Let f(x) = x^2/(1+x^2). Then, range (f) = ?

Question

Mark (√) against the correct answer in the following:

Let \(f(x)=\frac{x^2}{(1+x^2)}\) Then, range (f) = ?

A. [1, ∞) 

B. [0, 1) 

C. [ - 1, 1] 

D. (0, 1]

Answer 1

Correct Answer is (B) = [0, 1)

\(f(x)=\frac{x^2}{1+x^2}\)

⇒ \(y=\frac{x^2}{1+x^2}\)

⇒ y + yx² = x²

⇒ y = x² (1 - y)

⇒ x = \(\sqrt{\frac{y}{1-y}}\)

\(\frac{y}{1-y}\geq 0\)

⇒ y ≥ 0

And

1 - y > 0

⇒ y < 1

Taking intersection we get

range (f) = [0, 1)