Q100 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

If $\mathrm{f}(\mathrm{x})=-|\mathrm{x}|$, then $\left(f^{\prime}\right) f(0)(\mathrm{x})+(f \circ f \circ f)(-\mathrm{x})=$

  • A
    $-2 f(\mathrm{x})$
  • B
    $|f(\mathrm{x})|$
  • Correct answer
    C
    $2 f(\mathrm{x})$
  • D
    $-|f(\mathrm{x})|$

Solution

$f(x)=-|x|$ $\begin{aligned}(f \circ f \circ f)(x) & =f \circ f(-|x|) \\ & =f \circ(-|-| x \mid)=-|-|-|x||| \\ & =\left\{\begin{array}{cc}-x, & x \geq 0 \\ x, & x < 0\end{array}\right. \\ \text { fofof }(-x) & =f \circ f(-|-x|)\end{aligned}$ $\begin{gathered}\quad=f(-|-|-x \|)=-|-|-|-x| \mid \\ = \begin{cases}-x, & x \geq 0 \\ x, & x < 0\end{cases} \\ \Rightarrow \quad(\text { fofof })(x)+(\text { fofof })(-x)=\left\{\begin{array}{cl}-2 x, & x \geq 0 \\ 2 x & x < 0\end{array}\right. \\ \quad f(x)=\left\{\begin{array}{cl}-x, & x \geq 0 \\ x & x < 0\end{array}\right. \\ \therefore \quad(\text { fofof })(x)+(\text { fofof })(-x)=2 f(x) .\end{gathered}$

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

Relevant Questions