Continuity and Differentiability of Piecewise Function

To determine the correct options, we analyze the continuity and differentiability of the piecewise function $f(x)$ given by:

$$f(x) = \begin{cases} -x - \frac{\pi}{2}, & x \le -\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \le 0 \\ x - 1, & 0 < x \le 1 \\ \ln x, & x > 1 \end{cases}$$

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1. Analysis for Option (A): Continuity at $x = -\frac{\pi}{2}$

To test for continuity at $x = -\frac{\pi}{2}$, we compute the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the value of the function at $x = -\frac{\pi}{2}$:

• Left-Hand Limit (LHL):
  $\lim_{x \to -\frac{\pi}{2}^-} f(x) = \lim_{x \to -\frac{\pi}{2}^-} \left(-x - \frac{\pi}{2}\right) = -\left(-\frac{\pi}{2}\right) - \frac{\pi}{2} = 0$

• Value of the function:
  $f\left(-\frac{\pi}{2}\right) = -\left(-\frac{\pi}{2}\right) - \frac{\pi}{2} = 0$

• Right-Hand Limit (RHL):
  $\lim_{x \to -\frac{\pi}{2}^+} f(x) = \lim_{x \to -\frac{\pi}{2}^+} (-\cos x) = -\cos\left(-\frac{\pi}{2}\right) = 0$

Since $\text{LHL} = \text{RHL} = f\left(-\frac{\pi}{2}\right) = 0$, $f(x)$ is continuous at $x = -\frac{\pi}{2}$.
Thus, Option (A) is correct.

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2. Analysis for Option (B): Differentiability at $x = 0$

First, we check continuity at $x = 0$:

• $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-\cos x) = -\cos(0) = -1$
• $f(0) = -1$
• $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x - 1) = 0 - 1 = -1$

Since $\text{LHL} = \text{RHL} = f(0) = -1$, the function is continuous at $x = 0$.

Next, we calculate the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at $x = 0$:

• Left-Hand Derivative (LHD):
  $f'_{-}(0) = \left.\frac{d}{dx}(-\cos x)\right|_{x=0^-} = \left.\sin x\right|_{x=0} = 0$

• Right-Hand Derivative (RHD):
  $f'_{+}(0) = \left.\frac{d}{dx}(x - 1)\right|_{x=0^+} = 1$

Since $f'_{-}(0) \neq f'_{+}(0)$, $f(x)$ is not differentiable at $x = 0$.
Thus, Option (B) is correct.

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3. Analysis for Option (C): Differentiability at $x = 1$

First, we check continuity at $x = 1$:

• $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x - 1) = 0$
• $f(1) = 1 - 1 = 0$
• $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (\ln x) = \ln(1) = 0$

Since $\text{LHL} = \text{RHL} = f(1) = 0$, the function is continuous at $x = 1$.

Next, we calculate the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at $x = 1$:

• Left-Hand Derivative (LHD):
  $f'_{-}(1) = \left.\frac{d}{dx}(x - 1)\right|_{x=1^-} = 1$

• Right-Hand Derivative (RHD):
  $f'_{+}(1) = \left.\frac{d}{dx}(\ln x)\right|_{x=1^+} = \left.\frac{1}{x}\right|_{x=1} = 1$

Since $f'_{-}(1) = f'_{+}(1) = 1$, $f(x)$ is differentiable at $x = 1$.
Thus, Option (C) is correct.

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4. Analysis for Option (D): Differentiability at $x = -\frac{3}{2}$

Note that $-\frac{3}{2} = -1.5$ and $-\frac{\pi}{2} \approx -1.5708$.
Since $-1.5 > -1.5708$, we have $-\frac{3}{2} \in \left(-\frac{\pi}{2}, 0\right)$.

In the open interval $\left(-\frac{\pi}{2}, 0\right)$, $f(x)$ is defined solely by the expression $f(x) = -\cos x$, which is a standard trigonometric function differentiable everywhere in its domain.

Therefore, $f(x)$ is differentiable at $x = -\frac{3}{2}$, with derivative:
$f'\left(-\frac{3}{2}\right) = \sin\left(-\frac{3}{2}\right) = -\sin\left(\frac{3}{2}\right)$

Thus, Option (D) is correct.

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Conclusion

All four statements (A), (B), (C), and (D) are correct.

Correct Options: A, B, C, D