
Since gof = fog = IR ,
(gof) (x) = (fog) (x) = x ∀ x ∈ R
g (f (x)) = f (g (x)) = x
g (10x + 7) = x and

Question 2.
Let f : W → W be defined as f(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.
Let m, n ∈ W and m and n be 2 distinct elements.

Case I:
When both m and n are even
f(m) = m + 1     f (n) = n + 1
m ≠ n ⇒ m + 1 ≠ n + 1
⇒ f (m) ≠ f (n)

Case II:
When both m and n are odd
f (m) = m – 1 f(n) = n – 1
m ≠ n ⇒ m -1 ≠ n – 1
⇒ f (m) ≠ f (n)

Case III:
When m is odd and n is even
f(m) = m – 1       f (n) = n + 1
f(m) ≠ f (n)
so, in all cases m ≠ n ⇒ f (m) ≠ f (n)
Hence, f is a one-one function.
If n ∈ W is any element
f (n – 1) = n if n is odd (∵ n – liseven)
f(n + 1) = n if n is even (∵ n + 1 is odd)
so, every element of W is the f – image of some element in W.
Hence f is onto
Thus, f is both one-one and onto.
i.e. f is one-one correspondence. Consequently,
f is invertible.
f (n – 1) = n if n is odd
f (n + 1) = n if n is even
n – 1 = f-1 (n) if n is odd
n + 1 = f-1 (n) if n is even
\begin{aligned}&\text { i.e. } f^{-1}(n)=\left\{\begin{array}{ll}{n-1} & {\text { if } \quad n \text { is odd }} \\{n+1} & {\text { if } \quad n \text { is even }}\end{array}\right.\\&\text { thus } f^{-1}=f\end{aligned}

Question 3.
If f : R → R is defined by f(x) = x2 – 3x +2, find f (f(x)).
f (x) = x2 – 3x + 2
∴ f (f (x)) = f (y) where
y = x2 – 3x + 2
= y2 – 3y + 2
= (x2 – 3x + 2)2 – 3 (x2 – 3x + 2) + 2
= x4 + 9x2 + 4 – 6x3 + 4x2 – 12x – 3x2 + 9x – 6 + 2
= x4 – 6x3 + 10x2 – 3x.

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

hence f is one-one
similarly we can prove that f is one-one.
When x1 < 0 and x2> 0 and x, < 0 and x2 > 0
To prove f is onto

Question 5.
Show that the function f : R → R given by f (x) = x3 is injective.
Let x1 ≠ x2 be any 2 reals
∴  x1 – x2 ≠ 0.
Also,

Question 6.
Give examples of two functions f : N → Z and g : Z → Z such that g : Z → Z is injective but £ is not injective. (Hint: Consider f(x) = x and g (x) = |x|).
As g (x) = g (-x) = |x| for all x ∈ Z ,
∴ g is not one-one
i.e. y is not injective
As f : N →  Z and g : Z → Z gof : N → Z
let x1, x2, ∈ N such that gof (x1) = gof (x2)
⇒ g (x1) = g (x2)
⇒ | x1 | = | x2 | ⇒ x1 = X2
(both x1 ,x2 >0)
Hence g o f is injective.

Question 7.
Give examples of two functions f : N → N and g : N → N such that gof is onto but f is not onto. (Hint: Consider f (x) = x + 1 and
$$g(x)=\left\{\begin{array}{r}{x-1 \text { if } x>1} \\{1 \text { if } x=1}\end{array}\right.$$
f(x) = x+ lxl + 1 ≥ 1 + 1 ∀ x ∈ N
(∵ v ≥ 1 x ∈ N)
⇒ f(x) ≥ 2 ∀ x ∈ N . Rf ^ N as 1 t Rf
Hence f is not onto
gof : → ∼ → N such that
gof (x) = g (f (x)) = g (x + 1)
= (x+1) -1 [ ∵ ∀ x ∈ N,x+ 1 > 1]
⇒ gof (x) = x      V x ∈ N
∴ Range of gof = N as gof is the identity f Hence gof 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.
Since A C A ∀ A ∈ P (X)
R is reflexive.
Also, for A, B, C ∈ P(X),A R B & B R C
⇒ A C B and B C C
⇒ A ⊂ C ⇒ A R C
∴ R is transitive.
R is not symmetric as A C B ⇒ B C A
so ARB ⇒ BRA
Hence R is not an equivalence relation.

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 *.
Let E ∈ P (X) be an identity elements, then
A * E = E * A = A ∀ A ∈ P(X)
⇒ A ∩ E = E ∩ A = A ∀ A ∈ P(X)
⇒ X ∩ E = X as X ∈ P (X)
⇒ X C E
Also E C X as E ∈ P (X)
∴ E = X
Thus, X is the identity element.
Let A ∈ P(X) be invertible, then there exists
B ∈ P(X) such that A * B = B*A = X, the identity element.
⇒ A ∩ B = B ∩ A = X
X C A & X C B
Also A, B, ⊂ X as A, B ∈ P (X)
A = X = B
X is the only invertible element and – X1 = B = X.

Question 10.
Find the number of all onto functions from the set {1, 2, 3,…, n} to itself.
Let A = {1, 2, 3,……… n}
f : A → A is an onto function, then range of f
⇒ f is one-one f is one and onto the number of onto function is nPn = n!

Question 11.
Let S = {a, b, c} and T = {1, 2, 3}. Find F1 of the following functions F from S to T, if it exists.
(i) F = {(a, 3), (b, 2), (c, 1)}
(ii) F = {(a, 2),(i, 1), (c, 1)}
(i) Range of {1, 2, 3} = T
⇒ F is onto Also F is onto as different elements of S have different F – images.
∴ F-1 exist and F-1 = {(3, a), (2, b), (1, c)}
∴ F1 T → S as defined by
F1 (3) = a, F-1 (2) = b, F1 (1) = c.

(ii) F (b) = 1, F (c) = 1 hence F (n1) = F (n2) ≠ n1, = n2
⇒ F is not one-one .: hence not invertible.

Question 12.
Consider the binary operations * : R x R → R and o : R x R → R defined as a* b = |a-b| and a o b = a, ∀ a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a *(b o c) = (a * b) o (a * b). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over? Justify your answer.
∀ a, b ∈ R
a * b = |a – b| = |b – a| = b * a
hence commutative ∀ 3, 5 ,1 ∈ R
(3 * 5) * 7
|3 – 5| * 7 = 2 * 7= |2 – 7| = 5
3* (5 * 7) = 3* |5 – 7| =3 * 2= |3 – 2| = 1
a* (b * c) & (a * b) *c
∴ * is not associative.
a o b = a, b o a = b
hence a o b ≠ boa
hence not commutative
(a o b)o c : a o c = a
a o (b o c) = a o b = a
hence o is associative
a*(b o c) (a * b)=|a – b|
(a * b) o (a * c) = |a – b| o |a – c| = |a – b|
hence a* (b o c) = (a * b) o (a * c)
hence o is distributive over *.

Question 13.
Given a non-empty set X, let * : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ) is the identity for the operation * and all the elements A of P(X) are invertible with A-1 = A.
(Hint: (A – ϕ) ∪ (ϕ – A) = A and (A – A) ∪ (A – A) = A * A = ϕ
ϕ ∈ P (X) be an identity element.
To prove it let E ∈ P (X) be the identity element such that A * E = E * A = A
∀ A ∈ P (X)
(A – E) ∪ (E – A) = A ⇒ E = φ
i.e. (A – ϕ) ∪ (E – ϕ)) = A
A * ϕ = ϕ * A = A
* A = A, hence ϕ is the identity element.
Let Be P(X) be the inverse of A
∴ A* B = B * A = φ ∀ A e P(X)
(A – B) ∪ (B – A) = 0 ⇒ B = A
because A – B = ϕ    B – A = ϕ⇒ A = B
∴ ∀ A ∈ P(X),  A * A = ϕ
A is the invertible element of A
∴ A-1 = A

Question 14.
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as
$$\mathbf{a} * \mathbf{b}=\left\{\begin{array}{ll}{\mathbf{a}+\mathbf{b},} & {\text { if } \mathbf{a}+\mathbf{b}<\mathbf{6}} \\{\mathbf{a}+\mathbf{b}-\mathbf{6}} & {\text { if } \mathbf{a}+\mathbf{b} \geq \mathbf{6}}\end{array}\right.$$
Show that zero is the identity for this operation and each element a ±0 of the set is invertible with 6-a being the inverse of a.

 * 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4

From the composition table it is obvious that a
* 0 = a ∀ a ∈ {0, 1/2, 3,4, 5}
∴ 0 is the identity element
Also ∀ a ∈ {0, 1, 2, 3, 4, 5} ∃ (6 – a) ∈ {0,1,2, 3,4,5} such that
a * (6 – a) = (6 – a) * a = 0, a ≠ 0
hence 6 – a is the inverse of a .
However when a = 0, 6 – a g {0, 1, 2, 3,4, 5} hence 6 – a is the inverse of a when a ≠ 0.

Question 15.
Let A = {- 1, 0, 1, 2}, B = {- 4, – 2, 0, 2} and f, g : A → B be functions defined by
f (x) = x2 – x, x A and g (x) = -1, x A, Are f and g equal? Justify your answer.
(Hint: One may note that two functions f : A → B and g : A → B such that f(a) =
$$2\left|x-\frac{1}{2}\right|-1, x \in A A$$, are called equal functions).
f (-1) = (-1)2 + 1=2
f (0) = 02 – 0 = 0
f(1) = (1)2– 1 = 0
f(2) = (2)2 -2 = 2

Question 16.
Let A = {1, 2,3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1
(B) 2
(C) 3
(D) 4
R1 = {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (3, 1), (2, 1)} is the only relation which is reflexive symmetric but not transitive and is such that {1, 2], {1, 3} ∈ R1 Correct answer is “A”

Question 17.
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C )3
(D) 4
(i) {(1, 1), (1, 2), (2, 1), (2,2), (2, 3)}
(ii) {(1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1)} (2,3), (3,2)}
There are two equivalence relation

Question 18.
Let f : R → R be the Signum Function defined as
$$\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ll}{\mathbf{1},} {\mathbf{x}>\mathbf{0}} \\{\mathbf{0},} & {\mathbf{x}=\mathbf{0}} \\{-\mathbf{1},} & {\mathbf{x}<\mathbf{0}}\end{array}\right.$$
fog (n) = f g(n) = f (0) = 0, 0 ≤ n < 1
f (1) = n = 1
gof (n) = g [1] = 1

hence fog and gof does not coincide.

Question 19.
Number of binary operations on the set {a, b} are
(A) 10
(B) 16
(C) 20
(D) 8
Number of Binary operation on a set {a, b} is
n (A) = 2    ∴ n(A×A) = 4
number of binary operation is 24 = 16.

### 2nd PUC Maths Relations and Functions Miscellaneous Exercise Additional Questions and Answers

One Mark Questions:

Question 1.
f: R → R defined by $$\left(f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}\right.$$ find fof (x) [CBSE 2010]

Question 2.
If  $$\mathbf{f}(\mathbf{x})=\frac{1}{1-\mathbf{x}} \forall \mathbf{x} \in \mathbf{R}, \mathbf{n} \neq \mathbf{1}, \text { Find } \mathbf{f}[\mathbf{f}(\mathbf{f}(\mathbf{x}))]$$

Question 3.
Let A = (1, 2, 3}, B = (4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A → B. State whether f is one-one or not f (1) = 4, f(2) = 5, f (3) = 6 [CBSE 2011]
f (x1) = f(x2) → x1 = x2 ∀ n ∈ A
hence f is one-one.

Question 4.
* : R x R → R is a binary operation, a* b = 2a + b . Find (2* 3) * 4 [CBSE 2011]
(2 * 3) *4 = (4 + 7) *4 = 11 * 4 = 26.

Question 5.
If $$f(n)=\frac{4 n+3}{6 n-4}, n \neq \frac{2}{3}$$. show that f[f(n)]= n

Four Marks Questions:

Question 1.
Show that the function f: R → R defined by $$f(n)=\frac{2 n-1}{3}, n \in R$$ is one -one onto function .Also find the inverse of f.

Question 2.
Let f : R → R defined as f(n) = 10n + 7 . Find the function g : R→ R such the
gof = fog = IR

Question 3.
Show that f : A → B where A = (0,2 π) and B = [-1,1] such that fin) = sin n is not. invertible  [I I T]
$$\text { Let } n_{1}=\frac{\pi}{4}, \text { and } n_{2}=\frac{3 \pi}{4}$$
$$f\left(n_{1}\right)=\frac{1}{\sqrt{2}} f\left(n_{2}\right)=\frac{1}{\sqrt{2}}$$
f(n1) = f(n2) ⇒ n1≠ n2
hence it is not one-one
∀ y ∈ [-1, 1] ∃ n ∈ (0, 2π) such that f (n) = sin n
hence f is onto
since f is not one-one.
It is not invertible.

Question 4.
Is the function sin (sin1 n) bijective? [I I T] Which is defined on [-1, 1] to [-1, 1].
Let f (n) = sin (sin-1 n)
let n1 n2 ∈ [-1, 1]
f (n1) = f (n2) ⇒ sin (sin-1 n1) = sin [sin1 (n2)]
⇒ n1 = n2
hence f is one-one
let y = sin (sin-1 n) = n
∀ y ∈  [-1,1] ∃ [-1, 1] f(n) = n
hence f is onto
since f is one-one and onto it is bijective.

Question 5.
If f be greatest integer function and g be the absolute function. Find
$$fog\left( \cfrac { -3 }{ 2 } \right) +gof\left( \cfrac { 4 }{ 3 } \right)$$ [CBSE 2007]