The number of real values of $a$ for which $f(x) = x^a+\sin x-ax$ is a periodic function.

$\bf{My\; Try::}$ If function $f(x)$ is Periodic function , Then it must satisfy the

condition $f(x+T) = f(x)\;,$ Where $T$ is period of that function (Smallest positive value)

So Here $(x+T)^a+\sin (x+T)-a(x+T) = x^{a}+\sin x-ax$

So we get $(x+T)^a+\sin (x+T)-aT = \sin x$

From here we can see for $a=0$ and $a=1.$ We get $\sin(x+T)=\sin x.$

So it is a periodic function for $T=2\pi.$

But I did not understand how can we prove that there are only two values of $a$

  • 2
    $\begingroup$ the point is i think, that $g(x,a)=x^a-ax$ have to be a constant or has to vanish for $f(x)$ being periodic. the reason for that is that $g(x,a)$ is monotonically increasing/decreasing for $x>x_0(a)$ and therefore this part can not be periodic when the above condtions are not fullfiled $\endgroup$
    – tired
    Commented Nov 25, 2015 at 17:55

1 Answer 1


$$If \space f(x+T)=f(x) \space then \space f'(x+T)=f'(x)$$ $$f'(x)=ax^{a-1}+\cos x-a=a(x+T)^{a-1}+\cos (x+T)-a$$ i) $a=0$: $$\cos x = \cos (x+T)$$ $$T=2k\pi$$ ii) $a\neq0$: $$\left(\frac{x}{x+T}\right)^{a-1}=\frac{\cos x \cos T - \sin x \sin T}{\cos x}=\cos T - \sin T \tan x$$ $$\therefore a=1, T=2k\pi$$

  • 1
    $\begingroup$ But as the OP already stated, a=0 also works. $\endgroup$
    – tethernova
    Commented Nov 25, 2015 at 17:58
  • $\begingroup$ yeah, I added it. $\endgroup$
    – Kay K.
    Commented Nov 25, 2015 at 18:02

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