Revision Notes: Relations & Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download

Table: Domain and range of some standard functions

Inverse Function:

f⁻¹ exists iff f is both one–one and onto.
f⁻¹: B → A, f⁻¹(b) = a ⇒ f(a) = b

Even and Odd Function: A function is said to be
(a)Even function if f(x) = f(-x)
(b) Odd function if f(-x) = -f(x)

Properties of Even & Odd Function:
(a)The graph of an even function is always symmetric about the y-axis.
(b) The graph of an odd function is always symmetric about the origin.
(c) The product of two even or odd functions is an even function.
(d) The sum & difference of two even (odd) functions is an even (odd) function.
(e) The product of an even and odd function is an odd function.
(f) The sum of an even and odd function is neither even nor odd function.
(g)The zero function, i.e. f(x) = 0, is the only function which is both even and odd.
(h) If f(x) is an odd (even) function, then f'(x) is even (odd) function, provided f(x) is differentiable on R.
(i)A given function can be expressed as the sum of an even and odd function.
i.e. f(x) = (1/2) [ f(x) + f(-x) ] + (1/2) [ f(x) - f(-x) ] = even function + odd function

Increasing Function:A function f(x) is an increasing function in the domain D if the value of the function does not decrease by increasing the value of x.

Decreasing Function:A function f(x) is a decreasing function in the domain D if the value of the function does not increase by increasing the value of x.

Periodic Function:A function f(x) will be periodic if a positive real number T exists such that:
f(x + T) = f(x), ∀ x ∈ Domain.
There may be infinitely many such real numbers T that satisfy the above equality. The least positive number T is called the period of f(x).

(i)If a function f(x) has period T, then:

• The period of f(xn + a) = T / n
• The period of (x / n + a) = nT

(ii) If the period of f(x) is T₁ and g(x) has T₂, then the period of f(x) ± g(x) will be L.C.M. of T₁ and T₂, provided it satisfies the definition of a periodic function.

(iii)If the period of f(x) and g(x) are the same T, then the period of af(x) + bg(x) will also be T.

Solved Examples

Q1: ({x}, represents fractional part function)
(i) Domain of the function
(ii) Range of the function cos(2 sin x) is ______.
Ans:
(i)1 - {x} > 0 (Always true)

And 4 - x² ≥ 0
x² ≤ 4
x ∈ [-2,2]

(ii) Range of cos(2 sin x)
-1 ≤ sin x ≤ 1
-2 ≤ 2 sin x ≤ 2
y = cos(2 sin x) ∈ [cos 2, 1].

Q2: Let A = {a, b, c, d}. Examine which of the following relations is a function on A?
(i) f = {(a, a), (b, c), (c, d), (d, c)}
(ii) g = {(a, c), (b, d), (b, c)}
(iii) h = {(b, c), (d, a), (a, a)}
Ans: (i) Yes, it is because it covers all A and each corresponds to one value.
(ii) No, as it gives two values at x = b.
(iii) No, as x = c has no image.

Q3: Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a linear function from Z into Z and g(x) = x. Find f + g.
Ans: f = a×x + b (Suppose)
f(1) = 1 = a + b
f(2) = 3 = 2a + b
Solving for a and b:
a = 2, b = -1
Thus, f(x) = 2x - 1
g(x) = x
Now, f + g = (2x - 1) + x = 3x - 1
Calculating values:
(f + g)(1) = 3(1) - 1 = 2
(f + g)(2) = 3(2) - 1 = 5
(f + g)(0) = 3(0) - 1 = -1
(f + g)(-1) = 3(-1) - 1 = -4
So, f + g = {(1,2), (2,5), (0,-1), (-1,-4)}

Q4: The void relation on a set A is
(a) Reflexive
(b) Symmetric and transitive
(c) Reflexive and symmetric
(d) Reflexive and transitive
Ans: (b)
Void relation: A → A
It is also called an empty relation.
A relation R is a void relation if no element of set A is related to any element of A.
(x, y) ⇒ x is not related to y
∴ (y, x) ⇒ y is not related to x
∴ Symmetric
(x, x) ⇒ x is any, how will be related to itself
∴ So it can’t be reflexive
(x, y)(y, z)
⇒ x is not related to y
y is not related to z
∴ This is transitive.

Q5: If A, B, and C are these sets such that A ∩ B = A ∩ C and A ∪ B = A ∪ C, then
(a) A = B
(b) A = C
(c) B = C
(d) A ∩ B = ϕ
An: (c)
Given A ∩ B = A ∩ C … (i)
and A ∪ B = A ∪ C    … (ii)
We know that A ∪ B = A + B - A ∩ B    
From (i) and (ii):
A ∪ C = A + B - A ∩ C    … (iii)
But A ∪ C = A + C - A ∩ C    … (iv)
From (iii) - (iv):
0 = B - C
B = C

Q6: The minimum number of elements that must be added to the relation R = {(1, 2), (2, 3)} on the set {1, 2, 3}, so that it is an equivalence is
(a) 4
(b) 7
(c) 6
(d) 5
Ans:(b)
R = {(1, 2), (2, 3)} on the set A (assume)
So A = {1, 2, 3}
→ Reflexive: (x, x) ∈ R, x ∈ A
→ Symmetric: (x, y) ∈ R for (y, x) ∈ R
→ Transitive: (1, 2) ∈ R and (2, 3) ∈ R
Now, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1), (3, 2)}
Now for symmetric pair (1, 3), pair (3, 1) must be added
Total added pairs →
7 {(1,1), (2,2), (3,3), (2,1), (3,2), (1,3), (3,1)}

Q7: The graph of function f(x) is as shown, adjacently
The graph of

(a)

(b)

(c)

(d)

Ans: R = {(a, b) ∈ A × H: a is sister of b}
Domain and range both (a, b) are set {A (a, b) ∈ A × A}
And it is given that A is only boy’s school a and b both
are boys. So It is not possible that a is sister of b.

Q8: Ifdenoting the greatest integer function, then

(a) f(0) = 0

(b) 

(c) 

(D) f(π) = 0

Ans: (c)
f(x) cos 4x sin4x
f(0)=1

f(π) = 1

Q9: If the function f : [1, ∞) → [1, ∞) is defined by
f(x) = 2ˣ(ˣ⁻¹), then f⁻¹(x) is
(a)(b)

(c) 

(D) Not defined
Ans: (b)
Let y = 2ˣ(ˣ⁻¹), where y ≥ 1 as x ≥ 1
Taking log₂ on both sides, we get
log₂ y = log₂ 2ˣ(ˣ⁻¹)
⇒ log₂ y = x(x - 1)
⇒ x² - x - log₂ y = 0
For y ≥ 1, log₂ y ≥ 0⇒ 4 log₂ y ≥ 0
But x ≥ 1
Therefore,

Q10: If S = {x ∈ R : f(x) = f(-x)} ; then S:
(a) Contains exactly one element
(b) Contains exactly two elements.
(c) Contains more than two elements.
(D) Is an empty set.
Ans: (b)
S : f(x) = f(–x)
… (i)… (ii)
Now, f (x) = f (-x)
Exactly two elements.