Let $f_1:R\to R,f_2:\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\to R{f}_3:\left(-1,e^{\frac{\pi }{2}}-2\right)\to R$and $f_4:R\to R$be functions defined by\displaystyle{{f}_{{1}}{\left({x}\right)}}={\sin{{\left(\sqrt{{{1}-{e}^{{-{{x}}}}^{2}}}\right)}}},(ii) $f_2\left(x\right)=\{\frac{\left|\sin x\right|}{{\tan }^{-1}x}\text{if}x\ne 01\text{if}x=0,$where the inverse trigonometric function ${\tan }^{-1}x$assumes values in $\left(\frac{\pi }{2},\frac{\pi }{2}\right)$,(iii) $f_3\left(x\right)=\left[\sin \left({\left(\log \right)}_e\left(x+2\right)\right)\right]$, where, for $t\in R$, $\left[t\right]$denotes the greatest integer less than or equal to $t$,(iv) $f_4\left(x\right)=\{x^2\sin \left(\frac{1}{x}\right)\text{if}x\ne 00\text{if}x=0$.LIST-I LIST-IIP. The function $f_1$is 1. NOT continuous at $x=0$Q. The function $f_2$is 2. continuous at $x=0$and NOTR. The function $f_2$is differentiable at $x=0$S. The function $f_2$is 3. differentiable at $x=0$and itsis NOT continuous at $x=0$4. differentiable at $x=0$and itsderivative is continuous at $x=0$The correct option is$P\to 2;Q\to 3;R\to 1;S\to 4$(b) $P\to 4;Q\to 1;R\to 2;S\to 3$(c) $P\to 4;Q\to 2;R\to 1;S\to 3$(d) $P\to 2;Q\to 1;R\to 4;S\to 3$

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