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

The Correct Option is C

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