Q.Find the range of the following functions given by

Here are the ranges for the given functions:

(i) $f(x) = \dfrac{3}{2 - x^2}$

• Concept: Range of a rational function by solving for $x^2$.
• Reasoning:
  1. Let $y = f(x)$. Then $y = \frac{3}{2 - x^2}$.
  2. Rearrange to express $x^2$ in terms of $y$: $2 - x^2 = \frac{3}{y} \implies x^2 = 2 - \frac{3}{y} = \frac{2y - 3}{y}$.
  3. Since $x^2 \ge 0$, we must have $\frac{2y - 3}{y} \ge 0$. …

Finding the range of a function means determining all possible values that the function can output. This often involves understanding the fundamental properties of the operations involved (like squares, absolute values, or trigonometric functions) and how they restrict or expand the set of possible values. We will analyze each function individually.

(i) $f(x) = \dfrac{3}{2 - x^2}$

This is a rational function. The key to finding its range lies in understanding the behavior of the denominator $2 - x^2$.

1. Analyze the denominator $2 - x^2$:

   We know that for any real number $x$, $x^2 \ge 0$.

   Multiplying by $-1$ reverses the inequality: $-x^2 \le 0$.

   Adding $2$ to both sides: $2 - x^2 \le 2$.

   So, the denominator $2 - x^2$ can take any value less than or equal to $2$, except for $0$ (because division by zero is undefined).

   Thus, $2 - x^2 \in (-\infty, 0) \cup (0, 2]$.

2. Consider two cases for the denominator:

   • Case 1: $2 - x^2 > 0$

     This implies $x^2 < 2$, or $-\sqrt{2} < x < \sqrt{2}$.

     In this interval, $0 < 2 - x^2 \le 2$.

     Taking the reciprocal of each part (and reversing inequalities because all terms are positive):

     $\dfrac{1}{2 - x^2} \ge \dfrac{1}{2}$.

     Now, multiply by $3$:

     $f(x) = \dfrac{3}{2 - x^2} \ge \dfrac{3}{2}$.

     So, in this case, $f(x) \in \left[\frac{3}{2}, \infty\right)$.

   • Case 2: $2 - x^2 < 0$

     This implies $x^2 > 2$, or $x < -\sqrt{2}$ or $x > \sqrt{2}$.

     In this interval, $2 - x^2$ can take any negative value.

     As $x$ approaches $\pm \sqrt{2}$ from outside the interval $(-\sqrt{2}, \sqrt{2})$, $2 - x^2$ approaches $0$ from the negative side ($0^-$).

     Therefore, $\dfrac{3}{2 - x^2}$ approaches $-\infty$.

     As $x \to \pm \infty$, $x^2 \to \infty$, so $2 - x^2 \to -\infty$.

     Therefore, $\dfrac{3}{2 - x^2}$ approaches $0$ from the negative side ($0^-$).

     This means that when $2 - x^2 < 0$, $f(x)$ can take any negative value.

     So, in this case, $f(x) \in (-\infty, 0)$.

3. Combine the results:

   The range of $f(x)$ is the union of the ranges from both cases. …