Question

Let $f\left(x\right)$ be defined for all $x$ $ϵ$ $R$ and continuous. Let $f(x+y)-f(x-y)=4xy$ for all $x,y\in R$ and $f\left(0\right)=0$, then

• A. The minimum value of $f\left(x\right)$ is 0
• B. $f\left(x\right)=f\left(\frac{1}{x}\right)=f(x+\frac{1}{x})+2$
• C. $f\left(x\right)+f\left(\frac{1}{x}\right)=f(x-\frac{1}{x})+2$
• D. none of these

Answer 1

Answer: (A), (C)

The minimum value of $f\left(x\right)$ is 0
$f\left(x\right)+f\left(\frac{1}{x}\right)=f(x-\frac{1}{x})+2$

Substitute $y=x$ in $f(x+y)-f(x-y)=4xy$
$f\left(2x\right)-f\left(0\right)=4x^2$
However it is given that $f\left(0\right)=0$
Hence,$f\left(2x\right)=4x^2$
$f\left(2x\right)=(2x{)}^2$
Therefore
$f\left(x\right)=x^2$
The minimum value of $f\left(x\right)$ is $0.$
Which also satisfies the equation $f\left(x\right)+\frac{1}{f\left(x\right)}=f(x-\frac{1}{x})+2$
Hence, options 'A' and 'C' are correct.