Let $f_{1}: \mathrm{R} \rightarrow \mathrm{R}, f_{2}:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rig

Question

Let $f_{1}: \mathrm{R} \rightarrow \mathrm{R}, f_{2}:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathrm{R}, f_{3}:\left(-1, e^{\frac{\pi}{2}}-2\right) \rightarrow \mathrm{R}$, and $f_{4}: \mathrm{R} \rightarrow \mathrm{R}$ be functions defined by\ (i) $f_{1}(x)=\sin \left(\sqrt{1-e^{-x^{2}}}\right)$,\ (ii) $f_{2}(x)=\left\{\begin{array}{cc}\frac{|\sin x|}{\tan ^{-1} x}, & \text { if } x \neq 0 \ 1, & \text { if } x=0\end{array}\right.$, where the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ \(iii) $f_{3}(x)=\left[\sin \left(\log _{e}(x+2)\right)\right]$, where for $t \in \mathrm{R},[t]$ denotes the greatest integer less than or equal to $t$,\ (iv) $f_{4}(x)=\left\{\begin{array}{ll}x^{2} \sin \left(\frac{1}{x}\right), & \text { if } x \neq 0 \ 0 & \text {, if } x=0\end{array}\right.$.  The correct option is:

Professor PC Thomas Classes · Answer

(d) : (i) $f_{1}^{\prime}(0)=\lim _{h ightarrow 0} \frac{\sin \sqrt{1-e^{-h^{2}}}}{h}$\ $=\lim _{h ightarrow 0} \frac{\sin \sqrt{1-e^{-h^{2}}}}{\sqrt{1-e^{-h^{2}}}} \cdot \sqrt{\frac{1-e^{-h^{2}}}{h^{2}}} \cdot \frac{|h|}{h}$\ As $\lim _{h ightarrow 0} \frac{|h|}{h}$ doesn't exist, hence $f^{\prime}{ }_{1}(0)$ doesn't exist.\ So, $\mathrm{P} ightarrow 2$\ (ii) $\lim _{x ightarrow 0} f_{2}(x)=\lim _{x ightarrow 0} \frac{|\sin x|}{	an ^{-1} x}$\ $=\lim _{x ightarrow 0} \frac{|\sin x|}{|x|} \cdot \frac{x}{	an ^{-1} x} \cdot \frac{|x|}{x}=1 \cdot 1 \cdot \lim _{x ightarrow 0} \frac{|x|}{x}$\ As $\lim _{x ightarrow 0} \frac{|x|}{x}$ doesn't exist,\ hence $\lim _{x ightarrow 0} f_{2}(x)$ doesn't exist.\ So, $\mathrm{Q} ightarrow 1$\ (iii) For $1<x+2<e^{\pi / 2}$,\ we have $0<\log (x+2)<\pi / 2$\ $\Rightarrow \quad 0<\sin (\log (x+2))<1$\ $	herefore \quad[\sin (\log (x+2))]=0$\ Thus, $f_{3}(x)=0$. Hence $\mathrm{R} ightarrow 4$\ (iv) $f_{4}(x)=\left\{\begin{array}{cc}x^{2} \sin \left(\frac{1}{x}ight), & x 
eq 0 \ 0, & x=0\end{array}ight.$\ $\lim _{x ightarrow 0} f_{4}(x)=\lim _{x ightarrow 0}\left(x^{2} \sin \frac{1}{x}ight)=0$\ $f_{4}^{\prime}(x)=\lim _{h ightarrow 0} \frac{h^{2} \sin \frac{1}{h}}{h}=\lim _{h ightarrow 0} h \sin \frac{1}{h}=0$\ $	herefore f_{4}^{\prime}(x)=-\cos \frac{1}{x}+2 x \sin \frac{1}{x}$\ Now, $\lim _{x ightarrow 0} f_{4}^{\prime}(x)$ doesn't exist.\ Hence, $S ightarrow 3$\