Problem : A function f(x) satisfies equation given :

f x + 1 + f x - 1 = 3 f x f x + 1 + f x - 1 = 3 f x

Determine its period.

Solution : Here, main strategy is to replace “x” such that we get expression on RHS which is same as the expression on LHS. Replacing “x” by “x+1” and replacing “x” by “x-1” separately in given equation, we get two equations :

⇒ f x + 2 + f x = 3 f x + 1 ⇒ f x + 2 + f x = 3 f x + 1 ⇒ f x + f x - 2 = 3 f x - 1 ⇒ f x + f x - 2 = 3 f x - 1

Adding these two equations, we get term on RHS, which is same as LHS :

⇒ f x + 2 + f x - 2 + 2 f x = 3 { f x + 1 + f x - 1 } = 3 f x ⇒ f x + 2 + f x - 2 + 2 f x = 3 { f x + 1 + f x - 1 } = 3 f x ⇒ f x + 2 + f x - 2 = f x ⇒ f x + 2 + f x - 2 = f x

Replacing “x” by “x+2” in above equation, we have :

⇒ f x + 4 + f x = f x + 2 ⇒ f x + 4 + f x = f x + 2

Note that we have not replaced “x” by “x-2”, because that yields a related which has argument form of “x-4”. As definition of periodic function involves addition of a positive constant being added to independent variable, we opt to replace “x” by “x+2”. Now, adding two preceding equations,

⇒ f x + 4 + f x - 2 = 0 ⇒ f x + 4 + f x - 2 = 0

Replacing “x” by “x+2” so that one of two term becomes f(x), we have :

⇒ f x + 6 + f x = 0 ⇒ f x + 6 + f x = 0 ⇒ f x + 6 = - f x ⇒ f x + 6 = - f x

Replacing “x” by “x+6”, we have :

⇒ f x + 12 = - f x + 6 = f x ⇒ f x + 12 = - f x + 6 = f x ⇒ T = 12 ⇒ T = 12

Problem : Find period of function :

f x = 3 sin 2 { 3 x } + 5 cos 3 { 2 x } f x = 3 sin 2 { 3 x } + 5 cos 3 { 2 x }

where {} denotes fraction part function.

Solution : The period of {x} is 1. Therefore, period of function {3x} is 1/3. Domains of sin2x and FPF functions are R. Thus, domain of FPF is proper subset of domain of sin2x. Hence, period of 3sin2{3x} is 1/3. On the other hand, period of function {2x} is 1/2. Domains of cos3x and FPF functions are R. Hence, period of 5cos3{2x} is 1/2. LCM of 1/2 and 1/3 is :

⇒ L C M = LCM of 1,1 HCF of 2,3 = 1 1 = 1 ⇒ L C M = LCM of 1,1 HCF of 2,3 = 1 1 = 1

Problem : Find period of function :

f x = 3 sin 2 3 x + 2 cos 5 3 x f x = 3 sin 2 3 x + 2 cos 5 3 x

Solution : Period of 3 sin 2 3 x 3 sin 2 3 x is :

⇒ T 1 = 2 π 2 3 ⇒ T 1 = 2 π 2 3

Period of 2 cos 5 3 x 2 cos 5 3 x is :

⇒ T 2 = 2 π 5 3 ⇒ T 2 = 2 π 5 3

Two irrational periods are of same kind. Hence, period of given function is :

⇒ L C M = LCM of 2 π , 2 π HCF of 2 3, 5 3 = 2 π 3 ⇒ L C M = LCM of 2 π , 2 π HCF of 2 3, 5 3 = 2 π 3

Problem : Find period of function :

f x = sin 2 x + cos 3 x f x = sin 2 x + cos 3 x

Solution : Period of sin 2 x sin 2 x is :

⇒ T 1 = 2 π 2 ⇒ T 1 = 2 π 2

Period of cos 3 x cos 3 x is :

⇒ T 2 = 2 π 3 ⇒ T 2 = 2 π 3

Two irrational periods are not of same kind. Hence, function is not periodic.

Problem : Determine whether the function given below is periodic?

f x = cos x f x = cos x

If the function is periodic, then find its period.

Solution : We need to check periodicity applying definition of periodic function. According to definition of periodic function,

f x + T = f x f x + T = f x

⇒ cos x + T = cos x ⇒ cos x + T = cos x

⇒ x + T = 2 n π ± x , n ∈ Z ⇒ x + T = 2 n π ± x , n ∈ Z

Squaring both sides and solving for “T”, we have :

⇒ x + T = 2 n π ± x 2 ⇒ x + T = 2 n π ± x 2

⇒ T = 2 n π ± x 2 − x ⇒ T = 2 n π ± x 2 − x

If we expand the square term, then we find that the expression of “T” is not independent of “x”. Hence, given function is not a periodic function.

Problem : Find period of function :

f x = sin x + tan x 2 f x = sin x + tan x 2

Solution : The function is addition of two trigonometric functions. Each of the functions is periodic. We see that neither sine nor tangent function is even function. Therefore, we apply LCM rule to find the period of the function.

The period of “sin x” is “2π”. Hence,

⇒ T 1 = 2 π ⇒ T 1 = 2 π

We also know that period of g(ax+b) is equal to the period of g(x), divided by “|a|”. The period of second term of “f(x)”, therefore, is equal to the period of “tanx”, divided by “1/2”. Thus, period of “tanx” is “π”.

⇒ T 2 = π 1 2 = 2 π ⇒ T 2 = π 1 2 = 2 π

Now, LCM of “2π” and “2π” is “2π”. Hence,

⇒ T = 2 π ⇒ T = 2 π

Problem : Determine whether the function given below is periodic?

f x = x cos x f x = x cos x

If the function is periodic, then find its period.

Solution : The function is product of algebraic function “x” and trigonometric function “cosx”. We need to check periodicity applying definition of a periodic function.

Let the function be periodic. Then,

f x + T = f x f x + T = f x

⇒ x + T cos x + T = x cos x ⇒ x + T cos x + T = x cos x

⇒ T cos x + T = x cos x − x cos x + T = x { cos x − cos x + T } ⇒ T cos x + T = x cos x − x cos x + T = x { cos x − cos x + T }

We see that right hand side is product of algebraic function “x” and trigonometric function. On the left hand side, there is only trigonometric function (apart from “T”, which is a constant). There is no algebraic function in “x” on the left which can cancel "x" on the right. Thus, we conclude that “T” is not independent of “x” and as such give function is not periodic.

Problem : If the function f x = sin x + cos a x f x = sin x + cos a x be periodic, then prove that “a” is rational.

Solution :

Statement of the problem : The function is sum of two trigonometric functions. It is given that the function is a periodic function.

Let “T” be the period, then according to definition :

sin x + T + cos a x + T = sin x + cos a x sin x + T + cos a x + T = sin x + cos a x

Putting, x = 0, we have :

⇒ sin T + cos a T = 1 ⇒ sin T + cos a T = 1

Putting, x = -T, we have :

⇒ sin 0 + cos 0 = - sin T + cos T ⇒ sin 0 + cos 0 = - sin T + cos T

⇒ - sin T + cos a T = 1 ⇒ - sin T + cos a T = 1

Subtracting one equation from another,

sin T = 0 sin T = 0

⇒ T = n π , n ∈ Z ⇒ T = n π , n ∈ Z

and two equations :

cos a T = 1 cos a T = 1

⇒ a T = 2 m π , n ∈ Z ⇒ a T = 2 m π , n ∈ Z

Combining two results :

⇒ a = 2m n ⇒ a = 2m n

Hence, “a” is a rational number.

Problem : Find value of “n”, if period of given function is π/4.

f x = 4 sin 2 n x 1 + cos 2 n x ; n ∈ N f x = 4 sin 2 n x 1 + cos 2 n x ; n ∈ N

Solution : The numerator and denominator repeat simultaneously. Hence, period of given function is LCM of the periods of numerator and denominator. Now, period of 4sin2nx is :

⇒ T 1 = 2 π 2 n = π n ⇒ T 1 = 2 π 2 n = π n

Using trigonometric identity, the denominator of the given function is :

⇒ 1 + cos 2 n x = 1 + 1 + cos 2 n x 2 ⇒ 1 + cos 2 n x = 1 + 1 + cos 2 n x 2

The period of denominator is :

⇒ T 2 = 2 π 2 n = π n ⇒ T 2 = 2 π 2 n = π n

Hence, period of given function is LCM of π/n and π/n, which is π/n. According to question,

⇒ π n = π 4 ⇒ π n = π 4

⇒ n = 4 ⇒ n = 4

Problem : Find period of function :

f x = sin 3 x + sin 2 x + π 3 - cos x cos x + π 3 f x = sin 3 x + sin 2 x + π 3 - cos x cos x + π 3

Solution : The given function is not an even function as sin 3 - x = - sin 3 x sin 3 - x = - sin 3 x . We can use LCM rule to determine period of given function. The periods of sin 3 x sin 3 x and sin 2 x + π 3 sin 2 x + π 3 are 2π and π respectively. Using trigonometric identity,

cos x cos x + π 3 = 1 2 cos 2 x + π 3 + cos π 3 cos x cos x + π 3 = 1 2 cos 2 x + π 3 + cos π 3 ⇒ cos x cos x + π 3 = 1 2 cos 2 x + π 3 + 1 2 ⇒ cos x cos x + π 3 = 1 2 cos 2 x + π 3 + 1 2

The period of cos (2x+π/3) is 2π/2 = π. Now, periods of three terms of given function are 2π,π and π. Hence, period of given function is LCM of three terms i.e. 2π.