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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 = I_{R}.

Answer:

Since gof = fog = I_{R} ,

(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.

Answer:

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) = x^{2} – 3x +2, find f (f(x)).

Answer:

f (x) = x^{2} – 3x + 2

∴ f (f (x)) = f (y) where

y = x^{2} – 3x + 2

= y^{2} – 3y + 2

= (x^{2} – 3x + 2)^{2} – 3 (x^{2} – 3x + 2) + 2

= x^{4} + 9x^{2} + 4 – 6x^{3} + 4x^{2} – 12x – 3x^{2} + 9x – 6 + 2

= x^{4} – 6x^{3} + 10x^{2} – 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.

Answer:

hence f is one-one

similarly we can prove that f is one-one.

When x_{1} < 0 and x_{2}> 0 and x, < 0 and x_{2} > 0

To prove f is onto

Question 5.

Show that the function f : R → R given by f (x) = x^{3} is injective.

Answer:

Let x_{1} ≠ x_{2} be any 2 reals

∴ x_{1 }– x_{2} ≠ 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|).

Answer:

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 x_{1}, x_{2}, ∈ N such that gof (x_{1}) = gof (x_{2})

⇒ g (x_{1}) = g (x_{2})

⇒ | x_{1 }| = | x_{2} | ⇒ x_{1 }= X_{2
}(both x_{1 },x_{2} >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

\(g(x)=\left\{\begin{array}{r}{x-1 \text { if } x>1} \\{1 \text { if } x=1}\end{array}\right.\)

Answer:

f(x) = x+ lxl + 1 ≥ 1 + 1 ∀ x ∈ N

(∵ v ≥ 1 x ∈ N)

⇒ f(x) ≥ 2 ∀ x ∈ N . R_{f} ^ N as 1 t R_{f
}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.

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 *.

Answer:

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 – X^{1} = B = X.

Question 10.

Find the number of all onto functions from the set {1, 2, 3,…, n} to itself.

Answer:

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 ^{n}P_{n} = n!

Question 11.

Let S = {a, b, c} and T = {1, 2, 3}. Find F^{1 }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)}

Answer:

(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)}

∴ F^{1} T → S as defined by

F^{1} (3) = a, F^{-1} (2) = b, F^{1} (1) = c.

(ii) F (b) = 1, F (c) = 1 hence F (n_{1}) = F (n_{2}) ≠ n_{1}, = n_{2
}⇒ 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.

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

Answer:

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.

Answer:

* | 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) = x^{2} – 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).

Answer:

f (-1) = (-1)^{2} + 1=2

f (0) = 0^{2} – 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

Answer:

R_{1} = {(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} ∈ R_{1} 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

Answer:

(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

Correct answer is “B”.

Question 18.

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

\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ll}{\mathbf{1},} {\mathbf{x}>\mathbf{0}} \\{\mathbf{0},} & {\mathbf{x}=\mathbf{0}} \\{-\mathbf{1},} & {\mathbf{x}<\mathbf{0}}\end{array}\right.\)

Answer:

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

Answer:

Number of Binary operation on a set {a, b} is

n (A) = 2 ∴ n(A×A) = 4

number of binary operation is 2^{4} = 16.

correct answer is B.

### 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]

Answer:

Question 2.

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

Answer:

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]

Answer:

f (x_{1}) = f(x_{2}) → x_{1} = x_{2} ∀ 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]

Answer:

(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

Answer:

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.

Answer:

Question 2.

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

gof = fog = I_{R
}Answer:

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]

Answer:

\(\text { Let } n_{1}=\frac{\pi}{4}, \text { and } n_{2}=\frac{3 \pi}{4}\)

\(f\left(n_{1}\right)=\frac{1}{\sqrt{2}} f\left(n_{2}\right)=\frac{1}{\sqrt{2}}\)

f(n_{1}) = f(n_{2}) ⇒ n_{1}≠ n_{2
}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 (sin^{1} n) bijective? [I I T] Which is defined on [-1, 1] to [-1, 1].

Answer:

Let f (n) = sin (sin^{-1} n)

let n_{1} n_{2} ∈ [-1, 1]

f (n_{1}) = f (n_{2}) ⇒ sin (sin^{-1} n_{1}) = sin [sin^{1} (n_{2})]

⇒ n_{1 }= n_{2}

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]

Answer: