Q46 / 87 · NTA Abhyas (JEE Main) 2020 · Mathematics · Functions

If fx=x2-x21+x2-x2 (where . represents the greatest integer part of x ), then the range of fx is

  • A
    0,1
  • B
    -1,1
  • C
    0,∞
  • Correct answer
    D
    0,12

Solution

x2-x2=x2
∴fx=x21+x2=1+x2-11+x2=1-11+x2

∵0≤x2<1⇒1≤x2+1<2
12<1x2+1≤1⇒-12>-11+x2≥-1
12>1+-11+x2≥0⇒ Range of fx∈0,12

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

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