GATE 2021 AE – Question 41
For $f(x)=e^{-x}|\cos x|$, which statements hold?
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Correct answer: (B) Differentiable at π.; (D) Continuous at 2π.
Explanation
**Differentiability of $f(x)=e^{-x}|\cos x|$.**
The factor $e^{-x}$ is smooth, so the question is where $|\cos x|$ fails to be differentiable.
- $|\cos x|$ has a **corner** wherever $\cos x=0$ (the graph touches zero and bounces back), i.e. at odd multiples of $\pi/2$: $\pi/2,\ 3\pi/2,\ \dots$
- At $x=\pi$, $\cos\pi=-1\neq0$, so $|\cos x|=-\cos x$ near there, which is smooth.
Multiplying by the smooth, non-zero $e^{-x}$ does not remove the corners.
- **A. Differentiable at $\pi/2$.** No (corner). ✗
- **B. Differentiable at $\pi$.** Yes. ✓
- **C. Differentiable at $3\pi/2$.** No (corner). ✗
- **D. Continuous at $2\pi$.** Yes. The function is continuous everywhere. ✓
Answer **B and D**.