What is the range of the function \({\rm{y}} = \frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}\) where x ∈ R?

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This question was previously asked in

NDA (Held On: 17 April 2016) Maths Previous Year paper

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Answer 1

Concept:

For a quadratic equation ax² + bx +c = 0 to have real roots: D = b² – 4ac ≥ 0

Calculation:

\({\rm{y}} = \frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}\), Where x ∈ R

On solving,

y + yx² = x²

⇒ (y - 1) x² + y = 0

⇒(y - 1) x² + y = 0

Also for the quadratic equation to exist y – 1 = 0 is not possible

Which is a quadratic equation in x , on comparing ax² + bx +c = 0 we get a = (y - 1) , b = 0 , c = y.

For x ∈ R considering D ≥ 0 so b² – 4ac ≥ 0

0 – 4 y ( y – 1 ) ≥ 0

⇒ y ( y - 1 ) ≤ 0

Which gives the solution y ∈ [0, 1)