Question

If f(x) = sgn(x$^5$), then which of the following is/are false (where sgn denotes signum function):

• A. f'(0$^+$) = 1
• B. f'(0) = -1
• C. f is continuous but not differentiable at x = 0
• D. f is discontinuous at x = 0

Answer 1

Answer: (D)

(A), (B), (C)

Answer 2

The given function is $f(x) = \text{sgn}(x^5)$.

The signum function $\text{sgn}(u)$ is defined as:

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

Let's analyze $f(x)$ based on the value of $x^5$:

1. If $x > 0$, then $x^5 > 0$, so $f(x) = \text{sgn}(x^5) = 1$.
2. If $x = 0$, then $x^5 = 0$, so $f(x) = \text{sgn}(x^5) = 0$.
3. If $x < 0$, then $x^5 < 0$, so $f(x) = \text{sgn}(x^5) = -1$.

Thus, $f(x)$ can be simplified to:

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

This is precisely the definition of the signum function $\text{sgn}(x)$. So, $f(x) = \text{sgn}(x)$.

Now, let's evaluate the continuity and differentiability of $f(x)$ at $x=0$.

Continuity at $x=0$:

For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

1. Left-hand limit (LHL):
   $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1$.
2. Right-hand limit (RHL):
   $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1$.
3. Function value at $x=0$:
   $f(0) = 0$.

Since LHL ($\neq -1$) is not equal to RHL ($= 1$), the limit of $f(x)$ as $x \to 0$ does not exist. Therefore, $f(x)$ is discontinuous at $x=0$.

Differentiability at $x=0$:

A function must be continuous at a point to be differentiable at that point. Since $f(x)$ is discontinuous at $x=0$, it is not differentiable at $x=0$.

Now let's check each statement:

(A) f'(0$^+$) = 1

This refers to the right-hand derivative at $x=0$.

$f'(0^+) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}$

For $h > 0$, $f(h) = 1$ and $f(0) = 0$.

$f'(0^+) = \lim_{h \to 0^+} \frac{1 - 0}{h} = \lim_{h \to 0^+} \frac{1}{h} = \infty$.

Since $\infty \neq 1$, statement (A) is false.

(B) f'(0) = -1

As established, $f(x)$ is not differentiable at $x=0$. Therefore, $f'(0)$ does not exist.

Since $f'(0)$ does not exist, it cannot be equal to $-1$. Statement (B) is false.

(C) f is continuous but not differentiable at x = 0

We have determined that $f$ is discontinuous at $x=0$.

Therefore, the first part of the statement, "f is continuous at x = 0", is false. This makes the entire compound statement false.

(D) f is discontinuous at x = 0

We have determined that $f$ is discontinuous at $x=0$.

Therefore, statement (D) is true.

The question asks for the statements that are false. Based on our analysis, statements (A), (B), and (C) are false.