Question

Let ${{f}_{1}}\text{ }:\text{ }R\text{ }\to R,{{f}_{2}}\text{ }:\left( -\pi /2,\pi /2 \right)\to R,\text{ }{{f}_{2}}\text{ }:\text{ }\left( -1,\text{ }{{e}^{\dfrac{_{\pi }}{2}}}-2 \right)\to R$ and
${{f}_{4}}\text{ }:\text{ }R\to ~R\$ be functions defined by (i) ${{f}_{1}}(x)=\sin (\sqrt{1-{{e}^{-{{x}^{2}}}}})$$$$$ (ii) ${{f}_{2}}\left( x \right)=\left\\{ \begin{matrix} \dfrac{\left| \sin x \right|}{{{\tan }^{-1}}x},\text{if }x\ne 0, \\\ 1,\text{if }x=0 \\\ \end{matrix} \right.$ where the inverse trigonometric function ${{\tan }^{-1}}x$ assumes values $\left( -\pi /2,\pi /2 \right)$ .

(iii) ${{f}_{3}}\left( x \right)=\left[ \sin \left( {{\log }_{e}}\left( x\text{ }+\text{ }2 \right) \right) \right]$ where, for $t\in ~R,\text{ }\left[ t \right]$ denotes the greatest integer less than or equal to is $t$$$$$

(iv) {{f}_{4}}\left( x \right)=\left\\{ \begin{matrix} {{x}^{2}}\sin \left( \dfrac{1}{x} \right) & ,x\ne 0 \\\ 0 & ,x=0 \\\ \end{matrix} \right.$$$$$ List-I P. The function {{f}{1}} Q. The function ${{f}_{2}}
R. The function {{f}_{3}}$$$$$ S. The function {{f}{4}} List-II:$$$$ 1\. NOT continuous at $x=0
2. Continuous at $x=0$ and NOT differentiable at $x=0$
3. Differentiable at $x=0$ and its derivative is NOT continuous at $x=0$. 4\. Differentiable at $x=0$ its derivative is continuous at $x=0$.
Choose the correct option:$$$$
A.P\to 2,Q\to 3,R\to 1,S\to 4$$$$$ B. P\to 4,Q\to 1,R\to 2,S\to 3 C. $P\to 4,Q\to 2,R\to 1,S\to 3
D. $P\to 2,Q\to 1,R\to 4,S\to 3$$$$$

Answer 1

Use the definition of continuity and differentiability at any point on the basis of limits. Calculate the left hand limit, the right hand limit , the left hand derivative and right hand derivative of all the functions to reach the correct result.$$$$

Complete step-by-step answer:
We know that if a function $f\left( x \right)$ is continuous at any point $x=a$ then if and only if Left hand limit(LHL)= right hand limit(RHL)=the value of the function at $x=a$. In symbols,

& \text{LHL}=\text{RHL=}f\left( a \right) \\\ & \Rightarrow \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) \\\ \end{aligned}$$ The function $f\left( x \right)$ is differentiable at $x=a$ if and only if $f\left( x \right)$ is continuous and Left hand derivative (LHD)=Right Hand Derivative $$\begin{aligned} & \text{LHD}=\text{RHD} \\\ & \Rightarrow \underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( h \right)}{h}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( h \right)}{h} \\\ \end{aligned}$$ (i) The first function is given by ${{f}_{1}}(x)=\sin (\sqrt{1-{{e}^{-{{x}^{2}}}}})$. Testing for continuity at $x=0$, $$\text{LHL}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,(\sqrt{1-{{e}^{-{{x}^{2}}}}})=\text{RHL}=f\left( 0 \right)$$ So ${{f}_{1}}$ is continuous at $x=0$. Testing for differentiability at $x=0$, $$\begin{aligned} & \text{LHD}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin (\sqrt{1-{{e}^{-{{h}^{2}}}}})}{-h}=-1 \\\ & \text{RHD}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin (\sqrt{1-{{e}^{-{{h}^{2}}}}})}{h}=1 \\\ \end{aligned}$$ So ${{f}_{1}}$ is not differentiable at $x=0$. So $P\to 2$$$$$ (ii) The first function is given by$${{f}_{2}}\left( x \right)=\left\\{ \begin{matrix} \dfrac{\left| \sin x \right|}{{{\tan }^{-1}}x},\text{if }x\ne 0, \\\ 1,\text{if }x=0 \\\ \end{matrix} \right.$$. Testing for continuity at $x=0$, $$\begin{aligned} & \text{LHL}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left| \sin x \right|}{{{\tan }^{-1}}x}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left| \sin x \right|}{{{\tan }^{-1}}x}=\dfrac{-\sin \left( -h \right)}{{{\tan }^{-}}\left( -h \right)}=-1 \\\ & \text{RHL}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left| \sin x \right|}{{{\tan }^{-1}}x}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left| \sin x \right|}{{{\tan }^{-1}}x}=\dfrac{\sin \left( h \right)}{{{\tan }^{-}}\left( h \right)}=1 \\\ \end{aligned}$$ As LHL and RHL are not same , ${{f}_{2}}$ is not continuous at $x=0$. So $Q\to 2$.$$$$ (iii) The third function is given by $${{f}_{3}}\left( x \right)=\left[ \sin \left( {{\log }_{e}}\left( x\text{ }+\text{ }2 \right) \right) \right]$$ . Testing for continuity at $x=0$, $$\begin{aligned} & \text{LHL}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \sin \left( {{\log }_{e}}\left( x\text{ }+\text{ }2 \right) \right) \right]=0 \\\ & \text{RHL}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \sin \left( {{\log }_{e}}\left( x\text{ }+\text{ }2 \right) \right) \right]=0 \\\ & {{f}_{3}}\left( 0 \right)=\left[ \sin \left( {{\log }_{e}}\left( \text{0 }+\text{ }2 \right) \right) \right]=0 \\\ \end{aligned}$$ So ${{f}_{3}}$ is continuous at $x=0$. Testing for differentiability at $x=0$, $$\begin{aligned} & \text{LHD}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left[ \sin \left( {{\log }_{e}}\left( \text{-h }+\text{ }2 \right) \right) \right]}{-h}=0 \\\ & \text{RHD}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left[ \sin \left( {{\log }_{e}}\left( \text{h }+\text{ }2 \right) \right) \right]}{h}=0 \\\ \end{aligned}$$ So ${{f}_{3}}$ is differentiable at $x=0$ and also ${{f}_{3}}^{'}\left( x \right)$ is differentiable in neighbourhood differentiable $x=0$. So ${{f}_{3}}^{'}\left( x \right)$ is continuous at $x=0$. So $R\to 4$ $$$$ (iv) The last function is given by ${{f}_{4}}\left( x \right)=\left\\{ \begin{matrix} {{x}^{2}}\sin \left( \dfrac{1}{x} \right) & ,x\ne 0 \\\ 0 & ,x=0 \\\ \end{matrix} \right.$ . Testing for continuity at $x=0$, $$\begin{aligned} & \text{LHL}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=0 \\\ & \text{RHL}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)==0 \\\ & {{f}_{3}}\left( 0 \right)=0 \\\ \end{aligned}$$ So ${{f}_{3}}$ is continuous at $x=0$. Testing for differentiability at $x=0$, $$\begin{aligned} & \text{LHD}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{h}^{2}}\sin \left( -\dfrac{1}{h} \right)}{-h}=0 \\\ & \text{RHD}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{h}^{2}}\sin \left( \dfrac{1}{h} \right)}{h}=0 \\\ \end{aligned}$$ So ${{f}_{4}}$ is differentiable at $x=0$ and also ${{f}_{4}}^{'}\left( 0 \right)$ is not differentiable everywhere in the neighborhood of differentiable $x=0$. So ${{f}_{3}}^{'}\left( x \right)$ is not differentiable at $x=0$. So $R\to 4$ $$$$ **So, the correct answer is “Option D”.** **Note:** We need to be careful of calculation and substitution which will lead us to the correct result. We need to take care of the negative and positive signs while finding left and right hand derivatives because they are going to be critical if the modulus function is involved.