Question

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

Answer 1

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

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

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