Question

Consider the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = e^{-| \log_e x |}.$If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is

• A. 0
• B. 3
• C. 1
• D. 2

Answer 1

Answer: (C)

1

Answer 2

Rewrite $f(x)$ in terms of piecewise functions based on the value of $x$:

$f(x) = e^{-\lvert \ln x \rvert} = \begin{cases} e^{\ln x} = x & \text{for } x \geq 1 \\\ e^{-\ln x} = \frac{1}{x} & \text{for } 0 < x < 1 \end{cases}$

Check for continuity. The function $f(x)$ is continuous for $x > 0$ because:

• $f(x) = \frac{1}{x}$ for $0 < x < 1$, $f(x) = x$ for $x \geq 1$.
• At $x = 1$, $f(1) = 1$ from both the left and right limits.

Thus, $f(x)$ is continuous at $x = 1$ and everywhere else in $(0, \infty)$. So, $m = 0$.

Check for differentiability at $x = 1$. To check differentiability at $x = 1$, compute the left-hand derivative and the right-hand derivative at $x = 1$.

For $0 < x < 1$, $f(x) = \frac{1}{x}$, so:

$f'_{-}(1) = \lim_{x \to 1^{-}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{-}} \frac{\frac{1}{x} - 1}{x - 1} = -1.$

For $x \geq 1$, $f(x) = x$, so:

$f'_{+}(1) = \lim_{x \to 1^{+}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{+}} \frac{x - 1}{x - 1} = 1.$

Since $f'_{-}(1) \neq f'_{+}(1)$, $f(x)$ is not differentiable at $x = 1$. Therefore, $n = 1$.

Conclusion:

$m + n = 0 + 1 = 1$

Thus, the answer is: 1