Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise

Question 1.

Let R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that gof = fog = I_{R}.

Solution:

Let f : X → Y, where X ⊆ R, Y ⊆ R.

Consider an arbitrary element y of Y.

By definition, y = 10x + 7 for some x ∈ X.

⇒ x = \(\frac { y-7 }{ 10 }\).

Let us define g : Y → X by g(y) = \(\frac { y-7 }{ 10 }\)

= \(\frac { (10x+7)-7 }{ 10 }\) = x

Now gof(x) = g[f(x)] = \(\frac { f(x)-7 }{ 10 }\)

= \(\frac { (10x+7)-7 }{ 10 }\)

and fog(x) = f[g(x)] + 10g(y) + 7 = 10.\(\frac { y-7 }{ 10 }\) + 7 = y.

⇒ gof(x) = I_{R}, fog(y) = I_{R}

⇒ f is invertible and g : Y → X, such that g(y) = \(\frac { y-7 }{ 10 }\).

Question 2.

Let f : W → W be defined as f(n) = n -1, if n is odd and f(n) = n + 1, if n is even. Show that/is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Solution:

We are given, f : W → W, defined as

\(\left\{\begin{array}{ll} n-1, & n \text { is odd } \\ n+1, & n \text { is even } \end{array}\right.\)

Let f(n_{1}) =f(n_{2}). If n_{1} is odd and n_{2} is even,

then n_{1} – 1 = n_{2} + 1, i.e., n_{1} – n_{2} = 2, which is not possible as n_{1} is odd and n_{2} is even.

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

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

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

f(n_{1}) = f(n_{2}) ⇒ n_{1} + 1 = n_{2} + 1 ⇒ n_{1} = n_{2}.

This shows f is one-one.

Any odd number 2r + 1 in codomain is the image of 2r and any even number 2r in the codomain W is the image of 2r + 1.

⇒ f is one-one and onto.

∴ f is invertible.

Further, when n is odd, y = n – 1, n = y + 1, y is even.

When n is even, y = n + l, n = y y is odd.

Hence, f^{-1}(y) = g(y) defined as

g : W → W, such that g(y) = \(\left\{\begin{array}{ll}

y-1, & y \text { is odd } \\

y+1, & y \text { is even }

\end{array}\right.\)

⇒ The inverse of f is f itself.

Question 3.

If f : R → R is defined by f(x) = x² – 3x + 2, find f[f(x)].

Solution:

f(x) = x² – 3x + 2

f[f(x)] = [f(x)]² – 3[f(x)] + 2

= (x² – 3x + 2)² – 3(x² -3x + 2) + 2

= (x^{4} + 9x² + 4 – 6x³ – 12x + 4x²) – 3x² + 9x – 6 + 2

= x^{4} – 6x³ + 10x² – 3x.

Question 4.

Show that the function f : R → { x ∈ R : – 1 – x < 1} defined by f(x) = \(\frac{x}{1+|x|}\) x ∈ R is one-one and onto. Solution: (i) Let x >0, f(x) = \(\frac { x }{ 1+x }\)

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

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

or x_{1}(1 + x_{2}) = x_{2}(1 + x_{1})

or x_{1} + x_{1}x_{2} = x_{2} + x_{1}x_{2}

⇒ x_{1} = x_{2}

When x< 0, f(x) = \(\frac { x }{ 1-x }\)

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

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

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

or x_{1} – x_{1}x_{2} = x_{2} – x_{2}x_{1}

⇒ x_{1} = x_{2}

(ii) As -1 < x < 1, f(x) = \(\frac{x}{1+|x|}\) lies between and – \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 2 }\) When x >0, y = \(\frac { x }{ 1+x }\).

∴ (1 + x)y = x

or xy – x = – y

∴ x(y-1) = – y

or x = \(\frac { -y }{ y-1 }\) = \(\frac { 1 }{ 1-y }\).

When x < 0, f{x) = \(\frac { x }{ 1-x }\) = y

∴ x = y – xy

x(1 + y) = y

∴ x = \(\frac { y }{ 1+y }\)

In both cases, each value of y in codomain of f has a unique value in its domain.

Hence, f is onto.

Thus, f is one-one and onto.

Question 5.

Show that the function f : R → R given by f(x) = x³ is injective.

Solution:

f(x) = x³

⇒ f(x_{1}) = f(x_{2})

⇒ x³_{1} = x³_{1}

⇒ x_{1} = x_{2}

⇒ f is one-one, i.e., f is injective.

Question 6.

Give examples of two functions f : N → Z and g : Z → Z such that gof is injective but g is not injective.

Solution:

Example:

(i) Let f(x) = x

and g(x) = | x |.

Consider g(x) = | x | ⇒ g is not injective,

since – 1 and 1 have the same image 1.

But (gof) (x) = g[f(x)] = g(x) = | x |

But domain of gof is N.

For every natural number n, | n | = n, n ∈ N, gof has a unique image.

∴ gof is injective.

(ii) f(x) = 2x

and g(x) = x².

Question 7.

Give examples of two functions f : N → N and g : N → N such that gof is onto but f is not onto.

Solution:

Example:

(i) Let f(x) = x + 1

and g(x) = \(\left\{\begin{array}{cl}

x-1 & \text { if } x>1 \\

1 & \text { if } x=1

\end{array}\right.\)

Let f(x) = y – x + 1

∴ x = y – 1.

For y = 1, x = 0 (Not a natural number)

⇒ Function/is not onto.

When x > 1 ,(gof)(x) = g[f(x)] = g(x +1) = x + 1 – 1 = x

This is the identity function.

Hence, it is an onto function.

(ii) Let f(x) = x + 2

and g(x) = \(\left\{\begin{array}{cl}

x-2, & \text { if } x>2 \\

2, & \text { if } x=2

\end{array}\right.\)

Question 8.

Given a non-empty set X, consider P(X) which is the set of all such sets 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.

Solution:

(i) A⊂A ⇒ R is reflexive.

(ii) A ⊂ B, B ⊂ A ∴ R is not symmetric.

(iii) If A ⊂ B, B ⊂ C, then A ⊂ C ∴ R is transitive.

⇒ R is not an equivalence relation.

Question 9.

Given a non-empty set X, consider the binary operaion * : P(X) x P(X) → P(X), given by A * B = A ∩ B, for all 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 *.

Solution:

We have * : P(X) x P(X) → P(X)

given by A* B = A ∩B

∴ X * A = X ∩ A = A * X = A for all A.

∴ X is an identity element.

Let I is an other identity

⇒ I ∩ A = A ∩ I = A for all A

and x ∈ X, I ∩ {x} = {x}

x ∈ I ⇒ X ⊂ I and I ⊂ X

⇒ I = X.

Question 10.

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

Solution:

Let

Y: 1 2 3 4 …. n

X: 1 2 3 4 …. n

One of the elements of set Y (say 1) may have any one of the pre-images 1, 2, 3, … n, i.e., in n ways.

The second element say 2 will have the pre-image in (n – 1) ways.

∴ The number of ways, we can have the pre-images

n x (n – 1)(n – 2) …. 3. 2. 1 = n!

Thus, there are nl possible onto functions.

∈

Question 11.

Let S = {a, b, c] and T = {1, 2, 3}. Find F^{-1} of the following functions F from S to T_{1}, if it exists:

(i) F = {(a, 3), (b, 2), (c, 1)}

(ii) F = {(a, 2), (b, 1), (c, 1)}

Solution:

S = {a, b, c} and T = {1, 2, 3}

(i) F = [(a, 3), (b, 2), (c, 1)}

i.e., F(a) = 3, F(b) = 2, F(c) = 1

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

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

(ii) F = {(a, 2), (b, 1), (c, 1)}

F is not one-one since elements b and c have the same image I.

∴ It is not a one-one function. So, its inverse does not exist.

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 for all a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that for all a, b, c ∈ R:

a * ( b o c) = (a * b) o (a * c)

[If it is so, we say that operation * distributes over the operation o]. Does o distributes over *? Justify your answer.

Solution:

We have: a * b = | a – b | and a o b = a.

(i) (a) a * b = | a – b |

and b * a = | b – a | = | a – b |

⇒ Operation * is commutative.

(b) (a * b) * c = | a – b | * c = || a-b | – c |

and a * (b * c) = a * | b – c | = |a- |b – c |

Clearly, ||a – b | – c | ≠ |a – |b – c||

⇒ Operation * is not associative.

(ii) (a) a o b = a, b o a = b.

Thus, a o b ≠ ⇒ b o a

⇒ Operation o is not commutative.

(b) (a o b) o c = a o c = a

and a o (b o c) = a o b = a

⇒ Operation o is associative.

(iii) For a * (b o c) = (a * b) o (a * c):

L.H.S. = a * (b o c) = a * b [∵ b o c = b]

= | a – b |

R.H.S. = (a * b) o (a * c)

= | a -b | o | a – c | = | a – b |

⇒ L.H.S. = R.H.S.

Hence, verified,

i.e., a * (b o c) = (a * b) o (a * c)

(iv) a o (b * c) = (a o b) * (a o c)

L.H.S. = a o (b * c) = a o | b – c | = a.

R.H.S. = (a o b) * (a * c) = a * a

= | a – a | = 0.

⇒ L.H.S. ≠ R.H.S.

o is not distributive over *.

Question 13.

Given a non-empty set X, let * : P(X) x P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A) for all A, B ∈ P(X). Show that emptry set Φ is the identity for the operation * and all the elements of P(X) are invertible with A^{-1} = A.

Solution:

Φ is the empty set.

The operation * is defined as

A * B = (A – B) ∪ (B-A)

Putting B = Φ, we get

A * Φ = (A – Φ) ∪ (Φ – A) = A ∪ Φ = A

Φ * A = (Φ – A) ∪ (A – Φ) = Φ ∪ A = A

⇒ A * Φ = Φ * A

Also, A * A = (A – A) ∪ (A – A) = Φ ∪ Φ = Φ

⇒ is an identity element.

Also, A * A = Φ ⇒ A^{-1} = A.

Question 14.

Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as

a * b = \(\left\{\begin{array}{cl} a+b, & \text { if } a+b<6 \\ a+b-6, & \text { if } a+b \geq 6 \end{array}\right.\)

Show that 0 is the identity for this operation and each element of the set is invertible with 6 – a being the inverse of a.

Solution:

(i) e is the identity element, if a * e = e * a = a.

a * 0 = a + 0, 0 * a = 0 + a = a.

⇒ a * 0 = 0 * a = a.

∴ 0 is the identity for the operation.

(ii) b is the inverse of a if a * b = b * a = e.

Now, a * (6 – a) = a + (6 – a) – 6 = 0

and (6 – a) * a = (6 – a) + a – 6 = 0.

Hence, each element of a of the set is invertible with inverse 6 – a.

Question 15.

Let A = {-1, 0, 1, 2}, B = {- 4, – 2, 0, 2} and/and g: A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = 2 |x – \(\frac { 1 }{ 2 }\)| – 1, x ∈ A. Are f and g equal? Justify your answer.

Solution:

Thus, for each a ∈ A, f(a) = g(a).

⇒ f and g are equal functions.

Question 16.

Let A = {1,2,3}. Then, the number of relations containing {1, 2} and {1, 3} which are reflexive, symmetric but not transitive are

(A) 1

(B) 2

(C) 3

(D) 4

Solution:

There is only one relation containing (1, 2) and (1, 3), which is reflexive and symmetric but not transitive is {1, 1}.

Part (A) is the correct answer.

Question 17.

Let A = {1, 2, 3}. Then, number of equivalence relations containing (1, 2) is

(A) 1

(B) 2

(C) 3

(D) 4

Solution:

There are two equivalence relations containing (1, 2).

Part (B) is the correct answer.

Question 18.

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

f(x) = \(\left\{\begin{array}{ll}

1, & x>0 \\

0, & x=0 \\

-1, & x<0

\end{array}\right.\)

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

Solution:

f : R → R, the Signum Function is defined as

f(x) = \(\left\{\begin{array}{ll}

1, & x>0 \\

0, & x=0 \\

-1, & x<0

\end{array}\right.\)

Question 19.

Number of binary operations on the set (1, 2} having 1 as the identity element is

(A) 10

(B) 16

(C) 20

(D) 8

Solution:

Let * be a binary operation.

There are two elements in the set {1, 2}.

∴ Number of binary operations is 2^{4} = 16.

Part (B) is the correct answer.