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 address the problem, we will analyze the provided function:

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

Step 1: Continuity Analysis

The function \(f(x) = e^{-| \log_e x |}\) can be simplified. The absolute value of \(| \log_e x |\) is defined as:

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

Consequently, the function takes the following forms:

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

At \(x = 1\), the function evaluates to \(f(x) = e^{0} = 1\).

Step 2: Discontinuity Point Examination

The function \(f(x)\) is continuous on the intervals \((0, 1)\) and \((1, \infty)\) as it is composed of continuous functions. We must check the limits at \(x = 1\):

• The left-hand limit as \(x \to 1^{-}\) is \(\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} x = 1\).
• The right-hand limit as \(x \to 1^{+}\) is \(\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} x^{-1} = 1\).

Since both limits equal the function's value at \(x = 1\), the function is continuous at \(x = 1\). Therefore, there are no points of discontinuity, and \(m = 0\).

Step 3: Non-Differentiability Point Examination

We examine the derivatives on both sides of \(x = 1\):

• For \(x > 1\), the derivative of \(f(x) = x^{-1}\) is \(\frac{d}{dx}(x^{-1}) = -x^{-2}\).
• For \(0 < x < 1\), the derivative of \(f(x) = x\) is \(\frac{d}{dx}(x) = 1\).

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

• As \(x \to 1^{-}\), the derivative approaches \(1\).
• As \(x \to 1^{+}\), the derivative approaches \(-1^{-2} = -1\) (correction: the previous statement was likely a typo, it should be \( -1 \)).

The derivatives from the left and right do not match. Therefore, the function \(f(x)\) is not differentiable at \(x = 1\). Consequently, there is one point of non-differentiability, and \(n = 1\).

Conclusion

The sum \(m + n = 0 + 1 = 1\). The final answer is \(1\).