Q42 / 160 · TS EAMCET 2016 · Mathematics

The number of real linear functions $f(x)$ satisfying $f(f(x))=x+f(x)$ is

  • A
    0
  • B
    4
  • C
    5
  • Correct answer
    D
    2

Solution

Let $f(x)=a x+b$ Given, $$ \begin{aligned} & f(f(x))=x+f(x) \\ & +b)+b=x+(a x+b) \\ & +a b+b=x+a x+b \end{aligned} $$ $$ \begin{aligned} & \Rightarrow & a(a x+b)+b & =x+(a x+b) \\ & \Rightarrow & a^2 x+a b+b & =x+a x+b \\ & \Rightarrow & x\left(a^2-a-1\right)+a b & =0 \\ & \therefore & a^2-a-1 & =0 \text { and } a b=0 \\ & \Rightarrow & a & \neq 0, b=0 \\ & \therefore & a & =\frac{1 \pm \sqrt{5}}{2} \end{aligned} $$ Now, $\quad f(x)=\left(\frac{1 \pm \sqrt{5}}{2}\right) x+0=\left(\frac{1 \pm \sqrt{5}}{2}\right) x$ $$ f(x)=\left(\frac{1+\sqrt{5}}{2}\right) x,\left(\frac{1-\sqrt{5}}{2}\right) x $$ $\therefore$ It has two linear equation.

Hence, the correct answer is option: 'D'.

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