Question

Let $f\left(x\right)$ be a continuous even periodic function of period $a$ and $f\left(0\right)=0$ then

• A. $f\left(x\right)$ is monotonic in $[0,\frac{a}{2}]$
• B. $f(\frac{a}{2}-x)=f(\frac{a}{2}+x)$ for all $x$
• C. $f\left(x\right)$ is differentiable in $[0,\frac{a}{2}]$
• D. $f\left(x\right)$ has a maximum value

Answer 1

Answer: (B), (D)

The correct options are
B $f(\frac{a}{2}-x)=f(\frac{a}{2}+x)$ for all $x$
D $f\left(x\right)$ has a maximum value

Consider
$f\left(x\right)=1-cos\left(x\right)$
$f(-x)=1-cos(-x)$
$=1-cos\left(x\right)$
$=f\left(x\right)$
Hence $f\left(x\right)=f(-x)$ making f(x) a even function.
$f\left(0\right)=0$
Therefore $a=2\pi$
Now we know that cos(x) has a maximum value at $x=\frac{\pi }{2}$.
The maximum value being 1.
Now
$\to cos(\pi -x)=-cos\left(x\right)$
$=cos(\pi +x)$
Hence
$f(\frac{a}{2}-x)=f(\frac{a}{2}+x)$