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Question: The function which are aperiodic are:- A.\(Y = \left[ {x + 1} \right]\) B.\(y = \sin {x^2}\) C...

The function which are aperiodic are:-
A.Y=[x+1]Y = \left[ {x + 1} \right]
B.y=sin⁡x2y = \sin {x^2}
C.y=sin⁡2xy = {\sin ^2}x
D.y=sin⁡−1xy = {\sin ^{ - 1}}x

Explanation

Solution

Hint: A aperiodic function never repeats itself. Time period of aperiodic function is 0 or dependent on the value of xx. For aperiodic function, f(x+T)≠f(x)f\left( {x + T} \right) \ne f\left( x \right)

Complete step-by-step answer:
We have to find aperiodic functions, which implies the functions which are not periodic.
On substituting, x+1=tx + 1 = t in first option, Y=[t]Y = \left[ t \right], we get,
Y=[t]Y = \left[ t \right]
And, we know that the greatest integer function is always periodic with a fixed time period of 1.
Next, substitute x2=t{x^2} = t in y=sin⁡x2y = \sin {x^2} , we get,
y=sin⁡ty = \sin t
And sin⁡t\sin t is also periodic function with a fixed time period of 2π2\pi
Similarly y=sin⁡2xy = {\sin ^2}x is always a periodic function is periodic function with time-period of π\pi
For the given function,y=sin⁡−1xy = {\sin ^{ - 1}}x we can check its periodicity by trying to find the period.
Let us assume the TTbe the time period of y=sin⁡−1xy = {\sin ^{ - 1}}x.
Thus, sin⁡−1(x+T)=sin⁡−1x{\sin ^{ - 1}}\left( {x + T} \right) = {\sin ^{ - 1}}x
On further solving the equation sin⁡−1(x+T)=sin⁡−1x{\sin ^{ - 1}}\left( {x + T} \right) = {\sin ^{ - 1}}x, we get
sin⁡−1(x+T)−sin⁡−1x=0 ⇒sin⁡−1((x+T)1−x2−x1−(x+T)2)=0 ⇒(x+T)1−x2−x1−(x+T)2=0 ⇒(x+T)1−x2=x1−(x+T)2 ⇒(x+T)2(1−x2)=x2(1−(x+T)2) ⇒(x+T)2+x2(x+T)2=x2+x2(x+T)2 ⇒(x+T)2=x2 ⇒x2+2xT+T2=x2 ⇒T2+2xT=0 ⇒T(T+2x)=0 ⇒T=0,−2x  {\sin ^{ - 1}}\left( {x + T} \right) - {\sin ^{ - 1}}x = 0 \\\ \Rightarrow {\sin ^{ - 1}}\left( {\left( {x + T} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {x + T} \right)}^2}} } \right) = 0 \\\ \Rightarrow \left( {x + T} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {x + T} \right)}^2}} = 0 \\\ \Rightarrow \left( {x + T} \right)\sqrt {1 - {x^2}} = x\sqrt {1 - {{\left( {x + T} \right)}^2}} \\\ \Rightarrow {\left( {x + T} \right)^2}\left( {1 - {x^2}} \right) = {x^2}\left( {1 - {{\left( {x + T} \right)}^2}} \right) \\\ \Rightarrow {\left( {x + T} \right)^2} + {x^2}{\left( {x + T} \right)^2} = {x^2} +{x^2}{\left( {x + T} \right)^2} \\\ \Rightarrow {\left( {x + T} \right)^2} = {x^2} \\\ \Rightarrow {x^2} + 2xT + {T^2} = {x^2} \\\ \Rightarrow {T^2} + 2xT = 0 \\\ \Rightarrow T\left( {T + 2x} \right) = 0 \\\ \Rightarrow T = 0, - 2x \\\
Since the time period of the function y=sin⁡−1xy = {\sin ^{ - 1}}x is either 0 or is dependent on xx. No fixed time period is therefore possible for the time period y=sin⁡−1xy = {\sin ^{ - 1}}x.
Thus option D is the correct answer.

Note: For the equation sin⁡−1x=0{\sin ^{ - 1}}x = 0, the value of xxwill be zero, as the sin⁡−1{\sin ^{ - 1}} is defined on the principal range of −π2 - \dfrac{\pi }{2}to π2\dfrac{\pi }{2}. Thus the only possible solution for the equation sin⁡−1x=0{\sin ^{ - 1}}x = 0 is the x=0x = 0.