Correct Answer: D (The domain of $h(x)$ is different from the domain of $g(x)$ at the same point.)

To determine which statement is not true, we need to analyze the properties of the composite function $h(x) = sgn(g(x))$ based on the definitions of the functions involved.

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Step 1: Understand the Signum Function

The signum function, $sgn(u)$, is defined as:

$$ sgn(u) = \begin{cases} 1 & \text{if } u > 0 \\ 0 & \text{if } u = 0 \\ -1 & \text{if } u < 0 \end{cases} $$

Key properties of the signum function are:

• Domain: The set of all real numbers, $\mathbb{R}$.
• Continuity: It is continuous everywhere except at $u=0$, where it has a jump discontinuity.

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Step 2: Analyze the Domain and Continuity of $h(x)$

The function $h(x) = sgn(g(x))$ is a composition of $g$ and the signum function.

• Domain of $h(x)$: The domain of a composite function $(f \circ g)(x)$ is the set of all $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$. Here, the domain of $g$ is $\mathbb{R}$, and its output $g(x)$ is a real number. Since the domain of the signum function is also $\mathbb{R}$, it can accept any output from $g(x)$. Therefore, the domain of $h(x)$ is the same as the domain of $g(x)$, which is $\mathbb{R}$.
• Continuity of $h(x)$: A composite function can be discontinuous where the inner function is discontinuous, or where the inner function's value causes the outer function to be discontinuous. Since the signum function is only discontinuous at 0, the composite function $h(x)$ will be discontinuous at all values of $x$ for which $g(x) = 0$.

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Step 3: Evaluate Each Statement

Now let's check the truthfulness of each option:

1. The domain of continuity of h(x) equal the domain of continuity of g(x) - {$x \in \mathbb{R}, g(x) = 0$}. This statement says that $h(x)$ is continuous everywhere that $g(x)$ is continuous, except for the points where $g(x)=0$. This is correct because the signum function introduces discontinuities precisely at the points where its argument is zero. So, this statement is TRUE.
2. $h(x)$ is discontinuous at $g(x) = 0$. As established above, the signum function is discontinuous at 0. Therefore, $h(x)$ will be discontinuous for any $x$ that makes $g(x)=0$. This statement is TRUE.
3. The domain of $h(x)$ is the same as the domain of $g(x)$. As analyzed in Step 2, both functions are defined for all real numbers. The domain of $g(x)$ is $\mathbb{R}$, and the domain of $h(x)$ is also $\mathbb{R}$. This statement is TRUE.
4. The domain of $h(x)$ is different from the domain of $g(x)$ at the same point. This statement is poorly phrased but implies that the two domains are not the same set. Since we proved in (C) that the domains are identical, this statement is a direct contradiction. This statement is NOT TRUE.

The question asks to select the option which is not true, which is D.