Question

If $f(x) = \begin{cases} x + \{x\} + x \sin\{x\} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$

where {.} denotes the fractional part function, then:

• A. f is continuous & differentiable at x = 0
• B. f is continuous but not differentiable at x = 0
• C. f is continuous & differentiable at x = 2
• D. none of these

Answer 1

Answer: (D)

none of these

Answer 2

The function is given by $f(x) = x + \{x\} + x \sin\{x\}$ for $x \neq 0$ and $f(0) = 0$. The fractional part function $\{x\}$ is defined as $x - \lfloor x \rfloor$.

Continuity at $x=0$: For continuity at $x=0$, we require $\lim_{x \to 0} f(x) = f(0)$. We are given $f(0) = 0$.

The right-hand limit: As $x \to 0^+$, $\lfloor x \rfloor = 0$, so $\{x\} = x$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + x + x \sin x) = \lim_{x \to 0^+} (2x + x \sin x) = 0$ The left-hand limit: As $x \to 0^-$, $\lfloor x \rfloor = -1$, so $\{x\} = x - (-1) = x+1$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + (x+1) + x \sin(x+1)) = 0 + (0+1) + 0 \sin(0+1) = 1$ Since $\lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x)$, the limit $\lim_{x \to 0} f(x)$ does not exist. Thus, $f$ is not continuous at $x=0$.

Continuity at $x=2$: For continuity at $x=2$, we require $\lim_{x \to 2} f(x) = f(2)$. For $x=2$, $\lfloor 2 \rfloor = 2$, so $\{2\} = 2 - 2 = 0$. $f(2) = 2 + \{2\} + 2 \sin\{2\} = 2 + 0 + 2 \sin(0) = 2$

The right-hand limit: As $x \to 2^+$, $\lfloor x \rfloor = 2$, so $\{x\} = x-2$. $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x + (x-2) + x \sin(x-2)) = 2 + (2-2) + 2 \sin(2-2) = 2 + 0 + 0 = 2$ The left-hand limit: As $x \to 2^-$, $\lfloor x \rfloor = 1$, so $\{x\} = x-1$. $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + (x-1) + x \sin(x-1)) = 2 + (2-1) + 2 \sin(2-1) = 2 + 1 + 2 \sin(1) = 3 + 2 \sin(1)$ Since $\lim_{x \to 2^-} f(x) = 3 + 2 \sin(1) \neq f(2) = 2$, $f$ is not continuous at $x=2$.

Since $f$ is not continuous at $x=0$ and $x=2$, the first three options are false. The correct option is none of these.