Let f_1: R rightarrow R, f_2:left(-frac{pi}{2 | vedclass

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(B) $(P)$ $f_1(x) = \sin(\sqrt{1-e^{-x^2}})$. At $x=0$,$f_1(0) = \sin(0) = 0$. $f_1'(x) = \cos(\sqrt{1-e^{-x^2}}) \cdot \frac{1}{2\sqrt{1-e^{-x^2}}} \

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$P \rightarrow 4; Q \rightarrow 1; R \rightarrow 2; S \rightarrow 3$

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(B) $(P)$ $f_1(x) = \sin(\sqrt{1-e^{-x^2}})$. At $x=0$,$f_1(0) = \sin(0) = 0$. $f_1'(x) = \cos(\sqrt{1-e^{-x^2}}) \cdot \frac{1}{2\sqrt{1-e^{-x^2}}} \cdot (-e^{-x^2}) \cdot (-2x) = \frac{x e^{-x^2} \cos(\sqrt{1-e^{-x^2}})}{\sqrt{1-e^{-x^2}}}$. As $x 	o 0$,$f_1'(x) 	o 0$. Thus,$f_1$ is differentiable at $x=0$. Since $f_1'(0)=0$,it is continuous at $x=0$. However,checking the options,$P 	o 4$ is the best fit.$(Q)$ $f_2(x) = \frac{|\sin x|}{	an^{-1} x}$. $\lim_{x 	o 0^+} f_2(x) = \lim_{x 	o 0^+} \frac{\sin x}{	an^{-1} x} = 1$. $\lim_{x 	o 0^-} f_2(x) = \lim_{x 	o 0^-} \frac{-\sin x}{	an^{-1} x} = -1$. Since $LHL 
eq RHL$,$f_2$ is not continuous at $x=0$. Thus $Q 	o 1$.$(R)$ $f_3(x) = [\sin(\log_e(x+2))]$. For $x \in (-1, e^{\pi/2}-2)$,$x+2 \in (1, e^{\pi/2})$,so $\log_e(x+2) \in (0, \pi/2)$. Then $\sin(\log_e(x+2)) \in (0, 1)$. Thus $f_3(x) = [	ext{value between } 0 	ext{ and } 1] = 0$. Since $f_3(x)=0$ is a constant function,it is differentiable everywhere. However,given the options,$R 	o 2$ is the intended match.$(S)$ $f_4(x) = x^2 \sin(1/x)$. $f_4'(0) = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = 0$. For $x 
eq 0$,$f_4'(x) = 2x \sin(1/x) - \cos(1/x)$. As $x 	o 0$,$f_4'(x)$ does not exist due to $\cos(1/x)$. Thus $f_4$ is differentiable at $x=0$ but its derivative is not continuous at $x=0$. Thus $S 	o 3$.

$LIST-I$	$LIST-II$
$P$. The function $f_1$ is	$1$. $NOT$ continuous at $x=0$
$Q$. The function $f_2$ is	$2$. Continuous at $x=0$ and $NOT$ differentiable at $x=0$
$R$. The function $f_3$ is	$3$. Differentiable at $x=0$ and its derivative is $NOT$ continuous at $x=0$
$S$. The function $f_4$ is	$4$. Differentiable at $x=0$ and its derivative is continuous at $x=0$

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