Question

The number of real linear functions $f\left(x\right)$ satisfying $f\left\{f\right(x\left)\right\}=x+f\left(x\right)$

• A. $0$
• B. $4$
• C. $5$
• D. $2$

Answer 1

The correct option is D $2$

Let $f\left(x\right)=ax+b$

$f\left(f\right(x\left)\right)=a(ax+b)+b=a^{2}x+ab+b$

Given, $f\left(f\right(x\left)\right)=x+f\left(x\right)$

$⟹a^{2}x+ab+b=x+ax+b$

$⟹(a^2-a-1)x+ab=0$

Since above equation is valid for all values of x, we have

$a^2-a-1=0$ and $ab=0$

$a^2-a-1=0$

$a=\frac{1\pm \sqrt{1+4}}{2}$

$⟹a=\frac{1\pm \sqrt{5}}{2}$

$ab=0$

Since $a$ is non-zero,

$⟹b=0$

$f\left(x\right)=\frac{1+\sqrt{5}}{2}x$ and $f\left(x\right)=\frac{1-\sqrt{5}}{2}x$ satisfy the condition.

Hence the answer is option (D)