Question

An even polynomial function $f(x)$ satisfies a relation
$f(2x)\left(1-f\left(\frac{1}{2x}\right)\right)+f\left(16x^{2}y\right)=f(-2)-f(4xy)\forall x,y\in R-\{0\}$ and $f(4)=-255,f(0)=1$Which of the following hold(s) good?

• A. $f(x)$ has local maximum at $x=1$.
• B. $f(x)f\left(\frac{1}{x}\right)\le 0$
• C. Range of values of $k$ for which $|f(x)|=k-2$ has exactly four distinct solutions is $(2,3)$.
• D. ${\int }_0^{1}f(x)dx=\frac{3}{4}$.

Answer 1

Answer: (B), (C)

We have $f(2x)-f(2x)f\left(\frac{1}{2x}\right)+f\left(16x^{2}y\right)=f(-2)-f(4xy)$
Replacing y by $\frac{1}{8x^2}$, we get
$f(2x)-f(2x)f\left(\frac{1}{2x}\right)+f(2)=f(-2)-f\left(\frac{1}{2x}\right)$
$f(2x)+f\left(\frac{1}{2x}\right)=f(2x)f\left(\frac{1}{2x}\right)($ As $f(x)$ is even)
$\therefore f(2x)=1\pm (2x{)}^n$
$\Rightarrow f(x)=1\pm x^n$
Now $f(4)=1\pm 4^n=-255$ (Given)
Taking negative sign, we get $256=4^n\Rightarrow n=4$
Hence $f(x)=1-x^4$, which is even function.
Now $|f(x)|=k-2$
$\Rightarrow 0<k-2<1\Rightarrow 2<k<3$
Clearly $f(x)$ has local maximum at $x=0$.

Also $\underset{0}{\overset{1}{\int }}f(x)dx=\underset{0}{\overset{1}{\int }}\left(1-x^4\right)dx={\left(1-\frac{x^5}{5}\right)}_0^1=1-\frac{1}{5}=\frac{4}{5}$.