Class 11 Mathematics Function Notes
Complete typed notes covering ordered pairs, Cartesian product, relations, functions, inverse functions, composition of functions, domain, range and solved exercises.
1. Ordered Pair
A pair which consists of two elements is called an ordered pair.
Hence, (a, b) is an ordered pair where a is the first element and b is the second element.
The ordered pairs (a, b) and (c, d) are equal if and only if:
2. Cartesian Product
Let A and B be two non-empty sets. Then the Cartesian product of A and B is denoted by A × B and is defined by:
3. Relation
A subset of a Cartesian product is called a relation. It is denoted by R.
4. Exercise 3.1
1. Equality of ordered pairs
What do you mean by the equality of two ordered pairs? If:
then, by equality of ordered pairs:
2x = 2
x = 1
y = -6
Hence, x = 1 and y = -6.
2. Find x and y
Given:
From equality of ordered pairs:
y = x + 2y …(ii)
From (ii):
Substituting in (i):
-3y = 4
y = -4/3
Then:
3. Cartesian products
If A = {1, 2, 3} and B = {a, b}, find A × B and B × A. Are they equal?
A × B ≠ B × A.
4. Relation on A
Let A = {1, 2, 3, 4}. Find the relation on A determined by the condition x + y < 5.
= {(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)}
5. Relations from A to B
If A = {1,2,3,4} and B = {3,4,5}, find relations from A to B determined by the following conditions.
a. x + y = 6
b. x + y > 6
c. y = 2x
d. y = x
e. x + y = 3
6. Find the domain, range and inverse of the following relations
a.
Domain: {1,2,3,4}
Range: {2,3,4,5}
b.
Domain: {0,-1,2,3,…}
Range: {0,-1,-2,-3,…}
5. Function
A function is a relation between two sets A and B in which each element of the first set A is associated with exactly one element in B.
6. Types of Function
1) One-to-One Function (Injective Function)
A function f : A → B is said to be one-to-one if distinct elements of A have distinct images in B.
or, if f(x1) = f(x2), then x1 = x2
2) Onto Function (Surjective Function)
A function f : A → B is said to be onto if every element of B has at least one pre-image in A.
3) Bijective Function
If a function f : A → B is both one-to-one and onto, then it is called a bijective function.
7. Inverse Function
If a function f : A → B is bijective, then the inverse function f-1 : B → A associates each element of B with a unique element of A.
8. Composition and Algebra of Functions
9. Exercise 3.2
1. Area of a circle as a function of diameter d
A = πd2/4
2. Area of an equilateral triangle as a function of side x
3. If h(x) = √(49 – x²), find h(0), h(7) and h(9)
h(7) = √(49 – 49) = 0
h(9) = √(49 – 81) = √(-32)
Therefore, h(9) has no real value.
4. Function from a mapping diagram
The mapping diagram on page 8 shows:
1 → 4
2 → 5
3 → 6
5. Find the range
Let A = {-1, 0, 2, 4, 6} and f : A → R be defined by:
f(0) = 0
f(2) = 1/2
f(4) = 2/3
f(6) = 3/4
Range of f = {-1, 0, 1/2, 2/3, 3/4}
6. Piecewise function
The handwritten source defines:
f(x) = 2 + 3x, for 0 ≤ x < 1/2
f(x) = -2 – 3x, for x ≥ 1/2
Then:
f(0) = 2
f(1/2) = -7/2
f(2/3) = -4
For -1/2 ≤ h < 0:
= [2 – 3h – 2] / h
= -3
7. Another piecewise function
The source gives:
f(x) = |x|, for x < 0
f(0) = 1
f(1) = 0
The handwritten note says the requested value at x = 7 does not exist because 7 is outside the stated domain.
8. Function f(x) = (x – 1)/(x + 2)
The source uses A = {-1, 0, 1, 2, 3, 4}. The calculated values are:
f(0) = -1/2
f(1) = 0
f(2) = 1/4
f(3) = 2/5
f(4) = 1/2
Range: {-2, -1/2, 0, 1/4, 2/5, 1/2}
9. Determine whether each function is one-to-one, onto, both or neither
a. f : {-2,-1,0,1,2} → R defined by f(x) = x²
Since different inputs have the same image, the function is not one-to-one.
b. f : {0,1,2} → R defined by f(x) = x²
10. Show that f(x) = x³ is one-to-one and onto
For one-to-one:
x13 = x23
x1 = x2
Hence, f is one-to-one.
For onto:
y = x³ ⇒ x = ∛y ∈ R
Hence, f is onto.
11. Prove f(x) = ax + b, a ≠ 0, is bijective
For one-to-one:
ax1 + b = ax2 + b
ax1 = ax2
x1 = x2
For onto:
x = (y – b)/a ∈ R
Therefore f is bijective.
12. Find inverse functions
a. f(x) = x³
x = ∛y
f-1(x) = ∛x
b. h(x) = 3x – 6
y + 6 = 3x
x = (y + 6)/3
h-1(x) = (x + 6)/3
13. Inverse of f(x) = 1/x
x = 1/y
f-1(x) = 1/x
14. Composition of functions
Given:
g(x) = x + 1
a. g ∘ f
= x² + 2x + 2
b. f ∘ g
= x² + 4x + 4
c. f ∘ f
d. g ∘ g
15. Composition from ordered pairs
Given:
g = {(3,2), (5,1), (7,3)}
The source gives:
g ∘ f = {(1,2), (2,1), (3,3)}
18. Find f-1 and verify inverse identities
Given:
Let y = 2x + 1:
x = (y – 1)/2
Hence:
The handwritten solution verifies:
f-1(f(x)) = x
19. Piecewise function on integers
The source defines a function on integers with domain from -1 to 5:
f(x) = -3x + 1, for 0 < x ≤ 3
f(x) = -2x + 1, for 3 < x ≤ 6
From the values shown in the handwritten notes:
f(0) = 1
f(1) = -2
f(2) = -5
f(3) = -8
f(4) = -7
f(5) = -9
Domain: {-1,0,1,2,3,4,5}
Range: {-1,1,-2,-5,-8,-7,-9}
The source concludes that the function is one-to-one.
10. Domain and Range
20(a). y = x² + 1
Since this is a polynomial function:
Also, x² ≥ 0, hence:
20(b). y = 1/(x – 1)
The function is not defined at x = 1.
For the range:
y(x-1) = 1
x = (y+1)/y
Thus y ≠ 0.
20(c). y = (x – 3)/(x + 3)
The function is not defined at x = -3.
20(d). y = (x² – 9)/(x – 3)
The source notes that the function is not defined at x = 3.
Since x² – 9 = (x-3)(x+3), for x ≠ 3:
The handwritten solution gives the range as:
20(e). y = |x – 3|/(x – 3)
The function is not defined at x = 3.
The handwritten solution gives:
20(f). y = √(x – 3)
For the function to be defined:
x ≥ 3
R(f) = [0, ∞)
20(g). y = √(-x)
For the function to be defined:
x ≤ 0
R(f) = [0, ∞)
20(h). y = 1/√(x – 3)
The denominator must be positive:
x > 3
R(f) = (0, ∞)
20(i). y = √(6 – x – x²)
The radicand must be non-negative:
x² + x – 6 ≤ 0
(x + 3)(x – 2) ≤ 0
Completing the square:
y² = 25/4 – (x + 1/2)²
Therefore:
R(f) = [0, 5/2]
Note: This typed version follows the uploaded handwritten PDF closely. Where the handwriting is unclear, the wording has been kept conservative rather than adding unsupported material.
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