Question

If $y=f\left(x\right)$ satisfies the property $(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2),f\left(1\right)=1$, then the number of real roots of $f\left(x\right)=4$ will be

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given property can be considered as
$\frac{f(x+y)}{(x+y)}-\frac{f(x-y)}{(x-y)}=4xy$ ......(1)
Now if $g\left(x\right)=\frac{f\left(x\right)}{x}$
then from (1)
$\Rightarrow \underset{y\to x}{lim}\left[\frac{g(x+y)-g(x-y)}{(x+y)-(x-y)}\right]=2x$
$\Rightarrow g^{\prime }\left(x\right)=2x$
$\frac{d}{dx}\left[\frac{f\left(x\right)}{x}\right]=2x$
$\int d\left[\frac{f\left(x\right)}{x}\right]=2\int xdx$
$\Rightarrow \frac{f\left(x\right)}{x}=x^2+cf\left(x\right)=x^3+cx$ ......(2)
Substituting $x=0$ in the given property
$f\left(y\right)+f(-y)=0\Rightarrow f\left(0\right)=0$
$\because f\left(1\right)=1,c=0$
$\Rightarrow f\left(x\right)=x^3$
$x^3=4$ will have one real solution.