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26 votes
26 votes

\lfloor(x^2+1)/(10)\rfloor+\lfloor(10)/(x^2+1)\rfloor=1


\small\textrm{$\bullet\ \lfloor ()\rfloor$ denotes the greatest integer that does not exceed the number}

User Anikhan
by
2.8k points

1 Answer

8 votes
8 votes

Answer:

  • See below

Explanation:

This equation is solved if one of the following 2 conditions is met

1.

  • (x² + 1) / 10 is between 1 and 2

and

  • 10 / (x² + 1) is between 0 and 1

Solve this:

  • 1 < (x² + 1) / 10 < 2
  • 10 < x² + 1 < 20
  • 9 < x² < 19
  • 3 < |x| < √19
  • x ∈ (- √19, - 3) ∪ (3, √19)
  • 0 < 10 / (x² + 1) < 1
  • x² + 1 > 10
  • x² > 9
  • |x| > 3

Solution for this case is x ∈ (- √19, - 3) ∪ (3, √19)

2.

  • (x² + 1) / 10 is between 0 and 1

and

  • 10 / (x² + 1) is between 1 and 2

Solve this:

  • 0 < (x² + 1) / 10 < 1
  • 0 < x² + 1 < 10
  • - 1 < x² < 9
  • 0 ≤ |x| < 3
  • x ∈ ( - 3, 3)
  • 1 < 10 / (x² + 1) < 2
  • x² + 1 < 10 ⇒ x² < 9 ⇒ |x| < 3
  • x² + 1 > 5 ⇒ x² > 5 ⇒ |x| > √5

Solution for this case is x ∈ (- 3, - √5) ∪ (√5, 3)

User Krishn Patel
by
2.5k points
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