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EEJ MAIN Mathematics QUESTION #7486
Question 1
If the function \(f: \mathbb{R} - \{1,-1\} \to A\) defined by \(f(x) = \frac{x^2}{1-x^2}\) is surjective, then A is equal to:
  • \(\mathbb{R} - \{-1\}\)
  • \([0,\infty)\)
  • \(\mathbb{R} - [-1,0)\)✔️
  • \(\mathbb{R} - (-1,0)\)
Correct Answer Explanation
To find the range, let \(y = \frac{x^2}{1-x^2}\). Solving for \(x^2\): \(y(1-x^2) = x^2\), so \(y = x^2 + yx^2 = x^2(1+y)\), giving \(x^2 = \frac{y}{1+y}\). For real x, we need \(\frac{y}{1+y} \geq 0\). This holds when \(y \geq 0\) and \(1+y > 0\) (i.e., \(y > -1\)), or \(y < 0\) and \(1+y < 0\) (i.e., \(y < -1\)). Combining: \(y \in (-\infty, -1) \cup [0, \infty) = \mathbb{R} - [-1,0)\). So A = \(\mathbb{R} - [-1,0)\).