Question

Which of the following statement(s) is/are true about the function defined as \( f(x)= e^{-x} \lvert \cos x \rvert \) for \( x>0 \)?
 

• A. Differentiable at \( x = \dfrac{\pi}{2} \)
• B. Differentiable at \( x = \pi \)
• C. Differentiable at \( x = \dfrac{3\pi}{2} \)
• D. Continuous at \( x = 2\pi \)

Hint:

Functions with absolute values are non-differentiable where the inside term changes sign.

Answer 1

The Correct Option is B, D

The function is \[ f(x) = e^{-x} |\cos x|. \]

Step 1: Check continuity.
Since \( e^{-x} \) is continuous and \( |\cos x| \) is also continuous everywhere, the product is continuous for all \( x>0 \). This includes \( x = 2\pi \). Thus, (D) is true.

Step 2: Check differentiability.
Non-differentiability occurs where \( |\cos x| \) has a kink. This happens when \( \cos x = 0 \), i.e. at \[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \] Thus, function is not differentiable at \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). So (A) and (C) are false.
At \( x = \pi \), \(\cos(\pi) = -1\), and there is no kink. Both \( e^{-x} \) and \( |\cos x| \) are smooth here; hence the product is differentiable. So (B) is true.

Final Answer: (B), (D)