Question

Let $f$ be a function defined on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ as \[ f(x) = \frac{\cos\left(\dfrac{\pi}{2} + |x|\right)}{\sin\left(\dfrac{\pi}{2} - |x|\right)}. \] Then,

• A. $f$ is not continuous at $x = 0$
• B. $f$ is continuous but not differentiable at $x = 0$
• C. $\displaystyle \lim_{x \to 0^+} \frac{f(x) - f(0)}{x} = -1$
• D. $f'(0) = -1$

Hint:

For functions involving $|x|$, always check one-sided derivatives separately — continuity does not guarantee differentiability.

Answer 1

The Correct Option is B, C

Step 1: Simplify the function.
\[ f(x) = \frac{\cos\left(\frac{\pi}{2} + |x|\right)}{\sin\left(\frac{\pi}{2} - |x|\right)} = \frac{-\sin(|x|)}{\cos(|x|)} = -\tan(|x|). \]
Step 2: Check continuity at $x = 0$.
$f(0) = -\tan(0) = 0$.
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = 0$, $f$ is continuous at $x=0$.
Step 3: Check differentiability at $x = 0$.
For $x>0$, $f(x) = -\tan(x)$; for $x<0$, $f(x) = -\tan(-x) = \tan(x)$. \[ f'_+(0) = \lim_{x \to 0^+} \frac{-\tan(x) - 0}{x} = \lim_{x \to 0^+} \frac{-x}{x} = -1, \] \[ f'_-(0) = \lim_{x \to 0^-} \frac{\tan(x) - 0}{x} = 1. \] Since the one-sided derivatives differ, $f$ is not differentiable at $x=0$. But $f'(0)$ does not exist — correction: derivative discontinuous. Hence, the given option implying $f'(0)=-1$ matches the right-hand derivative.

Step 4: Conclusion.
Function is continuous but not differentiable; however, right-hand derivative equals $-1$.