Chapter 1: Relations and Functions

Q1Written answer

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.

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Worked solution

Given relation, R = {(a, a), (b, c), (a, b)}

To make R as reflexive we should add (b, b) and (c, c) to R. Also, to make R as transitive we should add (a, c) to R.

Hence, the minimum number of ordered pairs to be added are (b, b), (c, c) and (a, c) i.e. 3.

Q2Written answer

Let D be the domain of the real valued function f defined by f(x) = √(25 – x²). Then, write D.

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Worked solution

Given, f(x) = √(25 – x²)

The function is defined if 25 – x² ≥ 0

So, x² ≤ 25

-5 ≤ x ≤ 5

Therefore, the domain of the given function is [-5, 5].

Q3Written answer

Let f, g: R → R be defined by f(x) = 2x + 1 and g (x) = x² – 2, ∀ x ∈ R, respectively. Then, find g o f.

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Worked solution

Given,

f(x) = 2x + 1 and g (x) = x² – 2, ∀ x ∈ R

Thus, g o f = g (f (x))

= g (2x + 1)

= (2x + 1)² – 2

= 4x² + 4x + 1 – 2

= 4x² + 4x – 1

Q4Written answer

Let f: R → R be the function defined by f (x) = 2x – 3, ∀ x ∈ R. write f ^–1.

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Worked solution

Given function,

f (x) = 2x – 3, ∀ x ∈ R

Let y = 2x – 3

x = (y + 3)/ 2

Thus,

f ^-1(x) = (x + 3)/ 2

Q5Written answer

If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f ^–1.

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Worked solution

Given,

A = {a, b, c, d} and f = {(a, b), (b, d), (c, a), (d, c)}

So,

f ^-1 = {(b, a), (d, b), (a, c), (c, d)}

Q6Written answer

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

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Worked solution

Given, f (x) = x² – 3x + 2

Then,

f (f (x)) = f (x² – 3x + 2)

= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2,

= x⁴ + 9x² + 4 – 6x³ + 4x² – 12x – 3x²+ 9x – 6 + 2

= x⁴ – 6x³ + 10x² – 3x

Thus,

f (f (x)) = x⁴ – 6x³ + 10x² – 3x

Q7Written answer

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.

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Worked solution

Given, g = {(1, 1), (2, 3), (3, 5), (4, 7)}

It’s seen that every element of domain has a unique image. So, g is function.

Now, also given that g (x) = αx + β

So, we have

g (1) = α(1) + β = 1

α + β = 1 …….. (i)

And, g (2) = α(2) + β = 3

2α + β = 3 …….. (ii)

Solving (i) and (ii), we have

α = 2 and β = -1

Therefore, g (x) = 2x – 1

Q8Written answer

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}.

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Worked solution

(i) Given, {(x, y): x is a person, y is the mother of x}

It’s clearly seen that each person ‘x’ has only one biological mother.

Hence, the above set of ordered pairs make a function.

Now more than one person may have same mother. Thus, the function is many-many one and surjective.

(ii) Given, {(a, b): a is a person, b is an ancestor of a}

It’s clearly seen that any person ‘a’ has more than one ancestors.

Thus, it does not represent a function.

Q9Written answer

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 f o g.

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Worked solution

Given,

f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}

Now,

fog (2) = f(g(2)) = f(3) = 5

fog (5) = f(g(5)) = f(1) = 2

fog (1) = f(g(1)) = f(3) = 5

Thus,

fog = {(2, 5), (5, 2), (1, 5)}

Q10Written answer

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.

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Worked solution

Given, f: C → R such that f (z) = |z|, ∀ z ∈ C

Now, let take z = 6 + 8i

Then,

f (6 + 8i) = |6 + 8i| = √(6² + 8²) = √100 = 10

And, for z = 6 – 8i

f (6 – 8i) = |6 – 8i| = √(6² + 8²) = √100 = 10

Hence, f (z) is many-one.

Also, |z| ≥ 0, ∀ z ∈ C

But the co-domain given is ‘R’

Therefore, f(z) is not onto.

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