Q43 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Mathematics

[.] represents greatest integer function. Let $g(x)=1+x-[x]$ and $f(x)=\left\{\begin{array}{cl}-3, & x < 0 \\ 0, & x=0, \text { then } \\ 5, & x>0\end{array}\right.$ $f(g(x))$ is

  • A
    $f(x)$
  • B
    $-15$
  • Correct answer
    C
    $5$
  • D
    $-3$

Solution

We have, $g(x)=1+x-[x]$ $g(x)=1+\{x\} \quad[\because x-[x]=\{x\}]$ $\Rightarrow g(x) \in[1,2) \quad[\because\{x\} \in[0,1)]$ Now, $f(x)=\left\{\begin{array}{cc}-3, & x < 0 \\ 0, & x=0 \\ 5, & x>0\end{array}\right.$ $\therefore \quad f(g(x))=5 ; g(x) \geq 1$

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

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