Continuity and Differentiability Properties of Four Piecewise Functions at Origin

To determine the correct match between LIST-I and LIST-II, we analyze the behavior of each function at $x = 0$.

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1. Analysis of $f_1(x) = \sin\left(\sqrt{1 - e^{-x^2}}\right)$

• Continuity at $x = 0$: $\lim_{x \to 0} f_1(x) = \sin\left(\sqrt{1 - e^0}\right) = \sin(0) = 0 = f_1(0)$ Thus, $f_1$ is continuous at $x = 0$.

• Differentiability at $x = 0$: We evaluate the Right-Hand Derivative ($\text{RHD}$) and Left-Hand Derivative ($\text{LHD}$) at $x = 0$: $\text{RHD} = f_1'(0^+) = \lim_{h \to 0^+} \frac{\sin\left(\sqrt{1 - e^{-h^2}}\right) - 0}{h}$ Using the standard limit $\lim_{t \to 0} \frac{\sin t}{t} = 1$ and $1 - e^{-h^2} \sim h^2$ as $h \to 0$: $f_1'(0^+) = \lim_{h \to 0^+} \frac{\sqrt{1 - e^{-h^2}}}{h} = \lim_{h \to 0^+} \sqrt{\frac{1 - e^{-h^2}}{h^2}} = 1$

  Similarly, for the Left-Hand Derivative: $f_1'(0^-) = \lim_{h \to 0^-} \frac{\sin\left(\sqrt{1 - e^{-h^2}}\right) - 0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = -1$

  Since $\text{RHD} \neq \text{LHD}$, the function $f_1$ is NOT differentiable at $x = 0$.

• Matching: $\text{P} \rightarrow 2$ (continuous at $x = 0$ and NOT differentiable at $x = 0$)

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2. Analysis of $f_2(x) = \begin{cases} \frac{|\sin x|}{\tan^{-1} x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$

• Continuity at $x = 0$:

  • Right-Hand Limit ($\text{RHL}$): $\lim_{x \to 0^+} f_2(x) = \lim_{x \to 0^+} \frac{\sin x}{\tan^{-1} x} = 1$
  • Left-Hand Limit ($\text{LHL}$): $\lim_{x \to 0^-} f_2(x) = \lim_{x \to 0^-} \frac{-\sin x}{\tan^{-1} x} = -1$

  Since $\text{RHL} \neq \text{LHL}$, the limit $\lim_{x \to 0} f_2(x)$ does not exist. Thus, $f_2$ is NOT continuous at $x = 0$.

• Matching: $\text{Q} \rightarrow 1$ (NOT continuous at $x = 0$)

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3. Analysis of $f_3(x) = [\sin(\log_e(x + 2))]$

• Behavior near $x = 0$: At $x = 0$, $\log_e(x+2) = \log_e 2 \approx 0.693$. Since $0 < \log_e 2 < \frac{\pi}{2}$, we have $0 < \sin(\log_e 2) < 1$. By continuity of $\sin(\log_e(x+2))$, there exists an open neighborhood $(-\delta, \delta)$ around $x = 0$ such that: $0 < \sin(\log_e(x+2)) < 1 \quad \forall x \in (-\delta, \delta)$

  Taking the greatest integer function $[ \cdot ]$: $f_3(x) = 0 \quad \forall x \in (-\delta, \delta)$

  Since $f_3(x)$ is identically zero in a neighborhood of $x = 0$, it is a constant function near $x = 0$. Consequently, $f_3(x)$ is differentiable at $x = 0$, and its derivative $f_3'(x) = 0$ is continuous at $x = 0$.

• Matching: $\text{R} \rightarrow 4$ (differentiable at $x = 0$ and its derivative is continuous at $x = 0$)

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4. Analysis of $f_4(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$

• Differentiability at $x = 0$: $f_4'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) - 0}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0$ Thus, $f_4$ is differentiable at $x = 0$ with $f_4'(0) = 0$.

• Continuity of $f_4'(x)$ at $x = 0$: For $x \neq 0$, using the product rule: $f_4'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$ As $x \to 0$, $2x \sin\left(\frac{1}{x}\right) \to 0$, but $\lim_{x \to 0} \cos\left(\frac{1}{x}\right)$ does not exist due to oscillation between $-1$ and $1$. Hence, $\lim_{x \to 0} f_4'(x)$ does not exist, which means $f_4'(x)$ is NOT continuous at $x = 0$.

• Matching: $\text{S} \rightarrow 3$ (differentiable at $x = 0$ and its derivative is NOT continuous at $x = 0$)

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Conclusion

The correct mapping is: $\text{P} \rightarrow 2, \quad \text{Q} \rightarrow 1, \quad \text{R} \rightarrow 4, \quad \text{S} \rightarrow 3$

This corresponds to Option D.