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If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(2+x) = f(2-x)$ and $f(20-x) = f(x)\;\forall x\in \mathbb{R}$ and $f(2)\neq f(6)$

Then fundamental period of function $f(x)$ is

$\bf{My\; Try::}$ Given $f(2+x) = f(2-x)\;,$ Now replace $x\rightarrow (2-x)\;,$ We get

$f(4-x)=f(x)$ and given $f(20-x)=f(x)\;,$ So we get $f(4-x)=f(20-x)$

Now Replace $x\rightarrow (4-x)\;,$ We get $f(x)=f(x+16)$

So period of function $f(x)$ is $=16\;,$ But Options given as

$(a)\;\;\;\; 1\;\;\;\; (b)\;\;\;\; 8$ $(c)\;$ Period can not be $1\;\;\;\; (d)\; $ may be one

Help required, Thanks

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juantheron
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    $f(x)=sin({\frac \pi 4}x)$ is an example of function of period 8, respecting the required conditions. –  May 16 '16 at 06:22

1 Answers1

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Let $f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=1$ for $x=16k+2, k\in \mathbb{Z}$ and $f(x)=0$ otherwise.

First, let's notice this function is periodic of fundamental period 16.

Let's verify the conditions:

1) $f(2+x) = f(2-x)$ If $2 + x=16k+2$ then $2-x=2-16k$ so $1=1$ (because of periodicity), otherwise $0=0$

2) $f(20-x) = f(x)$ Because of periodicity this is equivalent to $f(4-x)=f(x)$ which is equivalent to 1).

3) $f(2)\neq f(6)$ because $f(2)=1 \neq f(6)=0$

The conclusion:

(a) false - because $f(2)\neq f(6)$

(b) false - we just prove it by a counter-example

(c) true

(d) false - because $f(2)\neq f(6)$

Taking $f(x)=sin({\frac \pi 4}x)$ we get un example of periodic function of fundamental period 8, satisfying the problem's requirements.

Therefore the problem's requirements are not strong enough to allow us determining the fundamental period.