Question

Let $f(x)$ be a polynomial of even degree satisfying $f(2x)\left(1-f\left(\frac{1}{2x}\right)\right)+f\left(16x^{2}y\right)=f(-2)-f(4xy)$ for all $x,y\in R\sim \{0\}$ and $f(4)=-255,f(0)=1$. Then the value of $\left|\frac{f(2)+1}{2}\right|$ is

• A. 4
• B. 5
• C. 7
• D. 6

Answer 1

Answer: (C)

Solution: Replacing $y$ by $\frac{1}{8x^2}$ in the given functional equation, we obtain
$f(2x)\left(1-f\left(\frac{1}{2x}\right)\right)+f(2)=f(-2)-f\left(\frac{1}{2x}\right).$
Since $f$ is an even function so $f(2)=f(-2)$,
$\text{ so }f(2x)-f(2x)f\left(\frac{1}{2x}\right)=-f\left(\frac{1}{2x}\right)$
$\Rightarrow f(2x)+f\left(\frac{1}{2x}\right)=f(2x)f\left(\frac{1}{2x}\right)$
Replacing $x$ by $\frac{x}{2}$, we have
$f(x)+f\left(\frac{1}{x}\right)=f(x)f\left(\frac{1}{x}\right)$
Since $f$ is a polynomial, so $f(x)=\pm x^n+1$ (see Example 88) But $-255=f(4)=\pm 4^n+1$.
nly negative sign is possible, thus $4^n=256\Rightarrow n=4$.
i.e. $f(x)=-x^4+1.f(2)=-2^4+1=-15$ $\left|\frac{f(2)+1}{2}\right|=\left|-\frac{14}{2}\right|=7$