Find domain of inverse sine function - jelet 2024 - jelet previous year question solution by sandip sir jelet academy

Problem: 1: Find domain of inverse sine function

Solution:

**Assumptions:**

*   The function is defined over real numbers.

*   Standard definitions of inverse trigonometric functions and logarithms are used.

1.  **Identify Domain Restrictions:**

    The function is $f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right)$.

    For $f(x)$ to be defined, two conditions must be met:

    *   The argument of the inverse sine function must be in the interval $[-1, 1]$.

    *   The argument of the logarithm function must be strictly positive.

2.  **Apply Logarithm Domain Condition:**

    The argument of $\log_2$ is $\frac{x^2}{2}$.

    $\frac{x^2}{2} > 0$

    This implies $x^2 > 0$, which means $x \ne 0$.

3.  **Apply Inverse Sine Domain Condition:**

    The argument of $\sin^{-1}$ is $\log_2 \left( \frac{x^2}{2} \right)$.

    $-1 \le \log_2 \left( \frac{x^2}{2} \right) \le 1$

    Since the base of the logarithm is $2$ (which is greater than $1$), we can exponentiate all parts of the inequality with base $2$ without changing the direction of the inequalities.

    $2^{-1} \le \frac{x^2}{2} \le 2^1$

    $\frac{1}{2} \le \frac{x^2}{2} \le 2$

4.  **Solve the Compound Inequality:**

    This compound inequality can be split into two separate inequalities:

    *   **Inequality 1:** $\frac{1}{2} \le \frac{x^2}{2}$

        Multiply by $2$: $1 \le x^2$

        Rearrange: $x^2 \ge 1$

        This inequality holds when $x \le -1$ or $x \ge 1$.

        In interval notation: $(-\infty, -1] \cup [1, \infty)$.

    *   **Inequality 2:** $\frac{x^2}{2} \le 2$

        Multiply by $2$: $x^2 \le 4$

        Take the square root of both sides: $\sqrt{x^2} \le \sqrt{4}$

        $|x| \le 2$

        This inequality holds when $-2 \le x \le 2$.

        In interval notation: $[-2, 2]$.

5.  **Combine All Conditions:**

    We need to find the intersection of all valid $x$ values:

    *   $x \ne 0$

    *   $(-\infty, -1] \cup [1, \infty)$

    *   $[-2, 2]$

    The intersection of $(-\infty, -1] \cup [1, \infty)$ and $[-2, 2]$ is:

    $([-2, 2]) \cap ((-\infty, -1] \cup [1, \infty))$

    $= ([-2, 2] \cap (-\infty, -1]) \cup ([-2, 2] \cap [1, \infty))$

    $= [-2, -1] \cup [1, 2]$

    The condition $x \ne 0$ is already satisfied by the interval $[-2, -1] \cup [1, 2]$, as $0$ is not included in this set.

The domain of the function $f(x)$ is:

$\boxed{[-2, -1] \cup [1, 2]}$