Question

The range of the function \[ f:\mathbb{R}-\{-1,1\}\to\mathbb{R}, \qquad f(x)=\frac{x^2}{1-x^2} \] is

• A. \((-\infty,-1]\cup(0,\infty)\)
• B. \((-\infty,-1]\cup[0,\infty)\)
• C. \((-\infty,-1)\cup(0,\infty)\)
• D. \((-\infty,-1)\cup[0,\infty)\)

Hint:

For rational functions, put \[ y=f(x) \] and solve for \(x\). Then use conditions such as \[ x^2\ge 0 \] to determine the range.

Answer 1

The Correct Option is B

Concept: To find the range, let \[ y=\frac{x^2}{1-x^2} \] and solve for \(x^2\). Since \[ x^2\ge 0, \] the obtained expression must satisfy the non-negativity condition.

Step 1: Express \(x^2\) in terms of \(y\). \[\begin{aligned} y &= \frac{x^2}{1-x^2} \end{aligned}\] \[\begin{aligned} y(1-x^2) &= x^2 \end{aligned}\] \[\begin{aligned} y &= x^2(1+y) \end{aligned}\] \[\begin{aligned} x^2 &= \frac{y}{1+y} \end{aligned}\]

Step 2: Use the condition \(x^2\ge 0\). \[ \frac{y}{1+y}\ge 0 \] Critical points are \[ y=0,\qquad y=-1 \] Sign analysis gives \[ y\in(-\infty,-1)\cup[0,\infty) \]

Step 3: Check boundary values. \[ y=0 \] is attained at \[ x=0 \] Hence \(0\) belongs to the range. \[ y=-1 \] is impossible because \[ \frac{y}{1+y} \] is undefined at \(y=-1\). Therefore \(-1\) is excluded. \[\begin{aligned} \boxed{ (-\infty,-1)\cup[0,\infty) } \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.

Note: The mathematically correct answer is option \(\mathbf{(D)}\). If the official key shows option (B), it is incorrect.