Relations and Functions can become challenging in JEE Mathematics when a question combines domain restrictions, composite functions, inverse functions, and graph transformations. Applying a formula without checking its conditions can lead to incorrect answers, particularly in questions involving logarithms, square roots, one-one functions, and periodicity. A clear understanding of the definitions and key results is essential for solving these problems accurately.
Physics Wallah supports JEE aspirants through dedicated online batches for Class 11, Class 12, and droppers. Its preparation resources include DPPs, mock and sectional tests, practice questions, and live and recorded classes to help you revise concepts, practise different question types, and strengthen your problem-solving skills.
For sets A and B, the Cartesian product A x B is the set of all ordered pairs (a, b), where a belongs to A and b belongs to B.
If |A| = m and |B| = n, then: |A x B| = mn
In general, A x B is not equal to B x A
because the order of coordinates changes. However:
|A x B| = |B x A|
A relation from A to B is any subset of A x B:
R is a subset of A x B
If |A| = m and |B| = n, the number of possible relations from A to B is:
2^(mn)
This follows because every relation is a subset of a Cartesian product containing mn ordered pairs.
For a relation from A to B:
Domain: Elements of A that actually occur as first components.
Range: Elements of B that actually occur as second components.
Codomain: The complete target set B.
The range is always a subset of the codomain:
Range is a subset of Codomain
The inverse relation is obtained by reversing every ordered pair: (a, b) becomes (b, a)
Therefore, the domain and range of a relation interchange in its inverse.
A relation R on A is reflexive if: (a, a) belongs to R for every a in A
An identity relation contains only self-pairs: I_A = {(a, a): a belongs to A}
Every identity relation is reflexive, but a reflexive relation may contain additional pairs.
A relation is symmetric if:
(a, b) belongs to R implies (b, a) belongs to R
For transitivity:
(a, b) and (b, c) belong to R implies (a, c) belongs to R
An equivalence relation must be:
Reflexive
Symmetric
Transitive
A function from A to B is a relation in which every element of A has exactly one image in B.
Multiple inputs may have the same output, but one input cannot have two different outputs.
One-One Function: Different inputs have different outputs.
Many-One Function: Multiple inputs may have the same output.
Onto Function: Range equals codomain.
Into Function: Range is a proper subset of codomain.
A function is bijective when it is both one-one and onto.
Horizontal-Line Test: A horizontal line cutting the graph more than once shows many-one behaviour.
Differentiation: If the derivative remains positive or negative throughout the domain, the function is strictly monotonic and one-one.
For a cubic function, differentiate and check the derivative’s discriminant:
D less than 0: one-one, when the derivative keeps one sign.
D greater than 0: many-one, when the derivative changes sign.
For functions f and g:
(f + g)(x) = f(x) + g(x)
(f - g)(x) = f(x) - g(x)
(fg)(x) = f(x)g(x)
(f/g)(x) = f(x)/g(x)
For addition, subtraction, and multiplication, the domain is the intersection of the domains of f and g.
For division, also exclude values where: g(x) = 0
For a reciprocal: 1/f(x)
Exclude all values where f(x) equals zero.
Square root: radicand must be greater than or equal to zero.
Logarithm: argument must be greater than zero.
Logarithm base: base must be positive and not equal to one.
Denominator: must not be zero.
Quotient under a square root: numerator and denominator must have the same sign, with a non-zero denominator.
For f(g(x)), f is the outer function, and g is the inner function. Replace x in f(x) with the complete expression g(x).
To determine the domain of a composite function:
Start with the innermost expression.
Apply its domain condition.
Apply the outer function’s condition.
Take the intersection of all restrictions.
For nested square roots, every radicand must be non-negative. For logarithmic compositions, every logarithmic argument must be positive.
For y = f(x):
y = f(x - a): shift right by a units.
y = f(x + a): shift left by a units.
y = f(x) + a: shift upward.
y = f(x) - a: shift downward.
y = f(-x): reflection in the y-axis.
y = -f(x): reflection in the x-axis.
For modulus graphs:
y = |f(x)|: reflect portions below the x-axis upward.
y = f(|x|): retain the right side and reflect it across the y-axis.
For a reciprocal graph y = 1/f(x), zeros of f(x) become vertical asymptote locations, and every point (a, b) becomes (a, 1/b), where b is not zero.
A function is periodic if:
f(x + T) = f(x)
where T is the least positive period.
|
Function |
Period |
|---|---|
|
Sine, cosine, secant, cosecant |
2 pi |
|
Tangent, cotangent |
pi |
|
Fractional part function |
1 |
For f(cx + d), the new period is:
Original period / absolute value of c
Constants outside the function do not change its period. Always simplify first if the domain remains unchanged. For example:
sin^2 x + cos^2 x = 1
The resulting constant function is periodic, but its period is not defined.
For even powers of sine, cosine, secant, and cosecant, the period generally becomes pi. For odd powers, it remains 2 pi.
Functional equations are often solved by substitution.
Replace x with 1 - x.
Replace x with 1/x.
Replace x with a suitable expression such as 24/x.
Add or subtract the resulting equations to eliminate one function term.
Important patterns include:
f(x + y) = f(x)f(y) suggests an exponential function.
Weighted-average relations suggest a linear function of the form px + q.
Periodic functional equations are solved by comparing arguments and finding the least positive shift that leaves the function unchanged.
An inverse exists only for a bijective function.
To find the inverse:
Write y = f(x).
Solve for x in terms of y.
Interchange x and y.
For example:
y = 1 + ln(x + 2)
gives:
x = exp(y - 1) - 2
Therefore:
f inverse of x = exp(x - 1) - 2
The graphs of a function and its inverse are symmetric about:
y = x
Hence, to solve:
f(x) = f inverse of x
it is unnecessary to calculate the inverse. Use:
f(x) = x
For f(x) = x^2 - 3x + 4:
x^2 - 3x + 4 = x
(x - 2)^2 = 0
Therefore: x = 2
Relations and Functions is an important foundation for Class 12 Mathematics and JEE preparation. A clear understanding of Cartesian products, relation properties, function types, domain restrictions, composite functions, graph transformations, periodicity, and inverses can help you approach a wider range of questions confidently. Physics Wallah supports JEE preparation through dedicated Class 11, Class 12, and dropper batches, DPPs, mock and sectional tests, practice questions, and live and recorded lessons. Combine concept revision with regular problem-solving to identify gaps and improve accuracy.
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