Periodic & Inverse Function - (Maths) for JEE Main & Advanced PDF Download

Periodic Function

Definition. A function f(x) is called periodic if there exists a positive number T > 0 such that f(x + T) = f(x) for all x for which both sides are defined. Any such T is called a period of f. The smallest positive period, when it exists, is called the fundamental or principal period.

Examples of common periodic functions

• sin x and cos x have fundamental period 2π.
• tan x has fundamental period π.
• A constant function is periodic (every shift leaves it unchanged) but it does not have a unique fundamental period.

Remarks and useful properties

• If f(x) has period p, then any translate f(x + b) also has period p.
• If f(x) has period T and a > 0, then f(ax + b) has period T/a.
• If f(x) has period T and g is defined on the range of f, then the composite g∘f has period T. If g is one-to-one on the range of f, then T is a period of g∘f and may be its fundamental period.
• If f has period T1 and g has period T2, then any common period of f and g is a multiple of both T1 and T2; a natural candidate for a common period is the least common multiple (L.C.M.) of T1 and T2 when this L.C.M. exists. The L.C.M. need not be the fundamental period of f ± g or f·g. If T1 and T2 are incommensurate (their ratio is irrational), then f ± g or f·g will be non-periodic in general.
• If f is periodic then its inverse need not exist because periodic functions are not one-to-one on their whole domain. Only a restriction to a one-to-one interval can be inverted.

Examples and Solutions - Periodic Functions

Example 1. Find the period of the following functions.

(ii) f(x) = {x} + sin x

(iii) f(x) = cos x · cos 3x

Solution.

(i) Period of sin(x/2) is 4π.
Period of cos(x/3) is 6π.
A common period is L.C.M.(4π, 6π) = 12π.
Therefore sin(x/2) + cos(x/3) is periodic with period 12π. The fundamental period may be a divisor of 12π depending on function symmetry; here 12π is a valid (not necessarily minimal) period.

(ii) Period of sin x is 2π.
Period of the fractional part function {x} is 1.
Since the ratio 2π : 1 is irrational, there is no positive T that is simultaneously an integer multiple of 1 and an integer multiple of 2π.
Therefore {x} + sin x is not periodic.

(iii) Consider f(x) = cos x · cos 3x.
Period of cos x is 2π.
Period of cos 3x is 2π/3.
Any common period is a multiple of both 2π and 2π/3. L.C.M.(2π, 2π/3) = 2π, so 2π is a period.
Check if a smaller positive period exists.
Evaluate f(x + π) = cos(x + π)·cos(3x + 3π) = (-cos x)·(-cos 3x) = cos x·cos 3x = f(x).
Thus π is a period.
Therefore the fundamental period of cos x · cos 3x is π.

Example 2. If f(x) = sin x + cos(ax) is periodic, show that a is a rational number.

Solution.

Assume f has a positive period T.
Then sin(x + T) = sin x and cos(a(x + T)) = cos(ax) for all x.
Therefore T is simultaneously a multiple of the fundamental period of sin x, which is 2π, and of cos(ax), which is 2π/a.
Hence there exist integers m, n > 0 such that T = 2π·m = (2π/a)·n.
Equate the two expressions: 2π m = (2π/a) n.
Simplify to get a = n/m.
Thus a is a rational number.

Example 3. A partial graph of an even periodic function f with period 8 is given. If [·] denotes greatest integer function, evaluate the expression:

f(-3) + 2|f(-1)| + [f(7/8)] + f(0) + arccos(f(-2)) + f(-7) + f(20)

Solution.

Because f is even, f(-x) = f(x).
Therefore f(-3) = f(3), and from the graph f(3) = 2, so f(-3) = 2.
Because f is even, f(-1) = f(1), and from the graph f(1) = -3, so f(-1) = -3.
Therefore 2|f(-1)| = 2·|-3| = 6.
From the graph -4 < f(7/8) < -3, so [f(7/8)] = -4 if the value is strictly between -4 and -3; verify graph value carefully and choose the correct integer part accordingly.
From the graph f(0) = 0.
f(-2) = f(2) and from the graph f(2) = 1, so arccos(f(-2)) = arccos(1) = 0.
Since the function has period 8, f(-7) = f(-7 + 8) = f(1) = -3.
Also f(20) = f(20 - 16) = f(4), and from the graph f(4) = 3.
Now sum the values: 2 + 6 + (integer part) + 0 + 0 - 3 + 3.
If [f(7/8)] = -3 (if graph shows value in (-3, -2] adjust accordingly), compute with that; with the indicated values in the original worked example the result was 5.

Example 4. If the periodic function f(x) satisfies the functional equation f(x + 1) + f(x - 1) = √3 · f(x) for all real x, then find a period of f.

Solution.

Given f(x + 1) + f(x - 1) = √3 f(x) for all x.
Replace x by x + 1 to obtain f(x + 2) + f(x) = √3 f(x + 1).
Replace x by x - 1 to obtain f(x) + f(x - 2) = √3 f(x - 1).
Multiply the first given equation and its shifted versions appropriately and eliminate intermediate terms to obtain relations among values spaced by 2 and 4 units.
From manipulations one can derive f(x + 2) + f(x - 2) = f(x).
Replace x by x + 2 to get f(x + 4) + f(x) = f(x + 2).
Combine these equations to eliminate terms and obtain a linear recurrence in shifts of f; continuing this process yields f(x + 12) = f(x).
Therefore f is periodic with period 12.

Inverse of a Function

Definition. Let f : A → B be a bijection (one-to-one and onto). Then there exists a unique function g : B → A such that f(x) = y ⇔ g(y) = x for x ∈ A and y ∈ B. This function g is called the inverse of f and is denoted f^-1.

Alternate description. The inverse can be written as f^-1 = {(f(x), x) | (x, f(x)) ∈ f}.

Properties of inverse function

• Uniqueness. The inverse of a bijection is unique and is itself a bijection.
• Composition. If f : A → B has inverse g, then f∘g = I_B and g∘f = I_A, where I denotes the identity map on the respective set.
• Graph symmetry. The graph of f^-1 is the reflection of the graph of f across the line y = x.
• Intersection with inverse. Points where f(x) = f^-1(x) lie on the line y = x, though intersections can occur at other points if both functions are defined there.
• Inverse of composition. If f : A → B and g : B → C are bijections, then (g∘f)^-1 = f^-1 ∘ g^-1.

Example - Inverse of a Logarithmic Function

Example 5. Find the inverse of the function f(x) = ln(x^2 + 3x + 1), where x ∈ [1, 3], assuming f is onto its image (so invertible on this domain).

Solution.

Let y = f(x) = ln(x^2 + 3x + 1).

Exponentiate both sides:
e^y = x^2 + 3x + 1.

Rearrange to a quadratic in x:
x^2 + 3x + 1 - e^y = 0.

Compute the discriminant D:
D = 3^2 - 4·1·(1 - e^y) = 9 - 4 + 4e^y = 5 + 4e^y.

Solutions for x:
x = [-3 ± √(5 + 4e^y)]/2.

Determine which sign gives x in the given domain [1, 3].
When x = 1, y = ln(1 + 3 + 1) = ln 5.
When x = 3, y = ln(9 + 9 + 1) = ln 19.
Evaluate the expression with the positive sign at y = ln 5:
√(5 + 4e^{ln 5}) = √(5 + 4·5) = √25 = 5.
x = (-3 + 5)/2 = 1, which matches the left endpoint.
Evaluate at y = ln 19:
√(5 + 4·19) = √(5 + 76) = √81 = 9.
x = (-3 + 9)/2 = 3, which matches the right endpoint.

Therefore the correct branch is with the plus sign, and the inverse function is:
f^-1(y) = [-3 + √(5 + 4e^y)] / 2,
with domain y ∈ [ln 5, ln 19] and range [1, 3].

Summary

Periodic functions repeat their values after a fixed shift; determining fundamental periods often requires checking for the smallest positive shift that leaves the function unchanged. The sum or product of periodic functions may or may not be periodic depending on whether their periods are commensurate. The inverse of a function exists only when the function is one-to-one and onto on the relevant domain; the inverse can be obtained by solving y = f(x) for x and choosing the correct branch consistent with the given domain or range.