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

To solve the problem, we need to analyze the function given:

Function: \(f(x) = e^{-| \log_e x |}\) where \( f : (0, \infty) \rightarrow \mathbb{R} \).

Step 1: Analyze Continuity

The function \(f(x) = e^{-| \log_e x |}\) can be rewritten as follows: Analyzing the absolute value, \(| \log_e x |\) can be expressed as:

• If \(x > 1\), then \(\log_e x > 0\), and hence \(| \log_e x | = \log_e x\).
• If \(0 < x < 1\), then \(\log_e x < 0\), and hence \(| \log_e x | = -\log_e x\).

Thus,

• For \(x > 1\), \(f(x) = e^{-\log_e x} = x^{-1}\).
• For \(0 < x < 1\), \(f(x) = e^{\log_e x} = x\).

At \(x = 1\), \(f(x) = e^{0} = 1\).

Step 2: Check Points of Discontinuity

The function \(f(x)\) is composed of continuous functions on intervals \((0, 1)\) and \((1, \infty)\). At \(x = 1\), the function's left-hand limit and right-hand limit should be checked:

• As \(x \to 1^{-}\), \(f(x) = x \to 1\).
• As \(x \to 1^{+}\), \(f(x) = x^{-1} \to 1\).

Since both limits are equal to the value of the function at \(x = 1\), the function is continuous at \(x = 1\). Thus, \(m = 0\).

Step 3: Check Points of Non-Differentiability

The derivatives on either side of \(x = 1\) should be checked:

• For \(x > 1\), \(f(x) = x^{-1}\) differentiates to \(-x^{-2}\).
• For \(0 < x < 1\), \(f(x) = x\) differentiates to \(1\).

The derivatives from the left and right at \(x = 1\) are:

• As \(x \to 1^{-}\), the derivative is \(1\).
• As \(x \to 1^{+}\), the derivative is \(-1\).

Since these derivatives do not match, the function \(f(x)\) is not differentiable at \(x = 1\). Thus, \(n = 1\).

Conclusion

The value of \(m + n = 0 + 1 = 1\). Therefore, the correct answer is \(1\).

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