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Periodic Function
A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the function such that f (x + T) = f(x), for all values of x and x + T within the domain of f(x). The least positive period is called the principal or fundamental period of f.
Example: The function sin x & cos x both are periodic over 2π & tan x is periodic over π

Remark:
(a) A constant function is always periodic, with no fundamental period.
(b) If f(x) has a period p, then 1/f(x) and √f(x) also has a period p.
(c) If f(x) has a period T then f(ax + b) has a period  T/a  (a > 0).
(d) If f(x) has a period T1 & g(x) also has a period T2 then period of f(x) ± g(x) or f(x)/g(x) s L.C.M of T1 & Tprovided their L.C.M. exists. However that L.C.M. (if exists) need not to be fundamental period. If L.C.M. does not exists then f(x) ± g(x) or f(x). g(x) or f(x)/g(x) is non-periodic e.g. |sin x| has the period p, |cos x| also has the period π |sin x| + |cos x| also has a period p. But the fundamental period of |sin x| + |cos x| is π/2.
(e) If g is a function such that gof is defined on the domain of f and f is periodic with T, then gof is also periodic with T as one of its periods. Further if # g is one-one, then T is the period of gof # g is also periodic with T' as the period and the range of f is a subset of [0, T'], then T is the period of gof
(f) Inverse of a periodic function does not exist.

Example 1. Find period of the following functions
Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced
(ii) f(x) = {x} + sin x
(iii) f(x) = cos x . cos 3x
Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced 
Solution.
(i) Period of sin x/2 is 4p while period of cos x/3 is 6p. Hence period of sin x/2 + cos x/3 is 12 π {L.C.M. or 4 & 6 is 12}
(ii) Period of sin x = 2π ; Period of {x} = 1; but L.C.M. of 2π & 1 is not possible ∴ it is a periodic
(iii) f(x) = cos x . cos 3x ; Period of f(x) is L.C.M. of (2π, 2π/3) = 2π but 2π may or may not be the fundamental period. The fundamental period can be 2π/n where  n ∈ Np may or may not be the fundamental period. The fundamental period can be f(π + x) = (–cos x) (– cos 3x) = f(x)
Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced

Example 2. If f(x) = sin x + cos ax is a periodic function, show that a is a rational number.
Solution. Given f(x) = sin x + cos ax
Period of sin x = 2π/1  and period of cos ax 2π/a
Hence period of f(x) = L.C.M.
or  
Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced
where k = H.C.F. of 1 and a
1/k = integer = q (say), (≠0) and k a/k = integer = p (say)
Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced
a = p/q
a = rational number

Example 3. Given below is a partial graph of an even periodic function f whose period is 8. If [*] denotes greatest integer function then find the value of the expression.
Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advancedf (-3) + 2 | f (-1) | + [f(7/8)] + f (0) + arc cos (f(-2)) + f (-7) + f (20)
Solution.
f (-3) = f (3) = 2 [f(x) is an even function, ∴ f(-x) = f (x)]
again f (-1) = f (1) = - 3
∴ 2 | f (-1) | = 2 | f (1) | = 2 | -3 | = 6
from the graph, -3 < f(7/8) < -3 ∴  [f(7/8)] = -3
f (0) = 0 (obviously from the graph)
cos -1 (f(-2)) = cos-1 (f (2)) = cos -1(1) = 0
f (-7) = f (-7 + 8) = f (1) = - 3 [f (x) has period 8]
f (20) = f (4 + 16) = f (4) = 3 [f (nT + X) = f (x)]
sum = 2 + 6 - 3 + 0 + 0 - 3 + 3 = 5

Example 4. If the periodic function f(x) satisfies the equation f(x + 1) + f(x - 1) = √3  f(x) ∀ x ∈ R then find the period of f(x)
Solution.
We have f(x + 1) + f(x – 1) = √3 f(x) ∀ x ∈ R
Replacing x by x – 1 and x + 1 in (1) then f(x) + f(x – 2) = √3 f(x – 1) ...(2)
Adding (2) and (3), we get 2f(x) + f(x – 2) + f(x + 2) = √3 (f (x – 1) + f(x + 1)) ...(3)
2f (x) + f(x – 2) + f(x + 2) = √3 . √3 f(x) [From (1)]
f(x + 2) + f(x – 2) = f(x) ...(4)
Replacing x by x + 2 in equation (4) then f (x + 4) + f (x) = f (x + 2) ...(5)
Adding equations (4) and (5), we get f(x + 4) + f (x - 2) = 0       ...(6)
Again replacing x by x + 6 in (6) then f (x + 10) + f (x + 4) = 0  ...(7)
Subtracting (6) from (7), we get f (x + 10) - f (x - 2) = 0     ...(8)
Replacing x by x + 2 in (8) then f (x + 12) - f(x) = 0 or f (x + 12) = f(x)
Hence f(x) is periodic function with period 12.

Inverse Of A Function
Let  f : A → B  be a one-one & onto function, then their exists a unique function
g :  B → A such that f(x) = y ⇔ g(y) = x, x ∈ A & y ∈ B. Then g is said to be inverse of f.
Thus g = f-1 :  B → A =  {(f(x), x) | (x,  f(x)) ∈ f}.

Properties of inverse function :
(i) The inverse of a bijection is unique, and it is also a bijection.
(ii) If f : A → B is a bijection & g : B → A is the inverse of f, then fog = IB and gof = IA, where IA & IB are identity functions on the sets A & B respectively.
(iii) The graphs of f & g are the mirror images of each other in the line y = x.
(iv) Normally points of intersection of f and f–1 lie on the straight line y = x. However it must be noted that f(x) and f–1(x) may intersect otherwise also.
(v) In general fog(x) and gof(x) are not equal. But if either f and g are inverse of each other or atleast one of f, g is an identity function, then gof = fog.
(vi) If f & g are two bijections  f : A → B ,  g : B → C then the inverse of gof exists and  (gof)-1 = f-1 og-1.

Example 5. Find the inverse of the function f(x) = ln(x2 + 3x +1); x ε [1, 3] and assuming it to be an onto function.
Solution.
Given f(x) = ln (x2 + 3x + 1)

Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced
which is a strictly increasing function. Thus f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Now let y = f(x) = ln (x2 + 3x + 1) then x = f–1 (y) ...(1)
and y = ln (x2 + 3x + 1) ⇒ ey = x 2 + 3x + 1 ⇒ x2 + 3x + 1 – ey = 0

Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced

Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced
From (1) and (2), we get

Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced

Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced

The document Periodic & Inverse Function | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on Periodic & Inverse Function - Mathematics (Maths) for JEE Main & Advanced

1. What is a periodic function?
A periodic function is a mathematical function that repeats its values at regular intervals or periods. In other words, a periodic function has a specific pattern that repeats itself infinitely. The period of a function is the smallest interval at which the function repeats.
2. Can you provide an example of a periodic function?
Certainly! A commonly known example of a periodic function is the sine function (sin(x)). It repeats its values every 2π radians or 360 degrees. Another example is the cosine function (cos(x)), which also repeats its values every 2π radians or 360 degrees.
3. How can you determine if a function is periodic?
To determine if a function is periodic, you need to check if there exists a constant value, such as "P," for which f(x + P) = f(x) holds true for all values of x. If such a constant exists, then the function is periodic, and P is the period of the function. If there is no such constant, the function is not periodic.
4. What is an inverse function?
An inverse function is a function that undoes the action of another function. In other words, if you have a function f(x), its inverse function, denoted as f^(-1)(x), will take the output of f(x) and give you back the original input. The inverse function undoes the effect of the original function.
5. How can you find the inverse of a function?
To find the inverse of a function, you can follow these steps: 1. Replace the function notation f(x) with y. 2. Swap the x and y variables. Now, you have x = f(y). 3. Solve the resulting equation for y to express it in terms of x. 4. Replace y with f^(-1)(x) to obtain the inverse function notation. It's important to note that not all functions have inverses. A function must be one-to-one (injective) for its inverse to exist.
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